Note. An earlier and incomplete version of this work appeared in two parts on Hexagon, as 2610.00446v1 (the spectral calculations on the Dwork and weighted Fermat pencils) and 2610.00493v1 (the mirror clause), with the archived versions doi:10.5281/zenodo.23567115, doi:10.5281/zenodo.23646340 and doi:10.5281/zenodo.23647986. This version adds Section 16: the residual Abelian factor is forced for every rank-five split bundle on a threefold with at most four Kähler parameters, the polystable locus of the fixed bundle is the volume ray, the perturbative selection rules are decided on eleven split data, and the extension branch of the eleven data is decided for its two smallest shapes; and Definition 9.1, which fixes the acceptance windows of the three phenomenological conditions. The programs that check the finite arithmetic are in the repository creelie/geometric-selection-conjecture on GitHub, release v3.1.0, concept DOI 10.5281/zenodo.23567114, version DOI 10.5281/zenodo.23688897.
Introduction
The physical problem has two distinct levels. One can calculate a spectrum and cubic form on a specified Calabi–Yau family. One can also ask for a single compactification satisfying the gauge, matter, mass, stabilization and supersymmetry requirements C1–C9. The first calculation is not a solution to the second existence problem.
This paper brings together the spectral calculations on the Dwork and weighted Fermat pencils, the mirror restriction of the selector, and the construction criteria for the conjecture, in one narrative with one bibliography. The three strands are kept apart where they must be: the Dwork mass matrix below is not substituted for a Yukawa matrix of the tetraquadric bundle later in the paper.
The fixed-candidate calculation is Theorems 13.2 and 14.2 : a single equation at the parameter θ = π \theta= \pi θ = π − 3 -3 − 3 [29 ] . The determinant and explicit seed identities make the chosen complex structure auditable without assuming it is generic. § 17 then shows that the two failures found at that point are not accidents of the choice: the residual Abelian factor is forced for every sum of five line bundles on a threefold with at most four Kähler parameters (Theorem 16.1 ), and the perturbative selection rules, decided by a flow criterion, exclude six of eleven split data on the tetraquadric and leave five (Proposition 16.4). The same section follows the extension branch away from the split locus: one non-split extension never removes the forced gauge boson (Proposition 16.8), while a polystable sum of two rank-two extensions and a line bundle does so for the fixed datum and five of the ten further data, on one side of a stability wall through the polystable point, and is excluded for the other four at every Kähler class or on the general tetraquadric (Theorem 16.10 ). Definition 9.1 fixes the acceptance windows of C7–C9, so that the conjecture is the definite statement GSC(W) of Conjecture 9.2 ; the negative results do not depend on the windows. § 18 tests the fixed datum against every condition. It records both verified statements and failures.
For comparison, consider the Dwork family
X ψ = { ∑ i = 1 5 x i 5 − 5 ψ ∏ i = 1 5 x i = 0 } ⊂ P 4 , ψ 5 ≠ 1. (1) X_{\psi}=\left\{\sum_{i=1}^{5}x_i^5-5\psi\prod_{i=1}^{5}x_i=0\right\}\subset\mathbb{P}^{4},\qquad\psi^5\ne1. \tag*{(1)} X ψ = { i = 1 ∑ 5 x i 5 − 5 ψ i = 1 ∏ 5 x i = 0 } ⊂ P 4 , ψ 5 = 1. (1) The standard-embedding E 6 E_6 E 6 27 27 27 E 6 E_6 E 6 Γ ≅ ( Z / 5 ) 3 \Gamma\cong(\mathbb{Z}/5)^3 Γ ≅ ( Z /5 ) 3
Theorem 1.1 (Labels). The hundred and one classes carry hundred and one distinct characters of Γ \Gamma Γ Γ \Gamma Γ ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) S 5 S_5 S 5
Theorem 1.2 (Couplings). Let ψ 5 ≠ 1 \psi^5\ne1 ψ 5 = 1 x 1 3 ⋯ x 5 3 x_1^3\cdots x_5^3 x 1 3 ⋯ x 5 3 a , b , c \mathbf{a},\mathbf{b},\mathbf{c} a , b , c ψ r \psi^r ψ r a + b + c = ( 3 − r ) 1 ( m o d 5 ) \mathbf{a}+\mathbf{b}+\mathbf{c}=(3-r)\mathbf{1}\pmod5 a + b + c = ( 3 − r ) 1 ( mod 5 ) r ∈ { 0 , 1 , 2 , 3 } r\in\{0,1,2,3\} r ∈ { 0 , 1 , 2 , 3 } r = 0 , 1 , 2 , 3 r=0,1,2,3 r = 0 , 1 , 2 , 3 ψ ≠ 0 \psi\ne0 ψ = 0
Theorem 1.3 (Mass map). The invariant class h = x 1 x 2 x 3 x 4 x 5 h=x_1x_2x_3x_4x_5 h = x 1 x 2 x 3 x 4 x 5 b \mathbf{b} b σ ( b ) \sigma(\mathbf{b}) σ ( b ) σ \sigma σ h h h b \mathbf{b} b σ ( b ) \sigma(\mathbf{b}) σ ( b ) ψ \psi ψ ( 3 , 1 , 1 , 0 , 0 ) (3,1,1,0,0) ( 3 , 1 , 1 , 0 , 0 ) ψ 2 \psi^2 ψ 2 ( 3 , 2 , 0 , 0 , 0 ) (3,2,0,0,0) ( 3 , 2 , 0 , 0 , 0 ) 27 labelled by h h h ψ ≠ 0 \psi\ne0 ψ = 0 ψ 5 ≠ 1 \psi^5\ne1 ψ 5 = 1 ψ 70 \psi^{70} ψ 70 ψ = 0 \psi=0 ψ = 0 Γ \Gamma Γ ∣ ψ ∣ r ( b ) ( g b g σ ( b ) ) − 1 / 2 |\psi|^{r(\mathbf{b})}(g_{\mathbf{b}}g_{\sigma(\mathbf{b})})^{-1/2} ∣ ψ ∣ r ( b ) ( g b g σ ( b ) ) − 1/2 g b g_{\mathbf{b}} g b
The same residue and character questions extend to three weighted families of Fermat type, the pencils
X ψ = { ∑ i = 1 5 x i n i − d ψ x 1 x 2 x 3 x 4 x 5 = 0 } ⊂ P ( w ) , n i = d w i , (2) X_\psi=\left\{\sum_{i=1}^{5}x_i^{n_i}-d\psi x_1x_2x_3x_4x_5=0\right\}\subset\mathbb{P}(w),\qquad n_i=\frac{d}{w_i}, \tag*{(2)} X ψ = { i = 1 ∑ 5 x i n i − d ψ x 1 x 2 x 3 x 4 x 5 = 0 } ⊂ P ( w ) , n i = w i d , (2) of sextics in P ( 1 , 1 , 1 , 1 , 2 ) \mathbb{P}(1,1,1,1,2) P ( 1 , 1 , 1 , 1 , 2 ) P ( 1 , 1 , 1 , 1 , 4 ) \mathbb{P}(1,1,1,1,4) P ( 1 , 1 , 1 , 1 , 4 ) P ( 1 , 1 , 1 , 2 , 5 ) \mathbb{P}(1,1,1,2,5) P ( 1 , 1 , 1 , 2 , 5 ) [17 ] , Klemm and Theisen [18 ] and Font [19 ] . Here d = ∑ i w i d=\sum_i w_i d = ∑ i w i ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 δ = ∏ i w i w i \delta=\prod_i w_i^{w_i} δ = ∏ i w i w i Γ \Gamma Γ
Theorem 1.4 (Weighted pencils). Let X ψ X_\psi X ψ ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 X ψ X_\psi X ψ h 1 , 1 = 1 h^{1,1}=1 h 1 , 1 = 1 h 2 , 1 = 103 h^{2,1}=103 h 2 , 1 = 103
(i) The generations are the monomials x b x^b x b d d d 0 ≤ b i ≤ n i − 2 0\le b_i\le n_i-2 0 ≤ b i ≤ n i − 2 Γ \Gamma Γ
(ii) The cubic form, normalised so that ∏ i x i n i − 2 \prod_i x_i^{n_i-2} ∏ i x i n i − 2 a , b , c a,b,c a , b , c ψ r ∏ i w i k i \psi^r\prod_i w_i^{k_i} ψ r ∏ i w i k i a i + b i + c i = n i − 2 − r + n i k i a_i+b_i+c_i=n_i-2-r+n_i k_i a i + b i + c i = n i − 2 − r + n i k i i i i r ∈ { 0 , … , d − 1 } r\in\{0,\ldots,d-1\} r ∈ { 0 , … , d − 1 } k i ≥ 0 k_i\ge0 k i ≥ 0 ψ ≠ 0 \psi\ne0 ψ = 0 Γ \Gamma Γ
(iii) For ψ ≠ 0 \psi\ne0 ψ = 0 ψ = 0 \psi=0 ψ = 0
(iv) No member, and no quotient of a member by a group acting freely, carries three families with the standard embedding.
These results are proved in Sections 2–8 and their appendices. They concern specified holomorphic maps and their stated metric normalization, not a completed phenomenological vacuum. The global family-index obstruction is strengthened in Section 12. The GSC construction problem is formulated in Section 9, and the one-candidate audit follows.
The dictionary
Throughout, X X X S U ( 3 ) SU(3) S U ( 3 ) X X X g g g Ω \Omega Ω h 1 , 0 = h 2 , 0 = 0 h^{1,0}=h^{2,0}=0 h 1 , 0 = h 2 , 0 = 0 T X TX T X S U ( 3 ) SU(3) S U ( 3 )
Spinors and the Dolbeault complex
On a Kähler manifold of complex dimension n n n S = ⨁ q Λ 0 , q T X ∗ S=\bigoplus_q\Lambda^{0,q}T_X^* S = ⨁ q Λ 0 , q T X ∗ D = 2 ( ∂ ˉ + ∂ ˉ ∗ ) D=\sqrt{2}(\bar{\partial}+\bar{\partial}^*) D = 2 ( ∂ ˉ + ∂ ˉ ∗ ) q q q [10 ] ). If V V V ( 1 , 1 ) (1,1) ( 1 , 1 ) V V V 2 ( ∂ ˉ V + ∂ ˉ V ∗ ) \sqrt{2}(\bar{\partial}_V+\bar{\partial}_V^*) 2 ( ∂ ˉ V + ∂ ˉ V ∗ ) Λ 0 , ∗ ⊗ V \Lambda^{0,*}\otimes V Λ 0 , ∗ ⊗ V ⨁ q H 0 , q ( X , V ) \bigoplus_q H^{0,q}(X,V) ⨁ q H 0 , q ( X , V )
The charged spectrum
Theorem 2.1 (Spectrum of the standard embedding). Let X X X E 8 × E 8 E_8\times E_8 E 8 × E 8 R 3 , 1 × X \mathbb{R}^{3,1}\times X R 3 , 1 × X E 6 E_6 E 6
h 2 , 1 ( X ) h^{2,1}(X) h 2 , 1 ( X ) 27 ⊕ h 1 , 1 ( X ) \mathbf{27}\ \oplus\ h^{1,1}(X) 27 ⊕ h 1 , 1 ( X ) 27 ‾ \overline{\mathbf{27}} 27
with the 27 \mathbf{27} 27 H 1 ( X , T X ) ≅ H 2 , 1 ( X ) H^1(X,T_X)\cong H^{2,1}(X) H 1 ( X , T X ) ≅ H 2 , 1 ( X ) 27 ‾ \overline{\mathbf{27}} 27 H 1 ( X , T X ∗ ) ≅ H 1 , 1 ( X ) H^1(X,T_X^*)\cong H^{1,1}(X) H 1 ( X , T X ∗ ) ≅ H 1 , 1 ( X ) ∣ h 2 , 1 − h 1 , 1 ∣ = 1 2 ∣ χ ( X ) ∣ \lvert h^{2,1}-h^{1,1}\rvert=\frac{1}{2}\lvert\chi(X)\rvert ∣ h 2 , 1 − h 1 , 1 ∣ = 2 1 ∣ χ ( X )∣
Proof. The ten-dimensional gaugino is a Majorana–Weyl spinor with values in the adjoint of E 8 E_8 E 8 E 8 ⊃ E 6 × S U ( 3 ) E_8 \supset E_6 \times SU(3) E 8 ⊃ E 6 × S U ( 3 )
248 = ( 78 , 1 ) ⊕ ( 1 , 8 ) ⊕ ( 27 , 3 ) ⊕ ( 27 ‾ , 3 ‾ ) , 248 = (78,1) \oplus(1,8) \oplus(27,3) \oplus(\overline{27},\overline{3}), 248 = ( 78 , 1 ) ⊕ ( 1 , 8 ) ⊕ ( 27 , 3 ) ⊕ ( 27 , 3 ) , and the standard embedding identifies the S U ( 3 ) SU(3) S U ( 3 ) T X TX T X 3 3 3 T X T_X T X 3 ‾ \overline{3} 3 T X ∗ T_X^* T X ∗ R 3 , 1 × X \mathbb{R}^{3,1} \times X R 3 , 1 × X 27 27 27 X X X T X T_X T X S U ( 3 ) SU(3) S U ( 3 ) 2 ( ∂ ˉ + ∂ ˉ ∗ ) \sqrt{2}(\bar{\partial}+\bar{\partial}^*) 2 ( ∂ ˉ + ∂ ˉ ∗ ) Λ 0 , ∙ ⊗ T X \Lambda^{0,\bullet}\otimes T_X Λ 0 , ∙ ⊗ T X H 0 , 1 ( X , T X ) ⊕ H 0 , 3 ( X , T X ) H^{0,1}(X,T_X) \oplus H^{0,3}(X,T_X) H 0 , 1 ( X , T X ) ⊕ H 0 , 3 ( X , T X ) H 0 , 0 ( X , T X ) ⊕ H 0 , 2 ( X , T X ) H^{0,0}(X,T_X) \oplus H^{0,2}(X,T_X) H 0 , 0 ( X , T X ) ⊕ H 0 , 2 ( X , T X ) Ω \Omega Ω Λ 3 T X ∗ \Lambda^3T_X^* Λ 3 T X ∗ Ω \Omega Ω H 0 , q ( X , T X ) ≅ H 2 , q ( X ) H^{0,q}(X,T_X) \cong H^{2,q}(X) H 0 , q ( X , T X ) ≅ H 2 , q ( X ) H 0 , q ( X , T X ∗ ) = H 1 , q ( X ) H^{0,q}(X,T_X^*) = H^{1,q}(X) H 0 , q ( X , T X ∗ ) = H 1 , q ( X ) h 2 , 0 = h 1 , 0 = 0 h^{2,0}=h^{1,0}=0 h 2 , 0 = h 1 , 0 = 0 H 0 , 0 ( T X ) = H 2 , 0 = 0 H^{0,0}(T_X)=H^{2,0}=0 H 0 , 0 ( T X ) = H 2 , 0 = 0 H 0 , 3 ( T X ) = H 2 , 3 = H 1 , 0 ∗ = 0 H^{0,3}(T_X)=H^{2,3}=H^{1,0*}=0 H 0 , 3 ( T X ) = H 2 , 3 = H 1 , 0 ∗ = 0 H 0 , 2 ( T X ) = H 2 , 2 ≅ H 1 , 1 H^{0,2}(T_X)=H^{2,2}\cong H^{1,1} H 0 , 2 ( T X ) = H 2 , 2 ≅ H 1 , 1 H 0 , 1 ( T X ) = H 2 , 1 H^{0,1}(T_X)=H^{2,1} H 0 , 1 ( T X ) = H 2 , 1 27 27 27 27 ‾ \overline{27} 27 h 2 , 1 h^{2,1} h 2 , 1 27 27 27 H 0 , 1 ( T X ) H^{0,1}(T_X) H 0 , 1 ( T X ) h 1 , 1 h^{1,1} h 1 , 1 27 ‾ \overline{27} 27 H 0 , 2 ( T X ) H^{0,2}(T_X) H 0 , 2 ( T X ) h 2 , 1 − h 1 , 1 h^{2,1}-h^{1,1} h 2 , 1 − h 1 , 1 − χ ( X ) / 2 -\chi(X)/2 − χ ( X ) /2 χ = 2 ( h 1 , 1 − h 2 , 1 ) \chi=2(h^{1,1}-h^{2,1}) χ = 2 ( h 1 , 1 − h 2 , 1 ) h 1 , 0 = h 2 , 0 = 0 h^{1,0}=h^{2,0}=0 h 1 , 0 = h 2 , 0 = 0
The cubic form
Let V = H 1 ( X , T X ) V=H^1(X,T_X) V = H 1 ( X , T X )
V ⊗ V ⊗ V ⟶ H 3 ( X , Λ 3 T X ) = H 3 ( X , K X − 1 ) ≅ H 3 ( X , O X ) ≅ C , V\otimes V\otimes V \longrightarrow H^3(X,\Lambda^3T_X)=H^3(X,K_X^{-1})\cong H^3(X,\mathcal{O}_X)\cong\mathbb{C}, V ⊗ V ⊗ V ⟶ H 3 ( X , Λ 3 T X ) = H 3 ( X , K X − 1 ) ≅ H 3 ( X , O X ) ≅ C , where the last two isomorphisms use Ω \Omega Ω Y Y Y V V V X X X ∇ \nabla ∇ Ω \Omega Ω
Y ( a , b , c ) = ∫ X Ω ∧ ∇ a ∇ b ∇ c Ω , (3) Y(a,b,c)=\int_X\Omega\wedge\nabla_a\nabla_b\nabla_c\Omega, \tag*{(3)} Y ( a , b , c ) = ∫ X Ω ∧ ∇ a ∇ b ∇ c Ω , (3) and this is the coupling of three 27 27 27 [3 ] . Changing Ω \Omega Ω f Ω f\Omega f Ω Y Y Y f 2 f^2 f 2 Y Y Y ⟨ Ω , Ω ‾ ⟩ \langle\Omega,\overline{\Omega}\rangle ⟨ Ω , Ω ⟩ [6 , 7 ] .
Mass terms
A vacuum value v v v 27 27 27 h ∈ V h\in V h ∈ V E 6 E_6 E 6 Y ( h , ⋅ , ⋅ ) Y(h,\cdot,\cdot) Y ( h , ⋅ , ⋅ ) v Y ( h , b , c ) vY(h,b,c) v Y ( h , b , c ) b b b c c c [2 ] . The physical mass matrix is
M b c = v Y ( h , b , c ) g b g c (4) M_{bc}=v\frac{Y(h,b,c)}{\sqrt{g_b g_c}} \tag*{(4)} M b c = v g b g c Y ( h , b , c ) (4) in an orthogonal basis of V V V g b = ⟨ ω b , ω ‾ b ⟩ g_b=\langle\omega_b,\overline{\omega}_b\rangle g b = ⟨ ω b , ω b ⟩ Y ( h , ⋅ , ⋅ ) Y(h,\cdot,\cdot) Y ( h , ⋅ , ⋅ ) h h h
Hypersurfaces
For a smooth hypersurface X = { F = 0 } ⊂ P 4 X=\{F=0\}\subset\mathbb{P}^4 X = { F = 0 } ⊂ P 4 J F = ( ∂ 1 F , … , ∂ 5 F ) J_F=(\partial_1F,\ldots,\partial_5F) J F = ( ∂ 1 F , … , ∂ 5 F ) R = C [ x 1 , … , x 5 ] / J F R=\mathbb{C}[x_1,\ldots,x_5]/J_F R = C [ x 1 , … , x 5 ] / J F [8 ] identifies H p r i m 3 − q , q ( X ) H_{\mathrm{prim}}^{3-q,q}(X) H prim 3 − q , q ( X ) R 5 ( q + 1 ) − 5 R_{5(q+1)-5} R 5 ( q + 1 ) − 5 P P P 5 q 5q 5 q
ω P = Res X P Ω 0 F q + 1 , Ω 0 = ∑ i ( − 1 ) i x i d x 1 ∧ ⋯ d x i ^ ⋯ ∧ d x 5 , \omega_P=\operatorname{Res}_X\frac{P\Omega_0}{F^{q+1}},\qquad
\Omega_0=\sum_i(-1)^i x_i\,dx_1\wedge\cdots\widehat{dx_i}\cdots\wedge dx_5, ω P = Res X F q + 1 P Ω 0 , Ω 0 = i ∑ ( − 1 ) i x i d x 1 ∧ ⋯ d x i ⋯ ∧ d x 5 , so H 3 , 0 = R 0 H^{3,0}=R_0 H 3 , 0 = R 0 H 2 , 1 = R 5 H^{2,1}=R_5 H 2 , 1 = R 5 H 1 , 2 = R 10 H^{1,2}=R_{10} H 1 , 2 = R 10 H 0 , 3 = R 15 H^{0,3}=R_{15} H 0 , 3 = R 15 R R R ( 1 + t + t 2 + t 3 ) 5 (1+t+t^2+t^3)^5 ( 1 + t + t 2 + t 3 ) 5 dim R 5 = 101 \dim R_5=101 dim R 5 = 101 dim R 15 = 1 \dim R_{15}=1 dim R 15 = 1 R 15 R_{15} R 15 [9 ] , says that the Hessian Hess F = det ( ∂ i ∂ j F ) F=\det(\partial_i\partial_jF) F = det ( ∂ i ∂ j F ) R 15 R_{15} R 15 R R R λ F : R 15 → C \lambda_F:R_{15}\to\mathbb{C} λ F : R 15 → C λ F ( Hess F ) = 1 \lambda_F(\operatorname{Hess}F)=1 λ F ( Hess F ) = 1 Ω \Omega Ω
Y ( ω P , ω Q , ω S ) = c λ F ( P Q S ) . (5) Y(\omega_P,\omega_Q,\omega_S)=c\lambda_F(PQS). \tag*{(5)} Y ( ω P , ω Q , ω S ) = c λ F ( P QS ) . (5) Figure 1 shows a curve on the Fermat quintic and the Hodge numbers of a quintic threefold: by Theorem 2.1 the hundred and one classes of type ( 2 , 1 ) (2,1) ( 2 , 1 ) ( 1 , 1 ) (1,1) ( 1 , 1 )
Figure 1. Left: the intersection of the Fermat quintic with the plane x 3 = x 4 = 0 x_3=x_4=0 x 3 = x 4 = 0 z 1 5 + z 2 5 = 1 z_1^5+z_2^5=1 z 1 5 + z 2 5 = 1 C 2 \mathbb{C}^2 C 2 [11 ] and projected to three dimensions along ( Re z 1 , Re z 2 , cos α Im z 1 + sin α Im z 2 ) (\operatorname{Re}z_1,\operatorname{Re}z_2,\cos\alpha\operatorname{Im}z_1+\sin\alpha\operatorname{Im}z_2) ( Re z 1 , Re z 2 , cos α Im z 1 + sin α Im z 2 ) Γ \Gamma Γ μ 5 × μ 5 \mu_5\times\mu_5 μ 5 × μ 5 h p , q h^{p,q} h p , q 1 + log 2 h p , q 1+\log_2 h^{p,q} 1 + log 2 h p , q h 2 , 1 = h 1 , 2 = 101 h^{2,1}=h^{1,2}=101 h 2 , 1 = h 1 , 2 = 101
Labels
Let ζ = e 2 π i / 5 \zeta=e^{2\pi i/5} ζ = e 2 π i /5 K = { a ∈ ( Z / 5 ) 5 : ∑ a i = 0 } K=\{a\in(\mathbb{Z}/5)^5:\sum a_i=0\} K = { a ∈ ( Z /5 ) 5 : ∑ a i = 0 } P 4 \mathbb{P}^4 P 4 x i ↦ ζ a i x i x_i\mapsto\zeta^{a_i}x_i x i ↦ ζ a i x i D D D P 4 \mathbb{P}^4 P 4 Γ = K / D ≅ ( Z / 5 ) 3 \Gamma=K/D\cong(\mathbb{Z}/5)^3 Γ = K / D ≅ ( Z /5 ) 3 F ψ = ∑ x i 5 − 5 ψ ∏ x i F_\psi=\sum x_i^5-5\psi\prod x_i F ψ = ∑ x i 5 − 5 ψ ∏ x i ψ \psi ψ Γ \Gamma Γ K K K D D D
Γ ^ = { b ∈ ( Z / 5 ) 5 : ∑ b i = 0 } / ⟨ 1 ⟩ , ∣ Γ ^ ∣ = 125 , \widehat{\Gamma}=\{b\in(\mathbb{Z}/5)^5:\sum b_i=0\}/\langle1\rangle,\qquad|\widehat{\Gamma}|=125, Γ = { b ∈ ( Z /5 ) 5 : ∑ b i = 0 } / ⟨ 1 ⟩ , ∣ Γ ∣ = 125 , with b b b a ↦ ζ a ⋅ b a\mapsto\zeta^{a\cdot b} a ↦ ζ a ⋅ b x b x^b x b a ∈ K a\in K a ∈ K ζ a ⋅ b \zeta^{a\cdot b} ζ a ⋅ b b b b
Let B B B b ∈ { 0 , 1 , 2 , 3 } 5 b\in\{0,1,2,3\}^5 b ∈ { 0 , 1 , 2 , 3 } 5 ∣ b ∣ = ∑ b i = 5 |b|=\sum b_i=5 ∣ b ∣ = ∑ b i = 5 x b x^b x b b ∈ B b\in B b ∈ B R 5 R_5 R 5 ψ 5 ≠ 1 \psi^5\ne1 ψ 5 = 1 F ψ F_\psi F ψ x i 4 − ψ ∏ j ≠ i x j x_i^4-\psi\prod_{j\ne i}x_j x i 4 − ψ ∏ j = i x j B B B dim R 5 \dim R_5 dim R 5 ( 9 4 ) = 126 \binom{9}{4}=126 ( 4 9 ) = 126
Proof of Theorem 1.1. Two elements b , b ′ ∈ B b,b'\in B b , b ′ ∈ B b ′ ≡ b + k 1 ( m o d 5 ) b'\equiv b+k1\pmod5 b ′ ≡ b + k 1 ( mod 5 ) k k k b ′ b' b ′ { 0 , … , 4 } \{0,\ldots,4\} { 0 , … , 4 } b ′ b' b ′ b + k 1 b+k1 b + k 1 b i ′ b_i' b i ′ b i + k b_i+k b i + k b i + k − 5 b_i+k-5 b i + k − 5 ∑ b i ′ = 5 \sum b_i'=5 ∑ b i ′ = 5 k = 1 k=1 k = 1 b i + 1 ≤ 4 b_i+1\le4 b i + 1 ≤ 4 b i = 4 b_i=4 b i = 4 b i + 1 b_i+1 b i + 1 10 ≠ 5 10\ne5 10 = 5 k = 2 k=2 k = 2 b i + 2 b_i+2 b i + 2 b i ≤ 2 b_i\le2 b i ≤ 2 b i − 3 b_i-3 b i − 3 b i = 3 b_i=3 b i = 3 15 − 5 m 15-5m 15 − 5 m m m m m = 2 m=2 m = 2 k = 3 k=3 k = 3 b i + 3 b_i+3 b i + 3 b i ≤ 1 b_i\leq1 b i ≤ 1 b i − 2 b_i-2 b i − 2 b i ≥ 2 b_i\geq2 b i ≥ 2 20 − 5 m ′ 20-5m' 20 − 5 m ′ m ′ m' m ′ m ′ = 3 m'=3 m ′ = 3 k = 4 k=4 k = 4 b i + 4 b_i+4 b i + 4 b i = 0 b_i=0 b i = 0 b i − 1 b_i-1 b i − 1 25 − 5 m ′ ′ 25-5m'' 25 − 5 m ′′ m ′ ′ m'' m ′′ m ′ ′ = 4 m''=4 m ′′ = 4 ( 2 , 1 , 1 , 1 , 0 ) (2,1,1,1,0) ( 2 , 1 , 1 , 1 , 0 ) b ′ = ( 1 , 0 , 0 , 0 , 4 ) \mathbf{b}'=(1,0,0,0,4) b ′ = ( 1 , 0 , 0 , 0 , 4 ) k = 0 k=0 k = 0 R 5 R_5 R 5
The group Γ ^ \widehat{\Gamma} Γ B B B ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) Γ ^ \widehat{\Gamma} Γ ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) B B B { 0 , … , 4 } \{0,\ldots,4\} { 0 , … , 4 } ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 )
Finally S 5 S_5 S 5 B B B { 1 , 1 , 1 , 1 , 1 } \{1,1,1,1,1\} { 1 , 1 , 1 , 1 , 1 } { 2 , 1 , 1 , 1 , 0 } \{2,1,1,1,0\} { 2 , 1 , 1 , 1 , 0 } { 3 , 2 , 0 , 0 , 0 } \{3,2,0,0,0\} { 3 , 2 , 0 , 0 , 0 } { 2 , 2 , 1 , 0 , 0 } \{2,2,1,0,0\} { 2 , 2 , 1 , 0 , 0 } { 3 , 1 , 1 , 0 , 0 } \{3,1,1,0,0\} { 3 , 1 , 1 , 0 , 0 }
Figure 2 shows the hundred and one labels.
Figure 2. The hundred and one generations of the Dwork quintic as points of the character group of Γ \Gamma Γ Γ ^ ≅ ( Z / 5 ) 3 \widehat{\Gamma} \cong(\mathbb{Z}/5)^3 Γ ≅ ( Z /5 ) 3 b \mathbf{b} b ( b 0 − b 4 , b 1 − b 4 , b 2 − b 4 ) (b_0-b_4,b_1-b_4,b_2-b_4) ( b 0 − b 4 , b 1 − b 4 , b 2 − b 4 ) S 5 S_5 S 5 ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) σ \sigma σ Theorem 1.3 , drawn as chords between paired labels, with the chord colour giving the power of ψ \psi ψ ψ \psi ψ ψ 2 \psi^2 ψ 2 h h h
The socle functional
Fix ψ \psi ψ ψ 5 ≠ 1 \psi^5\ne1 ψ 5 = 1 R = R F ψ R=R_{F_\psi} R = R F ψ m 0 = x 1 3 x 2 3 x 3 3 x 4 3 x 5 3 m_0=x_1^3x_2^3x_3^3x_4^3x_5^3 m 0 = x 1 3 x 2 3 x 3 3 x 4 3 x 5 3 x e x^{\mathbf{e}} x e e \mathbf{e} e r ∈ { 0 , 1 , 2 , 3 , 4 } r\in\{0,1,2,3,4\} r ∈ { 0 , 1 , 2 , 3 , 4 } e i ≡ 3 − r ( m o d 5 ) e_i\equiv3-r\pmod5 e i ≡ 3 − r ( mod 5 ) i i i e \mathbf{e} e
Theorem 4.1 (Socle functional). For ψ 5 ≠ 1 \psi^5 \ne1 ψ 5 = 1 m 0 ≠ 0 m_0 \ne0 m 0 = 0 R 15 R_{15} R 15 λ \lambda λ R 15 R_{15} R 15 λ ( m 0 ) = 1 \lambda(m_0)=1 λ ( m 0 ) = 1 x e x^{\mathbf e} x e
λ ( x e ) = { ψ r if e is in sector r , 0 otherwise. \lambda(x^{\mathbf e}) =
\begin{cases}
\psi^r & \text{if }\mathbf e\text{ is in sector }r,\\
0 & \text{otherwise.}
\end{cases} λ ( x e ) = { ψ r 0 if e is in sector r , otherwise. Moreover λ ( Hess F ψ ) = 3 200 000 ( 1 − ψ 5 ) \lambda(\operatorname{Hess} F_\psi)=3\,200\,000(1-\psi^5) λ ( Hess F ψ ) = 3 200 000 ( 1 − ψ 5 )
Proof. The relations x i 4 ≡ ψ ∏ j ≠ i x j x_i^4 \equiv\psi\prod_{j\ne i}x_j x i 4 ≡ ψ ∏ j = i x j R R R e \mathbf e e e i ≥ 4 e_i\ge4 e i ≥ 4
x e = ψ x e ′ , e ′ = e + 1 − 5 e i , (6) x^{\mathbf e}=\psi x^{\mathbf e'},\qquad\mathbf e'=\mathbf e+\mathbf1-5\mathbf e_i, \tag*{(6)} x e = ψ x e ′ , e ′ = e + 1 − 5 e i , (6) where e i \mathbf e_i e i i i i e \mathbf e e r r r e ′ \mathbf e' e ′ r − 1 r-1 r − 1 i i i e i − 4 ≡ 3 − ( r − 1 ) e_i-4\equiv3-(r-1) e i − 4 ≡ 3 − ( r − 1 ) e j + 1 ≡ 3 − ( r − 1 ) e_j+1\equiv3-(r-1) e j + 1 ≡ 3 − ( r − 1 ) e = ( 3 , 3 , 3 , 3 , 3 ) \mathbf e=(3,3,3,3,3) e = ( 3 , 3 , 3 , 3 , 3 ) e \mathbf e e ( 7 , 2 , 2 , 2 , 2 ) (7,2,2,2,2) ( 7 , 2 , 2 , 2 , 2 ) ψ m 0 \psi m_0 ψ m 0 e = ( 11 , 1 , 1 , 1 , 1 ) \mathbf e=(11,1,1,1,1) e = ( 11 , 1 , 1 , 1 , 1 ) ( 7 , 2 , 2 , 2 , 2 ) (7,2,2,2,2) ( 7 , 2 , 2 , 2 , 2 ) e \mathbf e e ( 15 , 0 , 0 , 0 , 0 ) (15,0,0,0,0) ( 15 , 0 , 0 , 0 , 0 ) ( 10 , 5 , 0 , 0 , 0 ) (10,5,0,0,0) ( 10 , 5 , 0 , 0 , 0 ) ( 5 , 5 , 5 , 0 , 0 ) (5,5,5,0,0) ( 5 , 5 , 5 , 0 , 0 ) r r r r r r x e ≡ ψ r m 0 x^{\mathbf e}\equiv\psi^r m_0 x e ≡ ψ r m 0 R R R
If e \mathbf e e x e x^{\mathbf e} x e Γ \Gamma Γ e \mathbf e e e ≡ c 1 \mathbf e\equiv c\mathbf1 e ≡ c 1 c c c e \mathbf e e J F ψ J_{F_\psi} J F ψ Γ \Gamma Γ R R R Γ \Gamma Γ x e x^{\mathbf e} x e R 15 R_{15} R 15 R 15 R_{15} R 15 m 0 m_0 m 0 m 0 ≠ 0 m_0\ne0 m 0 = 0 x e = 0 x^{\mathbf e}=0 x e = 0
It remains to show that m 0 ≠ 0 m_0\ne0 m 0 = 0 Hess F ψ \operatorname{Hess} F_\psi Hess F ψ Γ \Gamma Γ R R R m 0 m_0 m 0
Hess F ψ ≡ H ( ψ ) m 0 , \operatorname{Hess} F_\psi\equiv H(\psi)m_0, Hess F ψ ≡ H ( ψ ) m 0 , with H ( ψ ) H(\psi) H ( ψ ) ψ r \psi^r ψ r Hess F ψ ≠ 0 \operatorname{Hess} F_\psi\ne0 Hess F ψ = 0 R 15 R_{15} R 15 m 0 ≠ 0 m_0\ne0 m 0 = 0 H ( ψ ) ≠ 0 H(\psi)\ne0 H ( ψ ) = 0 20 x i 3 20x_i^3 20 x i 3 − 5 ψ ∏ k ≠ i , j x k -5\psi\prod_{k\ne i,j}x_k − 5 ψ ∏ k = i , j x k π \pi π π \pi π ∏ i ( entry ( i , π ( i ) ) ) \prod_i(\text{entry }(i,\pi(i))) ∏ i ( entry ( i , π ( i ))) π \pi π x a 3 x_a^3 x a 3 x a x_a x a m m m x m x_m x m m m m Table 1 lists the seven cycle types; the derivation of each row is in Appendix A. Summing the last column,
Table 1. The Hessian of F ψ F_{\psi} F ψ r r r H ( ψ ) H(\psi) H ( ψ ) x ≡ ψ r m 0 x \equiv\psi^{r}m_{0} x ≡ ψ r m 0
Table 1.
H ( ψ ) = 3 200 000 − ( 2 000 000 + 1 000 000 + 375 000 H(\psi)=3\,200\,000-(2\,000\,000+1\,000\,000+375\,000 H ( ψ ) = 3 200 000 − ( 2 000 000 + 1 000 000 + 375 000 + 75 000 − 187 500 − 62 500 ) ψ 5 = 3 200 000 ( 1 − ψ 5 ) . \begin{aligned}
\qquad+75\,000-187\,500-62\,500)\psi^5 \\
&=3\,200\,000(1-\psi^5).
\end{aligned} + 75 000 − 187 500 − 62 500 ) ψ 5 = 3 200 000 ( 1 − ψ 5 ) . Remark 4.2 (Normalisation). With the residue normalisation λ F ( Hess F ) = 1 \lambda_F(\operatorname{Hess} F)=1 λ F ( Hess F ) = 1 λ F ψ = λ / ( 3 200 000 ( 1 − ψ 5 ) ) \lambda_{F_\psi}=\lambda/(3\,200\,000(1-\psi^5)) λ F ψ = λ / ( 3 200 000 ( 1 − ψ 5 )) a = b = c = h \mathbf a=\mathbf b=\mathbf c=\mathbf h a = b = c = h x 31 = m 0 x^{31}=m_0 x 31 = m 0 Y ( h , h , h ) Y(h,h,h) Y ( h , h , h ) 1 / ( 1 − ψ 5 ) 1/(1-\psi^5) 1/ ( 1 − ψ 5 ) 5 / ( 1 − ψ 5 ) 5/(1-\psi^5) 5/ ( 1 − ψ 5 ) [5 ] in the gauge in which Ω \Omega Ω Ω 0 / F ψ \Omega_0/F_\psi Ω 0 / F ψ c c c ψ \psi ψ Theorem 1.2 is stated for λ \lambda λ
Couplings
Proof of Theorem 1.2. By (5) and Theorem 4.1 , Y ( x a , x b , x c ) = λ ( x a + b + c ) Y(x^{\mathbf a},x^{\mathbf b},x^{\mathbf c})=\lambda(x^{\mathbf a+\mathbf b+\mathbf c}) Y ( x a , x b , x c ) = λ ( x a + b + c ) a + b + c \mathbf a+\mathbf b+\mathbf c a + b + c ψ r \psi^{r} ψ r a + b + c ≡ ( 3 − r ) 1 \mathbf a+\mathbf b+\mathbf c\equiv(3-r)\mathbf1 a + b + c ≡ ( 3 − r ) 1 ψ ≠ 0 \psi\ne0 ψ = 0 § 5.1 . □
The count
A triple { a , b , c } \{\mathbf a,\mathbf b,\mathbf c\} { a , b , c } B B B a + b + c ≡ c 1 ( m o d 5 ) \mathbf a+\mathbf b+\mathbf c\equiv c\mathbf1\pmod5 a + b + c ≡ c 1 ( mod 5 ) c c c c ≡ 3 − r c\equiv3-r c ≡ 3 − r r r r c \mathbf c c a + b \mathbf a+\mathbf b a + b Γ ^ \widehat{\Gamma} Γ B B B ( a , b ) (\mathbf a,\mathbf b) ( a , b ) c \mathbf c c − ( a + b ) -(\mathbf a+\mathbf b) − ( a + b ) ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) a + b \mathbf a+\mathbf b a + b
Excluded pairs. The entries of a + b \mathbf a+\mathbf b a + b { 0 , … , 6 } \{0,\ldots,6\} { 0 , … , 6 } 0 , 1 , 2 , 3 , 4 0,1,2,3,4 0 , 1 , 2 , 3 , 4 a + b \mathbf a+\mathbf b a + b ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) k k k a + b \mathbf a+\mathbf b a + b k k k k = 0 , … , 4 k=0,\ldots,4 k = 0 , … , 4 p k p_k p k a \mathbf a a k k k 0 ≤ p k ≤ min ( k , 3 ) 0\le p_k\le\min(k,3) 0 ≤ p k ≤ min ( k , 3 ) k − p k ≤ 3 k-p_k\le3 k − p k ≤ 3 p 0 = 0 p_0=0 p 0 = 0 p 1 ∈ { 0 , 1 } p_1\in\{0,1\} p 1 ∈ { 0 , 1 } p 2 ∈ { 0 , 1 , 2 } p_2\in\{0,1,2\} p 2 ∈ { 0 , 1 , 2 } p 3 ∈ { 0 , 1 , 2 , 3 } p_3\in\{0,1,2,3\} p 3 ∈ { 0 , 1 , 2 , 3 } p 4 ∈ { 1 , 2 , 3 } p_4\in\{1,2,3\} p 4 ∈ { 1 , 2 , 3 } ∣ a ∣ = 5 |\mathbf a|=5 ∣ a ∣ = 5 p 1 + p 2 + p 3 + p 4 = 5 p_1+p_2+p_3+p_4=5 p 1 + p 2 + p 3 + p 4 = 5 t 5 t^5 t 5
( 1 + t ) ( 1 + t + t 2 ) ( 1 + t + t 2 + t 3 ) ( t + t 2 + t 3 ) , (1+t)(1+t+t^2)(1+t+t^2+t^3)(t+t^2+t^3), ( 1 + t ) ( 1 + t + t 2 ) ( 1 + t + t 2 + t 3 ) ( t + t 2 + t 3 ) , which is 16.2 101 2 − 1920 = 8281 101^2-1920=8281 10 1 2 − 1920 = 8281
Repetitions. An ordered triple ( a , a , a ) (\mathbf a,\mathbf a,\mathbf a) ( a , a , a ) 3 a ≡ c 1 3\mathbf a\equiv c\mathbf1 3 a ≡ c 1 a \mathbf a a { 0 , 1 , 2 , 3 } \{0,1,2,3\} { 0 , 1 , 2 , 3 } a = 1 \mathbf a=\mathbf1 a = 1 ( a , a , c ) (\mathbf a,\mathbf a,\mathbf c) ( a , a , c ) c \mathbf c c 2 a 2\mathbf a 2 a 2 a i 2a_i 2 a i a i = 0 , 1 , 2 , 3 a_i=0,1,2,3 a i = 0 , 1 , 2 , 3 0 , 2 , 4 , 1 0,2,4,1 0 , 2 , 4 , 1 a \mathbf a a a \mathbf a a c \mathbf c c c = a \mathbf c=\mathbf a c = a a = 1 \mathbf a=\mathbf1 a = 1 { a , a , c } \{\mathbf a,\mathbf a,\mathbf c\} { a , a , c } c ≠ a \mathbf{c} \ne\mathbf{a} c = a { 1 , 1 , 1 } \{1,1,1\} { 1 , 1 , 1 } 8281 − 300 − 1 = 7980 8281-300-1=7980 8281 − 300 − 1 = 7980
7980 6 + 100 + 1 = 1330 + 101 = 1431. \frac{7980}{6}+100+1=1330+101=1431. 6 7980 + 100 + 1 = 1330 + 101 = 1431. Sectors. In sector r r r e = a + b + c \mathbf{e}=\mathbf{a}+\mathbf{b}+\mathbf{c} e = a + b + c Theorem 4.1 with entries at most nine: ( 3 , 3 , 3 , 3 , 3 ) (3,3,3,3,3) ( 3 , 3 , 3 , 3 , 3 ) r = 0 r=0 r = 0 ( 7 , 2 , 2 , 2 , 2 ) (7,2,2,2,2) ( 7 , 2 , 2 , 2 , 2 ) r = 1 r=1 r = 1 ( 6 , 6 , 1 , 1 , 1 ) (6,6,1,1,1) ( 6 , 6 , 1 , 1 , 1 ) r = 2 r=2 r = 2 ( 5 , 5 , 5 , 0 , 0 ) (5,5,5,0,0) ( 5 , 5 , 5 , 0 , 0 ) r = 3 r=3 r = 3 O ( e ) O(\mathbf{e}) O ( e ) e \mathbf{e} e x 5 y 5 x^5y^5 x 5 y 5 ∏ i = 1 5 P e i ( x , y ) \prod_{i=1}^{5}P_{e_i}(x,y) ∏ i = 1 5 P e i ( x , y ) P e ( x , y ) = ∑ x a y b P_e(x,y)=\sum x^a y^b P e ( x , y ) = ∑ x a y b 0 ≤ a , b ≤ 3 0\leq a,b\leq3 0 ≤ a , b ≤ 3 0 ≤ e − a − b ≤ 3 0\leq e-a-b\leq3 0 ≤ e − a − b ≤ 3
O ( 3 , 3 , 3 , 3 , 3 ) = 3391 , O ( 7 , 2 , 2 , 2 , 2 ) = 480 , O ( 6 , 6 , 1 , 1 , 1 ) = 162 , O ( 5 , 5 , 5 , 0 , 0 ) = 87. \begin{aligned}
O(3,3,3,3,3)&=3391, & O(7,2,2,2,2)&=480,\\
O(6,6,1,1,1)&=162, & O(5,5,5,0,0)&=87.
\end{aligned} O ( 3 , 3 , 3 , 3 , 3 ) O ( 6 , 6 , 1 , 1 , 1 ) = 3391 , = 162 , O ( 7 , 2 , 2 , 2 , 2 ) O ( 5 , 5 , 5 , 0 , 0 ) = 480 , = 87. so the ordered counts by sector are 3391, 2400, 1620 and 870, with total 8281. The hundred unordered triples { a , a , c } \{a,a,c\} { a , a , c } c ≠ a c\ne a c = a 2 a + c 2a+c 2 a + c { 1 , 1 , 1 } \{1,1,1\} { 1 , 1 , 1 }
N r = O r − 3 D r − T r 6 + D r + T r = 566 , 425 , 280 , 160 , N_r=\frac{O_r-3D_r-T_r}{6}+D_r+T_r=566,\ 425,\ 280,\ 160, N r = 6 O r − 3 D r − T r + D r + T r = 566 , 425 , 280 , 160 , with O r O_r O r D r D_r D r T r T_r T r r r r § 26 .
The mass map
Proof of Theorem 1.3 . For b ∈ B \mathbf{b}\in B b ∈ B λ ( x b + c + 1 ) \lambda(x^{\mathbf{b}+\mathbf{c}+1}) λ ( x b + c + 1 ) b + c + 1 ≡ ( 3 − r ) 1 \mathbf{b}+\mathbf{c}+1\equiv(3-r)\mathbf{1} b + c + 1 ≡ ( 3 − r ) 1 c ≡ ( 2 − r ) 1 − b ( m o d 5 ) \mathbf{c}\equiv(2-r)\mathbf{1}-\mathbf{b}\pmod5 c ≡ ( 2 − r ) 1 − b ( mod 5 ) c ∈ B \mathbf{c}\in B c ∈ B r r r ( 2 − r ) 1 − b (2-r)\mathbf{1}-\mathbf{b} ( 2 − r ) 1 − b c \mathbf{c} c B B B { 0 , 1 , 2 , 3 } \{0,1,2,3\} { 0 , 1 , 2 , 3 } b i ≤ 2 b_i\leq2 b i ≤ 2 r = 0 r=0 r = 0 c = 21 − b ∈ { 0 , 1 , 2 } 5 \mathbf{c}=2\mathbf{1}-\mathbf{b}\in\{0,1,2\}^5 c = 2 1 − b ∈ { 0 , 1 , 2 } 5 r = 1 r=1 r = 1 1 − b \mathbf{1}-\mathbf{b} 1 − b 1 , 0 , 4 1,0,4 1 , 0 , 4 b i = 0 , 1 , 2 b_i=0,1,2 b i = 0 , 1 , 2 b \mathbf{b} b b ∈ { 0 , 1 } 5 \mathbf{b}\in\{0,1\}^5 b ∈ { 0 , 1 } 5 b = 1 \mathbf{b}=\mathbf{1} b = 1 r = 1 r=1 r = 1 r = 2 r=2 r = 2 − b -\mathbf{b} − b 0 , 4 , 3 0,4,3 0 , 4 , 3 b ∈ { 0 , 2 } 5 \mathbf{b}\in\{0,2\}^5 b ∈ { 0 , 2 } 5 r = 3 r=3 r = 3 − 1 − b -\mathbf{1}-\mathbf{b} − 1 − b 4 , 3 , 2 4,3,2 4 , 3 , 2 b ∈ { 1 , 2 } 5 \mathbf{b}\in\{1,2\}^5 b ∈ { 1 , 2 } 5 b = 1 \mathbf{b}=\mathbf{1} b = 1 21 − b 2\mathbf{1}-\mathbf{b} 2 1 − b { 1 , 1 , 1 , 1 , 1 } \{1,1,1,1,1\} { 1 , 1 , 1 , 1 , 1 } { 2 , 1 , 1 , 1 , 0 } \{2,1,1,1,0\} { 2 , 1 , 1 , 1 , 0 } { 2 , 2 , 1 , 0 , 0 } \{2,2,1,0,0\} { 2 , 2 , 1 , 0 , 0 } σ ( b ) = 21 − b \sigma(\mathbf{b})=2\mathbf{1}-\mathbf{b} σ ( b ) = 2 1 − b b = 1 \mathbf{b}=\mathbf{1} b = 1
If b \mathbf{b} b { 3 , 1 , 1 , 0 , 0 } \{3,1,1,0,0\} { 3 , 1 , 1 , 0 , 0 } r = 0 r=0 r = 0 r = 1 r=1 r = 1 3 , 0 , 0 , 1 , 1 3,0,0,1,1 3 , 0 , 0 , 1 , 1 c \mathbf{c} c b \mathbf{b} b b \mathbf{b} b b \mathbf{b} b ψ \psi ψ r = 2 r=2 r = 2 2 , 4 , … 2,4,\ldots 2 , 4 , … r = 3 r=3 r = 3 1 , 3 , 3 , 4 , 4 1,3,3,4,4 1 , 3 , 3 , 4 , 4 b \mathbf{b} b { 3 , 2 , 0 , 0 , 0 } \{3,2,0,0,0\} { 3 , 2 , 0 , 0 , 0 } r = 0 r=0 r = 0 r = 1 r=1 r = 1 3 , 4 , … 3,4,\ldots 3 , 4 , … r = 2 r=2 r = 2 2 , 3 , 0 , 0 , 0 2,3,0,0,0 2 , 3 , 0 , 0 , 0 c \mathbf{c} c b \mathbf{b} b ψ 2 \psi^2 ψ 2 r = 3 r=3 r = 3 1 , 2 , 4 , 4 , 4 1,2,4,4,4 1 , 2 , 4 , 4 , 4 σ \sigma σ 1 \mathbf{1} 1 M = ( λ ( x b + c + 1 ) ) b , c M=(\lambda(x^{\mathbf{b}+\mathbf{c}+1}))_{\mathbf{b},\mathbf{c}} M = ( λ ( x b + c + 1 ) ) b , c ψ \psi ψ ψ 2 \psi^2 ψ 2 ψ 30 + 40 = ψ 70 \psi^{30+40}=\psi^{70} ψ 30 + 40 = ψ 70 σ \sigma σ + 1 +1 + 1 ψ ≠ 0 \psi\ne0 ψ = 0 ψ = 0 \psi=0 ψ = 0
The Γ \Gamma Γ H 2 , 1 H^{2,1} H 2 , 1 Theorem 1.1 the components are the lines spanned by the x b x^{\mathbf{b}} x b g b g_{\mathbf{b}} g b ν \nu ν ψ r ( b ) ( g b g σ b ) − 1 / 2 \psi^{r(\mathbf{b})}(g_{\mathbf{b}}g_{\sigma\mathbf{b}})^{-1/2} ψ r ( b ) ( g b g σ b ) − 1/2 ( b , σ b ) (\mathbf{b},\sigma\mathbf{b}) ( b , σ b )
Figure 3. The mass spectrum of Theorem 1.3 along the pencil. Left: the three mass scales 1 1 1 ∣ ψ ∣ |\psi| ∣ ψ ∣ ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 ∣ ψ ∣ |\psi| ∣ ψ ∣ 0 ≤ ∣ ψ ∣ < 1 0 \leq|\psi| < 1 0 ≤ ∣ ψ ∣ < 1 h h h ψ \psi ψ ψ = 0 \psi= 0 ψ = 0 ψ 5 = 1 \psi^5 = 1 ψ 5 = 1
Remark 6.1 (Degenerations). The determinant ± ψ 70 \pm\psi^{70} ± ψ 70 X ψ X_\psi X ψ Γ ⋊ S 5 \Gamma\rtimes S_5 Γ ⋊ S 5 5 4 ⋅ 120 5^4 \cdot120 5 4 ⋅ 120 ψ = 0 \psi= 0 ψ = 0 1 1 1 ψ = 0 \psi= 0 ψ = 0 h h h b b b c c c 1 + b + c = ( 3 , 3 , 3 , 3 , 3 ) 1 + b + c = (3, 3, 3, 3, 3) 1 + b + c = ( 3 , 3 , 3 , 3 , 3 ) 21 − b 21 - b 21 − b ψ 5 = 1 \psi^5 = 1 ψ 5 = 1 M M M [5 ] for the period behavior.
The Fermat point
At ψ = 0 \psi= 0 ψ = 0
Proposition 7.1 (Hodge norms at the Fermat point). Let X 0 X_0 X 0 ω b \omega_b ω b x b x^b x b b ∈ B b \in B b ∈ B k i = b i + 1 k_i = b_i + 1 k i = b i + 1 γ ( t ) = Γ ( t ) / Γ ( 1 − t ) \gamma(t) = \Gamma(t)/\Gamma(1 - t) γ ( t ) = Γ ( t ) /Γ ( 1 − t ) C > 0 C > 0 C > 0 b b b
g b = ⟨ ω b , ω b ‾ ⟩ = C ∏ i = 1 5 γ ( k i / 5 ) . (7) g_b = \langle\omega_b,\overline{\omega_b}\rangle= C \prod_{i=1}^{5}\gamma(k_i/5). \tag*{(7)} g b = ⟨ ω b , ω b ⟩ = C i = 1 ∏ 5 γ ( k i /5 ) . (7) Proof. The full group μ 5 5 / μ 5 \mu_5^5/\mu_5 μ 5 5 / μ 5 X 0 X_0 X 0 ω b \omega_b ω b k = ( k i ) \mathbf{k} = (k_i) k = ( k i ) 1 ≤ k i ≤ 4 1 \leq k_i \leq4 1 ≤ k i ≤ 4 ∑ k i = 10 \sum k_i = 10 ∑ k i = 10 g b g_b g b ω b \omega_b ω b P P P P k P_{\mathbf{k}} P k k \mathbf{k} k H 3 ( X 0 , C ) H_3(X_0,\mathbb{C}) H 3 ( X 0 , C ) g b = ∣ ∫ P ω b ∣ 2 / ∣ P k ⋅ P − k ∣ g_b = \left|\int_P \omega_b\right|^2/\left|P_{\mathbf{k}}\cdot P_{-\mathbf{k}}\right| g b = ∫ P ω b 2 / ∣ P k ⋅ P − k ∣
@page 10/49 starts_mid=1 ends_mid=1
Figure 3 (PDF p. 10). The mass spectrum of Theorem 1.3 along the pencil. Left: the three mass scales 1 1 1 ∣ ψ ∣ |\psi| ∣ ψ ∣ ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 ∣ ψ ∣ |\psi| ∣ ψ ∣ 0 ≤ ∣ ψ ∣ < 1 0 \leq|\psi| < 1 0 ≤ ∣ ψ ∣ < 1 h h h ψ \psi ψ ψ = 0 \psi= 0 ψ = 0 ψ 5 = 1 \psi^5 = 1 ψ 5 = 1
symmetric monomial matrix with an involution are the absolute values of its entries, each two-cycle giving a doubly degenerate pair and the fixed point a singlet; this is the stated spectrum. □
Remark 6.1 (Degenerations). The determinant ± ψ 70 \pm\psi^{70} ± ψ 70 X ψ X_\psi X ψ Γ ⋊ S 5 \Gamma\rtimes S_5 Γ ⋊ S 5 5 4 ⋅ 120 5^4 \cdot120 5 4 ⋅ 120 ψ = 0 \psi= 0 ψ = 0 1 1 1 ψ = 0 \psi= 0 ψ = 0 h h h b b b c c c 1 + b + c = ( 3 , 3 , 3 , 3 , 3 ) 1 + b + c = (3, 3, 3, 3, 3) 1 + b + c = ( 3 , 3 , 3 , 3 , 3 ) 21 − b 21 - b 21 − b ψ 5 = 1 \psi^5 = 1 ψ 5 = 1 M M M [5 ] for the period behavior.
The Fermat point
At ψ = 0 \psi= 0 ψ = 0
Proposition 7.1 (Hodge norms at the Fermat point). Let X 0 X_0 X 0 ω b \omega_b ω b x b x^b x b b ∈ B b \in B b ∈ B k i = b i + 1 k_i = b_i + 1 k i = b i + 1 γ ( t ) = Γ ( t ) / Γ ( 1 − t ) \gamma(t) = \Gamma(t)/\Gamma(1 - t) γ ( t ) = Γ ( t ) /Γ ( 1 − t ) C > 0 C > 0 C > 0 b b b
g b = ⟨ ω b , ω b ‾ ⟩ = C ∏ i = 1 5 γ ( k i / 5 ) . (8) g_b = \langle\omega_b,\overline{\omega_b}\rangle= C \prod_{i=1}^{5}\gamma(k_i/5). \tag*{(8)} g b = ⟨ ω b , ω b ⟩ = C i = 1 ∏ 5 γ ( k i /5 ) . (8) Proof. The full group μ 5 5 / μ 5 \mu_5^5/\mu_5 μ 5 5 / μ 5 X 0 X_0 X 0 ω b \omega_b ω b k = ( k i ) \mathbf{k} = (k_i) k = ( k i ) 1 ≤ k i ≤ 4 1 \leq k_i \leq4 1 ≤ k i ≤ 4 ∑ k i = 10 \sum k_i = 10 ∑ k i = 10 g b g_b g b ω b \omega_b ω b P P P P k P_{\mathbf{k}} P k k \mathbf{k} k H 3 ( X 0 , C ) H_3(X_0,\mathbb{C}) H 3 ( X 0 , C ) g b = ∣ ∫ P ω b ∣ 2 / ∣ P k ⋅ P − k ∣ g_b = \left|\int_P \omega_b\right|^2/\left|P_{\mathbf{k}}\cdot P_{-\mathbf{k}}\right| g b = ∫ P ω b 2 / ∣ P k ⋅ P − k ∣ 1 − ζ k i 1-\zeta^{k_i} 1 − ζ k i Γ ( t ) Γ ( 1 − t ) = π / sin π t \Gamma(t)\Gamma(1-t)=\pi/\sin\pi t Γ ( t ) Γ ( 1 − t ) = π / sin π t [4 ] on the Fermat quintic and, in the language of the Gepner model, of Distler and Greene [13 ] ; the steps are written out in Appendix C. □
The proposition fixes the ratios of the Hodge norms inside H 2 , 1 H^{2,1} H 2 , 1 γ ( 1 − t ) = 1 / γ ( t ) \gamma(1-t)=1/\gamma(t) γ ( 1 − t ) = 1/ γ ( t ) γ ( 1 / 5 ) = 3.94325 \gamma(1/5)=3.94325 γ ( 1/5 ) = 3.94325 γ ( 2 / 5 ) = 1.48951 \gamma(2/5)=1.48951 γ ( 2/5 ) = 1.48951 k = ( 2 , 2 , 2 , 2 , 2 ) \mathbf{k}=(2,2,2,2,2) k = ( 2 , 2 , 2 , 2 , 2 ) g 1 = C γ ( 2 / 5 ) 5 g_1=C\gamma(2/5)^5 g 1 = C γ ( 2/5 ) 5 { 2 , 1 , 1 , 1 , 0 } \{2,1,1,1,0\} { 2 , 1 , 1 , 1 , 0 } k \mathbf{k} k ( 3 , 2 , 2 , 2 , 1 ) (3,2,2,2,1) ( 3 , 2 , 2 , 2 , 1 ) g b = C γ ( 2 / 5 ) 2 γ ( 1 / 5 ) g_b=C\gamma(2/5)^2\gamma(1/5) g b = C γ ( 2/5 ) 2 γ ( 1/5 ) { 2 , 2 , 1 , 0 , 0 } \{2,2,1,0,0\} { 2 , 2 , 1 , 0 , 0 } k \mathbf{k} k ( 3 , 3 , 2 , 1 , 1 ) (3,3,2,1,1) ( 3 , 3 , 2 , 1 , 1 ) g b = C γ ( 1 / 5 ) 2 / γ ( 2 / 5 ) g_b=C\gamma(1/5)^2/\gamma(2/5) g b = C γ ( 1/5 ) 2 / γ ( 2/5 ) Γ ( 3 / 5 ) 3 Γ ( 1 / 5 ) / Γ ( 2 / 5 ) 3 Γ ( 4 / 5 ) = 1.19324 \Gamma(3/5)^3\Gamma(1/5)/\Gamma(2/5)^3\Gamma(4/5)=1.19324 Γ ( 3/5 ) 3 Γ ( 1/5 ) /Γ ( 2/5 ) 3 Γ ( 4/5 ) = 1.19324 κ = 1.09236 \kappa=1.09236 κ = 1.09236 [13 ] expressed the normalised couplings of the Gepner model (see [12, eq. (4.3)]). The norm of Ω \Omega Ω k = ( 1 , 1 , 1 , 1 , 1 ) \mathbf{k}=(1,1,1,1,1) k = ( 1 , 1 , 1 , 1 , 1 ) [5 ] , is
G ψ ψ ˉ ( 0 ) = 25 Γ ( 2 / 5 ) 5 Γ ( 4 / 5 ) 5 Γ ( 1 / 5 ) 5 Γ ( 3 / 5 ) 5 = 0.19226 , G_{\psi\bar{\psi}}(0)=25\frac{\Gamma(2/5)^5\Gamma(4/5)^5}{\Gamma(1/5)^5\Gamma(3/5)^5}=0.19226, G ψ ψ ˉ ( 0 ) = 25 Γ ( 1/5 ) 5 Γ ( 3/5 ) 5 Γ ( 2/5 ) 5 Γ ( 4/5 ) 5 = 0.19226 , the value that Butbaia and collaborators [12 ] reproduce with numerical Ricci-flat metrics.3
Inserting (7) into Theorem 1.3 gives the fifty-one nonzero masses at the Fermat point. Each label of the types { 2 , 1 , 1 , 1 , 0 } \{2,1,1,1,0\} { 2 , 1 , 1 , 1 , 0 } { 2 , 2 , 1 , 0 , 0 } \{2,2,1,0,0\} { 2 , 2 , 1 , 0 , 0 } k i k_i k i g b = g σ b g_b=g_{\sigma b} g b = g σ b v / g b v/g_b v / g b v / g 1 v/g_1 v / g 1 Table 2 lists the values. In units of the singlet mass the two doublet masses are 0.83805 (twenty labels, ten doublets) and 0.70234 (thirty labels, fifteen doublets). The fifty generations with an exponent three are massless there, and acquire masses proportional to ∣ ψ ∣ |\psi| ∣ ψ ∣ ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2
Table 2. Hodge norms and masses at the Fermat point, from (7) and Theorem 1.3 . Norms are in units of C C C v / C v/C v / C ψ = 0 \psi=0 ψ = 0
Weighted Fermat pencils
The pencils
8.1. The pencils. In this section ( d ; w ) (d;w) ( d ; w ) ( 6 ; 1 , 1 , 1 , 1 , 2 ) (6;1,1,1,1,2) ( 6 ; 1 , 1 , 1 , 1 , 2 ) ( 8 ; 1 , 1 , 1 , 1 , 4 ) (8;1,1,1,1,4) ( 8 ; 1 , 1 , 1 , 1 , 4 ) ( 10 ; 1 , 1 , 1 , 2 , 5 ) (10;1,1,1,2,5) ( 10 ; 1 , 1 , 1 , 2 , 5 ) d = ∑ w i d=\sum w_i d = ∑ w i n i = d / w i n_i=d/w_i n i = d / w i n = ( 6 , 6 , 6 , 6 , 3 ) \mathbf{n}=(6,6,6,6,3) n = ( 6 , 6 , 6 , 6 , 3 ) ( 8 , 8 , 8 , 8 , 2 ) (8,8,8,8,2) ( 8 , 8 , 8 , 8 , 2 ) ( 10 , 10 , 10 , 5 , 2 ) (10,10,10,5,2) ( 10 , 10 , 10 , 5 , 2 ) F ψ = ∑ i x i n i − d ψ P F_\psi=\sum_i x_i^{n_i}-d\psi P F ψ = ∑ i x i n i − d ψ P P = x 1 x 2 x 3 x 4 x 5 P=x_1x_2x_3x_4x_5 P = x 1 x 2 x 3 x 4 x 5 d d d deg x i = w i \deg x_i=w_i deg x i = w i X ψ ⊂ P ( w ) X_\psi\subset\mathbb{P}(w) X ψ ⊂ P ( w )
δ = ∏ i w i w i ∈ { 4 , 256 , 12500 } , \delta=\prod_i w_i^{w_i}\in\{4,256,12500\}, δ = i ∏ w i w i ∈ { 4 , 256 , 12500 } , and use the letters i , j , k , l i,j,k,l i , j , k , l Table 3 collects the numbers proved in this section, with the quintic for comparison.4
Table 3. The four pencils of Fermat type. The rows give the degree and weights, the constant δ = ∏ i w i w i \delta=\prod_iw_i^{w_i} δ = ∏ i w i w i ψ d δ = 1 \psi^d\delta=1 ψ d δ = 1 L L L h 2 , 1 h^{2,1} h 2 , 1 Γ \Gamma Γ Γ \Gamma Γ ψ ≠ 0 \psi\ne0 ψ = 0 ψ = 0 \psi=0 ψ = 0
Lemma 8.1. If ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 X ψ X_\psi X ψ P ( w ) \mathbb{P}(w) P ( w ) h 1 , 1 = 1 h^{1,1}=1 h 1 , 1 = 1 L = c 1 ( O X ( 1 ) ) L=c_1(\mathcal{O}_X(1)) L = c 1 ( O X ( 1 )) H 2 ( X , Z ) H^2(X,\mathbb{Z}) H 2 ( X , Z )
L 3 = d ∏ i w i , c 2 ⋅ L = σ 2 ( w ) L 3 , L^3=\frac{d}{\prod_iw_i},\qquad c_2\cdot L=\sigma_2(w)L^3, L 3 = ∏ i w i d , c 2 ⋅ L = σ 2 ( w ) L 3 , {#eq:8}
with σ k \sigma_k σ k ψ d δ = 1 \psi^d\delta=1 ψ d δ = 1 X ψ X_\psi X ψ
Proof. The Euler relations x i ∂ i F ψ = n i x i n i − d ψ P x_i\partial_iF_\psi=n_ix_i^{n_i}-d\psi P x i ∂ i F ψ = n i x i n i − d ψ P x i n i = w i ψ P x_i^{n_i}=w_i\psi P x i n i = w i ψ P i i i P = 0 P=0 P = 0 x = 0 x=0 x = 0 i i i w i w_i w i ∏ i x i n i w i = P d \prod_i x_i^{n_iw_i}=P^d ∏ i x i n i w i = P d P d = ψ d δ P d P^d=\psi^d\delta P^d P d = ψ d δ P d ψ d δ = 1 \psi^d\delta=1 ψ d δ = 1 X ψ X_\psi X ψ ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 ψ = 2 − 1 / 3 \psi=2^{-1/3} ψ = 2 − 1/3 ψ = 1 / 2 \psi=1/2 ψ = 1/2 ψ = 2 − 1 / 5 5 − 1 / 2 \psi=2^{-1/5}5^{-1/2} ψ = 2 − 1/5 5 − 1/2 ( 1 , 1 , 1 , 1 , 2 1 / 3 ) (1,1,1,1,2^{1/3}) ( 1 , 1 , 1 , 1 , 2 1/3 ) ( 1 , 1 , 1 , 1 , 2 ) (1,1,1,1,2) ( 1 , 1 , 1 , 1 , 2 ) ( 1 , 1 , 1 , 2 1 / 5 , 5 1 / 2 ) (1,1,1,2^{1/5},5^{1/2}) ( 1 , 1 , 1 , 2 1/5 , 5 1/2 ) x i n i = w i ψ P = w i x_i^{n_i}=w_i\psi P=w_i x i n i = w i ψ P = w i ψ d δ = 1 \psi^d\delta=1 ψ d δ = 1 ζ ψ \zeta\psi ζ ψ ζ d = 1 \zeta^d=1 ζ d = 1 x 1 ↦ ζ − 1 x 1 x_1\mapsto\zeta^{-1}x_1 x 1 ↦ ζ − 1 x 1 X ζ ψ X_{\zeta\psi} X ζ ψ X ψ X_\psi X ψ n 1 = d n_1=d n 1 = d
The singular points of P ( w ) \mathbb{P}(w) P ( w ) ( 0 : 0 : 0 : 0 : 1 ) (0:0:0:0:1) ( 0 : 0 : 0 : 0 : 1 ) μ 2 \mu_2 μ 2 μ 4 \mu_4 μ 4 ( 0 : 0 : 0 : 1 : 0 ) (0:0:0:1:0) ( 0 : 0 : 0 : 1 : 0 ) ( 0 : 0 : 0 : 0 : 1 ) (0:0:0:0:1) ( 0 : 0 : 0 : 0 : 1 ) μ 2 \mu_2 μ 2 μ 5 \mu_5 μ 5 x i n i x_i^{n_i} x i n i F ψ F_\psi F ψ F ψ F_\psi F ψ X ψ X_\psi X ψ P ( w ) \mathbb{P}(w) P ( w ) K X = O X ( d − ∑ i w i ) = O X K_X=\mathcal{O}_X(d-\sum_iw_i)=\mathcal{O}_X K X = O X ( d − ∑ i w i ) = O X [21 ] gives h 1 , 0 = h 2 , 0 = 0 h^{1,0}=h^{2,0}=0 h 1 , 0 = h 2 , 0 = 0 h 1 , 1 = 1 h^{1,1}=1 h 1 , 1 = 1
The tangent bundle of P ( w ) \mathbb{P}(w) P ( w ) ∏ i ( 1 + w i H ) \prod_i(1+w_iH) ∏ i ( 1 + w i H ) X X X O X ( d ) \mathcal{O}_X(d) O X ( d ) c ( T X ) = ∏ i ( 1 + w i L ) ( 1 + c(T_X)=\prod_i(1+w_iL)(1+ c ( T X ) = ∏ i ( 1 + w i L ) ( 1 + d L ) − 1 dL)^{-1} d L ) − 1 σ 1 ( w ) = d \sigma_1(w)=d σ 1 ( w ) = d c 1 = 0 c_1=0 c 1 = 0 c 2 = σ 2 L 2 c_2=\sigma_2L^2 c 2 = σ 2 L 2 c 3 = ( σ 3 − d σ 2 ) L 3 c_3=(\sigma_3-d\sigma_2)L^3 c 3 = ( σ 3 − d σ 2 ) L 3 L 3 = d ∫ P ( w ) H 4 = d / ∏ w i L^3=d\int_{\mathbb{P}(w)}H^4=d/\prod w_i L 3 = d ∫ P ( w ) H 4 = d / ∏ w i ( σ 2 , σ 3 ) = ( 14 , 16 ) (\sigma_2,\sigma_3)=(14,16) ( σ 2 , σ 3 ) = ( 14 , 16 ) ( 22 , 28 ) (22,28) ( 22 , 28 ) ( 34 , 52 ) (34,52) ( 34 , 52 ) Table 3 . If L = m L ′ L=mL' L = m L ′ L ′ L' L ′ m 3 m^3 m 3 L 3 ∈ { 3 , 2 , 1 } L^3\in\{3,2,1\} L 3 ∈ { 3 , 2 , 1 } m = ± 1 m=\pm1 m = ± 1 H 2 ( X , Q ) H^2(X,\mathbb{Q}) H 2 ( X , Q ) L L L L L L H 2 ( X , Z ) H^2(X,\mathbb{Z}) H 2 ( X , Z )
Labels and characters
The Jacobian ring R = C [ x 1 , … , x 5 ] / ( ∂ 1 F ψ , … , ∂ 5 F ψ ) R=\mathbb{C}[x_1,\ldots,x_5]/(\partial_1F_\psi,\ldots,\partial_5F_\psi) R = C [ x 1 , … , x 5 ] / ( ∂ 1 F ψ , … , ∂ 5 F ψ ) deg x i = w i \deg x_i=w_i deg x i = w i d − w i d-w_i d − w i R R R ∏ i ( 1 − t d − w i ) / ( 1 − t w i ) \prod_i(1-t^{d-w_i})/(1-t^{w_i}) ∏ i ( 1 − t d − w i ) / ( 1 − t w i ) ∏ i ( n i − 1 ) \prod_i(n_i-1) ∏ i ( n i − 1 ) ∑ i ( d − 2 w i ) = 3 d \sum_i(d-2w_i)=3d ∑ i ( d − 2 w i ) = 3 d R 3 d R_{3d} R 3 d Hess F ψ \operatorname{Hess}F_\psi Hess F ψ [20 ] and Dolgachev [21 ] : with
Ω 0 = ∑ i ( − 1 ) i w i x i d x 1 ∧ ⋯ d x i ^ ⋯ ∧ d x 5 , \Omega_0=\sum_i(-1)^iw_ix_i\,dx_1\wedge\cdots\widehat{dx_i}\cdots\wedge dx_5, Ω 0 = i ∑ ( − 1 ) i w i x i d x 1 ∧ ⋯ d x i ⋯ ∧ d x 5 , the residue along X X X P Ω 0 / F ψ q + 1 P\Omega_0/F_\psi^{q+1} P Ω 0 / F ψ q + 1 P P P q d qd q d R q d R_{qd} R q d H 3 − q , q ( X ) H^{3-q,q}(X) H 3 − q , q ( X ) H 2 , 1 = R d H^{2,1}=R_d H 2 , 1 = R d R 3 d R_{3d} R 3 d R R R λ \lambda λ R 3 d R_{3d} R 3 d [9 ] uses only the residue map and the residue pairing of R R R
Let B B B b ∈ Z 5 \mathbf{b}\in\mathbb{Z}^5 b ∈ Z 5 0 ≤ b i ≤ n i − 2 0\le b_i\le n_i-2 0 ≤ b i ≤ n i − 2 ∑ i w i b i = d \sum_iw_ib_i=d ∑ i w i b i = d
Lemma 8.2. Let ψ δ ≠ 1 \psi^\delta\ne1 ψ δ = 1 x e x^{\mathbf e} x e d d d x e ≡ c ψ r x b x^{\mathbf e}\equiv c\psi^r x^{\mathbf b} x e ≡ c ψ r x b R R R b ∈ B \mathbf b\in B b ∈ B c c c r ≤ 1 r\le1 r ≤ 1 r ≤ 2 r\le2 r ≤ 2 x b x^{\mathbf b} x b b ∈ B \mathbf b\in B b ∈ B R d = H 2 , 1 R_d=H^{2,1} R d = H 2 , 1 ∣ B ∣ = h 2 , 1 |B|=h^{2,1} ∣ B ∣ = h 2 , 1
Proof. The relations ∂ i F ψ / n i = 0 \partial_iF_\psi/n_i=0 ∂ i F ψ / n i = 0
x i n i − 1 = w i ψ ∏ j ≠ i x j . (9) x_i^{n_i-1}=w_i\psi\prod_{j\ne i}x_j. \tag*{(9)} x i n i − 1 = w i ψ j = i ∏ x j . (9) In a monomial of degree d d d e i ≥ n i − 1 e_i\ge n_i-1 e i ≥ n i − 1 2 d − w i − w j > d 2d-w_i-w_j>d 2 d − w i − w j > d e i ≥ n i − 1 e_i\ge n_i-1 e i ≥ n i − 1 e i ∈ { n i − 1 , n i } e_i\in\{n_i-1,n_i\} e i ∈ { n i − 1 , n i } x e x^{\mathbf e} x e w i ψ x e ′ w_i\psi x^{\mathbf e'} w i ψ x e ′ e i ′ = e i − n i + 1 e'_i=e_i-n_i+1 e i ′ = e i − n i + 1 e j ′ = e j + 1 e'_j=e_j+1 e j ′ = e j + 1 j ≠ i j\ne i j = i e i = n i e_i=n_i e i = n i x e = x i n i x^{\mathbf e}=x_i^{n_i} x e = x i n i e ′ = 1 \mathbf e'=\mathbf1 e ′ = 1 e i = n i − 1 e_i=n_i-1 e i = n i − 1 e i ′ = 0 e'_i=0 e i ′ = 0 w i w_i w i e j ′ ≥ n j − 1 e'_j\ge n_j-1 e j ′ ≥ n j − 1 w j e j ≥ d − 2 w j w_je_j\ge d-2w_j w j e j ≥ d − 2 w j w i + 2 w j ≥ d w_i+2w_j\ge d w i + 2 w j ≥ d i ≠ j i\ne j i = j w i + 2 w j ≥ 6 w_i+2w_j\ge6 w i + 2 w j ≥ 6 1 ∈ B \mathbf1\in B 1 ∈ B n j ≥ 3 n_j\ge3 n j ≥ 3 j j j w i + 2 w j ≥ d w_i+2w_j\ge d w i + 2 w j ≥ d i ≠ j i\ne j i = j j = 5 j=5 j = 5 w 5 = d / 2 w_5=d/2 w 5 = d /2 n 5 = 2 n_5=2 n 5 = 2 e j ′ = 1 e'_j=1 e j ′ = 1 x 5 x_5 x 5 x 5 x_5 x 5 e 5 ′ ′ = 0 e''_5=0 e 5 ′′ = 0 w 5 + 2 w k ≥ d w_5+2w_k\ge d w 5 + 2 w k ≥ d w k ≥ d / 4 w_k\ge d/4 w k ≥ d /4 k ≤ 4 k\le4 k ≤ 4 e i ≤ n i − 2 e_i\le n_i-2 e i ≤ n i − 2
At ψ = 0 \psi=0 ψ = 0 ⨂ i C [ x i ] / ( x i n i − 1 ) \bigotimes_i\mathbb{C}[x_i]/(x_i^{n_i-1}) ⨂ i C [ x i ] / ( x i n i − 1 ) e i ≤ n i − 2 e_i\le n_i-2 e i ≤ n i − 2 ∣ B ∣ = dim R d |B|=\dim R_d ∣ B ∣ = dim R d ψ \psi ψ R d R_d R d ψ \psi ψ
The symmetry group. Let K K K ε ∈ ∏ i μ n i \varepsilon\in\prod_i\mu_{n_i} ε ∈ ∏ i μ n i ∏ i ε i = 1 \prod_i\varepsilon_i=1 ∏ i ε i = 1 x i ↦ ε i x i x_i\mapsto\varepsilon_ix_i x i ↦ ε i x i F ψ F_\psi F ψ ψ \psi ψ ( λ w 1 , … , λ w 5 ) (\lambda^{w_1},\ldots,\lambda^{w_5}) ( λ w 1 , … , λ w 5 ) λ d = 1 \lambda^d=1 λ d = 1 K K K P ( w ) \mathbb{P}(w) P ( w ) μ d \mu_d μ d Γ = K / μ d \Gamma=K/\mu_d Γ = K / μ d lcm ( n i ) = d \operatorname{lcm}(n_i)=d lcm ( n i ) = d ∏ i μ n i → μ d \prod_i\mu_{n_i}\to\mu_d ∏ i μ n i → μ d ∣ Γ ∣ = ∏ i n i / d 2 |\Gamma|=\prod_i n_i/d^2 ∣Γ∣ = ∏ i n i / d 2 ∏ i μ n i \prod_i\mu_{n_i} ∏ i μ n i b ∈ ⨁ i Z / n i \mathbf b\in\bigoplus_i\mathbb{Z}/n_i b ∈ ⨁ i Z / n i ε ↦ ∏ i ε i b i \varepsilon\mapsto\prod_i\varepsilon_i^{b_i} ε ↦ ∏ i ε i b i K K K μ d \mu_d μ d ∑ i w i b i ≡ 0 ( m o d d ) \sum_i w_i b_i \equiv0 \pmod{d} ∑ i w i b i ≡ 0 ( mod d ) w i n i = d w_i n_i=d w i n i = d
Γ ^ = { b ∈ ⨁ i Z / n i : ∑ i w i b i ≡ 0 ( m o d d ) } / ⟨ 1 ⟩ . \widehat{\Gamma}=\left\{\mathbf{b}\in\bigoplus_i\mathbb{Z}/n_i:\sum_i w_i b_i\equiv0\pmod{d}\right\}/\langle\mathbf{1}\rangle. Γ = { b ∈ i ⨁ Z / n i : i ∑ w i b i ≡ 0 ( mod d ) } / ⟨ 1 ⟩ . A monomial x e \mathbf{x}^{\mathbf{e}} x e d d d [ e ] [\mathbf{e}] [ e ] Ω 0 \Omega_0 Ω 0 ∏ i ϵ i = 1 \prod_i\epsilon_i=1 ∏ i ϵ i = 1 R q d ≅ H 3 − q , q R_{qd}\cong H^{3-q,q} R q d ≅ H 3 − q , q Γ \Gamma Γ R R R Γ ^ \widehat{\Gamma} Γ ∂ i ∂ j F ψ \partial_i\partial_jF_\psi ∂ i ∂ j F ψ ϵ i − 1 ϵ j − 1 \epsilon_i^{-1}\epsilon_j^{-1} ϵ i − 1 ϵ j − 1 ∏ i ϵ i − 2 = 1 \prod_i\epsilon_i^{-2}=1 ∏ i ϵ i − 2 = 1 R 3 d R_{3d} R 3 d F ψ F_\psi F ψ
Lemma 8.3 (Wrapping). Two labels b ≠ b ′ \mathbf{b}\ne\mathbf{b}' b = b ′ carry the same character if and only if there is a shift s ∈ { 1 , … , d − 1 } s\in\{1,\ldots,d-1\} s ∈ { 1 , … , d − 1 } such that, with a i = ⌊ ( b i + s ) / n i ⌋ a_i=\lfloor(b_i+s)/n_i\rfloor a i = ⌊( b i + s ) / n i ⌋
b i ′ = b i + s − n i a i , ∑ i a i = s , b_i'=b_i+s-n_i a_i,\qquad\sum_i a_i=s, b i ′ = b i + s − n i a i , i ∑ a i = s , b i + s ≢ n i − 1 ( m o d n i ) for all i . b_i+s\not\equiv n_i-1\pmod{n_i}\qquad\text{for all }i. b i + s ≡ n i − 1 ( mod n i ) for all i . The shift from b ′ \mathbf{b}' b ′ back to b \mathbf{b} b is then d − s d-s d − s
Proof. Both labels satisfy ∑ i w i b i = d ≡ 0 \sum_i w_i b_i=d\equiv0 ∑ i w i b i = d ≡ 0 b ′ ≡ b + s 1 \mathbf{b}'\equiv\mathbf{b}+s\mathbf{1} b ′ ≡ b + s 1 ( n 1 , … , n 5 ) (n_1,\ldots,n_5) ( n 1 , … , n 5 ) s s s d d d 0 ≤ b i ′ ≤ n i − 2 0\le b_i'\le n_i-2 0 ≤ b i ′ ≤ n i − 2 b i ′ b_i' b i ′ b i + s b_i+s b i + s { 0 , … , n i − 1 } \{0,\ldots,n_i-1\} { 0 , … , n i − 1 } n i − 1 n_i-1 n i − 1
∑ i w i b i ′ = ∑ i w i b i + s d − d ∑ i a i , \sum_i w_i b_i'=\sum_i w_i b_i+sd-d\sum_i a_i, i ∑ w i b i ′ = i ∑ w i b i + s d − d i ∑ a i , which equals d d d ∑ i a i = s \sum_i a_i=s ∑ i a i = s □ \square □
Proposition 8.4. Let ψ d ≠ 1 \psi^{d}\ne1 ψ d = 1
(a) Sextic: the 4 pairs { x i 4 x 5 , x j 2 x k 2 x l 2 } \{x_i^4x_5,x_j^2x_k^2x_l^2\} { x i 4 x 5 , x j 2 x k 2 x l 2 } { x i 3 x j 3 , x k 3 x l 3 } \{x_i^3x_j^3,x_k^3x_l^3\} { x i 3 x j 3 , x k 3 x l 3 }
(b) Octic: the 12 pairs { x i 6 x j 2 , x k 4 x l 2 x m 2 } \{x_i^6x_j^2,x_k^4x_l^2x_m^2\} { x i 6 x j 2 , x k 4 x l 2 x m 2 } { x i 6 x j x k , x j 3 x k 3 x l 2 } \{x_i^6x_jx_k,x_j^3x_k^3x_l^2\} { x i 6 x j x k , x j 3 x k 3 x l 2 } { x i 4 x j 4 , x k 4 x l 4 } \{x_i^4x_j^4,x_k^4x_l^4\} { x i 4 x j 4 , x k 4 x l 4 }
(c) Dectic: the three triples { x i 8 x 4 , x j 2 x k 2 x 3 2 , x j 2 x k 4 x 4 2 } \{x_i^8x_4,x_j^2x_k^2x_3^2,x_j^2x_k^4x_4^2\} { x i 8 x 4 , x j 2 x k 2 x 3 2 , x j 2 x k 4 x 4 2 } { x 1 c 1 + 2 x 2 c 2 + 2 x 3 c 3 + 2 x 4 3 , c 1 + c 2 + c 3 = 4 } \{x_1^{c_1+2}x_2^{c_2+2}x_3^{c_3+2}x_4^3,c_1+c_2+c_3=4\} { x 1 c 1 + 2 x 2 c 2 + 2 x 3 c 3 + 2 x 4 3 , c 1 + c 2 + c 3 = 4 } c \mathbf{c} c ( 2 , 2 , 0 ) (2,2,0) ( 2 , 2 , 0 ) { x i 8 x j 2 , x j 4 x k 2 x l 2 } \{x_i^8x_j^2,x_j^4x_k^2x_l^2\} { x i 8 x j 2 , x j 4 x k 2 x l 2 } { x i 8 x j x k , x j 3 x k 3 x l 2 } \{x_i^8x_jx_k,x_j^3x_k^3x_l^2\} { x i 8 x j x k , x j 3 x k 3 x l 2 } { x i 6 x j 2 x 4 , x j 6 x 4 4 } \{x_i^6x_j^2x_4,x_j^6x_4^4\} { x i 6 x j 2 x 4 , x j 6 x 4 4 } { x i 6 x j x k x 4 , x j 5 x 5 5 } \{x_i^6x_jx_kx_4,x_j^5x_5^5\} { x i 6 x j x k x 4 , x j 5 x 5 5 } { x i 7 x j x 4 , x i x j 5 x 4 4 } \{x_i^7x_jx_4,x_ix_j^5x_4^4\} { x i 7 x j x 4 , x i x j 5 x 4 4 } { x i 6 x 4 2 , x j 4 x k 4 x 4 } \{x_i^6x_4^2,x_j^4x_k^4x_4\} { x i 6 x 4 2 , x j 4 x k 4 x 4 } Γ \Gamma Γ
The characters carried by no label are the 12 classes of ( π ( 0 , 1 , 3 , 4 ) , 2 ) (\pi(0,1,3,4),2) ( π ( 0 , 1 , 3 , 4 ) , 2 ) ( π ( 0 , 2 , 4 , 6 ) , 1 ) (\pi(0,2,4,6),1) ( π ( 0 , 2 , 4 , 6 ) , 1 ) π \pi π χ ↦ − χ \chi\mapsto-\chi χ ↦ − χ
Proof. We use Lemma 8.3 and treat the shifts s ≤ d / 2 s\le d/2 s ≤ d /2 b i + s ≤ 9 < 12 b_i+s\le9<12 b i + s ≤ 9 < 12 i ≤ 4 i\le4 i ≤ 4 a i = 1 a_i=1 a i = 1 b i ≥ 6 − s b_i\ge6-s b i ≥ 6 − s n i − 1 = 5 n_i-1=5 n i − 1 = 5 b i = 5 − s b_i=5-s b i = 5 − s x 5 x_5 x 5 b 5 ≤ 1 b_5\le1 b 5 ≤ 1 a 5 = ⌊ ( b 5 + s ) / 3 ⌋ a_5=\lfloor(b_5+s)/3\rfloor a 5 = ⌊( b 5 + s ) /3 ⌋ b 5 + s ≡ 2 ( m o d 3 ) b_5+s\equiv2\pmod3 b 5 + s ≡ 2 ( mod 3 ) s = 1 s=1 s = 1 a 5 = 0 a_5=0 a 5 = 0 b i ≥ 5 b_i\ge5 b i ≥ 5 s = 2 s=2 s = 2 b 5 = 0 b_5=0 b 5 = 0 b 5 = 1 b_5=1 b 5 = 1 a 5 = 1 a_5=1 a 5 = 1 b i ≥ 4 b_i\ge4 b i ≥ 4 b = ( 4 , 0 , 0 , 1 , 1 ) \mathbf{b}=(4,0,0,1,1) b = ( 4 , 0 , 0 , 1 , 1 ) ( 0 , 2 , 2 , 2 , 0 ) (0,2,2,2,0) ( 0 , 2 , 2 , 2 , 0 ) s = 3 s=3 s = 3 a 5 = 1 a_5=1 a 5 = 1 b i ≥ 3 b_i\ge3 b i ≥ 3 b = ( 3 , 3 , 0 , 0 , 0 ) \mathbf{b}=(3,3,0,0,0) b = ( 3 , 3 , 0 , 0 , 0 ) ( 0 , 0 , 3 , 3 , 0 ) (0,0,3,3,0) ( 0 , 0 , 3 , 3 , 0 ) 103 − 7 = 96 103-7=96 103 − 7 = 96 n 5 = 2 n_5=2 n 5 = 2 b 5 = 0 b_5=0 b 5 = 0 a 5 = s / 2 a_5=s/2 a 5 = s /2 x 5 x_5 x 5 b i ≤ n i − 2 b_i\le n_i-2 b i ≤ n i − 2 a i a_i a i a 4 ∈ { 0 , 1 , 2 } a_4\in\{0,1,2\} a 4 ∈ { 0 , 1 , 2 }
The characters not carried by a label are 108 − 96 = 12 108-96=12 108 − 96 = 12 128 − 122 = 6 128-122=6 128 − 122 = 6 100 − 100 = 0 100-100=0 100 − 100 = 0 s 1 s1 s 1 ( π ( 0 , 1 , 3 , 4 ) , 2 ) (\pi(0,1,3,4),2) ( π ( 0 , 1 , 3 , 4 ) , 2 ) 5 5 5 s ≢ 0 ( m o d 3 ) s\not\equiv0\pmod3 s ≡ 0 ( mod 3 ) ( π ′ ( 0 , 1 , 3 , 4 ) , 2 ) (\pi'(0,1,3,4),2) ( π ′ ( 0 , 1 , 3 , 4 ) , 2 ) s = 3 s=3 s = 3 n 5 − 1 n_5-1 n 5 − 1 n i − 1 n_i-1 n i − 1 ( π ( 0 , 1 , 3 , 4 ) , 2 ) (\pi(0,1,3,4),2) ( π ( 0 , 1 , 3 , 4 ) , 2 ) s 1 s1 s 1 ( π ( 0 , 2 , 4 , 6 ) , 1 ) (\pi(0,2,4,6),1) ( π ( 0 , 2 , 4 , 6 ) , 1 ) 7 7 7 s s s 1 = n 5 − 1 1=n_5-1 1 = n 5 − 1 s s s − ( π ( 0 , 1 , 3 , 4 ) , 2 ) = ( π ( 0 , 5 , 3 , 2 ) , 1 ) -(\pi(0,1,3,4),2)=(\pi(0,5,3,2),1) − ( π ( 0 , 1 , 3 , 4 ) , 2 ) = ( π ( 0 , 5 , 3 , 2 ) , 1 ) ( π ( 1 , 0 , 4 , 3 ) , 2 ) (\pi(1,0,4,3),2) ( π ( 1 , 0 , 4 , 3 ) , 2 ) 1 1 1 − ( π ( 0 , 2 , 4 , 6 ) , 1 ) = ( π ( 0 , 6 , 4 , 2 ) , 1 ) -(\pi(0,2,4,6),1)=(\pi(0,6,4,2),1) − ( π ( 0 , 2 , 4 , 6 ) , 1 ) = ( π ( 0 , 6 , 4 , 2 ) , 1 )
Figure 4 compares the four pencils.
Figure 4. Left: for the quintic in P 4 \mathbb{P}^{4} P 4 P ( 1 , 1 , 1 , 1 , 2 ) \mathbb{P}(1,1,1,1,2) P ( 1 , 1 , 1 , 1 , 2 ) P ( 1 , 1 , 1 , 1 , 4 ) \mathbb{P}(1,1,1,1,4) P ( 1 , 1 , 1 , 1 , 4 ) P ( 1 , 1 , 1 , 2 , 5 ) \mathbb{P}(1,1,1,2,5) P ( 1 , 1 , 1 , 2 , 5 ) ψ ≠ 0 \psi\ne0 ψ = 0 ψ = 0 \psi=0 ψ = 0 ψ \psi ψ ψ = 0 \psi=0 ψ = 0 12 500 ψ 10 = 1 12\,500\psi^{10}=1 12 500 ψ 10 = 1
The socle functional
Let m 0 = x n − 21 m_0=x^{n-2\mathbf{1}} m 0 = x n − 2 1 n i − 2 n_i-2 n i − 2 ∑ i w i ( n i − 2 ) = 3 d \sum_i w_i(n_i-2)=3d ∑ i w i ( n i − 2 ) = 3 d x e x^e x e 3 d 3d 3 d r ∈ { 0 , … , d − 1 } r\in\{0,\ldots,d-1\} r ∈ { 0 , … , d − 1 } e i ≡ n i − 2 − r ( m o d n i ) e_i\equiv n_i-2-r\pmod{n_i} e i ≡ n i − 2 − r ( mod n i ) i i i
e i = n i − 2 − r + n i k i ; (10) e_i=n_i-2-r+n_i k_i; \tag*{(10)} e i = n i − 2 − r + n i k i ; (10) each k i k_i k i k i ≤ − 1 k_i\le-1 k i ≤ − 1 e i ≤ − 2 − r e_i\le-2-r e i ≤ − 2 − r ∑ i k i = r \sum_i k_i=r ∑ i k i = r ∑ i w i e i = 5 d − 2 d − r d + d ∑ i k i = 3 d \sum_i w_i e_i=5d-2d-rd+d\sum_i k_i=3d ∑ i w i e i = 5 d − 2 d − r d + d ∑ i k i = 3 d
Theorem 8.5 (Weighted socle functional). Let ψ d / δ ≠ 1 \psi^{d/\delta}\ne1 ψ d / δ = 1 m 0 m_0 m 0 R 3 d R_{3d} R 3 d λ \lambda λ R 3 d R_{3d} R 3 d λ ( m 0 ) = 1 \lambda(m_0)=1 λ ( m 0 ) = 1 x e x^e x e 3 d 3d 3 d
λ ( x e ) = { ψ r ∏ i w i k i if e is in sector r , with k as in (10) , 0 otherwise. \lambda(x^e)=
\begin{cases}
\psi^r\prod_i w_i^{k_i} & \text{if }e\text{ is in sector }r,\text{ with }k\text{ as in (10)},\\
0 & \text{otherwise.}
\end{cases} λ ( x e ) = { ψ r ∏ i w i k i 0 if e is in sector r , with k as in (10) , otherwise. A monomial of degree 3 d 3d 3 d
Proof. The character of x e x^{\mathbf e} x e e ≡ c 1 \mathbf e \equiv c\mathbf1 e ≡ c 1 ( n 1 , … , n 5 ) (n_1,\ldots,n_5) ( n 1 , … , n 5 ) c c c r ≡ − 2 − c ( m o d d ) r \equiv-2-c \pmod d r ≡ − 2 − c ( mod d ) n i ∣ d n_i\mid d n i ∣ d e i ≡ n i − 2 − r e_i \equiv n_i-2-r e i ≡ n i − 2 − r R 3 d R_{3d} R 3 d R R R
Let e \mathbf e e r ≥ 1 r\geq1 r ≥ 1 i i i n i k i ≥ r + 1 n_i k_i\geq r+1 n i k i ≥ r + 1 k i ≤ r / n i = r w i / d k_i\leq r/n_i=rw_i/d k i ≤ r / n i = r w i / d i i i r = ∑ i k i ≤ r ∑ i w i / d = r r=\sum_i k_i\leq r\sum_i w_i/d=r r = ∑ i k i ≤ r ∑ i w i / d = r n i k i = r n_i k_i=r n i k i = r i i i n i n_i n i r r r d d d 1 ≤ r < d 1\leq r<d 1 ≤ r < d i i i e i ≥ n i − 1 e_i\geq n_i-1 e i ≥ n i − 1 x e ≡ w i ψ x e ′ x^{\mathbf e}\equiv w_i\psi x^{\mathbf e'} x e ≡ w i ψ x e ′ e i ′ = e i − n i + 1 e'_i=e_i-n_i+1 e i ′ = e i − n i + 1 e j ′ = e j + 1 e'_j=e_j+1 e j ′ = e j + 1 j ≠ i j\ne i j = i e ′ \mathbf e' e ′ r − 1 r-1 r − 1 k i ′ = k i − 1 k'_i=k_i-1 k i ′ = k i − 1 k j k_j k j r r r k i k_i k i e = n − 21 \mathbf e=\mathbf n-2\mathbf1 e = n − 2 1 x e ≡ ψ r ∏ i w i k i m 0 x^{\mathbf e}\equiv\psi^r\prod_i w_i^{k_i}m_0 x e ≡ ψ r ∏ i w i k i m 0 R 3 d R_{3d} R 3 d m 0 m_0 m 0 R 3 d ≠ 0 R_{3d}\ne0 R 3 d = 0 m 0 m_0 m 0 R 3 d R_{3d} R 3 d
Corollary 8.6. Let ψ d δ ≠ 1 \psi^{d}\delta\ne1 ψ d δ = 1 a , b , c \mathbf a,\mathbf b,\mathbf c a , b , c λ ( x a + b + c ) \lambda(x^{\mathbf a+\mathbf b+\mathbf c}) λ ( x a + b + c ) λ \lambda λ ψ r ∏ i w i k i \psi^r\prod_i w_i^{k_i} ψ r ∏ i w i k i a + b + c \mathbf a+\mathbf b+\mathbf c a + b + c r r r x a + b + c x^{\mathbf a+\mathbf b+\mathbf c} x a + b + c ψ ≠ 0 \psi\ne0 ψ = 0 Γ \Gamma Γ
The mass map
The only label with trivial character is h = 1 h=1 h = 1 h = ( 2 , 2 , 2 , 2 , 0 ) h=(2,2,2,2,0) h = ( 2 , 2 , 2 , 2 , 0 ) b ≡ c 1 \mathbf b\equiv c\mathbf1 b ≡ c 1 b i = c b_i=c b i = c R d R_d R d x h x^h x h ψ ≠ 0 \psi\ne0 ψ = 0 ∂ ψ F ψ = − d P \partial_\psi F_\psi=-dP ∂ ψ F ψ = − d P P = x h P=x^h P = x h P = x 1 ≡ w 5 ψ x h P=x^1\equiv w_5\psi x^h P = x 1 ≡ w 5 ψ x h x 5 x_5 x 5 n 5 − 1 n_5-1 n 5 − 1
M b c = λ ( x h + b + c ) , b , c ∈ B . M_{\mathbf b\mathbf c}=\lambda(x^{h+\mathbf b+\mathbf c}),\qquad\mathbf b,\mathbf c\in B. M bc = λ ( x h + b + c ) , b , c ∈ B . Since h h h M b c ≠ 0 M_{\mathbf b\mathbf c}\ne0 M bc = 0 [ c ] = − [ b ] [\mathbf c]=-[\mathbf b] [ c ] = − [ b ] χ \chi χ B χ B_\chi B χ M M M B χ ∪ B − χ B_\chi\cup B_{-\chi} B χ ∪ B − χ { χ , − χ } \{\chi,-\chi\} { χ , − χ } − χ -\chi − χ χ \chi χ
Lemma 8.7. Let ψ ≠ 0 \psi\ne0 ψ = 0 ψ d δ ≠ 1 \psi^{d}\delta\ne1 ψ d δ = 1 ψ r ∏ i w i k i ≠ 0 \psi^r\prod_i w_i^{k_i}\ne0 ψ r ∏ i w i k i = 0 b , b ′ ∈ B χ \mathbf b,\mathbf b'\in B_\chi b , b ′ ∈ B χ c , c ′ ∈ B − χ \mathbf c,\mathbf c'\in B_{-\chi} c , c ′ ∈ B − χ r , r ′ , ρ , ρ ′ r,r',\rho,\rho' r , r ′ , ρ , ρ ′ h + b + c h+\mathbf b+\mathbf c h + b + c h + b ′ + c ′ h+\mathbf b'+\mathbf c' h + b ′ + c ′ h + b + c ′ h+\mathbf b+\mathbf c' h + b + c ′ h + b ′ + c h+\mathbf b'+\mathbf c h + b ′ + c Δ = r + r ′ − ρ − ρ ′ \Delta=r+r'-\rho-\rho' Δ = r + r ′ − ρ − ρ ′ Δ ∈ { − d , 0 , d } \Delta\in\{-d,0,d\} Δ ∈ { − d , 0 , d }
M b c M b ′ c ′ − M b c ′ M b ′ c = M b c ′ M b ′ c ( ( ψ d δ ) Δ / d − 1 ) , M_{\mathbf b\mathbf c}M_{\mathbf b'\mathbf c'}-M_{\mathbf b\mathbf c'}M_{\mathbf b'\mathbf c}
=M_{\mathbf b\mathbf c'}M_{\mathbf b'\mathbf c}\left((\psi^d\delta)^{\Delta/d}-1\right), M bc M b ′ c ′ − M b c ′ M b ′ c = M b c ′ M b ′ c ( ( ψ d δ ) Δ/ d − 1 ) , which vanishes exactly when Δ = 0 \Delta=0 Δ = 0
Proof. The first statement is Theorem 8.5, since each h + b + c h+\mathbf b+\mathbf c h + b + c ( h + b + c ) + ( h + b ′ + c ′ ) (h+\mathbf b+\mathbf c)+(h+\mathbf b'+\mathbf c') ( h + b + c ) + ( h + b ′ + c ′ ) ( h + b + c ′ ) + ( h + b ′ + c ) (h+\mathbf b+\mathbf c')+(h+\mathbf b'+\mathbf c) ( h + b + c ′ ) + ( h + b ′ + c ) k , k ′ , κ , κ ′ k,k',\kappa,\kappa' k , k ′ , κ , κ ′ n i ( k i + κ i ′ − k i ′ − κ i ) = Δ n_i(k_i+\kappa'_i-k'_i-\kappa_i)=\Delta n i ( k i + κ i ′ − k i ′ − κ i ) = Δ i i i n i n_i n i Δ \Delta Δ d d d ∣ Δ ∣ < 2 d |\Delta|<2d ∣Δ∣ < 2 d { 0 , … , d − 1 } \{0,\ldots,d-1\} { 0 , … , d − 1 } ψ Δ ∏ i w i Δ w i / d = ( ψ d δ ) Δ / d \psi^{\Delta}\prod_i w_i^{\Delta w_i/d}=(\psi^d\delta)^{\Delta/d} ψ Δ ∏ i w i Δ w i / d = ( ψ d δ ) Δ/ d
Proposition 8.8. Let ψ ≠ 0 \psi\ne0 ψ = 0 ψ d δ ≠ 1 \psi^{d}\delta\ne1 ψ d δ = 1 Table 4 . The mass map has rank 99, 125 and 109 for the sextic, octic and dectic, its kernel has dimension 4, 24 and 36 and lies in the span of the labels that share a character, and the rank is the same at every number with ψ ≠ 0 \psi\ne0 ψ = 0 ψ d δ ≠ 1 \psi^{d}\delta\ne1 ψ d δ = 1 ψ = 0 \psi=0 ψ = 0 b ↦ n − 21 − h − b \mathbf b\mapsto\mathbf n-2\mathbf1-h-\mathbf b b ↦ n − 2 1 − h − b b ≤ n − 21 − h \mathbf b\leq\mathbf n-2\mathbf1-h b ≤ n − 2 1 − h
Table 4. The blocks of the mass map of the invariant direction that involve labels sharing a character. Each row gives the labels of B χ B_{\chi} B χ B − χ B_{-\chi} B − χ M b c M_{bc} M b c b ∈ B χ b\in B_{\chi} b ∈ B χ ψ ≠ 0 \psi\ne0 ψ = 0 − χ = χ -\chi=\chi − χ = χ χ = − χ \chi=-\chi χ = − χ
Proof. The classes of Proposition 8.4 and their negatives give the rows of Table 4 : for one representative of each orbit the character − χ -\chi − χ k k k i = 1 i=1 i = 1 h + ( 3 , 1 , 1 , 1 , 0 ) + ( 4 , 0 , 0 , 0 , 1 ) = ( 8 , 2 , 2 , 2 , 2 ) h+(3,1,1,1,0)+(4,0,0,0,1)=(8,2,2,2,2) h + ( 3 , 1 , 1 , 1 , 0 ) + ( 4 , 0 , 0 , 0 , 1 ) = ( 8 , 2 , 2 , 2 , 2 ) k = ( 1 , 0 , 0 , 0 , 1 ) k=(1,0,0,0,1) k = ( 1 , 0 , 0 , 0 , 1 ) 2 ψ 2 2\psi^2 2 ψ 2 h + ( 3 , 1 , 1 , 1 , 0 ) + ( 0 , 2 , 2 , 2 , 0 ) = ( 4 , 4 , 4 , 4 , 1 ) = n − 21 h+(3,1,1,1,0)+(0,2,2,2,0)=(4,4,4,4,1)=n-21 h + ( 3 , 1 , 1 , 1 , 0 ) + ( 0 , 2 , 2 , 2 , 0 ) = ( 4 , 4 , 4 , 4 , 1 ) = n − 21 B χ , B − χ B_{\chi},B_{-\chi} B χ , B − χ
( 0 A A T 0 ) \begin{pmatrix} 0 & A \\ A^{\mathsf T} & 0 \end{pmatrix} ( 0 A T A 0 ) with A A A ∣ B χ ∣ + ∣ B − χ ∣ − 2 |B_{\chi}|+|B_{-\chi}|-2 ∣ B χ ∣ + ∣ B − χ ∣ − 2 2 × 2 2\times2 2 × 2 A A A Δ = 0 \Delta=0 Δ = 0 6 + 0 − 2 − 4 = 0 6+0-2-4=0 6 + 0 − 2 − 4 = 0 A A A Δ = ± d \Delta=\pm d Δ = ± d 4 ψ 6 − 1 4\psi^6-1 4 ψ 6 − 1 256 ψ 8 − 1 256\psi^8-1 256 ψ 8 − 1 12500 ψ 10 − 1 12500\psi^{10}-1 12500 ψ 10 − 1 χ = − χ \chi=-\chi χ = − χ Table 4 : 4 ⋅ 1 = 4 4\cdot1=4 4 ⋅ 1 = 4 12 ⋅ 1 + 6 ⋅ 2 = 24 12\cdot1+6\cdot2=24 12 ⋅ 1 + 6 ⋅ 2 = 24 3 + 6 + 3 + 6 + 3 ⋅ 2 + 3 ⋅ 2 + 3 ⋅ 2 = 36 3+6+3+6+3\cdot2+3\cdot2+3\cdot2=36 3 + 6 + 3 + 6 + 3 ⋅ 2 + 3 ⋅ 2 + 3 ⋅ 2 = 36 ψ \psi ψ ψ d / δ − 1 \psi^{d/\delta}-1 ψ d / δ − 1
At ψ = 0 \psi=0 ψ = 0 r ≥ 1 r\ge1 r ≥ 1 M b c ≠ 0 M_{bc}\ne0 M b c = 0 h + b + c = n − 21 h+b+c=n-21 h + b + c = n − 21 c = n − 21 − h − b c=n-21-h-b c = n − 21 − h − b 3 d − 2 d = d 3d-2d=d 3 d − 2 d = d n i − 2 n_i-2 n i − 2 b ≤ n − 21 − h b\le n-21-h b ≤ n − 21 − h n − 21 − h = ( 3 , 3 , 3 , 3 , 0 ) , ( 4 , 4 , 4 , 4 , 0 ) n-21-h=(3,3,3,3,0),(4,4,4,4,0) n − 21 − h = ( 3 , 3 , 3 , 3 , 0 ) , ( 4 , 4 , 4 , 4 , 0 ) ( 6 , 6 , 6 , 1 , 0 ) (6,6,6,1,0) ( 6 , 6 , 6 , 1 , 0 ) t 6 t^6 t 6 ( 1 + t + t 2 + t 3 ) 4 (1+t+t^2+t^3)^4 ( 1 + t + t 2 + t 3 ) 4 t 8 t^8 t 8 ( 1 + t + ⋯ + t 4 ) 4 (1+t+\cdots+t^4)^4 ( 1 + t + ⋯ + t 4 ) 4 36 + 36 = 72 36+36=72 36 + 36 = 72 b 1 + b 2 + b 3 + 2 b 4 = 10 b_1+b_2+b_3+2b_4=10 b 1 + b 2 + b 3 + 2 b 4 = 10 b i ≤ 6 b_i\le6 b i ≤ 6 b 4 ≤ 1 b_4\le1 b 4 ≤ 1
Three families
Proposition 8.9. Let X X X ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 H H H X X X X / H X/H X / H
Proof. By Theorem 2.1 the number of net families on Y = X / H Y=X/H Y = X / H ∣ χ ( Y ) ∣ / 2 = ∣ χ ( X ) ∣ / ( 2 ∣ H ∣ ) |\chi(Y)|/2=|\chi(X)|/(2|H|) ∣ χ ( Y ) ∣/2 = ∣ χ ( X ) ∣/ ( 2∣ H ∣ ) ∣ H ∣ = ∣ χ ( X ) ∣ / 6 |H|=|\chi(X)|/6 ∣ H ∣ = ∣ χ ( X ) ∣/6 296 / 6 296/6 296/6 H H H π : X → Y \pi:X\to Y π : X → Y ∣ H ∣ |H| ∣ H ∣ χ ( O Y ) = χ ( O X ) / ∣ H ∣ = 0 \chi(\mathcal{O}_Y)=\chi(\mathcal{O}_X)/|H|=0 χ ( O Y ) = χ ( O X ) /∣ H ∣ = 0 c 1 ( Y ) c_1(Y) c 1 ( Y ) H 2 ( X , Q ) = Q L H^2(X,\mathbb{Q})=\mathbb{Q}L H 2 ( X , Q ) = Q L L L L H 2 ( X , Z ) H^2(X,\mathbb{Z}) H 2 ( X , Z ) Lemma 8.1 ), every line bundle D D D Y Y Y π ∗ D ≡ m L \pi^*D\equiv mL π ∗ D ≡ m L m m m m m m Z \mathbb{Z} Z ∣ H ∣ |H| ∣ H ∣ ⨂ g ∈ H g ∗ O X ( 1 ) \bigotimes_{g\in H}g^*\mathcal{O}_X(1) ⨂ g ∈ H g ∗ O X ( 1 ) Y Y Y ∣ H ∣ L |H|L ∣ H ∣ L H H H H 2 ( X , Q ) H^2(X,\mathbb{Q}) H 2 ( X , Q ) m 0 > 0 m_0>0 m 0 > 0 D D D m 0 m_0 m 0 ∣ H ∣ |H| ∣ H ∣
D 3 = m 0 3 L 3 ∣ H ∣ , c 2 ( Y ) ⋅ D = m 0 c 2 ⋅ L ∣ H ∣ , D^3=\frac{m_0^3L^3}{|H|},\qquad c_2(Y)\cdot D=\frac{m_0c_2\cdot L}{|H|}, D 3 = ∣ H ∣ m 0 3 L 3 , c 2 ( Y ) ⋅ D = ∣ H ∣ m 0 c 2 ⋅ L , χ ( Y , D ) = D 3 6 + c 2 ( Y ) ⋅ D 12 , \chi(Y,D)=\frac{D^3}{6}+\frac{c_2(Y)\cdot D}{12}, χ ( Y , D ) = 6 D 3 + 12 c 2 ( Y ) ⋅ D , the last by Riemann–Roch with c 1 ( Y ) = 0 c_1(Y)=0 c 1 ( Y ) = 0 χ ( O Y ) = 0 \chi(\mathcal{O}_Y)=0 χ ( O Y ) = 0 D 3 = 3 m 0 3 / 34 D^3=3m_0^3/34 D 3 = 3 m 0 3 /34 34 ∣ m 0 34\mid m_0 34 ∣ m 0 m 0 = 34 m_0=34 m 0 = 34 D 3 = 3468 D^3=3468 D 3 = 3468 c 2 ( Y ) ⋅ D = 42 c_2(Y)\cdot D=42 c 2 ( Y ) ⋅ D = 42 χ ( Y , D ) = 578 + 7 / 2 \chi(Y,D)=578+7/2 χ ( Y , D ) = 578 + 7/2 D 3 = m 0 3 / 48 D^3=m_0^3/48 D 3 = m 0 3 /48 12 ∣ m 0 12\mid m_0 12 ∣ m 0 c 2 ( Y ) ⋅ D = 17 m 0 / 24 c_2(Y)\cdot D=17m_0/24 c 2 ( Y ) ⋅ D = 17 m 0 /24 24 ∣ m 0 24\mid m_0 24 ∣ m 0 m 0 ∈ { 24 , 48 } m_0\in\{24,48\} m 0 ∈ { 24 , 48 } χ ( Y , D ) = 48 + 17 / 12 \chi(Y,D)=48+17/12 χ ( Y , D ) = 48 + 17/12 384 + 17 / 6 384+17/6 384 + 17/6
Proof of Theorem 1.4. Smoothness and h 2 , 1 h^{2,1} h 2 , 1 Lemma 8.1 , part (i) is Proposition 8.4 , part (ii) is Corollary 8.6 , part (iii) is Proposition 8.8, and part (iv) is Proposition 8.9 together with ∣ χ ( X ) ∣ / 2 ∈ { 102 , 148 , 144 } |\chi(X)|/2\in\{102,148,144\} ∣ χ ( X ) ∣/2 ∈ { 102 , 148 , 144 } X X X
Remark 8.10 (Why the quintic is different). On the quintic all n i n_i n i ∑ i a i = s \sum_i a_i=s ∑ i a i = s Lemma 8.3 has no solution: this is the proof of Theorem 1.1 . When the n i n_i n i n i n_i n i n 5 = 3 n_5=3 n 5 = 3 n 5 = 2 n_5=2 n 5 = 2 Proposition 8.4 are exactly these balances. The mass map then has blocks larger than 1 × 1 1\times1 1 × 1 Δ = 0 \Delta=0 Δ = 0
The simultaneous construction problem
The GSC formulation in [27 , 28 ] concerns a landscape L = ∐ τ M τ \mathcal{L}=\coprod_\tau M_\tau L = ∐ τ M τ μ \mu μ D D D S : L → [ 0 , ∞ ) S:\mathcal{L}\to[0,\infty) S : L → [ 0 , ∞ ) μ \mu μ μ \mu μ [26 ] . We use the following short restatement, without supplying new numerical meanings for the physical acceptance conditions.
Table 12.
These are nine simultaneous predicates on one complete datum, not nine alternative cases. Let A i A_i A i C i C_i C i ⋂ i = 1 9 A i ≠ ∅ \bigcap_{i=1}^{9}A_i\ne\varnothing ⋂ i = 1 9 A i = ∅ A i A_i A i { 1 , 2 } \{1,2\} { 1 , 2 } { 0 , 2 } \{0,2\} { 0 , 2 }
For a complete datum one must specify the threefold, visible and hidden bundles, polarization, free action and equivariant structures when used, Wilson line, effective Kähler potential and superpotential, and the domain in which omitted corrections are controlled. C5 and C7–C9 also need explicit acceptance predicates. Without them the conjecture is not a mathematical statement, since a threshold “fixed by comparison with measurement” is not a predicate on a datum. We fix one choice. The numbers are conventions, and a different choice gives a different conjecture; the failures proved in this paper for C7 and C8 hold for every choice, as Remark 9.3 records.
Definition 9.1 (The acceptance windows W W W W W W
C 7 W C_{7W} C 7 W m 0 = 3 × 10 4 m_0 = 3 \times10^4 m 0 = 3 × 1 0 4 [39 , 40 ] .
C 8 W C_{8W} C 8 W μ = 2 × 10 16 \mu= 2 \times10^{16} μ = 2 × 1 0 16
m u m t = 5.3 × 10 − 6 , m c m t = 2.6 × 10 − 3 , m d m b = 8.9 × 10 − 4 , \frac{m_u}{m_t} = 5.3 \times10^{-6}, \qquad
\frac{m_c}{m_t} = 2.6 \times10^{-3}, \qquad
\frac{m_d}{m_b} = 8.9 \times10^{-4}, m t m u = 5.3 × 1 0 − 6 , m t m c = 2.6 × 1 0 − 3 , m b m d = 8.9 × 1 0 − 4 , m s m b = 1.6 × 10 − 2 , m e m μ = 2.8 × 10 − 4 , m μ m τ = 5.9 × 10 − 2 , \frac{m_s}{m_b} = 1.6 \times10^{-2}, \qquad
\frac{m_e}{m_\mu} = 2.8 \times10^{-4}, \qquad
\frac{m_\mu}{m_\tau} = 5.9 \times10^{-2}, m b m s = 1.6 × 1 0 − 2 , m μ m e = 2.8 × 1 0 − 4 , m τ m μ = 5.9 × 1 0 − 2 , the ratios of the running masses at that scale in the minimal supersymmetric standard model with tan β = 10 \tan\beta= 10 tan β = 10 [38 Table V] ; the Cabibbo element satisfies 0.15 ≤ ∣ V u s ∣ ≤ 0.30 0.15 \leq\lvert V_{us}\rvert\leq0.30 0.15 ≤ ∣ V u s ∣ ≤ 0.30 tan β \tan\beta tan β [38 ] and the threshold corrections at the matching scales.
C 9 W C_{9W} C 9 W [ 10 3 , 10 6 ] [10^3,10^6] [ 1 0 3 , 1 0 6 ] m 0 m_0 m 0 [40 ] .
Conjecture 9.2 (GSC(W)). There is a complete datum, in the sense of the list above, satisfying C1–C6 and C 7 W C_{7W} C 7 W C 9 W C_{9W} C 9 W
Remark 9.3 (Dependence on the windows). Only a positive certificate depends on W W W § 16 , violates C 7 W C_{7W} C 7 W m 0 > 0 m_0 > 0 m 0 > 0 C 8 W C_{8W} C 8 W
A marked-point selection construction
The functional conditions have a clean differential-topological answer. They do not manufacture the point that is to be selected.
Theorem 10.1 (Selection with a supplied point). Let M M M μ \mu μ D D D p ∈ M p \in M p ∈ M
(i) p ∉ D p \notin D p ∈ / D μ p = p \mu p = p μ p = p
(ii) There is a nonnegative smooth globally proper function S S S M M M p p p S ∘ μ = S S \circ\mu= S S ∘ μ = S D D D
The function can be chosen to have Hessian 2 g p 2g_p 2 g p p p p
Proof. An invariant function with unique zero must fix that zero under every symmetry defined there. This proves necessity.
Every second-countable smooth manifold has a nonnegative smooth proper exhaustion ρ \rho ρ K j ⊂ int K j + 1 K_j \subset\operatorname{int} K_{j+1} K j ⊂ int K j + 1 K j K_j K j j j j K j K_j K j [31 ] .
On a component preserved by μ \mu μ ρ \rho ρ ρ + ρ ∘ μ \rho+\rho\circ\mu ρ + ρ ∘ μ ρ \rho ρ p p p
Choose a normal ball B r ( p ) B_r(p) B r ( p ) f ( x ) = d ( p , x ) 2 f(x)=d(p,x)^2 f ( x ) = d ( p , x ) 2 p ∈ D p\in D p ∈ D f f f θ : [ 0 , ∞ ) → [ 0 , 1 ] \theta:[0,\infty)\to[0,1] θ : [ 0 , ∞ ) → [ 0 , 1 ] t ≤ r 2 / 4 t\le r^2/4 t ≤ r 2 /4 t ≥ r 2 / 2 t\ge r^2/2 t ≥ r 2 /2 p p p
S = θ ( f ) f + ( 1 − θ ( f ) ) ( 1 + ρ ) , S=\theta(f)f+(1-\theta(f))(1+\rho), S = θ ( f ) f + ( 1 − θ ( f )) ( 1 + ρ ) , extending the expression by 1 + ρ 1+\rho 1 + ρ θ ( f ) \theta(f) θ ( f ) p p p p p p f f f p p p 2 g p 2g_p 2 g p
On the remaining components use their exhaustion plus the positive orbit offset. For a fixed sublevel only finitely many offsets occur. On the distinguished component a sublevel is a closed subset of B r ( p ) ∪ { ρ ≤ t − 1 } B_r(p)\cup\{\rho\le t-1\} B r ( p ) ∪ { ρ ≤ t − 1 } S S S
Remark 10.2 (What the point supplies). The construction uses the marked point p p p p p p p p p
Proposition 10.3 (An intrinsic selector must respect all isometries). Suppose a rule assigns S g S_g S g S g S_g S g p p p p p p
Proof. Naturality for an isometry h h h S g ( h x ) = S g ( x ) S_g(hx)=S_g(x) S g ( h x ) = S g ( x ) h p hp h p p p p
This is an obstruction to a particular notion of metric-only selection, not a proof that the actual heterotic moduli metric is homogeneous. It also shows why invariance under an additional compact group cannot be discarded merely because the partial mirror map is undefined at p p p
Index constraints and the mirror restriction
Proposition 11.1 (Index and finite descent). Let X X X V V V c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0
χ ( X , V ) = 1 2 ∫ X c 3 ( V ) . \chi(X,V)=\frac{1}{2}\int_X c_3(V). χ ( X , V ) = 2 1 ∫ X c 3 ( V ) . If π : X → Y \pi:X\to Y π : X → Y d d d V = π ∗ V Y V=\pi^*V_Y V = π ∗ V Y χ ( X , V ) = d χ ( Y , V Y ) \chi(X,V)=d\chi(Y,V_Y) χ ( X , V ) = d χ ( Y , V Y ) ∣ ∫ X c 3 ( V ) ∣ = 6 d \left|\int_X c_3(V)\right|=6d ∫ X c 3 ( V ) = 6 d
Proof. In Hirzebruch–Riemann–Roch, the degree-six term of ch ( V ) td ( X ) \operatorname{ch}(V)\operatorname{td}(X) ch ( V ) td ( X ) ch 3 ( V ) + c 1 ( V ) c 2 ( X ) / 12 + rk ( V ) c 1 ( X ) c 2 ( X ) / 24 \operatorname{ch}_3(V)+c_1(V)c_2(X)/12+\operatorname{rk}(V)c_1(X)c_2(X)/24 ch 3 ( V ) + c 1 ( V ) c 2 ( X ) /12 + rk ( V ) c 1 ( X ) c 2 ( X ) /24 ch 3 ( V ) = c 3 ( V ) / 2 \operatorname{ch}_3(V)=c_3(V)/2 ch 3 ( V ) = c 3 ( V ) /2 d d d
For a quintic, expansion of ( 1 + H ) 5 / ( 1 + 5 H ) (1+H)^5/(1+5H) ( 1 + H ) 5 / ( 1 + 5 H ) c 3 ( T X ) = − 40 H 3 c_3(T_X)=-40H^3 c 3 ( T X ) = − 40 H 3 ∫ X H 3 = 5 \int_X H^3=5 ∫ X H 3 = 5 ∫ c 3 ( T X ) = − 200 \int c_3(T_X)=-200 ∫ c 3 ( T X ) = − 200 100 = 3 d 100=3d 100 = 3 d
Under the standard-type partial mirror convention of [28 ] , a fixed point has h 1 , 1 = h 2 , 1 h^{1,1}=h^{2,1} h 1 , 1 = h 2 , 1 χ ( X ) = 0 \chi(X)=0 χ ( X ) = 0 ∫ c 3 ( V ) = χ ( X ) \int c_3(V)=\chi(X) ∫ c 3 ( V ) = χ ( X ) Proposition 11.1 gives zero net families. This proves that a selected viable point cannot be a mirror-fixed point in that specified domain. It does not exclude nonstandard data outside the domain, nor establish a mirror map on all ( 0 , 2 ) (0,2) ( 0 , 2 )
The Dwork pencil
Proposition 11.2. (i) For ψ ≠ 0 \psi\ne0 ψ = 0 ψ 5 ≠ 1 \psi^5\ne1 ψ 5 = 1 h h h X ψ X_\psi X ψ 1 1 1 ψ \psi ψ ψ 2 \psi^2 ψ 2 Theorem 1.3 are ∣ ψ ∣ r ( b ) |\psi|^{r(b)} ∣ ψ ∣ r ( b ) ( g b g σ b ) − 1 / 2 (g_bg_{\sigma b})^{-1/2} ( g b g σ b ) − 1/2
(ii) Let X X X H H H X X X X / H X/H X / H 100 / ∣ H ∣ 100/|H| 100/∣ H ∣ X / H X/H X / H H H H ( X / H , T X / H ) (X/H,T_{X/H}) ( X / H , T X / H )
(iii) On the sextic, octic and dectic pencils of § 9 , with ψ ≠ 0 \psi\ne0 ψ = 0 ψ δ ≠ 1 \psi^\delta\ne1 ψ δ = 1
Proof. Part (i) is Theorem 1.3 . For (ii), a smooth quintic has h 1 , 1 = 1 h^{1,1}=1 h 1 , 1 = 1 h 2 , 1 = 101 h^{2,1}=101 h 2 , 1 = 101 χ ( X ) = − 200 \chi(X)=-200 χ ( X ) = − 200 1 / ∣ H ∣ 1/|H| 1/∣ H ∣ χ ( X / H ) = − 200 / ∣ H ∣ \chi(X/H)=-200/|H| χ ( X / H ) = − 200/∣ H ∣ Theorem 2.1 the net number of families is ∣ χ ( X / H ) ∣ / 2 = 100 / ∣ H ∣ |\chi(X/H)|/2=100/|H| ∣ χ ( X / H ) ∣/2 = 100/∣ H ∣ X / H X/H X / H 100 / ∣ H ∣ = 3 100/|H|=3 100/∣ H ∣ = 3 8.9 .
Free quotients of the one-parameter families
The four pencils are four of the fourteen one-parameter families of Calabi–Yau threefolds with h 1 , 1 = 1 h^{1,1}=1 h 1 , 1 = 1 [24 ] . Thirteen of them are complete intersections X = X d 1 , … , d c ⊂ P ( w ) X=X_{d_1,\ldots,d_c}\subset\mathbb{P}(w) X = X d 1 , … , d c ⊂ P ( w ) d 1 , … , d c d_1,\ldots,d_c d 1 , … , d c ∑ j d j = ∑ i w i \sum_j d_j=\sum_i w_i ∑ j d j = ∑ i w i P ( w ) \mathbb{P}(w) P ( w ) Table 5 . The fourteenth, X 2 , 12 ⊂ P ( 1 , 1 , 1 , 1 , 4 , 6 ) X_{2,12}\subset\mathbb{P}(1,1,1,1,4,6) X 2 , 12 ⊂ P ( 1 , 1 , 1 , 1 , 4 , 6 ) P ( 4 , 6 ) \mathbb{P}(4,6) P ( 4 , 6 ) Proposition 11.2 becomes a statement about every free quotient, through one arithmetic invariant.
Table 5. The thirteen one-parameter complete intersections. The columns give the type, the weights, the intersection numbers, the Euler characteristic, the values χ ( X , k L ) = k 3 L 3 / 6 + k c 2 ⋅ L / 12 \chi(X,kL)=k^3L^3/6+k\,c_2\cdot L/12 χ ( X , k L ) = k 3 L 3 /6 + k c 2 ⋅ L /12 k = 1 , 2 , 3 k=1,2,3 k = 1 , 2 , 3 g X g_X g X ∣ χ ∣ / 6 |\chi|/6 ∣ χ ∣/6
Table 5. The thirteen one-parameter complete intersections. The columns give the type, the weights, the intersection numbers, the Euler characteristic, the values χ ( X , k L ) = k 3 L 3 / 6 + k c 2 ⋅ L / 12 \chi(X,kL)=k^3L^3/6+k\,c_2\cdot L/12 χ ( X , k L ) = k 3 L 3 /6 + k c 2 ⋅ L /12 k = 1 , 2 , 3 k=1,2,3 k = 1 , 2 , 3 g X g_X g X ∣ χ ∣ / 6 |\chi|/6 ∣ χ ∣/6
Lemma 12.1 (Primes of a free action). Let X X X K X ≅ O X K_X\cong\mathcal{O}_X K X ≅ O X h 1 , 0 = h 2 , 0 = 0 h^{1,0}=h^{2,0}=0 h 1 , 0 = h 2 , 0 = 0 b 2 = 1 b_2=1 b 2 = 1 π 1 ( X ) = 0 \pi_1(X)=0 π 1 ( X ) = 0 L L L X X X g g g p p p p p p χ ( X , k L ) \chi(X,kL) χ ( X , k L ) k ≥ 1 k\ge1 k ≥ 1
Proof. Since X X X H 2 ( X , Z ) H^2(X,\mathbb{Z}) H 2 ( X , Z ) h 0 , 1 = h 0 , 2 = 0 h^{0,1}=h^{0,2}=0 h 0 , 1 = h 0 , 2 = 0 Pic X \operatorname{Pic}X Pic X H 2 ( X , Z ) ≅ Z H^2(X,\mathbb{Z})\cong\mathbb{Z} H 2 ( X , Z ) ≅ Z A A A L = A a L=A^a L = A a a ≥ 1 a\ge1 a ≥ 1 g ∗ A g^*A g ∗ A g ∗ A ≅ A g^*A\cong A g ∗ A ≅ A G = ⟨ g ⟩ G=\langle g\rangle G = ⟨ g ⟩ A A A H 2 ( G , C ∗ ) H^2(G,\mathbb{C}^*) H 2 ( G , C ∗ ) A A A G G G G G G H q ( X , A m ) H^q(X,A^m) H q ( X , A m ) m ≥ 1 m\geq1 m ≥ 1 A m = K X ⊗ A m A^m=K_X\otimes A^m A m = K X ⊗ A m A m A^m A m 1 ≤ j < p 1\leq j<p 1 ≤ j < p g j g^j g j [22 ] gives ∑ q ( − 1 ) q tr ( g j ∣ H q ( X , A m ) ) = 0 \sum_q(-1)^q\operatorname{tr}\left(g^j\mid H^q(X,A^m)\right)=0 ∑ q ( − 1 ) q tr ( g j ∣ H q ( X , A m ) ) = 0 tr ( g j ∣ H 0 ( X , A m ) ) = 0 \operatorname{tr}\left(g^j\mid H^0(X,A^m)\right)=0 tr ( g j ∣ H 0 ( X , A m ) ) = 0 Z / p \mathbb{Z}/p Z / p p p p h 0 ( X , A m ) = χ ( X , A m ) h^0(X,A^m)=\chi(X,A^m) h 0 ( X , A m ) = χ ( X , A m ) m = k a m=ka m = k a
Proposition 12.2. Let X ⊂ P ( w ) X\subset\mathbb{P}(w) X ⊂ P ( w ) Table 5 that misses the singular points of P ( w ) \mathbb{P}(w) P ( w ) H H H X X X ∣ H ∣ |H| ∣ H ∣ g X g_X g X H H H X 6 X_6 X 6 X 10 X_{10} X 10 X 3 , 2 , 2 X_{3,2,2} X 3 , 2 , 2 X 6 , 6 X_{6,6} X 6 , 6 X 4 , 3 X_{4,3} X 4 , 3 X 6 , 4 X_{6,4} X 6 , 4 X 6 , 2 X_{6,2} X 6 , 2 X 5 X_5 X 5 X 3 , 3 X_{3,3} X 3 , 3 X 8 X_8 X 8 X 4 , 2 X_{4,2} X 4 , 2 X 2 , 2 , 2 , 2 X_{2,2,2,2} X 2 , 2 , 2 , 2 X 4 , 4 X_{4,4} X 4 , 4 X X X
Proof. As in Lemma 8.1, X X X K X ≅ O X ( ∑ j d j − ∑ i w i ) = O X K_X\cong\mathcal{O}_X(\sum_jd_j-\sum_iw_i)=\mathcal{O}_X K X ≅ O X ( ∑ j d j − ∑ i w i ) = O X [21 ] gives h 1 , 0 = h 2 , 0 = 0 h^{1,0}=h^{2,0}=0 h 1 , 0 = h 2 , 0 = 0 b 2 = 1 b_2=1 b 2 = 1 π 1 \pi_1 π 1 C ⊂ C N C\subset\mathbb{C}^N C ⊂ C N X X X c c c C C C [23 ] its link is 2-connected; the weighted action of R > 0 \mathbb{R}_{>0} R > 0 C ∖ { 0 } C\setminus\{0\} C ∖ { 0 } X X X C ∗ \mathbb{C}^* C ∗ C ∖ { 0 } C\setminus\{0\} C ∖ { 0 } X X X π 1 ( X ) = 0 \pi_1(X)=0 π 1 ( X ) = 0
Let p p p ∣ H ∣ |H| ∣ H ∣ g ∈ H g\in H g ∈ H p p p L = O X ( 1 ) L=\mathcal{O}_X(1) L = O X ( 1 ) p p p χ ( X , k L ) \chi(X,kL) χ ( X , k L ) k ≥ 1 k\geq1 k ≥ 1 c 1 ( X ) = 0 c_1(X)=0 c 1 ( X ) = 0 χ ( O X ) = 0 \chi(\mathcal{O}_X)=0 χ ( O X ) = 0 χ ( X , k L ) = k 3 L 3 / 6 + k c 2 ⋅ L / 12 \chi(X,kL)=k^3L^3/6+k\,c_2\cdot L/12 χ ( X , k L ) = k 3 L 3 /6 + k c 2 ⋅ L /12 α 1 ( k 1 ) + α 2 ( k 2 ) + α 3 ( k 3 ) \alpha_1\binom{k}{1}+\alpha_2\binom{k}{2}+\alpha_3\binom{k}{3} α 1 ( 1 k ) + α 2 ( 2 k ) + α 3 ( 3 k ) α i \alpha_i α i k ≥ 1 k\geq1 k ≥ 1 α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α 1 , α 2 , α 3 k = 1 , 2 , 3 k=1,2,3 k = 1 , 2 , 3 g X g_X g X c ( T X ) = ∏ i ( 1 + w i L ) ∏ j ( 1 + d j L ) − 1 c(T_X)=\prod_i(1+w_iL)\prod_j(1+d_jL)^{-1} c ( T X ) = ∏ i ( 1 + w i L ) ∏ j ( 1 + d j L ) − 1 L 3 = ∏ j d j / ∏ i w i L^3=\prod_jd_j/\prod_iw_i L 3 = ∏ j d j / ∏ i w i Theorem 2.1 the quotient X / H X/H X / H ∣ χ ( X ) ∣ / ( 2 ∣ H ∣ ) |\chi(X)|/(2|H|) ∣ χ ( X ) ∣/ ( 2∣ H ∣ ) ∣ H ∣ = ∣ χ ( X ) ∣ / 6 |H|=|\chi(X)|/6 ∣ H ∣ = ∣ χ ( X ) ∣/6 X 5 X_5 X 5 X 8 X_8 X 8 X 4 , 2 X_{4,2} X 4 , 2 X 2 , 2 , 2 , 2 X_{2,2,2,2} X 2 , 2 , 2 , 2 X 6 , 2 X_{6,2} X 6 , 2 g X g_X g X
The proposition contains Theorem 1.4(iv) and Proposition 11.2(ii) and strengthens them: the sextic and dectic pencils admit no free action at all, and on a quintic every free action is by a 5-group, so that only the orders 5 and 25 divide ∣ χ ∣ = 200 |\chi|=200 ∣ χ ∣ = 200 Z / 5 × Z / 5 \mathbb{Z}/5 \times\mathbb{Z}/5 Z /5 × Z /5 [1 ] . For the five complete intersections in ordinary projective space the free actions were classified by Braun [25 ] with a computer search; the proposition uses only the holomorphic Lefschetz formula and Riemann–Roch.5
One fixed nonstandard threefold
We now replace a generic existence assertion by a single specified complex structure. The line-bundle data are the example of Buchbinder, Constantin and Lukas [29, Section 5.1]; neither the bundle nor its phenomenological interpretation is claimed here as a new discovery. The purpose of fixing the equation is to make every subsequent calculation refer to the same geometric datum.
In A = ( P 1 ) 4 A=(\mathbb{P}^{1})^{4} A = ( P 1 ) 4 [ x i : y i ] [x_i:y_i] [ x i : y i ] 1 ≤ i ≤ 4 1\leq i\leq4 1 ≤ i ≤ 4 S ⊂ { 1 , 2 , 3 , 4 } S\subset\{1,2,3,4\} S ⊂ { 1 , 2 , 3 , 4 }
m S = ∏ i ∈ S y i 2 ∏ i ∉ S x i 2 , ( c 0 , c 1 , c 2 , c 3 , c 4 ) = ( 1 , 2 , 5 , 2 , 1 ) , F 0 = ∑ S c ∣ S ∣ m S . m_S=\prod_{i\in S}y_i^2\prod_{i\notin S}x_i^2,\qquad(c_0,c_1,c_2,c_3,c_4)=(1,2,5,2,1),\qquad F_0=\sum_S c_{|S|}m_S. m S = i ∈ S ∏ y i 2 i ∈ / S ∏ x i 2 , ( c 0 , c 1 , c 2 , c 3 , c 4 ) = ( 1 , 2 , 5 , 2 , 1 ) , F 0 = S ∑ c ∣ S ∣ m S . Write r i = x i y i r_i=x_i y_i r i = x i y i e i = x i 2 + y i 2 e_i=x_i^2+y_i^2 e i = x i 2 + y i 2 ( 12 ) , ( 13 ) , ( 14 ) , ( 23 ) , ( 24 ) , ( 34 ) (12),(13),(14),(23),(24),(34) ( 12 ) , ( 13 ) , ( 14 ) , ( 23 ) , ( 24 ) , ( 34 ) ( a i j ) = ( 1 , 2 , 3 , 5 , 7 , 11 ) (a_{ij})=(1,2,3,5,7,11) ( a ij ) = ( 1 , 2 , 3 , 5 , 7 , 11 )
G = ∑ i < j a i j r i r j e k e l + 13 r 1 r 2 r 3 r 4 , F θ = F 0 + θ G , X = X π = { F π = 0 } , (11) G=\sum_{i<j}a_{ij}r_i r_j e_k e_l+13r_1r_2r_3r_4,\qquad F_\theta=F_0+\theta G,\qquad X=X_\pi=\{F_\pi=0\}, \tag*{(11)} G = i < j ∑ a ij r i r j e k e l + 13 r 1 r 2 r 3 r 4 , F θ = F 0 + θ G , X = X π = { F π = 0 } , (11) where { k , l } \{k,l\} { k , l } { i , j } \{i,j\} { i , j } π \pi π
Lemma 13.1 (Smooth seed). The hypersurface F 0 = 0 F_0=0 F 0 = 0 C \mathbb{C} C
Proof. On a coordinate stratum set the remaining nonzero affine coordinates t i = y i / x i t_i=y_i/x_i t i = y i / x i u i = t i 2 u_i=t_i^2 u i = t i 2 k k k 4 − k − r 4-k-r 4 − k − r
f k , r ( u ) = ∑ j = 0 r c k + j e j ( u 1 , … , u r ) , 0 ≤ k , k + r ≤ 4 , f_{k,r}(u)=\sum_{j=0}^{r}c_{k+j}e_j(u_1,\ldots,u_r),\qquad0\leq k,\quad k+r\leq4, f k , r ( u ) = j = 0 ∑ r c k + j e j ( u 1 , … , u r ) , 0 ≤ k , k + r ≤ 4 , with e j e_j e j r r r ∂ / ∂ t i = 2 t i ∂ / ∂ u i \partial/\partial t_i=2t_i\partial/\partial u_i ∂ / ∂ t i = 2 t i ∂ / ∂ u i t i ≠ 0 t_i\ne0 t i = 0 f k , r f_{k,r} f k , r
For r = 0 r=0 r = 0 r = 1 r=1 r = 1 r = 2 r=2 r = 2 c k c k + 2 − c k + 1 2 c_kc_{k+2}-c_{k+1}^2 c k c k + 2 − c k + 1 2 1 , − 21 , 1 1,-21,1 1 , − 21 , 1
For r = 3 , k = 0 r=3,k=0 r = 3 , k = 0 ( u j − u i ) ( 5 + 2 u l ) (u_j-u_i)(5+2u_l) ( u j − u i ) ( 5 + 2 u l ) − 5 / 2 -5/2 − 5/2 − 21 / 2 -21/2 − 21/2 s s s ( u j − u i ) ( 2 + u l ) (u_j-u_i)(2+u_l) ( u j − u i ) ( 2 + u l ) − 2 -2 − 2
q 0 , 3 = 1 + 6 s + 15 s 2 + 2 s 3 , q 1 , 3 = 2 + 15 s + 6 s 2 + s 3 . q_{0,3}=1+6s+15s^2+2s^3,\qquad q_{1,3}=2+15s+6s^2+s^3. q 0 , 3 = 1 + 6 s + 15 s 2 + 2 s 3 , q 1 , 3 = 2 + 15 s + 6 s 2 + s 3 . Both are relatively prime to their derivatives, as the explicit Bézout identities in Appendix § 30 show.
For r = 4 r=4 r = 4 f = f 0 , 4 f=f_{0,4} f = f 0 , 4 Q ( x , y ) = 5 + 2 ( x + y ) + x y = ( x + 2 ) ( y + 2 ) + 1 Q(x,y)=5+2(x+y)+xy=(x+2)(y+2)+1 Q ( x , y ) = 5 + 2 ( x + y ) + x y = ( x + 2 ) ( y + 2 ) + 1
f i − f j = ( u j − u i ) Q ( u k , u l ) . f_i-f_j=(u_j-u_i)Q(u_k,u_l). f i − f j = ( u j − u i ) Q ( u k , u l ) . Classify the multiplicities of four coordinates. If all are equal, the polynomial is q 0 , 4 = 1 + 8 s + 30 s 2 + 8 s 3 + s 4 q_{0,4}=1+8s+30s^2+8s^3+s^4 q 0 , 4 = 1 + 8 s + 30 s 2 + 8 s 3 + s 4 ( 3 , 1 ) (3,1) ( 3 , 1 ) ( x , x , x , y ) (x,x,x,y) ( x , x , x , y ) x ≠ y x\ne y x = y Q ( x , x ) = x 2 + 4 x + 5 = 0 Q(x,x)=x^2+4x+5=0 Q ( x , x ) = x 2 + 4 x + 5 = 0 2 + 10 x + 2 x 2 + y Q ( x , x ) 2+10x+2x^2+yQ(x,x) 2 + 10 x + 2 x 2 + y Q ( x , x ) 2 ( x − 4 ) 2(x-4) 2 ( x − 4 ) Q ( x , x ) Q(x,x) Q ( x , x ) Q ( 4 , 4 ) = 37 Q(4,4)=37 Q ( 4 , 4 ) = 37
For multiplicities ( 2 , 1 , 1 ) (2,1,1) ( 2 , 1 , 1 ) Q ( x , x ) = Q ( x , z ) = 0 Q(x,x)=Q(x,z)=0 Q ( x , x ) = Q ( x , z ) = 0 x ≠ z x\ne z x = z ( x − z ) ( x + 2 ) = 0 (x-z)(x+2)=0 ( x − z ) ( x + 2 ) = 0 Q ( − 2 , − 2 ) = 1 Q(-2,-2)=1 Q ( − 2 , − 2 ) = 1 Q ( x , y ) = Q ( x , z ) = 0 Q(x,y)=Q(x,z)=0 Q ( x , y ) = Q ( x , z ) = 0 x = − 2 x=-2 x = − 2 Q ( x , y ) = 1 Q(x,y)=1 Q ( x , y ) = 1
For multiplicities ( 2 , 2 ) (2,2) ( 2 , 2 ) ( x , x , y , y ) (x,x,y,y) ( x , x , y , y ) x ≠ y x\ne y x = y Q ( x , y ) = 0 Q(x,y)=0 Q ( x , y ) = 0 y ≠ − 2 y\ne-2 y = − 2 x = − ( 2 y + 5 ) / ( y + 2 ) x=-(2y+5)/(y+2) x = − ( 2 y + 5 ) / ( y + 2 ) f f f
g = y 2 − 8 y − 21 , h = y 4 − 16 y 3 − 66 y 2 − 16 y + 89. g=y^2-8y-21,\qquad h=y^4-16y^3-66y^2-16y+89. g = y 2 − 8 y − 21 , h = y 4 − 16 y 3 − 66 y 2 − 16 y + 89. Their denominators are y + 2 y+2 y + 2 ( y + 2 ) 2 (y+2)^2 ( y + 2 ) 2 § 30 proves they cannot both vanish. These cases exhaust the strata and partitions. □
Let a a a [ x : y ] ↦ [ x : − y ] [x:y]\mapsto[x:-y] [ x : y ] ↦ [ x : − y ] b b b [ x : y ] ↦ [ y : x ] [x:y]\mapsto[y:x] [ x : y ] ↦ [ y : x ] PGL 2 \operatorname{PGL}_2 PGL 2 Γ ≅ ( Z / 2 ) 2 \Gamma\cong(\mathbb{Z}/2)^2 Γ ≅ ( Z /2 ) 2 F 0 F_0 F 0 G G G
Theorem 13.2 (The fixed quotient). The threefold X π X_\pi X π Γ \Gamma Γ Y = X π / Γ Y=X_\pi/\Gamma Y = X π /Γ − 32 -32 − 32 ( h 1 , 1 , h 2 , 1 ) = ( 4 , 20 ) (h^{1,1},h^{2,1})=(4,20) ( h 1 , 1 , h 2 , 1 ) = ( 4 , 20 )
Proof. The locus of singular fibres of the projective pencil (11) is a closed algebraic subset of the parameter line, defined over Q \mathbb{Q} Q A × A 1 A\times A^1 A × A 1 Lemma 13.1 , so it is finite and all its points are algebraic over Q \mathbb{Q} Q π \pi π
Each of a , b , a b a,b,ab a , b , ab { 0 , ∞ } \{0,\infty\} { 0 , ∞ } { 1 , − 1 } \{1,-1\} { 1 , − 1 } { i , − i } \{i,-i\} { i , − i } F 0 F_0 F 0 1 , 2 , 5 1,2,5 1 , 2 , 5 48 48 48 16 16 16 G G G F 0 + π G F_0+\pi G F 0 + π G G = 0 G=0 G = 0 π \pi π
Adjunction gives K X = O X K_X=\mathcal{O}_X K X = O X − 1 -1 − 1 1 1 1 h 1 , 0 ( Y ) = h 2 , 0 ( Y ) = 0 h^{1,0}(Y)=h^{2,0}(Y)=0 h 1 , 0 ( Y ) = h 2 , 0 ( Y ) = 0 c 3 ( T X ) = − 16 ∑ i < j < k J i J j J k c_3(T_X)=-16\sum_{i<j<k}J_iJ_jJ_k c 3 ( T X ) = − 16 ∑ i < j < k J i J j J k − 128 -128 − 128 χ ( Y ) = − 32 \chi(Y)=-32 χ ( Y ) = − 32 H 2 ( X , Q ) H^2(X,\mathbb{Q}) H 2 ( X , Q ) h 1 , 1 ( Y ) = 4 h^{1,1}(Y)=4 h 1 , 1 ( Y ) = 4 χ ( Y ) = 2 ( h 1 , 1 − h 2 , 1 ) \chi(Y)=2(h^{1,1}-h^{2,1}) χ ( Y ) = 2 ( h 1 , 1 − h 2 , 1 ) h 2 , 1 ( Y ) = 20 h^{2,1}(Y)=20 h 2 , 1 ( Y ) = 20
The bundle, its spectrum and the fixed Wilson line
On this X X X V = ⨁ a = 1 5 L a V=\bigoplus_{a=1}^{5}L_a V = ⨁ a = 1 5 L a L a = O X ( k a ) L_a=\mathcal{O}_X(k_a) L a = O X ( k a )
K = ( k a i ) = ( − 1 − 1 0 1 1 0 − 3 1 1 1 0 2 − 1 − 1 0 1 2 0 − 1 − 2 ) (12) K=(k_a^i)=
\begin{pmatrix}
-1 & -1 & 0 & 1 & 1\\
0 & -3 & 1 & 1 & 1\\
0 & 2 & -1 & -1 & 0\\
1 & 2 & 0 & -1 & -2
\end{pmatrix} \tag*{(12)} K = ( k a i ) = − 1 0 0 1 − 1 − 3 2 2 0 1 − 1 0 1 1 − 1 − 1 1 1 0 − 2 (12) Fix the Kähler class to be J = 100 ( J 1 + J 2 + J 3 + J 4 ) J = 100(J_1 + J_2 + J_3 + J_4) J = 100 ( J 1 + J 2 + J 3 + J 4 ) a , b a,b a , b
Proposition 14.1 (Chern classes and slopes). The bundle has c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 is slope-polystable at J J J and has
( ∫ X c 2 ( V ) J i ) = ( 24 , 8 , 20 , 12 ) , ∫ X c 3 ( V ) = − 24 , χ ( X , V ) = − 12. \left(\int_X c_2(V)J_i\right)=(24,8,20,12), \qquad\int_X c_3(V)=-24,\quad\chi(X,V)=-12. ( ∫ X c 2 ( V ) J i ) = ( 24 , 8 , 20 , 12 ) , ∫ X c 3 ( V ) = − 24 , χ ( X , V ) = − 12. The cover anomaly class is the effective class 6 J 1 J 3 + 2 J 1 J 4 6J_1J_3+2J_1J_4 6 J 1 J 3 + 2 J 1 J 4 The bundle descends to V Y V_Y V Y on Y Y Y with χ ( Y , V Y ) = − 3 \chi(Y,V_Y)=-3 χ ( Y , V Y ) = − 3
Proof. In the ambient ring J i 2 = 0 J_i^2=0 J i 2 = 0 2 ∑ i J i 2\sum_i J_i 2 ∑ i J i ∫ X J i J j J k = 2 \int_X J_iJ_jJ_k=2 ∫ X J i J j J k = 2 c 2 ( T X ) = 4 ∑ i < j J i J j c_2(T_X)=4\sum_{i<j}J_iJ_j c 2 ( T X ) = 4 ∑ i < j J i J j J i J_i J i c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 12 ⋅ 100 2 ∑ i k a i = 0 12\cdot100^2\sum_i k_a^i=0 12 ⋅ 10 0 2 ∑ i k a i = 0 l a = ∑ i k a i J i l_a=\sum_i k_a^iJ_i l a = ∑ i k a i J i c 2 ( V ) = − 1 2 ∑ a l a 2 c_2(V)=-\frac{1}{2}\sum_a l_a^2 c 2 ( V ) = − 2 1 ∑ a l a 2 c 3 ( V ) = 1 3 ∑ a l a 3 c_3(V)=\frac{1}{3}\sum_a l_a^3 c 3 ( V ) = 3 1 ∑ a l a 3
χ ( O X ( k ) ) = 2 ∑ i < j < l k i k j k l + 2 ∑ i k i \chi(\mathcal{O}_X(k))=2\sum_{i<j<l}k_ik_jk_l+2\sum_i k_i χ ( O X ( k )) = 2 i < j < l ∑ k i k j k l + 2 i ∑ k i are 0 , − 8 , 0 , 0 , − 4 0,-8,0,0,-4 0 , − 8 , 0 , 0 , − 4 ( 0 , 16 , 4 , 12 ) (0,16,4,12) ( 0 , 16 , 4 , 12 ) 6 J 1 J 3 + 2 J 1 J 4 6J_1J_3+2J_1J_4 6 J 1 J 3 + 2 J 1 J 4 H 4 ( X , Z ) H^4(X,\mathbb{Z}) H 4 ( X , Z ) H 2 H^2 H 2
The lifts of a , b a,b a , b O ( 1 ) \mathcal{O}(1) O ( 1 ) O ( k a ) \mathcal{O}(k_a) O ( k a )
( − 1 ) ∑ i k a i = 1 ; (-1)^{\sum_i k_a^i}=1; ( − 1 ) ∑ i k a i = 1 ; their squares are one. These are commuting equivariant structures. Their product determinant on V V V ∑ a k a = 0 \sum_a k_a=0 ∑ a k a = 0 étale descent gives χ ( Y , V Y ) = χ ( X , V ) / 4 = − 3 \chi(Y,V_Y)=\chi(X,V)/4=-3 χ ( Y , V Y ) = χ ( X , V ) /4 = − 3
Effectiveness of the anomaly on the cover proves rational effectiveness downstairs by pushforward and division by four. It does not by itself certify a quantized hidden sector or the integral torsion part of the anomaly on Y Y Y
An exact cohomology calculation
Let R Γ R_\Gamma R Γ 1 \mathbf{1} 1 ( a , b ) (a,b) ( a , b ) ( + + ) , ( + − ) , ( − + ) , ( − − ) (++),(+-),(-+),(--) ( + + ) , ( + − ) , ( − + ) , ( − − )
Theorem 14.2 (Cohomology at the specified complex structure). For the fixed choices above,
H 1 ( X , V ) = 3 R Γ , H 2 ( X , V ) = 0 , H^1(X,V)=3R_\Gamma,\qquad H^2(X,V)=0, H 1 ( X , V ) = 3 R Γ , H 2 ( X , V ) = 0 , H 1 ( X , Λ 2 V ) = 3 R Γ + ( R Γ − 1 ) , H 2 ( X , Λ 2 V ) = R Γ − 1. H^1(X,\Lambda^2V)=3R_\Gamma+(R_\Gamma-\mathbf{1}),\qquad H^2(X,\Lambda^2V)=R_\Gamma-\mathbf{1}. H 1 ( X , Λ 2 V ) = 3 R Γ + ( R Γ − 1 ) , H 2 ( X , Λ 2 V ) = R Γ − 1 . Here R Γ − 1 R_\Gamma-\mathbf{1} R Γ − 1 is the actual sum of the three nontrivial characters. The only nonzero individual dimensions are
h 1 ( L 2 ) = 8 , h 1 ( L 5 ) = 4 , h^1(L_2)=8,\qquad h^1(L_5)=4, h 1 ( L 2 ) = 8 , h 1 ( L 5 ) = 4 , h 1 ( L 2 L 4 ) = 4 , h 1 ( L 4 L 5 ) = 8 , h^1(L_2L_4)=4,\qquad h^1(L_4L_5)=8, h 1 ( L 2 L 4 ) = 4 , h 1 ( L 4 L 5 ) = 8 , h 1 ( L 2 L 5 ) = h 2 ( L 2 L 5 ) = 3. h^1(L_2L_5)=h^2(L_2L_5)=3. h 1 ( L 2 L 5 ) = h 2 ( L 2 L 5 ) = 3. Proof. For any multidegree k k k
0 ⟶ O A ( k − 2 ) → F π O A ( k ) ⟶ O X ( k ) ⟶ 0. 0\longrightarrow\mathcal{O}_A(k-2)\xrightarrow{F_\pi}\mathcal{O}_A(k)\longrightarrow\mathcal{O}_X(k)\longrightarrow0. 0 ⟶ O A ( k − 2 ) F π O A ( k ) ⟶ O X ( k ) ⟶ 0. On P 1 \mathbb{P}^1 P 1 H 0 ( O ( d ) ) H^0(\mathcal{O}(d)) H 0 ( O ( d )) d + 1 d+1 d + 1 d ≥ 0 d\geq0 d ≥ 0 H 1 ( O ( d ) ) H^1(\mathcal{O}(d)) H 1 ( O ( d )) − d − 1 -d-1 − d − 1 d ≤ − 2 d\leq-2 d ≤ − 2 d = − 1 d=-1 d = − 1 k k k L 2 L_2 L 2 H 2 H^2 H 2 L 5 L_5 L 5 H 1 H^1 H 1 L 2 L 4 L_2L_4 L 2 L 4 H 1 H^1 H 1 L 4 L 5 L_4L_5 L 4 L 5 H 2 H^2 H 2 L 2 L 5 = O X ( 0 , − 2 , 2 , 0 ) L_2L_5 = \mathcal{O}_X(0,-2,2,0) L 2 L 5 = O X ( 0 , − 2 , 2 , 0 ) H 1 H^1 H 1 H 3 H^3 H 3
The remaining product is L 1 L 2 = O X ( − 2 , − 3 , 2 , 3 ) L_1L_2 = \mathcal{O}_X(-2,-3,2,3) L 1 L 2 = O X ( − 2 , − 3 , 2 , 3 ) H 2 H^2 H 2 M ( θ ) = M 0 + θ M G M(\theta) = M_0 + \theta M_G M ( θ ) = M 0 + θ M G
det M ( 1 ) = 393460125696 ≠ 0. (13) \det M(1) = 393460125696 \ne0. \tag*{(13)} det M ( 1 ) = 393460125696 = 0. (13) This is a certificate that det M ( θ ) ∈ Z [ θ ] \det M(\theta) \in\mathbb{Z}[\theta] det M ( θ ) ∈ Z [ θ ] π \pi π M ( π ) M(\pi) M ( π ) L 1 L 2 L_1L_2 L 1 L 2 X 1 X_1 X 1
The character calculation is equally explicit. On a factor represent H 0 ( O ( d ) ) H^0(\mathcal{O}(d)) H 0 ( O ( d )) x d − v y v x^{d-v}y^v x d − v y v 0 ≤ v ≤ d 0 \le v \le d 0 ≤ v ≤ d H 1 ( O ( d ) ) H^1(\mathcal{O}(d)) H 1 ( O ( d )) d + 1 ≤ v ≤ − 1 d+1 \le v \le-1 d + 1 ≤ v ≤ − 1 a a a ( − 1 ) v (-1)^v ( − 1 ) v b b b v v v d − v d-v d − v H 1 H^1 H 1 2 R Γ 2R_\Gamma 2 R Γ R Γ R_\Gamma R Γ R Γ R_\Gamma R Γ 2 R Γ 2R_\Gamma 2 R Γ L 2 L 5 L_2L_5 L 2 L 5 ( 0 , 1 , 1 , 1 ) (0,1,1,1) ( 0 , 1 , 1 , 1 )
Projection of the charged matter
Fix
ρ ( a ) = diag ( 1 , 1 , 1 , − 1 , − 1 ) ∈ S U ( 5 ) , ρ ( b ) = I . \rho(a) = \operatorname{diag}(1,1,1,-1,-1) \in SU(5), \qquad\rho(b) = I. ρ ( a ) = diag ( 1 , 1 , 1 , − 1 , − 1 ) ∈ S U ( 5 ) , ρ ( b ) = I . The centralizer inside S U ( 5 ) SU(5) S U ( 5 ) S ( U ( 3 ) × U ( 2 ) ) S(U(3) \times U(2)) S ( U ( 3 ) × U ( 2 )) Γ \Gamma Γ R Γ − 1 R_\Gamma- 1 R Γ − 1 H 1 ( Λ 2 V ) H^1(\Lambda^2 V) H 1 ( Λ 2 V ) ( − , + ) (-,+) ( − , + ) ( + , + ) (+,+) ( + , + ) H 2 H^2 H 2 10 + 5 ‾ 10 + \overline{5} 10 + 5
A residual gauge factor at this same point
The preceding centralizer computation concerns only the S U ( 5 ) SU(5) S U ( 5 ) S ( U ( 1 ) 5 ) S(U(1)^5) S ( U ( 1 ) 5 ) K K K
q = ( − 4 , 1 , 6 , − 4 , 1 ) , K q = 0 , ∑ a q a = 0. (14) q = (-4,1,6,-4,1), \qquad Kq = 0, \qquad\sum_a q_a = 0. \tag*{(14)} q = ( − 4 , 1 , 6 , − 4 , 1 ) , K q = 0 , a ∑ q a = 0. (14) Thus one traceless Abelian combination is not removed by that map. A universal-axion row of the form β i k a i \beta_i k_a^i β i k a i q q q 10 10 10 L 2 , L 5 L_2,L_5 L 2 , L 5 − 3 -3 − 3 5 ‾ \overline{5} 5 L 2 L 4 , L 4 L 5 L_2L_4,L_4L_5 L 2 L 4 , L 4 L 5 ± 2 \pm2 ± 2
The Yukawa and stabilization tests at that point
This section keeps (11), (12), J J J
Proposition 15.1 (A perturbative rank obstruction). At the split-bundle point, the renormalizable perturbative up-type Yukawa matrix has rank at most 2, and the down-type matrix is zero. These statements hold after any invertible matter normalization.
Proof. Let e a e_a e a e 2 e_2 e 2 e 5 e_5 e 5 − e 2 − e 5 -e_2-e_5 − e 2 − e 5
Y u = ( 0 0 A 0 0 B A B 0 ) , Y_u =
\begin{pmatrix}
0 & 0 & A \\
0 & 0 & B \\
A & B & 0
\end{pmatrix}, Y u = 0 0 A 0 0 B A B 0 , which has rank at most two regardless of the values of A , B A,B A , B e 2 + e 4 e_2+e_4 e 2 + e 4 e 4 + e 5 e_4+e_5 e 4 + e 5 e 2 + e 5 e_2+e_5 e 2 + e 5 10 5 ‾ H d 10\,\overline{5}\,H_d 10 5 H d ( 1 , 1 , 1 , 1 , 1 ) (1,1,1,1,1) ( 1 , 1 , 1 , 1 , 1 ) S ( U ( 1 ) 5 ) S(U(1)^5) S ( U ( 1 ) 5 ) Y d = 0 Y_d=0 Y d = 0
Nonzero singlet expectation values, a non-Abelian bundle deformation, or computed nonperturbative contributions could change these matrices. None is part of the fixed point just audited. Their possible existence cannot turn its rank-deficient matrices into a verified C8 certificate.
For a leading-order unfluxed Calabi–Yau description, take vanishing classical superpotential on the holomorphic locus and vanishing charged expectation values. The slopes remain zero along J = λ i J i J=\lambda^iJ_i J = λ i J i λ > 0 \lambda>0 λ > 0 W = 0 W=0 W = 0
A targeted deformation beyond the split locus
The charge calculation indicates a concrete next construction, rather than a change of title. Consider the extensions
0 ⟶ L 2 ⟶ E ⟶ L 1 ⊕ L 3 ⟶ 0 , V ′ = E ⊕ L 4 ⊕ L 5 . 0 \longrightarrow L_2 \longrightarrow E \longrightarrow L_1 \oplus L_3 \longrightarrow0,\qquad V' = E \oplus L_4 \oplus L_5. 0 ⟶ L 2 ⟶ E ⟶ L 1 ⊕ L 3 ⟶ 0 , V ′ = E ⊕ L 4 ⊕ L 5 . Their parameters lie in H 1 ( L 2 L 1 − 1 ) ⊕ H 1 ( L 2 L 3 − 1 ) H^1(L_2L_1^{-1}) \oplus H^1(L_2L_3^{-1}) H 1 ( L 2 L 1 − 1 ) ⊕ H 1 ( L 2 L 3 − 1 ) H 1 H^1 H 1 H 2 H^2 H 2
The charges of these singlets under (14) are 5 and − 5 -5 − 5 E E E Λ 2 E \Lambda^2E Λ 2 E E ⊗ L i E\otimes L_i E ⊗ L i L 1 L_1 L 1 L 3 L_3 L 3 L 1 L 3 L_1L_3 L 1 L 3 L 1 L 2 L_1L_2 L 1 L 2 L__2L_3 and the indicated products with L 4 , L 5 L_4,L_5 L 4 , L 5
This does not give a replacement nine-condition certificate. At the original polarization all three L 1 , L 2 , L 3 L_1,L_2,L_3 L 1 , L 2 , L 3 L 2 L_2 L 2 κ i = ∫ X J i J 2 \kappa_i=\int_X J_iJ^2 κ i = ∫ X J i J 2 L 4 L_4 L 4 L 5 L_5 L 5 κ 3 = κ 4 \kappa_3=\kappa_4 κ 3 = κ 4 κ 1 + κ 2 = 2 κ 4 \kappa_1+\kappa_2=2\kappa_4 κ 1 + κ 2 = 2 κ 4 κ = ( c + s , c − s , c , c , c ) \kappa=(c+s,c-s,c,c,c) κ = ( c + s , c − s , c , c , c ) L 1 , L 3 , L 2 L_1,L_3,L_2 L 1 , L 3 , L 2 − s , − s , 2 s -s,-s,2s − s , − s , 2 s s < 0 s<0 s < 0
A stronger obstruction for analytic singlet insertions
The extension route above does not, by itself, repair the missing down-type matrix. In fact the charge obstruction persists for a larger specified class of local operators.
Proposition 15.2 (Analytic singlet obstruction). Consider a perturbative superpotential holomorphic near the split point in the singlet coordinates H 1 ( X , End V ) H^1(X,\operatorname{End} V) H 1 ( X , End V )
Proof. A charged singlet in H 1 ( L a L b − 1 ) H^1(L_aL_b^{-1}) H 1 ( L a L b − 1 ) e a − e b e_a-e_b e a − e b a = 1 a=1 a = 1 a = 3 a=3 a = 3 b ≠ a b\ne a b = a H 1 = 0 H^1=0 H 1 = 0 ( h 0 , h 1 , h 2 , h 3 ) (h^0,h^1,h^2,h^3) ( h 0 , h 1 , h 2 , h 3 )
Table 13 (PDF p. 28).
Every other cohomology degree in this table is zero. These facts follow from the ambient Koszul sequence without any multiplication-rank assumption: a degree − 1 -1 − 1
The charge Q Q Q 10 5 ‾ H d 10\,\overline{5}\,H_d 10 5 H d Q 1 = Q 3 = 0 Q_1=Q_3=0 Q 1 = Q 3 = 0 ∑ a Q a = 5 \sum_a Q_a=5 ∑ a Q a = 5 ( 1 , 1 , 1 , 1 , 1 ) (1,1,1,1,1) ( 1 , 1 , 1 , 1 , 1 )
This is an all-orders statement about the indicated analytic perturbative singlet expansion , not about every string contribution. Nonperturbative factors with axionic charges, a singular change of effective description, or a different topological bundle are outside its hypotheses. A new construction using them must calculate their actual coefficients and stabilize the resulting point. In particular the existence of an extension bundle alone cannot close C8 here.
The split branch on the tetraquadric
The ledger of the next section records two failures at the fixed datum that do not depend on the complex structure: the residual Abelian factor of Section 14.3 and the Yukawa deficiency of Proposition 15.1. This section asks whether another choice of five line bundles on the same manifold could avoid them while staying at the split locus. The first failure cannot be avoided by any sum of line bundles on any Calabi–Yau threefold with at most four Kähler parameters; the second is a property of the charge matrix, and the selection rules decide it in finitely many steps for each datum.
The residual Abelian factor is forced
Let X X X h = h 1 , 1 ( X ) h=h^{1,1}(X) h = h 1 , 1 ( X ) J 1 , … , J h J_1,\ldots,J_h J 1 , … , J h H 1 , 1 ( X , Q ) H^{1,1}(X,\mathbb{Q}) H 1 , 1 ( X , Q ) V = ⨁ a = 1 n L a V=\bigoplus_{a=1}^{n}L_a V = ⨁ a = 1 n L a c 1 ( L a ) = ∑ i k a i J i c_1(L_a)=\sum_i k_a^iJ_i c 1 ( L a ) = ∑ i k a i J i K = ( k a i ) ∈ Z h × n K=(k_a^i)\in\mathbb{Z}^{h\times n} K = ( k a i ) ∈ Z h × n k a k_a k a J J J κ i ( J ) = ∫ X J i J 2 \kappa_i(J)=\int_XJ_iJ^2 κ i ( J ) = ∫ X J i J 2
Theorem 16.1 (The residual factor). Suppose c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 V V V J J J
rank K ≤ h − 1 \operatorname{rank}K\le h-1 rank K ≤ h − 1
(ii) the traceless kernel { q ∈ Q n : K q = 0 , ∑ a q a = 0 } \{q \in\mathbb{Q}^{n}: Kq = 0,\ \sum_a q_a = 0\} { q ∈ Q n : K q = 0 , ∑ a q a = 0 } n − h n-h n − h
(iii) at the split locus, with vanishing expectation values of all charged singlets, exactly n − 1 − rank K ≥ n − h n-1-\operatorname{rank} K \ge n-h n − 1 − rank K ≥ n − h n − 1 n-1 n − 1 S ( U ( 1 ) n ) S(U(1)^n) S ( U ( 1 ) n )
In particular, for n = 5 n=5 n = 5 h ≤ 4 h \le4 h ≤ 4 X X X
Proof. The slope of L a L_a L a μ J ( L a ) = ∫ X c 1 ( L a ) J 2 = ∑ i k a i κ i \mu_J(L_a)=\int_X c_1(L_a)J^2=\sum_i k_a^i\kappa_i μ J ( L a ) = ∫ X c 1 ( L a ) J 2 = ∑ i k a i κ i μ J ( V ) \mu_J(V) μ J ( V ) μ J ( V ) = ∫ X c 1 ( V ) J 2 = 0 \mu_J(V)=\int_X c_1(V)J^2=0 μ J ( V ) = ∫ X c 1 ( V ) J 2 = 0 K T κ = 0 K^T\kappa=0 K T κ = 0 κ \kappa κ ∑ i t i κ i = ∫ X J 3 > 0 \sum_i t_i\kappa_i=\int_X J^3>0 ∑ i t i κ i = ∫ X J 3 > 0 J = ∑ i t i J i J=\sum_i t_iJ_i J = ∑ i t i J i K K K rank K ≤ h − 1 \operatorname{rank} K\le h-1 rank K ≤ h − 1 c 1 ( V ) = ∑ a c 1 ( L a ) = 0 c_1(V)=\sum_a c_1(L_a)=0 c 1 ( V ) = ∑ a c 1 ( L a ) = 0 1 = ( 1 , … , 1 ) \mathbf{1}=(1,\ldots,1) 1 = ( 1 , … , 1 ) ker K \ker K ker K n − rank K ≥ n − h + 1 n-\operatorname{rank} K\ge n-h+1 n − rank K ≥ n − h + 1 ∑ a q a = 0 \sum_a q_a=0 ∑ a q a = 0 1 \mathbf{1} 1 ker K \ker K ker K n − h n-h n − h
For (iii) we use the Green–Schwarz masses of the Abelian gauge bosons as computed by Anderson, Gray, Lukas and Palti [33 Section 2.7] . The axions of the h h h U ( 1 ) U(1) U ( 1 ) k a i k_a^i k a i β i k a i \beta_i k_a^i β i k a i A a A_a A a n n n M a b = k a T G ~ k b M_{ab}=k_a^T\widetilde{G}k_b M ab = k a T G k b G ~ = G + β β T / ( 4 s 2 ) \widetilde{G}=G+\beta\beta^T/(4s^2) G = G + β β T / ( 4 s 2 ) G G G s s s G ~ \widetilde{G} G q ↦ q T M q = ( K q ) T G ~ ( K q ) q\mapsto q^TMq=(Kq)^T\widetilde{G}(Kq) q ↦ q T M q = ( K q ) T G ( K q ) ker K \ker K ker K ∑ a q a A a \sum_a q_aA_a ∑ a q a A a K q = 0 Kq=0 K q = 0 S ( U ( 1 ) n ) S(U(1)^n) S ( U ( 1 ) n ) n − 1 − rank K n-1-\operatorname{rank} K n − 1 − rank K [33 eq. (2.31)] , and (i) gives the lower bound n − h n-h n − h [33 eq. (2.32)] that fewer than five Kähler moduli leave at least one U ( 1 ) U(1) U ( 1 ) q q q ker K \ker K ker K
Remark 16.2 (Reading the fixed datum). For the matrix (12), rank K = 3 = h − 1 \operatorname{rank} K=3=h-1 rank K = 3 = h − 1 ker K \ker K ker K 1 \mathbf{1} 1 q q q § 15.3 is therefore not a defect of the choice (12); it is the minimum that the tetraquadric, with h 1 , 1 = 4 h^{1,1}=4 h 1 , 1 = 4 K T κ = 0 K^T\kappa=0 K T κ = 0 κ \kappa κ h − rank K = 1 h-\operatorname{rank} K=1 h − rank K = 1 ( 1 , 1 , 1 , 1 ) (1,1,1,1) ( 1 , 1 , 1 , 1 ) J = ∑ i t i J i J=\sum_i t_iJ_i J = ∑ i t i J i t i > 0 t_i>0 t i > 0 Proposition 14.1 give κ i = 4 e 2 ( t j : j ≠ i ) = 4 e 2 ( t ) − t i ( e 1 ( t ) − t i ) \kappa_i=4e_2(t_j:j\ne i)=4e_2(t)-t_i(e_1(t)-t_i) κ i = 4 e 2 ( t j : j = i ) = 4 e 2 ( t ) − t i ( e 1 ( t ) − t i ) e 1 , e 2 e_1,e_2 e 1 , e 2 κ i \kappa_i κ i t i ( e 1 ( t ) − t i ) t_i(e_1(t)-t_i) t i ( e 1 ( t ) − t i ) t i t_i t i x ( e 1 − x ) = c x(e_1-x)=c x ( e 1 − x ) = c t i ∈ { c ′ , e 1 − c ′ } t_i\in\{c',e_1-c'\} t i ∈ { c ′ , e 1 − c ′ } c ′ c' c ′ m m m c ′ c' c ′ 4 − m 4-m 4 − m e 1 − c ′ e_1-c' e 1 − c ′ ( m − 3 ) e 1 = ( 2 m − 4 ) c ′ (m-3)e_1=(2m-4)c' ( m − 3 ) e 1 = ( 2 m − 4 ) c ′ t i > 0 t_i>0 t i > 0 m = 0 m=0 m = 0 m = 4 m=4 m = 4 J = λ ∑ i J i J=\lambda\sum_iJ_i J = λ ∑ i J i § 16 is the whole locus of Kähler classes at which (12) is polystable, and the flat classical direction of C7 and the massless generator of C2 are two readings of the one equality 4 − rank K = 1 4-\operatorname{rank} K=1 4 − rank K = 1 n = h + 1 n=h+1 n = h + 1 n − 1 − rank K n-1-\operatorname{rank} K n − 1 − rank K h − rank K h-\operatorname{rank} K h − rank K
Selection rules on eleven split data
Let X ⊂ ( P 1 ) 4 X\subset(\mathbb{P}^1)^4 X ⊂ ( P 1 ) 4 V = ⨁ a = 1 5 O X ( k a ) V=\bigoplus_{a=1}^{5}\mathcal{O}_X(k_a) V = ⨁ a = 1 5 O X ( k a ) c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 S ( U ( 1 ) 5 ) S(U(1)^5) S ( U ( 1 ) 5 ) § 15.3 , in the coordinates e 1 , … , e 5 e_1,\ldots,e_5 e 1 , … , e 5 1 \mathbf{1} 1 Table 6 . A term of the perturbative superpotential is a product of charged fields times a monomial in the singlets, and it is gauge invariant exactly when its total charge is a multiple of 1 \mathbf{1} 1 10 a 5 ‾ b c 5 ‾ d e 10_a\overline{5}_{bc}\overline{5}_{de} 1 0 a 5 b c 5 d e Q = e a + e b + e c + e d + e e Q=e_a+e_b+e_c+e_d+e_e Q = e a + e b + e c + e d + e e 1 − Q 1-Q 1 − Q 10 a 10 b 5 ‾ d e 10_a10_b\overline{5}_{de} 1 0 a 1 0 b 5 d e e d + e e − e a − e b e_d+e_e-e_a-e_b e d + e e − e a − e b X X X 0 → O A ( k − 21 ) → O A ( k ) → O X ( k ) → 0 0\to\mathcal{O}_A(k-2\mathbf{1})\to\mathcal{O}_A(k)\to\mathcal{O}_X(k)\to0 0 → O A ( k − 2 1 ) → O A ( k ) → O X ( k ) → 0
Table 6. The charged fields of a rank-five split datum under S ( U ( 1 ) 5 ) S(U(1)^5) S ( U ( 1 ) 5 )
h i ( X , O X ( k ) ) ≤ h i ( A , O A ( k ) ) + h i + 1 ( A , O A ( k − 21 ) ) . (15) h^i(X,\mathcal{O}_X(k))\le h^i(A,\mathcal{O}_A(k))+h^{i+1}(A,\mathcal{O}_A(k-2\mathbf{1})). \tag*{(15)} h i ( X , O X ( k )) ≤ h i ( A , O A ( k )) + h i + 1 ( A , O A ( k − 2 1 )) . (15) with the ambient groups given by the Künneth formula on A = ( P 1 ) 4 A=(\mathbb{P}^1)^4 A = ( P 1 ) 4 possibly present when the bound (15) for its group is positive. A coupling forbidden with every possibly present singlet is forbidden on every smooth tetraquadric in the class; a coupling allowed by the charges is only not excluded by them.
Lemma 16.3 (Flow criterion). Let E E E ( p , q ) (p,q) ( p , q ) N = { 1 , … , n } N=\{1,\ldots,n\} N = { 1 , … , n } v ∈ Z n v\in\mathbb{Z}^n v ∈ Z n ∑ p v p = 0 \sum_p v_p=0 ∑ p v p = 0
(a) v = ∑ ( p , q ) ∈ E m p q ( e p − e q ) v=\sum_{(p,q)\in E}m_{pq}(e_p-e_q) v = ∑ ( p , q ) ∈ E m pq ( e p − e q ) m p q ≥ 0 m_{pq}\ge0 m pq ≥ 0
(b) For every B ⊂ N B\subset N B ⊂ N E E E p ∈ B p\in B p ∈ B ( p , q ) ∈ E (p,q)\in E ( p , q ) ∈ E q ∈ B q\in B q ∈ B ∑ p ∈ B v p ≤ 0 \sum_{p\in B}v_p\le0 ∑ p ∈ B v p ≤ 0
Proof. A pair ( p , q ) (p,q) ( p , q ) e p − e q e_p-e_q e p − e q B B B 1 1 1 p ∈ B ∌ q p\in B\not\ni q p ∈ B ∋ q − 1 -1 − 1 q ∈ B ∌ p q\in B\not\ni p q ∈ B ∋ p 0 0 0 B B B ∑ B v ≤ 0 \sum_Bv\le0 ∑ B v ≤ 0
Conversely, form the network with a source s s s t t t N N N s → p s\to p s → p v p v_p v p v p > 0 v_p>0 v p > 0 q → t q\to t q → t − v q -v_q − v q v q < 0 v_q<0 v q < 0 p → q p\to q p → q ( p , q ) ∈ E (p,q)\in E ( p , q ) ∈ E S ∋ s S\ni s S ∋ s T ∋ t T\ni t T ∋ t B = S ∩ N B=S\cap N B = S ∩ N E E E B B B
∑ p ∉ B , v p > 0 v p + ∑ q ∈ B , v q < 0 ( − v q ) = ∑ v p > 0 v p − ∑ p ∈ B v p ≥ ∑ v p > 0 v p \sum_{\substack{p\notin B,\ v_p>0}}v_p+\sum_{\substack{q\in B,\ v_q<0}}(-v_q)=\sum_{v_p>0}v_p-\sum_{p\in B}v_p\ge\sum_{v_p>0}v_p p ∈ / B , v p > 0 ∑ v p + q ∈ B , v q < 0 ∑ ( − v q ) = v p > 0 ∑ v p − p ∈ B ∑ v p ≥ v p > 0 ∑ v p by (b). The cut S = { s } S=\{s\} S = { s } ∑ v p > 0 v p \sum_{v_p>0}v_p ∑ v p > 0 v p [34 ] gives an integral flow of that value, because all capacities are integers. It saturates every arc at s s s t t t m p q m_{pq} m pq p → q p\to q p → q p ∈ N p\in N p ∈ N ∑ q m p q − ∑ q m q p \sum_qm_{pq}-\sum_qm_{qp} ∑ q m pq − ∑ q m q p s s s t t t v p v_p v p
:::
The criterion is decided by inspecting the 2 5 2^5 2 5 6
Proposition 16.4 (Eleven split data). Consider the matrix (12) and the ten rank-five sums of line bundles with entries in [ − 2 , 2 ] [-2,2] [ − 2 , 2 ] Table 7 , which are the classes, up to permutation of the four factors of ( P 1 ) 4 (\mathbb{P}^1)^4 ( P 1 ) 4 [41 ] to satisfy c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 − 3 ∣ G ∣ -3|G| − 3∣ G ∣ G G G § 15 . With all possibly present singlets inserted to every order:
(i) for (12) every down-type coupling vanishes on every smooth tetraquadric, while the up-type couplings 10 2 10 5 5 25 10_2 10_5 5_{25} 1 0 2 1 0 5 5 25 e 5 − e 2 e_5-e_2 e 5 − e 2 10 2 10 2 5 25 10_2 10_2 5_{25} 1 0 2 1 0 2 5 25 10 5 10 5 5 25 10_5 10_5 5_{25} 1 0 5 1 0 5 5 25
Table 7. The eleven split data. Rows of K K K k a k_a k a L a L_a L a H 1 ( L a ) H^1(L_a) H 1 ( L a ) B B B p → q p\to q p → q e p − e q e_p-e_q e p − e q
(ii) for the classes 9 and 10 every up-type and every down-type coupling vanishes; for the classes 3, 4 and 5 every up-type coupling vanishes;
(iii) for the classes 1, 2, 6, 7 and 8 both couplings are allowed by the charges, with the singlet insertions of Table 7 .
Hence six of the eleven split data fail C8 to all orders of the perturbative singlet expansion on every smooth tetraquadric, and the five others are not excluded by the selection rules.
Proof. For each datum the possibly present fields follow from (15) and the Künneth formula h 0 ( P 1 , O ( m ) ) = m + 1 h^0(\mathbb{P}^1,\mathcal{O}(m))=m+1 h 0 ( P 1 , O ( m )) = m + 1 m ≥ 0 m\geq0 m ≥ 0 h 1 ( P 1 , O ( m ) ) = − m − 1 h^1(\mathbb{P}^1,\mathcal{O}(m))=-m-1 h 1 ( P 1 , O ( m )) = − m − 1 m ≤ − 2 m\leq-2 m ≤ − 2 H 1 ( L 2 ) H^1(L_2) H 1 ( L 2 ) H 1 ( L 5 ) H^1(L_5) H 1 ( L 5 ) 5 ‾ \overline{5} 5 H 1 H^1 H 1 L 1 L 2 L_1L_2 L 1 L 2 L 2 L 4 L_2L_4 L 2 L 4 L 2 L 5 L_2L_5 L 2 L 5 L 4 L 5 L_4L_5 L 4 L 5 5 5 5 H 2 H^2 H 2 L 1 L 2 L_1L_2 L 1 L 2 L 2 L 5 L_2L_5 L 2 L 5 e p − e q e_p-e_q e p − e q p ∈ { 2 , 4 , 5 } p\in\{2,4,5\} p ∈ { 2 , 4 , 5 } ( p , q ) (p,q) ( p , q ) ( 2 , 1 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 4 , 2 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) (2,1),(2,3),(2,4),(2,5),(4,2),(5,1),(5,2),(5,3) ( 2 , 1 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 4 , 2 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) p = 1 p=1 p = 1 p = 3 p=3 p = 3 B = { 1 } B=\{1\} B = { 1 } B = { 3 } B=\{3\} B = { 3 } v = 1 − Q v=1-Q v = 1 − Q v 3 = 1 − Q 3 = 1 v_3=1-Q_3=1 v 3 = 1 − Q 3 = 1 e 3 e_3 e 3 B = { 3 } B=\{3\} B = { 3 } Q = e a + e b − e 2 − e 5 Q=e_a+e_b-e_2-e_5 Q = e a + e b − e 2 − e 5 a , b ∈ { 2 , 5 } a,b\in\{2,5\} a , b ∈ { 2 , 5 } Q = 0 Q=0 Q = 0 { a , b } = { 2 , 5 } \{a,b\}=\{2,5\} { a , b } = { 2 , 5 } a = b = 2 a=b=2 a = b = 2 ( 5 , 2 ) (5,2) ( 5 , 2 ) e 5 − e 2 = − Q e_5-e_2=-Q e 5 − e 2 = − Q a = b = 5 a=b=5 a = b = 5 ( 2 , 5 ) (2,5) ( 2 , 5 )
For (ii) and (iii) the same two steps are carried out for each row of Table 7 . Where the table gives a set B B B ( p , q ) (p,q) ( p , q ) p ∈ B p\in B p ∈ B q ∉ B q\notin B q ∈ / B ∑ B v > 0 \sum_B v>0 ∑ B v > 0 L 1 L_1 L 1 5 ‾ \overline{5} 5 v 1 = 1 v_1=1 v 1 = 1 B B B 5 5 5 ∑ B v = 2 > 0 \sum_Bv=2>0 ∑ B v = 2 > 0 B B B 5 5 5 1 − Q 1-Q 1 − Q − Q -Q − Q Q = 1 Q=1 Q = 1
Remark 16.5 (What the proposition does and does not say). The proposition decides a necessary condition. For the six excluded data the conclusion is unconditional within the perturbative singlet expansion: no complex structure, no polarization and no singlet expectation values restore the missing matrices, since the bound (15) holds on every smooth member and Lemma 16.3(b) is violated. For the five remaining data nothing is asserted beyond the absence of this obstruction. Their down-type couplings arise only at second or third order in the singlets, the singlet groups may vanish on a given member, the free action and its projection of the spectrum have not been computed, and the stability of the bundle after the singlets acquire expectation values, which is the same extension problem as in Section 15, is open. By Theorem 16.1 every one of the eleven data keeps a massless Abelian gauge boson at its split point, so the five classes are candidates for the extension branch, not for the split locus. The next subsection decides, for each of the eleven data, whether the smallest non-Abelian deformations of the split bundle remove that gauge boson.
The extension branch
Throughout this subsection X ⊂ ( P 1 ) 4 X \subset(\mathbb{P}^{1})^{4} X ⊂ ( P 1 ) 4 V = ⨁ a = 1 5 L a V = \bigoplus_{a=1}^{5} L_a V = ⨁ a = 1 5 L a Table 7 other than the class 3, so that rank K = 3 \operatorname{rank} K = 3 rank K = 3 u ∈ Z 5 u \in\mathbb{Z}^{5} u ∈ Z 5 { q : K q = 0 , ∑ a q a = 0 } \{q : Kq = 0, \sum_a q_a = 0\} { q : K q = 0 , ∑ a q a = 0 } Theorem 16.1 (ii); the vectors u u u Table 8 . For a Kähler class J = ∑ i t i J i J = \sum_i t_i J_i J = ∑ i t i J i t i > 0 t_i > 0 t i > 0 μ a ( J ) = μ J ( L a ) = ∑ i κ i a κ i ( J ) \mu_a(J) = \mu_J(L_a) = \sum_i \kappa_i^a \kappa_i(J) μ a ( J ) = μ J ( L a ) = ∑ i κ i a κ i ( J ) μ ( J ) = ( μ 1 , … , μ 5 ) = K κ ( J ) \mu(J) = (\mu_1,\ldots,\mu_5) = K\kappa(J) μ ( J ) = ( μ 1 , … , μ 5 ) = K κ ( J ) μ ( J ) \mu(J) μ ( J ) K K K ∑ a μ a = 0 \sum_a \mu_a = 0 ∑ a μ a = 0
Table 8. The realised shapes of Theorem 16.10(ii): the two extensions, the line bundle left alone, the vector w w w W W W f > 0 f>0 f > 0 H 1 ( L s u b L q u o t − 1 ) H^1(L_{\mathrm{sub}}L_{\mathrm{quot}}^{-1}) H 1 ( L sub L quot − 1 ) u u u t t t J 0 = ∑ i t i J i J_0=\sum_i t_iJ_i J 0 = ∑ i t i J i r = 1.42764 … r=1.42764\ldots r = 1.42764 … 9 r 4 − 36 r 2 + 14 r + 16 9r^4-36r^2+14r+16 9 r 4 − 36 r 2 + 14 r + 16 ( 1.4 , 1.5 ) (1.4,1.5) ( 1.4 , 1.5 ) t 1 = ( 18 r 3 + 27 r 2 − 39 r − 28 ) / 5 = 4.7454 … t_1=(18r^3+27r^2-39r-28)/5=4.7454\ldots t 1 = ( 18 r 3 + 27 r 2 − 39 r − 28 ) /5 = 4.7454 … t 2 = ( 9 r 3 + 6 r 2 − 22 r − 4 ) / 10 = 0.3008 … t_2=(9r^3+6r^2-22r-4)/10=0.3008\ldots t 2 = ( 9 r 3 + 6 r 2 − 22 r − 4 ) /10 = 0.3008 …
The bundles that replace V V V § 16 acquire expectation values are the polystable bundles E = ⨁ j E j E = \bigoplus_j E_j E = ⨁ j E j E j E_j E j L a L_a L a a ∈ B j a \in B_j a ∈ B j { B j } \{B_j\} { B j } { 1 , … , 5 } \{1,\ldots,5\} { 1 , … , 5 } V V V shape of E E E [35 , 36 ] : an extension class in H 1 ( L p L q − 1 ) = Ext 1 ( L q , L p ) H^1(L_pL_q^{-1}) = \operatorname{Ext}^1(L_q,L_p) H 1 ( L p L q − 1 ) = Ext 1 ( L q , L p ) e p − e q e_p-e_q e p − e q
Lemma 16.6 (The Abelian factor on a branch). Let E = ⨁ j = 1 m E j E = \bigoplus_{j=1}^{m} E_j E = ⨁ j = 1 m E j be polystable of slope zero with the shape { B j } \{B_j\} { B j } The structure group of E E E reduces to S ( ∏ j U ( ∣ B j ∣ ) ) S\left(\prod_j U(\lvert B_j\rvert)\right) S ( ∏ j U (∣ B j ∣) ) whose centre is parametrized by the traceless vectors q ∈ Q 5 q \in\mathbb{Q}^{5} q ∈ Q 5 constant on every block. The Abelian gauge boson of such a generator q q q is massless at E E E if and only if K q = 0 Kq = 0 K q = 0 Consequently E E E keeps a massless Abelian gauge boson beyond the S U ( 5 ) SU(5) S U ( 5 ) factor exactly when u u u is constant on every block of its shape.
Proof. Each E j E_j E j ∣ B j ∣ \lvert B_j\rvert ∣ B j ∣ E E E ∏ j U ( ∣ B j ∣ ) \prod_j U(\lvert B_j\rvert) ∏ j U (∣ B j ∣) c 1 ( E ) = c 1 ( V ) = 0 c_1(E) = c_1(V) = 0 c 1 ( E ) = c 1 ( V ) = 0 q q q Theorem 16.1 : the axions shift under the generator q q q ∑ j q B j c 1 ( E j ) = ∑ a q a c 1 ( L a ) \sum_j q_{B_j}c_1(E_j) = \sum_a q_a c_1(L_a) ∑ j q B j c 1 ( E j ) = ∑ a q a c 1 ( L a ) K q Kq K q q q q ( K q ) T G ~ ( K q ) (Kq)^T\widetilde{G}(Kq) ( K q ) T G ( K q ) G ~ \widetilde{G} G [33 Section 2.7] , and q q q K q = 0 Kq = 0 K q = 0 K q = 0 Kq = 0 K q = 0 u u u u u u [36 ] : the extension class of a block is a singlet of charge e p − e q e_p-e_q e p − e q p , q p,q p , q q q q q p ≠ q q q_p \ne q_q q p = q q □ \square □
Lemma 16.7 (One side of a wall). Let Y Y Y be a smooth projective threefold, L , L ′ L,L' L , L ′ line bundles on Y Y Y and J 0 ∗ a n a m p l e c l a s s w i t h ∗ J_0 *an ample class with* J 0 ∗ anam pl ec l a ss w i t h ∗ . ∗ L e t ∗ $ 0 → L → E → L ′ → 0 . *Let* \$0 \to L \to E \to L' \to0 . ∗ L e t ∗ $0 → L → E → L ′ → 0 be a non-split extension. There is a neighbourhood U U U of J 0 J_0 J 0 in the ample cone such that E E E is μ J \mu_J μ J stable for every J ∈ U J \in U J ∈ U with μ J ( L ) < μ J ( L ′ ) \mu_J(L) < \mu_J(L') μ J ( L ) < μ J ( L ′ )
Proof. A saturated subsheaf M ⊂ E M \subset E M ⊂ E Y Y Y [37 Section 1.1] ; it suffices to show μ J ( M ) < μ J ( E ) = 1 2 ( μ J ( L ) + μ J ( L ′ ) ) \mu_J(M) < \mu_J(E) = \frac{1}{2}\bigl(\mu_J(L) + \mu_J(L')\bigr) μ J ( M ) < μ J ( E ) = 2 1 ( μ J ( L ) + μ J ( L ′ ) ) M M M M → L ′ M \to L' M → L ′ M ⊂ L M \subset L M ⊂ L M = L ( − D ) M = L(-D) M = L ( − D ) D D D μ J ( M ) ≤ μ J ( L ) < μ J ( E ) \mu_J(M) \le\mu_J(L) < \mu_J(E) μ J ( M ) ≤ μ J ( L ) < μ J ( E ) M → L ′ M \to L' M → L ′ M = L ′ ( − D ) M = L'(-D) M = L ′ ( − D ) D D D D ≠ 0 D \ne0 D = 0 D = 0 D = 0 D = 0 δ J = 1 2 ( μ J ( L ′ ) − μ J ( L ) ) \delta_J = \frac{1}{2}\bigl(\mu_J(L') - \mu_J(L)\bigr) δ J = 2 1 ( μ J ( L ′ ) − μ J ( L ) ) μ J ( M ) < μ J ( E ) \mu_J(M) < \mu_J(E) μ J ( M ) < μ J ( E ) J 2 ⋅ D > δ J J^2 \cdot D > \delta_J J 2 ⋅ D > δ J H H H J 0 − H J_0-H J 0 − H [37 Lemma 1.7.9] the saturated subsheaves M ⊂ E M\subset E M ⊂ E μ H ( M ) ≥ μ H ( L ′ ) − 1 \mu_H(M)\geq\mu_H(L')-1 μ H ( M ) ≥ μ H ( L ′ ) − 1 H 2 ⋅ D ≤ 1 H^2\cdot D\leq1 H 2 ⋅ D ≤ 1 D 1 , … , D m ∈ H 2 ( Y , Z ) D_1,\ldots,D_m\in H^2(Y,\mathbb{Z}) D 1 , … , D m ∈ H 2 ( Y , Z ) D i D_i D i J 0 2 ⋅ D i > 0 J_0^2\cdot D_i>0 J 0 2 ⋅ D i > 0 c 1 = min i J 0 2 ⋅ D i > 0 c_1=\min_i J_0^2\cdot D_i>0 c 1 = min i J 0 2 ⋅ D i > 0 U U U J 0 J_0 J 0 J − H J-H J − H J 2 ⋅ D i > c 1 / 2 J^2\cdot D_i>c_1/2 J 2 ⋅ D i > c 1 /2 i i i δ J < min ( c 1 / 2 , 1 ) \delta_J<\min(c_1/2,1) δ J < min ( c 1 /2 , 1 ) δ J 0 = 0 \delta_{J_0}=0 δ J 0 = 0 J ∈ U J\in U J ∈ U M M M J 2 ⋅ D > c 1 / 2 > δ J J^2\cdot D>c_1/2>\delta_J J 2 ⋅ D > c 1 /2 > δ J M M M H 2 ⋅ D > 1 H^2\cdot D>1 H 2 ⋅ D > 1 J = H + ( J − H ) J=H+(J-H) J = H + ( J − H ) J 2 ⋅ D ≥ H 2 ⋅ D > 1 > δ J J^2\cdot D\geq H^2\cdot D>1>\delta_J J 2 ⋅ D ≥ H 2 ⋅ D > 1 > δ J H ⋅ ( J − H ) ⋅ D H\cdot(J-H)\cdot D H ⋅ ( J − H ) ⋅ D ( J − H ) 2 ⋅ D (J-H)^2\cdot D ( J − H ) 2 ⋅ D D D D μ J ( M ) < μ J ( E ) \mu_J(M)<\mu_J(E) μ J ( M ) < μ J ( E )
Proposition 16.8 (One block does not suffice). No polystable bundle of slope zero whose shape has exactly one block of size two, E = E 1 ⊕ L r ⊕ L s ⊕ L o E=E_1\oplus L_r\oplus L_s\oplus L_o E = E 1 ⊕ L r ⊕ L s ⊕ L o E 1 E_1 E 1 L q L_q L q L p L_p L p u u u u p = u q u_p=u_q u p = u q u u u
Proof. Polystability of slope zero requires μ r = μ s = μ o = 0 \mu_r=\mu_s=\mu_o=0 μ r = μ s = μ o = 0 κ ( J ) ⊥ k r , k s , k o \kappa(J)\perp k_r,k_s,k_o κ ( J ) ⊥ k r , k s , k o { K T κ = 0 } \{K^T\kappa=0\} { K T κ = 0 } 4 − rank K = 1 4-\operatorname{rank}K=1 4 − rank K = 1 μ p = μ q = 0 \mu_p=\mu_q=0 μ p = μ q = 0 E 1 E_1 E 1 L p L_p L p v ∈ ker K v\in\ker K v ∈ ker K p p p q q q ker K = Q 1 ⊕ Q u \ker K=\mathbb{Q}1\oplus\mathbb{Q}u ker K = Q 1 ⊕ Q u v = α 1 + β u v=\alpha1+\beta u v = α 1 + β u α + β u p = α + β u q = 0 \alpha+\beta u_p=\alpha+\beta u_q=0 α + β u p = α + β u q = 0 β ≠ 0 \beta\neq0 β = 0 u p = u q u_p=u_q u p = u q
Two blocks of size two can do what one cannot. For a shape { p , q } , { r , s } , { o } \{p,q\},\{r,s\},\{o\} { p , q } , { r , s } , { o }
W = { μ ∈ Im K : μ o = 0 , μ p + μ q = 0 } , W=\{\mu\in\operatorname{Im}K:\mu_o=0,\ \mu_p+\mu_q=0\}, W = { μ ∈ Im K : μ o = 0 , μ p + μ q = 0 } , the slopes allowed to a polystable bundle of that shape (the third condition μ r + μ s = 0 \mu_r+\mu_s=0 μ r + μ s = 0 ∑ a μ a = 0 \sum_a\mu_a=0 ∑ a μ a = 0 W W W Q w \mathbb{Q}w Q w J J J μ ( J ) = f ( J ) w \mu(J)=f(J)w μ ( J ) = f ( J ) w f ( J ) ≠ 0 f(J)\neq0 f ( J ) = 0 f f f f w a < 0 f w_a<0 f w a < 0 w w w f > 0 f>0 f > 0 realised at J J J H 1 ( L s u b L q u o t − 1 ) H^1(L_{\mathrm{sub}}L_{\mathrm{quot}}^{-1}) H 1 ( L sub L quot − 1 )
Lemma 16.9 (A vanishing on the general tetraquadric). For the class 1 of Table 7 , h 1 ( X , L 2 L 5 − 1 ) = h 1 ( X , L 5 L 2 − 1 ) = 0 h^1(X,L_2L_5^{-1})=h^1(X,L_5L_2^{-1})=0 h 1 ( X , L 2 L 5 − 1 ) = h 1 ( X , L 5 L 2 − 1 ) = 0 X X X
Proof. Here k 2 − k 5 = ( − 3 , − 2 , 2 , 3 ) k_2-k_5=(-3,-2,2,3) k 2 − k 5 = ( − 3 , − 2 , 2 , 3 ) k 5 − k 2 = ( 3 , 2 , − 2 , − 3 ) k_5-k_2=(3,2,-2,-3) k 5 − k 2 = ( 3 , 2 , − 2 , − 3 ) k = ( − 3 , − 2 , 2 , 3 ) k=(-3,-2,2,3) k = ( − 3 , − 2 , 2 , 3 ) h 1 ( A , O A ( k ) ) = 0 h^1(A,\mathcal{O}_A(k))=0 h 1 ( A , O A ( k )) = 0 h 2 ( A , O A ( k ) ) = h 2 ( A , O A ( k − 21 ) ) = 2 ⋅ 1 ⋅ 3 ⋅ 4 = 24 h^2(A,\mathcal{O}_A(k))=h^2(A,\mathcal{O}_A(k-21))=2\cdot1\cdot3\cdot4=24 h 2 ( A , O A ( k )) = h 2 ( A , O A ( k − 21 )) = 2 ⋅ 1 ⋅ 3 ⋅ 4 = 24 H 1 ⊗ H 1 ⊗ H 0 ⊗ H 0 H^1\otimes H^1\otimes H^0\otimes H^0 H 1 ⊗ H 1 ⊗ H 0 ⊗ H 0 h 1 ( X , O X ( k ) ) = dim ker ( s : H 2 ( A , O A ( k − 21 ) ) → H 2 ( A , O A ( k ) ) ) h^1(X,\mathcal{O}_X(k))=\dim\ker(s:H^2(A,\mathcal{O}_A(k-21))\to H^2(A,\mathcal{O}_A(k))) h 1 ( X , O X ( k )) = dim ker ( s : H 2 ( A , O A ( k − 21 )) → H 2 ( A , O A ( k ))) s s s X X X H 1 ( P 1 , O ( m ) ) H^1(\mathbb{P}^1,\mathcal{O}(m)) H 1 ( P 1 , O ( m )) x a y b x^a y^b x a y b a , b ≤ − 1 a,b\leq-1 a , b ≤ − 1 a + b = m a+b=m a + b = m H 0 ( P 1 , O ( m ) ) H^0(\mathbb{P}^1,\mathcal{O}(m)) H 0 ( P 1 , O ( m )) a , b ≥ 0 a,b\geq0 a , b ≥ 0
s 0 = x 1 2 x 2 2 x 3 2 x 4 2 + y 1 2 y 2 2 x 3 2 x 4 2 + y 1 2 x 2 2 x 3 2 y 4 2 + x 1 2 y 2 2 y 3 2 y 4 2 + y 1 2 ( x 2 y 2 ) ( x 3 y 3 ) y 4 2 + x 1 2 ( x 2 y 2 ) ( x 3 y 3 ) x 4 2 s_0=x_1^2x_2^2x_3^2x_4^2+y_1^2y_2^2x_3^2x_4^2+y_1^2x_2^2x_3^2y_4^2+x_1^2y_2^2y_3^2y_4^2+y_1^2(x_2y_2)(x_3y_3)y_4^2+x_1^2(x_2y_2)(x_3y_3)x_4^2 s 0 = x 1 2 x 2 2 x 3 2 x 4 2 + y 1 2 y 2 2 x 3 2 x 4 2 + y 1 2 x 2 2 x 3 2 y 4 2 + x 1 2 y 2 2 y 3 2 y 4 2 + y 1 2 ( x 2 y 2 ) ( x 3 y 3 ) y 4 2 + x 1 2 ( x 2 y 2 ) ( x 3 y 3 ) x 4 2 each of the 24 monomials of the source is carried by exactly one of the six terms to a nonzero monomial of the target, and the 24 images are distinct, so the map is a bijection of the two bases; the same holds for k = ( 3 , 2 , − 2 , − 3 ) k=(3,2,-2,-3) k = ( 3 , 2 , − 2 , − 3 ) s s s 7
Theorem 16.10 (Two blocks). Let V V V Table 7 other than the class 3.
(i) For the classes 4, 5 and 9 there is no Kähler class J J J
(ii) For the datum (12) and the classes 2, 6, 7, 8, and 10 there are a shape, listed in Table 8 , and a Kähler class J 0 J_0 J 0 V V V S = { J : μ o ( J ) = 0 , μ p ( J ) + μ q ( J ) = 0 } S=\{J:\mu_o(J)=0,\ \mu_p(J)+\mu_q(J)=0\} S = { J : μ o ( J ) = 0 , μ p ( J ) + μ q ( J ) = 0 } J 0 J_0 J 0 μ = f w \mu=fw μ = f w f f f J 0 J_0 J 0 { f = 0 } \{f=0\} { f = 0 } J ∈ S J\in S J ∈ S f ( J ) > 0 f(J)>0 f ( J ) > 0 J 0 J_0 J 0 E 1 ⊕ E 2 ⊕ L o E_1\oplus E_2\oplus L_o E 1 ⊕ E 2 ⊕ L o V V V S ( U ( 2 ) × U ( 2 ) × U ( 1 ) ) S(U(2)\times U(2)\times U(1)) S ( U ( 2 ) × U ( 2 ) × U ( 1 )) S U ( 5 ) SU(5) S U ( 5 )
Proof. (i) A polystable bundle of slope zero and shape { p , q } , { r , s } , { o } \{p,q\},\{r,s\},\{o\} { p , q } , { r , s } , { o } μ ( J ) ∈ W \mu(J)\in W μ ( J ) ∈ W Table 9 lists the shapes with W ≠ 0 W\ne0 W = 0 W W W μ ( J ) = f w \mu(J)=fw μ ( J ) = f w f ≠ 0 f\ne0 f = 0 f f f H 1 ( L s u b L q u o t − 1 ) H^1(L_{\mathrm{sub}}L_{\mathrm{quot}}^{-1}) H 1 ( L sub L quot − 1 ) H 1 ( L 2 L 5 − 1 ) H^1(L_2L_5^{-1}) H 1 ( L 2 L 5 − 1 ) H 1 ( L 5 L 2 − 1 ) H^1(L_5L_2^{-1}) H 1 ( L 5 L 2 − 1 ) L 3 ≅ L 2 L_3\cong L_2 L 3 ≅ L 2
Table 9. Theorem 16.10(i): every shape with two blocks of size two whose space W W W W = 0 W = 0 W = 0 W W W w w w f f f H 1 ( L s u b L q u o t − 1 ) H^1(L_{\mathrm{sub}}L_{\mathrm{quot}}^{-1}) H 1 ( L sub L quot − 1 )
Table 9.
(ii) Fix a row of Table 8 and write { p , q } = { s u b 1 , q u o t 1 } \{p,q\}=\{\mathrm{sub}_1,\mathrm{quot}_1\} { p , q } = { sub 1 , quot 1 } { r , s } \{r,s\} { r , s } J 0 J_0 J 0 t = ( 1 , 1 , 1 , 1 ) t=(1,1,1,1) t = ( 1 , 1 , 1 , 1 ) κ i \kappa_i κ i K K K J 0 J_0 J 0 μ a ( t ) \mu_a(t) μ a ( t ) S S S J 0 J_0 J 0 μ o \mu_o μ o μ p + μ q \mu_p+\mu_q μ p + μ q i ≠ j i\ne j i = j ∂ κ i / ∂ t j = 4 ( e 1 ( t ) − t i − t j ) \partial\kappa_i/\partial t_j=4(e_1(t)-t_i-t_j) ∂ κ i / ∂ t j = 4 ( e 1 ( t ) − t i − t j ) ∂ κ i / ∂ t i = 0 \partial\kappa_i/\partial t_i=0 ∂ κ i / ∂ t i = 0 t = ( 1 , 1 , 1 , 1 ) t=(1,1,1,1) t = ( 1 , 1 , 1 , 1 ) d μ a = − 8 ∑ j k j a d t j d\mu_a=-8\sum_j k_j^a\,dt_j d μ a = − 8 ∑ j k j a d t j − 8 -8 − 8 k o k_o k o k p + k q k_p+k_q k p + k q ( 1 , 1 , − 1 , − 1 ) (1,1,-1,-1) ( 1 , 1 , − 1 , − 1 ) ( − 2 , − 3 , 2 , 3 ) (-2,-3,2,3) ( − 2 , − 3 , 2 , 3 ) S S S W = Q w W=\mathbb{Q}w W = Q w μ = f w \mu=fw μ = f w f = μ q / w q f=\mu_q/w_q f = μ q / w q S S S J 0 J_0 J 0 k 1 k_1 k 1 k 4 k_4 k 4 k 1 + k 2 k_1+k_2 k 1 + k 2 f f f S S S J 0 J_0 J 0
For J ∈ S J \in S J ∈ S f ( J ) > 0 f(J)>0 f ( J ) > 0 f w s u b < 0 f w_{\mathrm{sub}}<0 f w sub < 0 f w q u o t > 0 f w_{\mathrm{quot}}>0 f w quot > 0 μ o = 0 \mu_o=0 μ o = 0 μ s u b + μ q u o t = 0 \mu_{\mathrm{sub}}+\mu_{\mathrm{quot}}=0 μ sub + μ quot = 0 H 1 ( A , O A ( k − 21 ) ) H^1(A,\mathcal{O}_A(k-21)) H 1 ( A , O A ( k − 21 )) H 2 ( A , O A ( k ) ) H^2(A,\mathcal{O}_A(k)) H 2 ( A , O A ( k )) h 1 ( X , O X ( k ) ) = h 1 ( A , O A ( k ) ) + h 2 ( A , O A ( k − 21 ) ) h^1(X,\mathcal{O}_X(k))=h^1(A,\mathcal{O}_A(k))+h^2(A,\mathcal{O}_A(k-21)) h 1 ( X , O X ( k )) = h 1 ( A , O A ( k )) + h 2 ( A , O A ( k − 21 )) J 0 J_0 J 0 U U U J 0 J_0 J 0 E 1 E_1 E 1 E 2 E_2 E 2 J J J f ( J ) > 0 f(J)>0 f ( J ) > 0 L o L_o L o E = E 1 ⊕ E 2 ⊕ L o E=E_1\oplus E_2\oplus L_o E = E 1 ⊕ E 2 ⊕ L o J ∈ S ∩ U J\in S\cap U J ∈ S ∩ U f ( J ) > 0 f(J)>0 f ( J ) > 0 V V V c 1 ( E 1 ) = c 1 ( L p ) c_1(E_1)=c_1(L_p) c 1 ( E 1 ) = c 1 ( L p ) c 1 ( E 2 ) c_1(E_2) c 1 ( E 2 ) S ( U ( 2 ) × U ( 2 ) × U ( 1 ) ) S(U(2)\times U(2)\times U(1)) S ( U ( 2 ) × U ( 2 ) × U ( 1 )) u u u u s u b ≠ u q u o t u_{\mathrm{sub}}\ne u_{\mathrm{quot}} u sub = u quot Table 8 shows, so by Lemma 16.6 no Abelian gauge boson beyond S U ( 5 ) SU(5) S U ( 5 ) E E E
Remark 16.11 (What the branch does and does not give). Theorem 16.10(ii) removes the obstruction of Theorem 16.1 for six of the eleven data, including the fixed datum, at the cost of leaving the split locus: the polystable Kähler classes of E E E S S S X X X X / Γ X/\Gamma X /Γ Γ \Gamma Γ H 1 ( L 2 L 1 − 1 ) H^1(L_2L_1^{-1}) H 1 ( L 2 L 1 − 1 ) H 1 ( L 5 L 3 − 1 ) H^1(L_5L_3^{-1}) H 1 ( L 5 L 3 − 1 ) S S S Lemma 16.6 remove u u u u u u
Figure 5. Left: the singlet charges e p − e q e_p-e_q e p − e q p → q p\to q p → q e 1 e_1 e 1 e 3 e_3 e 3 Y d Y_d Y d Table 7 , with every possibly present singlet inserted.
Nine-condition ledger for the one fixed datum
The table refers only to X π / Γ X_\pi/\Gamma X π /Γ J = 100 ∑ J i J=100\sum J_i J = 100 ∑ J i § 16 . “Verified” below always carries this stated mathematical or perturbative scope. Table 10 is the result. Thus the fixed datum is not a viable nine-condition vacuum. The construction proves explicit geometry and charged-spectrum statements, and it proves definite failures of some tests. It neither proves nor disproves existence of a different vacuum passing all nine.
Table 10. The nine conditions at the fixed datum.
An estimate for coupling the stabilization sectors
Independent stabilization calculations cannot be joined without bounding their mixed derivatives. The following elementary estimate states one usable quantitative requirement.
Proposition 18.1 (Mixed-sector Hessian bound). Let the real symmetric Hessian have blocks
H = ( A B B T C ) , H=\begin{pmatrix} A & B \\ B^{T} & C \end{pmatrix}, H = ( A B T B C ) , with A ⪰ α I A\succeq\alpha I A ⪰ α I C ⪰ γ I C\succeq\gamma I C ⪰ γ I and ∥ B ∥ op ≤ β \lVert B\rVert_{\operatorname{op}}\le\beta ∥ B ∥ op ≤ β where α , γ > 0 \alpha,\gamma>0 α , γ > 0 If β 2 < α γ \beta^2<\alpha\gamma β 2 < α γ then
H ⪰ α + γ − ( α − γ ) 2 + 4 β 2 2 I > 0. H\succeq\frac{\alpha+\gamma-\sqrt{(\alpha-\gamma)^2+4\beta^2}}{2}I>0. H ⪰ 2 α + γ − ( α − γ ) 2 + 4 β 2 I > 0. Proof. For vectors v , w v,w v , w α ∥ v ∥ 2 − 2 β ∥ v ∥ ∥ w ∥ + γ ∥ w ∥ 2 \alpha\lVert v\rVert^2-2\beta\lVert v\rVert\lVert w\rVert+\gamma\lVert w\rVert^2 α ∥ v ∥ 2 − 2 β ∥ v ∥ ∥ w ∥ + γ ∥ w ∥ 2
This is a sufficient criterion with inputs to be measured or proved on a compactification chart. It does not provide those inputs for X π X_\pi X π
Why holomorphic masses do not close C8
After fixing conventions and the Higgs factor, a physical Yukawa matrix has the form
Y p h y s = c Z L − T / 2 Y h o l Z R − 1 / 2 , c > 0 , (16) Y_{\mathrm{phys}}=cZ_L^{-T/2}Y_{\mathrm{hol}}Z_R^{-1/2},\qquad c>0, \tag*{(16)} Y phys = c Z L − T /2 Y hol Z R − 1/2 , c > 0 , (16) where the matter metrics are positive definite. Their invertibility preserves rank but does not preserve singular-value ratios.
Proposition 19.1 (A normalization counterexample). For 0 < ϵ < 1 0<\epsilon<1 0 < ϵ < 1
Y h o l = diag ( ϵ 8 , ϵ 4 , 1 ) , Z L = I , Z R = diag ( ϵ 16 , ϵ 8 , 1 ) , c = 1. Y_{\mathrm{hol}}=\operatorname{diag}(\epsilon^8,\epsilon^4,1),\qquad Z_L=I,\qquad Z_R=\operatorname{diag}(\epsilon^{16},\epsilon^8,1),\qquad c=1. Y hol = diag ( ϵ 8 , ϵ 4 , 1 ) , Z L = I , Z R = diag ( ϵ 16 , ϵ 8 , 1 ) , c = 1. All metrics are positive definite and Y p h y s = I Y_{\mathrm{phys}}=I Y phys = I
Proof. The positive inverse square root of Z R Z_R Z R diag ( ϵ − 8 , ϵ − 4 , 1 ) \operatorname{diag}(\epsilon^{-8},\epsilon^{-4},1) diag ( ϵ − 8 , ϵ − 4 , 1 ) (16) gives the identity. □ \square □
The example does not claim that these metrics occur in the tetraquadric model. It refutes the inference based on a holomorphic matrix alone. It does not refute a hierarchy proved with regular nondegenerate metrics at the limiting point, as the following bound makes precise.
Proposition 19.2 (Bounded normalization preserves orders). Suppose m I ⪯ Z L , Z R ⪯ M I mI\preceq Z_L,Z_R\preceq MI m I ⪯ Z L , Z R ⪯ M I 0 < m ≤ M 0<m\leq M 0 < m ≤ M 0 < c − ≤ c ≤ c + 0<c_-\leq c\leq c_+ 0 < c − ≤ c ≤ c +
c − M σ i ( Y h o l ) ≤ σ i ( Y p h y s ) ≤ c + m σ i ( Y h o l ) . \frac{c_-}{M}\sigma_i(Y_{\mathrm{hol}})\leq\sigma_i(Y_{\mathrm{phys}})\leq\frac{c_+}{m}\sigma_i(Y_{\mathrm{hol}}). M c − σ i ( Y hol ) ≤ σ i ( Y phys ) ≤ m c + σ i ( Y hol ) . If these bounds are uniform as a parameter tends to zero, any nonzero power-law orders of the holomorphic singular values are preserved.
Proof. For invertible A A A B B B σ i ( A Y B ) ≤ ∥ A ∥ o p ∥ B ∥ o p σ i ( Y ) \sigma_i(AYB) \le\lVert A\rVert_{\mathrm{op}}\lVert B\rVert_{\mathrm{op}}\sigma_i(Y) σ i ( A Y B ) ≤ ∥ A ∥ op ∥ B ∥ op σ i ( Y ) Y = A − 1 ( A Y B ) B − 1 Y=A^{-1}(AYB)B^{-1} Y = A − 1 ( A Y B ) B − 1 m − 1 / 2 m^{-1/2} m − 1/2 M 1 / 2 M^{1/2} M 1/2 c c c
Continuity and positive definiteness at a regular limiting point give such bounds locally. The preceding counterexample loses that lower bound as ϵ → 0 \epsilon\to0 ϵ → 0 [30 ] ; that paper leaves stabilization and supersymmetry breaking untreated. This is a limitation of the result cited, not an assertion about every heterotic model in the literature.
A quantitative simultaneous-persistence criterion
The following theorem identifies bounds that would make a local construction robust. All its assumptions must be checked on the same compactification chart and in the same normalization.
Theorem 20.1 (A certified interior minimum). Let B = B ‾ r ( u 0 ) ⊂ R n B=\overline{B}_r(u_0)\subset\mathbb{R}^n B = B r ( u 0 ) ⊂ R n V 0 V_0 V 0 Δ \Delta Δ C 2 C^2 C 2 B B B B B B
D 2 V 0 ⪰ λ I , ∥ D 2 Δ ∥ o p ≤ b < λ , ∥ ∇ V 0 ( u 0 ) ∥ ≤ η , ∥ ∇ Δ ( u 0 ) ∥ ≤ a . D^2V_0 \succeq\lambda I,\qquad\lVert D^2\Delta\rVert_{\mathrm{op}}\le b<\lambda,\qquad\lVert\nabla V_0(u_0)\rVert\le\eta,\qquad\lVert\nabla\Delta(u_0)\rVert\le a. D 2 V 0 ⪰ λ I , ∥ D 2 Δ ∥ op ≤ b < λ , ∥ ∇ V 0 ( u 0 )∥ ≤ η , ∥ ∇Δ ( u 0 )∥ ≤ a . Put κ = λ − b \kappa=\lambda-b κ = λ − b η + a < κ r \eta+a<\kappa r η + a < κ r V = V 0 + Δ V=V_0+\Delta V = V 0 + Δ u ∗ u_* u ∗ B B B D 2 V ( u ∗ ) ⪰ κ I D^2V(u_*)\succeq\kappa I D 2 V ( u ∗ ) ⪰ κ I
∥ u ∗ − u 0 ∥ ≤ δ ≔ η + a κ < r . (17) \lVert u_*-u_0\rVert\le\delta\coloneqq\frac{\eta+a}{\kappa}<r. \tag*{(17)} ∥ u ∗ − u 0 ∥ ≤ δ : = κ η + a < r . (17) Proof. The Hessian of V V V κ I \kappa I κ I
( ∇ V ( x ) − ∇ V ( y ) ) ⋅ ( x − y ) ≥ κ ∥ x − y ∥ 2 . \bigl(\nabla V(x)-\nabla V(y)\bigr)\cdot(x-y)\ge\kappa\lVert x-y\rVert^2. ( ∇ V ( x ) − ∇ V ( y ) ) ⋅ ( x − y ) ≥ κ ∥ x − y ∥ 2 . For x x x v = x − u 0 v=x-u_0 v = x − u 0 ∇ V ( x ) ⋅ v ≥ κ r 2 − ( η + a ) r > 0 \nabla V(x)\cdot v\ge\kappa r^2-(\eta+a)r>0 ∇ V ( x ) ⋅ v ≥ κ r 2 − ( η + a ) r > 0 x = u ∗ x=u_* x = u ∗ y = u 0 y=u_0 y = u 0 ∇ V ( u ∗ ) = 0 \nabla V(u_*)=0 ∇ V ( u ∗ ) = 0 κ ∥ u ∗ − u 0 ∥ 2 ≤ ( η + a ) ∥ u ∗ − u 0 ∥ \kappa\lVert u_*-u_0\rVert^2\le(\eta+a)\lVert u_*-u_0\rVert κ ∥ u ∗ − u 0 ∥ 2 ≤ ( η + a ) ∥ u ∗ − u 0 ∥
Proposition 20.2 (Preserving masses and inequalities). In Theorem 20.1, suppose Y Y Y 3 × 3 3\times3 3 × 3 ∥ Y ( u ) − Y ( u 0 ) ∥ o p ≤ L ∥ u − u 0 ∥ \lVert Y(u)-Y(u_0)\rVert_{\mathrm{op}}\le L\lVert u-u_0\rVert ∥ Y ( u ) − Y ( u 0 ) ∥ op ≤ L ∥ u − u 0 ∥ B B B s i s_i s i u 0 u_0 u 0 u ∗ u_* u ∗ s i s_i s i e = L δ e=L\delta e = L δ s min > e s_{\min}>e s m i n > e s j > e s_j>e s j > e
max ( 0 , s i − e ) s j + e ≤ s i ( u ∗ ) s j ( u ∗ ) ≤ s i + e s j − e . \frac{\max(0,s_i-e)}{s_j+e}\le\frac{s_i(u_*)}{s_j(u_*)}\le\frac{s_i+e}{s_j-e}. s j + e max ( 0 , s i − e ) ≤ s j ( u ∗ ) s i ( u ∗ ) ≤ s j − e s i + e . If a real acceptance function q q q L q L_q L q q r e q q_{\mathrm{req}} q req q ( u 0 ) > q r e q + L q δ q(u_0)>q_{\mathrm{req}}+L_q\delta q ( u 0 ) > q req + L q δ q ( u ∗ ) > q r e q q(u_*)>q_{\mathrm{req}} q ( u ∗ ) > q req
Proof. The min–max formula for singular values and ∥ A v ∥ − ∥ B v ∥ ≤ ∥ ( A − B ) v ∥ \lVert Av\rVert-\lVert Bv\rVert\le\lVert(A-B)v\rVert ∥ A v ∥ − ∥ B v ∥ ≤ ∥( A − B ) v ∥ ∣ σ i ( A ) − σ i ( B ) ∣ ≤ ∥ A − B ∥ o p \lvert\sigma_i(A)-\sigma_i(B)\rvert\le\lVert A-B\rVert_{\mathrm{op}} ∣ σ i ( A ) − σ i ( B )∣ ≤ ∥ A − B ∥ op
Thus a controlled ball preserving the discrete spectrum and satisfying strict mass, hierarchy and consistency margins would turn an approximate minimum into a simultaneous certificate. If the real kinetic matrix obeys 0 < G ( u ∗ ) ⪯ Λ I 0<G(u_*)\preceq\Lambda I 0 < G ( u ∗ ) ⪯ Λ I G ( u ∗ ) − 1 / 2 D 2 V ( u ∗ ) G ( u ∗ ) − 1 / 2 G(u_*)^{-1/2}D^2V(u_*)G(u_*)^{-1/2} G ( u ∗ ) − 1/2 D 2 V ( u ∗ ) G ( u ∗ ) − 1/2 κ / Λ \kappa/\Lambda κ /Λ G − 1 / 2 v G^{-1/2}v G − 1/2 v
An explicit local effective-theory construction and its limit
It is useful to separate a solvable analytic problem from the missing string realization. Consider a classical global N = 1 N=1 N = 1 S , Z 1 , … , Z m S, Z_1,\ldots,Z_m S , Z 1 , … , Z m a > 0 a>0 a > 0 F ≠ 0 F\ne0 F = 0 m j ≠ 0 m_j\ne0 m j = 0
K = ∣ S ∣ 2 − a ∣ S ∣ 4 + ∑ j ∣ Z j ∣ 2 , W = F S + 1 2 ∑ j m j Z j 2 . K = |S|^2-a|S|^4+\sum_j |Z_j|^2,\qquad W=FS+\frac{1}{2}\sum_j m_j Z_j^2. K = ∣ S ∣ 2 − a ∣ S ∣ 4 + j ∑ ∣ Z j ∣ 2 , W = F S + 2 1 j ∑ m j Z j 2 . We work on ∣ S ∣ 2 < 1 / ( 4 a ) |S|^2<1/(4a) ∣ S ∣ 2 < 1/ ( 4 a )
Proposition 21.1 (A stabilized local model). For this specified classical global effective theory,
V = ∣ F ∣ 2 1 − 4 a ∣ S ∣ 2 + ∑ j ∣ m j ∣ 2 ∣ Z j ∣ 2 V=\frac{|F|^2}{1-4a|S|^2}+\sum_j |m_j|^2|Z_j|^2 V = 1 − 4 a ∣ S ∣ 2 ∣ F ∣ 2 + j ∑ ∣ m j ∣ 2 ∣ Z j ∣ 2 has a unique minimum at the origin of the stated domain. Supersymmetry is broken there. In canonical real coordinates its Hessian has diagonal entries 4 a ∣ F ∣ 2 4a|F|^2 4 a ∣ F ∣ 2 4 a ∣ F ∣ 2 4a|F|^2 4 a ∣ F ∣ 2 ∣ m j ∣ 2 |m_j|^2 ∣ m j ∣ 2 ∣ m j ∣ 2 |m_j|^2 ∣ m j ∣ 2 j j j
Proof. K S S ˉ = 1 − 4 a ∣ S ∣ 2 K_{S\bar S}=1-4a|S|^2 K S S ˉ = 1 − 4 a ∣ S ∣ 2 K Z i Z ˉ j = δ i j K_{Z_i\bar Z_j}=\delta_{ij} K Z i Z ˉ j = δ ij K A B ˉ W A W ‾ B K^{A\bar B}W_A\overline{W}_B K A B ˉ W A W B ∣ S ∣ 2 |S|^2 ∣ S ∣ 2 S S S W S = F W_S=F W S = F S = ( x + i y ) / 2 S=(x+iy)/\sqrt{2} S = ( x + i y ) / 2 V = ∣ F ∣ 2 + 2 a ∣ F ∣ 2 ( x 2 + y 2 ) + ∑ j ∣ m j ∣ 2 ( u j 2 + v j 2 ) / 2 + O ( ∣ S ∣ 4 ) V=|F|^2+2a|F|^2(x^2+y^2)+\sum_j |m_j|^2(u_j^2+v_j^2)/2+O(|S|^4) V = ∣ F ∣ 2 + 2 a ∣ F ∣ 2 ( x 2 + y 2 ) + ∑ j ∣ m j ∣ 2 ( u j 2 + v j 2 ) /2 + O ( ∣ S ∣ 4 )
This proposition supplies no Calabi–Yau bundle realizing its K K K W W W
Proposition 21.2 (A tree-level obstruction). Let a supergravity model have
K = K 0 ( z , z ˉ ) − ∑ i = 1 r n i log ( T i + T ˉ i ) , n i > 0 , ∑ i n i = 3 , W = W ( z ) . K=K_0(z,\bar z)-\sum_{i=1}^{r}n_i\log(T_i+\bar T_i),\qquad n_i>0,\qquad\sum_i n_i=3,\qquad W=W(z). K = K 0 ( z , z ˉ ) − i = 1 ∑ r n i log ( T i + T ˉ i ) , n i > 0 , i ∑ n i = 3 , W = W ( z ) . At a point with D z W = 0 D_zW=0 D z W = 0 T i T_i T i
Proof. The T T T K i = − n i / ( T i + T ˉ i ) K_i=-n_i/(T_i+\bar T_i) K i = − n i / ( T i + T ˉ i ) K i i ˉ = n i / ( T i + T ˉ i ) 2 K_{i\bar i}=n_i/(T_i+\bar T_i)^2 K i i ˉ = n i / ( T i + T ˉ i ) 2 ∑ i K i i ˉ K i K i ˉ = ∑ i n i = 3 \sum_i K^{i\bar i}K_iK_{\bar i}=\sum_i n_i=3 ∑ i K i i ˉ K i K i ˉ = ∑ i n i = 3 D i W = K i W D_iW=K_iW D i W = K i W z z z − 3 ∣ W ∣ 2 -3|W|^2 − 3∣ W ∣ 2 V F = e K ( K A B ˉ D A W D B W ‾ − 3 ∣ W ∣ 2 ) V_F=e^K(K^{A\bar B}D_AW\overline{D_BW}-3|W|^2) V F = e K ( K A B ˉ D A W D B W − 3∣ W ∣ 2 ) T T T D D D
Conclusion and verification scope
The algebraic spectral calculation and the existence of a viable vacuum are separate statements. The mirror argument restricts a selector on its specified domain; it does not supply a point outside that domain. The explicit tetraquadric construction supplies a free quotient, index and projected charged spectrum at one fixed complex structure. Its residual gauge generator and deficient perturbative Yukawa matrices are exact obstructions to using that split point as the sought vacuum. The extension branch changes the bundle and therefore requires its own stability, operator and potential calculations before replacing the audit. § 17 shows that this branch is the only one left on the tetraquadric: no sum of five line bundles on a threefold with at most four Kähler parameters passes the gauge condition at its split locus, and on the tetraquadric the selection rules already remove six of the eleven three-family split data from the Yukawa condition. On the branch itself a single extension never removes the forced gauge boson, while two rank-two extensions do so for the fixed datum and for five of the ten further data, on the cover and on one side of a stability wall, and cannot do so for the other four. What the branch has not been shown to do is to descend to the quotient, to keep three families with the required Yukawa matrices, and to lift the flat directions of C7W; the four classes closed at the two smallest shapes could only pass C3 on a shape with a block of rank three or more. With the windows of Definition 9.1 the conjecture is a definite statement, and every failure recorded here is independent of them.
The remaining task is to specify a single quantized compactification with its hidden sector and controlled effective potential, and to verify all nine predicates at its stabilized point. The estimates in this paper state sufficient analytic conditions for parts of that task. Their inputs have not been established for the fixed example. Consequently this manuscript does not prove the full Geometric Selection Conjecture.
Data and declarations
The spectral sections and their five appendices present the work of the two authors on the Dwork and weighted Fermat pencils [27 ] ; the mirror argument is that of [28 ] under its standard-type partial-mirror convention; the marked-point and effective-theory criteria reorganize [32 ] . The bundle example is the published one of [29 ] , and the ten further line-bundle sums of § 17.2 were produced by the scan of [41 ] .
Python programs recompute the smooth-seed polynomial identities, the fixed-pencil determinant and character cohomology, the Dwork label, coupling and mass-map counts, the thirteen-row Chern and index arithmetic, the selection rules of Proposition 16.4 with their flow certificates, and the slope spaces, orientations, extension dimensions and polystable points of Theorem 16.10 with the permutation matrices of Lemma 16.9. Each program emits a machine-readable result with its scope. They are in the repository creelie/geometric-selection-conjecture on GitHub (release v3.1.0, concept DOI 10.5281/zenodo.23567114). Finite algebra checks do not certify string corrections, stabilize moduli, or supply a proof of all nine conditions; no statement of the paper rests on them alone.
Funding. This research received no external funding.
Conflicts of interest. The authors declare no conflict of interest.
Appendix A. The Hessian in the socle
The Hessian matrix of F ψ F_{\psi} F ψ
H i i = 20 x i 3 , H i j = − 5 ψ ∏ k ≠ i , j x k ( i ≠ j ) . H_{ii}=20x_i^3,\qquad H_{ij}=-5\psi\prod_{k\ne i,j}x_k\quad(i\ne j). H ii = 20 x i 3 , H ij = − 5 ψ k = i , j ∏ x k ( i = j ) . For a permutation π \pi π { 1 , … , 5 } \{1,\ldots,5\} { 1 , … , 5 } m m m f = 5 − m f=5-m f = 5 − m ∏ i H i π ( i ) \prod_i H_{i\pi(i)} ∏ i H iπ ( i ) 20 f ( − 5 ψ ) m 20^f(-5\psi)^m 2 0 f ( − 5 ψ ) m a a a 3 3 3 1 1 1 m m m ( i , π ( i ) ) (i,\pi(i)) ( i , π ( i )) i i i π ( i ) \pi(i) π ( i ) c c c i = c i=c i = c π ( i ) = c \pi(i)=c π ( i ) = c π ( c ) ≠ c \pi(c)\ne c π ( c ) = c 1 1 1 m − 2 m-2 m − 2 π \pi π
∏ a f i x e d x a 3 + m ∏ c m o v e d x c m − 2 , \prod_{a\ \mathrm{fixed}}x_a^{3+m}\prod_{c\ \mathrm{moved}}x_c^{m-2}, a fixed ∏ x a 3 + m c moved ∏ x c m − 2 , of degree f ( 3 + m ) + m ( m − 2 ) = 15 f(3+m)+m(m-2)=15 f ( 3 + m ) + m ( m − 2 ) = 15 m m m
m = 0 m=0 m = 0 m 0 = ( 3 , 3 , 3 , 3 , 3 ) m_0=(3,3,3,3,3) m 0 = ( 3 , 3 , 3 , 3 , 3 )
m = 2 m=2 m = 2 ( 5 , 5 , 5 , 0 , 0 ) (5,5,5,0,0) ( 5 , 5 , 5 , 0 , 0 ) λ = ψ 3 \lambda=\psi^3 λ = ψ 3
m = 3 m=3 m = 3 ( 6 , 6 , 1 , 1 , 1 ) (6,6,1,1,1) ( 6 , 6 , 1 , 1 , 1 ) λ = ψ 2 \lambda=\psi^2 λ = ψ 2
m = 4 m=4 m = 4 ( 7 , 2 , 2 , 2 , 2 ) (7,2,2,2,2) ( 7 , 2 , 2 , 2 , 2 ) λ = ψ \lambda=\psi λ = ψ
m = 5 m=5 m = 5 ( 3 , 3 , 3 , 3 , 3 ) (3,3,3,3,3) ( 3 , 3 , 3 , 3 , 3 )
The contribution of a permutation to H ( ψ ) = λ ( Hess F ψ ) H(\psi)=\lambda(\operatorname{Hess} F_\psi) H ( ψ ) = λ ( Hess F ψ ) sgn ( π ) 20 f ( − 5 ψ ) m ψ r ( m ) \operatorname{sgn}(\pi)\,20^{f(-5\psi)}m\psi^{r(m)} sgn ( π ) 2 0 f ( − 5 ψ ) m ψ r ( m ) r ( 0 ) = r ( 5 ) = 0 r(0)=r(5)=0 r ( 0 ) = r ( 5 ) = 0 r ( 2 ) = 3 r(2)=3 r ( 2 ) = 3 r ( 3 ) = 2 r(3)=2 r ( 3 ) = 2 r ( 4 ) = 1 r(4)=1 r ( 4 ) = 1 m + r ( m ) = 5 m+r(m)=5 m + r ( m ) = 5 m ≥ 2 m\ge2 m ≥ 2 ψ 5 \psi^5 ψ 5
1 5 : 1 permutation , m = 0 , + 20 5 = 3 200 000 ; 2 1 3 : 10 , odd , m = 2 : 10 ⋅ ( − 1 ) ⋅ 20 3 ⋅ 25 = − 2 000 000 ; 3 1 2 : 20 , even , m = 3 : 20 ⋅ 20 2 ⋅ ( − 125 ) = − 1 000 000 ; 4 1 : 30 , odd , m = 4 : 30 ⋅ ( − 1 ) ⋅ 20 ⋅ 625 = − 375 000 ; 2 2 1 : 15 , even , m = 4 : 15 ⋅ 20 ⋅ 625 = 187 500 ; 5 : 24 , even , m = 5 : 24 ⋅ ( − 3125 ) = − 75 000 ; 3 2 : 20 , odd , m = 5 : 20 ⋅ ( − 1 ) ⋅ ( − 3125 ) = 62 500. \begin{aligned}
1^5 &: \ 1\text{ permutation},\ m=0,\ +20^5=3\,200\,000;\\
2\,1^3 &: \ 10,\ \text{odd},\ m=2:\ 10\cdot(-1)\cdot20^3\cdot25=-2\,000\,000;\\
3\,1^2 &: \ 20,\ \text{even},\ m=3:\ 20\cdot20^2\cdot(-125)=-1\,000\,000;\\
4\,1 &: \ 30,\ \text{odd},\ m=4:\ 30\cdot(-1)\cdot20\cdot625=-375\,000;\\
2^2\,1 &: \ 15,\ \text{even},\ m=4:\ 15\cdot20\cdot625=187\,500;\\
5 &: \ 24,\ \text{even},\ m=5:\ 24\cdot(-3125)=-75\,000;\\
3\,2 &: \ 20,\ \text{odd},\ m=5:\ 20\cdot(-1)\cdot(-3125)=62\,500.
\end{aligned} 1 5 2 1 3 3 1 2 4 1 2 2 1 5 3 2 : 1 permutation , m = 0 , + 2 0 5 = 3 200 000 ; : 10 , odd , m = 2 : 10 ⋅ ( − 1 ) ⋅ 2 0 3 ⋅ 25 = − 2 000 000 ; : 20 , even , m = 3 : 20 ⋅ 2 0 2 ⋅ ( − 125 ) = − 1 000 000 ; : 30 , odd , m = 4 : 30 ⋅ ( − 1 ) ⋅ 20 ⋅ 625 = − 375 000 ; : 15 , even , m = 4 : 15 ⋅ 20 ⋅ 625 = 187 500 ; : 24 , even , m = 5 : 24 ⋅ ( − 3125 ) = − 75 000 ; : 20 , odd , m = 5 : 20 ⋅ ( − 1 ) ⋅ ( − 3125 ) = 62 500. where the last six numbers are the coefficients of ψ 5 \psi^5 ψ 5 1 + 10 + 20 + 30 + 15 + 24 + 20 1+10+20+30+15+24+20 1 + 10 + 20 + 30 + 15 + 24 + 20 ψ 5 \psi^5 ψ 5 − 3 200 000 -3\,200\,000 − 3 200 000 H ( ψ ) = 3 200 000 ( 1 − ψ 5 ) H(\psi)=3\,200\,000(1-\psi^5) H ( ψ ) = 3 200 000 ( 1 − ψ 5 )
The coupling count
Ordered triples by sector
B.1. Ordered triples by sector. For e \mathbf{e} e O ( e ) O(\mathbf{e}) O ( e ) ( a , b , c ) ∈ B 3 (\mathbf{a},\mathbf{b},\mathbf{c})\in\mathcal{B}^3 ( a , b , c ) ∈ B 3 a + b + c = e \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{e} a + b + c = e c ∈ B \mathbf{c}\in\mathcal{B} c ∈ B c ≤ e \mathbf{c}\le\mathbf{e} c ≤ e s = e − c \mathbf{s}=\mathbf{e}-\mathbf{c} s = e − c a ∈ B \mathbf{a}\in\mathcal{B} a ∈ B s − a ∈ B \mathbf{s}-\mathbf{a}\in\mathcal{B} s − a ∈ B
max ( 0 , s i − 3 ) ≤ a i ≤ min ( 3 , s i ) for all i , ∑ a i = 5. \max(0,s_i-3)\le a_i\le\min(3,s_i)\quad\text{for all }i,\qquad\sum a_i=5. max ( 0 , s i − 3 ) ≤ a i ≤ min ( 3 , s i ) for all i , ∑ a i = 5. This number N ( s ) N(\mathbf{s}) N ( s ) t 5 t^5 t 5 ∏ i ( t max ( 0 , s i − 3 ) + ⋯ + t min ( 3 , s i ) ) \prod_i\left(t^{\max(0,s_i-3)}+\cdots+t^{\min(3,s_i)}\right) ∏ i ( t m a x ( 0 , s i − 3 ) + ⋯ + t m i n ( 3 , s i ) ) [ n ] = 1 + t + ⋯ + t n − 1 [n]=1+t+\cdots+t^{n-1} [ n ] = 1 + t + ⋯ + t n − 1
Sector 0, e = ( 3 , 3 , 3 , 3 , 3 ) \mathbf{e}=(3,3,3,3,3) e = ( 3 , 3 , 3 , 3 , 3 ) Here s = 3 1 − c \mathbf{s}=\mathbf{3\!1}-\mathbf{c} s = 3 1 − c { 0 , … , 3 } \{0,\ldots,3\} { 0 , … , 3 } N ( s ) N(\mathbf{s}) N ( s ) t 5 t^5 t 5 ∏ i [ s i + 1 ] \prod_i[s_i+1] ∏ i [ s i + 1 ] s \mathbf{s} s c \mathbf{c} c
Table 1 (PDF p. 41).
The values of N N N { 3 , 3 , 3 , 1 , 0 } \{3,3,3,1,0\} { 3 , 3 , 3 , 1 , 0 } [ 4 ] 3 = 1 + 3 t + 6 t 2 + 10 t 3 + 12 t 4 + 12 t 5 + ⋯ [4]^3=1+3t+6t^2+10t^3+12t^4+12t^5+\cdots [ 4 ] 3 = 1 + 3 t + 6 t 2 + 10 t 3 + 12 t 4 + 12 t 5 + ⋯ [ 2 ] [2] [ 2 ] 12 + 12 = 24 12+12=24 12 + 12 = 24 { 3 , 3 , 2 , 2 , 0 } \{3,3,2,2,0\} { 3 , 3 , 2 , 2 , 0 } [ 4 ] 2 = ( 1 , 2 , 3 , 4 , 3 , 2 , 1 ) [4]^2=(1,2,3,4,3,2,1) [ 4 ] 2 = ( 1 , 2 , 3 , 4 , 3 , 2 , 1 ) [ 3 ] 2 = ( 1 , 2 , 3 , 2 , 1 ) [3]^2=(1,2,3,2,1) [ 3 ] 2 = ( 1 , 2 , 3 , 2 , 1 ) 2 + 6 + 12 + 6 + 2 = 28 2+6+12+6+2=28 2 + 6 + 12 + 6 + 2 = 28 { 3 , 3 , 2 , 1 , 1 } \{3,3,2,1,1\} { 3 , 3 , 2 , 1 , 1 } [ 3 ] [ 2 ] 2 = ( 1 , 3 , 4 , 3 , 1 ) [3][2]^2=(1,3,4,3,1) [ 3 ] [ 2 ] 2 = ( 1 , 3 , 4 , 3 , 1 ) 2 + 9 + 16 + 9 + 2 = 38 2+9+16+9+2=38 2 + 9 + 16 + 9 + 2 = 38 { 3 , 2 , 2 , 2 , 1 } \{3,2,2,2,1\} { 3 , 2 , 2 , 2 , 1 } [ 3 ] 3 = ( 1 , 3 , 6 , 7 , 6 , 3 , 1 ) [3]^3=(1,3,6,7,6,3,1) [ 3 ] 3 = ( 1 , 3 , 6 , 7 , 6 , 3 , 1 ) [ 4 ] [ 2 ] = ( 1 , 2 , 2 , 2 , 1 ) [4][2]=(1,2,2,2,1) [ 4 ] [ 2 ] = ( 1 , 2 , 2 , 2 , 1 ) 3 + 12 + 14 + 12 + 3 = 44 3+12+14+12+3=44 3 + 12 + 14 + 12 + 3 = 44 { 2 , 2 , 2 , 2 , 2 } \{2,2,2,2,2\} { 2 , 2 , 2 , 2 , 2 } t 5 t^5 t 5 ( 1 + t + t 2 ) 5 (1+t+t^2)^5 ( 1 + t + t 2 ) 5
O ( 3 , 3 , 3 , 3 , 3 ) = 20 ⋅ 24 + 30 ⋅ 28 + 30 ⋅ 38 + 20 ⋅ 44 + 51 = 3391. O(3,3,3,3,3)=20\cdot24+30\cdot28+30\cdot38+20\cdot44+51=3391. O ( 3 , 3 , 3 , 3 , 3 ) = 20 ⋅ 24 + 30 ⋅ 28 + 30 ⋅ 38 + 20 ⋅ 44 + 51 = 3391. (The table for e = ( 3 , 3 , 3 , 3 , 3 ) \mathbf{e}=(3,3,3,3,3) e = ( 3 , 3 , 3 , 3 , 3 ) c ∈ B \mathbf{c}\in\mathcal{B} c ∈ B s = 3 1 − c \mathbf{s}=\mathbf{3\!1}-\mathbf{c} s = 3 1 − c a \mathbf{a} a
Sector 1 , e = ( 7 , 2 , 2 , 2 , 2 ) \mathbf{e}=(7,2,2,2,2) e = ( 7 , 2 , 2 , 2 , 2 ) a 1 + b 1 ≤ 6 a_1+b_1\le6 a 1 + b 1 ≤ 6 c 1 ≥ 1 c_1\ge1 c 1 ≥ 1 c i ≤ 2 c_i\le2 c i ≤ 2 i ≥ 2 i\ge2 i ≥ 2 s = e − c \mathbf{s}=\mathbf{e}-\mathbf{c} s = e − c s 1 = 7 − c 1 s_1=7-c_1 s 1 = 7 − c 1 a 1 ≥ 4 − c 1 a_1\ge4-c_1 a 1 ≥ 4 − c 1 s i = 2 − c i ≤ 2 s_i=2-c_i\le2 s i = 2 − c i ≤ 2 a i ≤ s i a_i\le s_i a i ≤ s i c \mathbf{c} c c 1 c_1 c 1 ( c 2 , … , c 5 ) (c_2,\ldots,c_5) ( c 2 , … , c 5 )
For instance, when c 1 = 3 c_1=3 c 1 = 3 ( c 2 , … , c 5 ) = ( 2 , 0 , 0 , 0 ) (c_2,\ldots,c_5)=(2,0,0,0) ( c 2 , … , c 5 ) = ( 2 , 0 , 0 , 0 ) a 1 a_1 a 1 { 1 , 2 , 3 } \{1,2,3\} { 1 , 2 , 3 } a i ≤ ( 0 , 2 , 2 , 2 ) a_i\le(0,2,2,2) a i ≤ ( 0 , 2 , 2 , 2 ) 5 − a 1 5-a_1 5 − a 1 t 4 t^4 t 4 t 3 t^3 t 3 t 2 t^2 t 2 [ 3 ] 3 [3]^3 [ 3 ] 3 O ( 7 , 2 , 2 , 2 , 2 ) = 480 O(7,2,2,2,2)=480 O ( 7 , 2 , 2 , 2 , 2 ) = 480
Sector 2 , e = ( 6 , 6 , 1 , 1 , 1 ) \mathbf{e}=(6,6,1,1,1) e = ( 6 , 6 , 1 , 1 , 1 ) c 3 , c 4 , c 5 ≤ 1 c_3,c_4,c_5\le1 c 3 , c 4 , c 5 ≤ 1
and O ( 6 , 6 , 1 , 1 , 1 ) = 162 O(6,6,1,1,1)=162 O ( 6 , 6 , 1 , 1 , 1 ) = 162
Sector 3 , e = ( 5 , 5 , 5 , 0 , 0 ) \mathbf{e}=(5,5,5,0,0) e = ( 5 , 5 , 5 , 0 , 0 ) c 4 = c 5 = 0 c_4=c_5=0 c 4 = c 5 = 0 c 1 + c 2 + c 3 = 5 c_1+c_2+c_3=5 c 1 + c 2 + c 3 = 5 { c 1 , c 2 , c 3 } = { 2 , 2 , 1 } \{c_1,c_2,c_3\}=\{2,2,1\} { c 1 , c 2 , c 3 } = { 2 , 2 , 1 } s \mathbf{s} s ( 3 , 3 , 4 ) (3,3,4) ( 3 , 3 , 4 ) a i ≤ 3 a_i\le3 a i ≤ 3 1 ≤ a 3 ≤ 3 1\le a_3\le3 1 ≤ a 3 ≤ 3 t 5 t^5 t 5 [ 4 ] 2 ( t + t 2 + t 3 ) [4]^2(t+t^2+t^3) [ 4 ] 2 ( t + t 2 + t 3 ) 3 + 4 + 3 = 10 3+4+3=10 3 + 4 + 3 = 10 { 3 , 1 , 1 } \{3,1,1\} { 3 , 1 , 1 } ( 2 , 4 , 4 ) (2,4,4) ( 2 , 4 , 4 ) t 5 t^5 t 5 [ 3 ] ( t + t 2 + t 3 ) 2 [3](t+t^2+t^3)^2 [ 3 ] ( t + t 2 + t 3 ) 2 2 + 3 + 2 = 7 2+3+2=7 2 + 3 + 2 = 7 { 3 , 2 , 0 } \{3,2,0\} { 3 , 2 , 0 } ( 2 , 3 , 5 ) (2,3,5) ( 2 , 3 , 5 ) t 5 t^5 t 5 [ 3 ] [ 4 ] ( t 2 + t 3 ) [3][4](t^2+t^3) [ 3 ] [ 4 ] ( t 2 + t 3 ) 3 + 3 = 6 3+3=6 3 + 3 = 6 c \mathbf{c} c O ( 5 , 5 , 5 , 0 , 0 ) = 30 + 21 + 36 = 87 O(5,5,5,0,0)=30+21+36=87 O ( 5 , 5 , 5 , 0 , 0 ) = 30 + 21 + 36 = 87
The ordered counts by sector are O 0 = 3391 O_0=3391 O 0 = 3391 O 1 = 5 ⋅ 480 = 2400 O_1=5\cdot480=2400 O 1 = 5 ⋅ 480 = 2400 O 2 = 10 ⋅ 162 = 1620 O_2=10\cdot162=1620 O 2 = 10 ⋅ 162 = 1620 O 3 = 10 ⋅ 87 = 870 O_3=10\cdot87=870 O 3 = 10 ⋅ 87 = 870 § 5.1 .
Repeated labels
For a ∈ B \mathbf{a}\in B a ∈ B c \mathbf{c} c 2 a + c 2\mathbf{a}+\mathbf{c} 2 a + c Theorem 1.3 . For a = 1 \mathbf{a}=1 a = 1 c = 1 \mathbf{c}=1 c = 1 a \mathbf{a} a { 2 , 1 , 1 , 1 , 0 } \{2,1,1,1,0\} { 2 , 1 , 1 , 1 , 0 } c i ≡ 2 − 2 a i c_i\equiv2-2a_i c i ≡ 2 − 2 a i c \mathbf{c} c a \mathbf{a} a a \mathbf{a} a { 3 , 1 , 1 , 0 , 0 } \{3,1,1,0,0\} { 3 , 1 , 1 , 0 , 0 } a \mathbf{a} a { 3 , 2 , 0 , 0 , 0 } \{3,2,0,0,0\} { 3 , 2 , 0 , 0 , 0 } { 2 , 2 , 1 , 0 , 0 } \{2,2,1,0,0\} { 2 , 2 , 1 , 0 , 0 } { a , a , c } \{\mathbf{a},\mathbf{a},\mathbf{c}\} { a , a , c } c ≠ a \mathbf{c}\ne\mathbf{a} c = a D r = 0 , 50 , 20 , 30 D_r=0,50,20,30 D r = 0 , 50 , 20 , 30 § 5.1 .
Appendix C. Hodge norms at the Fermat point
This appendix proves Proposition 7.1. Write G = μ 5 5 / μ 5 G=\mu_5^5/\mu_5 G = μ 5 5 / μ 5 P 4 \mathbb{P}^4 P 4 X 0 X_0 X 0 k ∈ { 1 , 2 , 3 , 4 } 5 \mathbf{k}\in\{1,2,3,4\}^5 k ∈ { 1 , 2 , 3 , 4 } 5 ∑ k i ≡ 0 ( m o d 5 ) \sum k_i\equiv0\pmod5 ∑ k i ≡ 0 ( mod 5 ) χ k ( a ) = ζ a ⋅ k \chi_{\mathbf{k}}(a)=\zeta^{\mathbf{a}\cdot\mathbf{k}} χ k ( a ) = ζ a ⋅ k
Eigenlines
The character χ k \chi_{\mathbf{k}} χ k H 3 ( X 0 , C ) H^3(X_0,\mathbb{C}) H 3 ( X 0 , C ) k \mathbf{k} k 1 5 ∑ j = 0 4 ( ∑ k = 1 4 ζ j k ) 5 = 1 5 ( 4 5 − 4 ) = 204 = b 3 ( X 0 ) \frac{1}{5}\sum_{j=0}^{4}\left(\sum_{k=1}^{4}\zeta^{jk}\right)^5=\frac{1}{5}(4^5-4)=204=b_3(X_0) 5 1 ∑ j = 0 4 ( ∑ k = 1 4 ζ j k ) 5 = 5 1 ( 4 5 − 4 ) = 204 = b 3 ( X 0 ) Res ( x k − 1 Ω 0 / F q + 1 ) \operatorname{Res}(x^{\mathbf{k}-1}\Omega_0/F^{q+1}) Res ( x k − 1 Ω 0 / F q + 1 ) 5 ( q + 1 ) = ∑ k i 5(q+1)=\sum k_i 5 ( q + 1 ) = ∑ k i χ k \chi_{\mathbf{k}} χ k x k − 1 x^{\mathbf{k}-1} x k − 1 C [ x ] / ( x 1 4 , … , x 5 4 ) \mathbb{C}[x]/(x_1^4,\ldots,x_5^4) C [ x ] / ( x 1 4 , … , x 5 4 ) ω b \omega_{\mathbf{b}} ω b k = b + 1 \mathbf{k}=\mathbf{b}+1 k = b + 1
The bilinear relation
Let Q Q Q H 3 ( X 0 , Z ) H_3(X_0,\mathbb{Z}) H 3 ( X 0 , Z ) H 3 H^3 H 3 H 3 H_3 H 3 D D D ∫ c ω = Q ( c , D ω ) \int_c\omega=Q(c,D\omega) ∫ c ω = Q ( c , D ω ) ∫ X 0 ω ∧ η = Q ( D ω , D η ) \int_{X_0}\omega\wedge\eta=Q(D\omega,D\eta) ∫ X 0 ω ∧ η = Q ( D ω , D η ) Q Q Q G G G L L L χ \chi χ L ′ L' L ′ χ ‾ \overline{\chi} χ e e e u u u L ′ L' L ′ L L L u ‾ \overline{u} u e e e D ω b = α u ‾ D\omega_{\mathbf{b}}=\alpha\overline{u} D ω b = α u u ‾ ≠ 0 \overline{u}\ne0 u = 0
∫ e ω b = Q ( e , α u ‾ ) = α Q ( u , u ‾ ) , ∫ X 0 ω b ∧ ω b ‾ = ∣ α ∣ 2 Q ( u ‾ , u ) , \int_e\omega_{\mathbf{b}}=Q(e,\alpha\overline{u})=\alpha Q(u,\overline{u}),\qquad
\int_{X_0}\omega_{\mathbf{b}}\wedge\overline{\omega_{\mathbf{b}}}=|\alpha|^2Q(\overline{u},u), ∫ e ω b = Q ( e , α u ) = α Q ( u , u ) , ∫ X 0 ω b ∧ ω b = ∣ α ∣ 2 Q ( u , u ) , and therefore
g b = ∣ ∫ X 0 ω b ∧ ω b ‾ ∣ = ∣ ∫ e ω b ∣ 2 ∣ Q ( u , u ‾ ) ∣ . (18) g_{\mathbf{b}}=\left|\int_{X_0}\omega_{\mathbf{b}}\wedge\overline{\omega_{\mathbf{b}}}\right|=\frac{\left|\int_e\omega_{\mathbf{b}}\right|^2}{|Q(u,\overline{u})|}. \tag*{(18)} g b = ∫ X 0 ω b ∧ ω b = ∣ Q ( u , u ) ∣ ∫ e ω b 2 . (18) It remains to compute both sides for one cycle.
The cycle of Pham
On the affine chart x 5 = 1 x_5=1 x 5 = 1 y i = x i / x 5 y_i=x_i/x_5 y i = x i / x 5 X 0 X_0 X 0 U = { f = 0 } U=\{f=0\} U = { f = 0 } f = y 1 5 + y 2 5 + y 3 5 + y 4 5 + 1 f=y_1^5+y_2^5+y_3^5+y_4^5+1 f = y 1 5 + y 2 5 + y 3 5 + y 4 5 + 1 μ 5 4 \mu_5^4 μ 5 4 y i ↦ ζ a i y i y_i\mapsto\zeta^{a_i}y_i y i ↦ ζ a i y i k \mathbf{k} k G G G ( k 1 , k 2 , k 3 , k 4 ) (k_1,k_2,k_3,k_4) ( k 1 , k 2 , k 3 , k 4 ) μ 5 4 \mu_5^4 μ 5 4 k 5 ≡ − ( k 1 + k 2 + k 3 + k 4 ) k_5\equiv-(k_1+k_2+k_3+k_4) k 5 ≡ − ( k 1 + k 2 + k 3 + k 4 ) [14 ] showed that U U U − 1 -1 − 1 H 3 ( U , Z ) H_3(U,\mathbb{Z}) H 3 ( U , Z ) Z [ μ 5 4 ] \mathbb{Z}[\mu_5^4] Z [ μ 5 4 ]
e = ( 1 − T 1 ) ( 1 − T 2 ) ( 1 − T 3 ) ( 1 − T 4 ) Δ , e=(1-T_1)(1-T_2)(1-T_3)(1-T_4)\Delta, e = ( 1 − T 1 ) ( 1 − T 2 ) ( 1 − T 3 ) ( 1 − T 4 ) Δ , where Δ \Delta Δ { y i = ϵ v i , v i ≥ 0 , ∑ i v i 5 = 1 } \{y_i=\epsilon v_i,\ v_i\ge0,\ \sum_i v_i^5=1\} { y i = ϵ v i , v i ≥ 0 , ∑ i v i 5 = 1 } ϵ 5 = − 1 \epsilon^5=-1 ϵ 5 = − 1 T i T_i T i y i y_i y i ζ \zeta ζ
The period. By Griffiths’ residue formula, ∫ e ω b = 1 2 π i ∫ τ ( e ) x b Ω 0 / F 2 \int_e\omega_{\mathbf{b}}=\frac{1}{2\pi i}\int_{\tau(e)}x^{\mathbf{b}}\Omega_0/F^2 ∫ e ω b = 2 π i 1 ∫ τ ( e ) x b Ω 0 / F 2 τ \tau τ y b ′ d y / f 2 y^{\mathbf{b}'}dy/f^2 y b ′ d y / f 2 b ′ = ( b 1 , … , b 4 ) \mathbf{b}'=(b_1,\ldots,b_4) b ′ = ( b 1 , … , b 4 ) f s = ∑ y i 5 + s f_s=\sum y_i^5+s f s = ∑ y i 5 + s y b ′ d y / f s 2 = − ∂ s ( y b ′ d y / f s ) y^{\mathbf{b}'}dy/f_s^2=-\partial_s(y^{\mathbf{b}'}dy/f_s) y b ′ d y / f s 2 = − ∂ s ( y b ′ d y / f s ) 1 2 π i ∫ τ ( e ) y b ′ d y / f s = ∫ e s y b ′ d y / d f s \frac{1}{2\pi i}\int_{\tau(e)}y^{\mathbf{b}'}dy/f_s=\int_{e_s}y^{\mathbf{b}'}dy/df_s 2 π i 1 ∫ τ ( e ) y b ′ d y / f s = ∫ e s y b ′ d y / d f s y = s 1 / 5 u y=s^{1/5}u y = s 1/5 u s A P s^A P s A P A = 1 5 ∑ i ≤ 4 k i − 1 A=\frac{1}{5}\sum_{i\le4}k_i-1 A = 5 1 ∑ i ≤ 4 k i − 1 P P P s = 1 s=1 s = 1 s = 1 s=1 s = 1
∫ e ω b = − A P . \int_e\omega_{\mathbf{b}}=-AP. ∫ e ω b = − A P . On Δ \Delta Δ t i = v i 5 t_i=v_i^5 t i = v i 5 5 − 4 ∏ i t i 1 / 5 − 1 d t 5^{-4}\prod_i t_i^{1/5-1}\,dt 5 − 4 ∏ i t i 1/5 − 1 d t ∑ t i = 1 \sum t_i=1 ∑ t i = 1 y b ′ = ϵ ∣ b ′ ∣ ∏ t i b i / 5 y^{\mathbf{b}'}=\epsilon^{|\mathbf{b}'|}\prod t_i^{b_i/5} y b ′ = ϵ ∣ b ′ ∣ ∏ t i b i /5 ∫ Δ = ϵ ∣ b ′ ∣ + 4 5 − 4 ∏ i ≤ 4 Γ ( k i / 5 ) / Γ ( ∑ i ≤ 4 k i / 5 ) \int_\Delta=\epsilon^{|\mathbf{b}'|+4}5^{-4}\prod_{i\le4}\Gamma(k_i/5)/\Gamma(\sum_{i\le4}k_i/5) ∫ Δ = ϵ ∣ b ′ ∣ + 4 5 − 4 ∏ i ≤ 4 Γ ( k i /5 ) /Γ ( ∑ i ≤ 4 k i /5 ) 1 − T i 1-T_i 1 − T i 1 − ζ k i 1-\zeta^{k_i} 1 − ζ k i ∑ i ≤ 4 k i = 10 − k 5 \sum_{i\le4}k_i=10-k_5 ∑ i ≤ 4 k i = 10 − k 5 A = 1 − k 5 / 5 A=1-k_5/5 A = 1 − k 5 /5 Γ ( A + 1 ) = A Γ ( A ) \Gamma(A+1)=A\Gamma(A) Γ ( A + 1 ) = A Γ ( A )
∣ ∫ e ω b ∣ = 5 − 4 ∏ i = 1 4 ∣ 1 − ζ k i ∣ ∏ i = 1 4 Γ ( k i / 5 ) Γ ( 1 − k 5 / 5 ) . (19) \left|\int_e\omega_{\mathbf{b}}\right|=5^{-4}\prod_{i=1}^{4}|1-\zeta^{k_i}|\frac{\prod_{i=1}^{4}\Gamma(k_i/5)}{\Gamma(1-k_5/5)}. \tag*{(19)} ∫ e ω b = 5 − 4 i = 1 ∏ 4 ∣1 − ζ k i ∣ Γ ( 1 − k 5 /5 ) ∏ i = 1 4 Γ ( k i /5 ) . (19) The intersection number. For one variable, the reduced homology of the fibre { y 5 = − 1 } \{y^5=-1\} { y 5 = − 1 } I I I Z [ t ] / ( t 5 − 1 ) \mathbb{Z}[t]/(t^5-1) Z [ t ] / ( t 5 − 1 ) t t t t − 1 t-1 t − 1 L ( x , z ) = ⟨ x , Var − 1 z ⟩ L(x,z)=\langle x,\operatorname{Var}^{-1}z\rangle L ( x , z ) = ⟨ x , Var − 1 z ⟩ x ^ ( k ) = x ( ζ k ) \widehat{x}(k)=x(\zeta^k) x ( k ) = x ( ζ k )
L ( x , z ) = 1 5 ∑ k = 1 4 x ^ ( k ) z ^ ( − k ) ζ − k − 1 . L(x,z)=\frac{1}{5}\sum_{k=1}^{4}\frac{\widehat{x}(k)\widehat{z}(-k)}{\zeta^{-k}-1}. L ( x , z ) = 5 1 k = 1 ∑ 4 ζ − k − 1 x ( k ) z ( − k ) . By the theorem of Sebastiani and Thom [15 ] , in the form of Sakamoto [16 ] for Seifert forms, the Seifert form of y 1 5 + ⋯ + y 4 5 y_1^5+\cdots+y_4^5 y 1 5 + ⋯ + y 4 5 Q = − L + L T Q=-L+L^{T} Q = − L + L T ( k 1 , … , k 4 ) (k_{1},\ldots,k_{4}) ( k 1 , … , k 4 )
Q ( x , z ) = ± 1 5 4 x ^ ( k ) z ^ ( − k ) ( ∏ i ≤ 4 1 ζ k i − 1 − ∏ i ≤ 4 1 ζ − k i − 1 ) . Q(x,z)=\pm\frac{1}{5^{4}}\hat{x}(\mathbf{k})\hat{z}(-\mathbf{k})\left(\prod_{i\leq4}\frac{1}{\zeta^{k_i}-1}-\prod_{i\leq4}\frac{1}{\zeta^{-k_i}-1}\right). Q ( x , z ) = ± 5 4 1 x ^ ( k ) z ^ ( − k ) ( i ≤ 4 ∏ ζ k i − 1 1 − i ≤ 4 ∏ ζ − k i − 1 1 ) . With four factors, ∏ ( ζ k i − 1 ) = ζ ∑ k i ∏ ( ζ − k i − 1 ) \prod(\zeta^{k_i}-1)=\zeta^{\sum k_i}\prod(\zeta^{-k_i}-1) ∏ ( ζ k i − 1 ) = ζ ∑ k i ∏ ( ζ − k i − 1 )
∣ 1 − ζ k 1 + k 2 + k 3 + k 4 ∣ ∏ i ≤ 4 ∣ 1 − ζ k i ∣ 2 = ∣ 1 − ζ k 5 ∣ ∏ i ≤ 4 ∣ 1 − ζ k i ∣ 2 . \frac{\left|1-\zeta^{k_{1}+k_{2}+k_{3}+k_{4}}\right|}{\prod_{i\leq4}\left|1-\zeta^{k_i}\right|^{2}}=\frac{\left|1-\zeta^{k_{5}}\right|}{\prod_{i\leq4}\left|1-\zeta^{k_i}\right|^{2}}. ∏ i ≤ 4 ∣ 1 − ζ k i ∣ 2 1 − ζ k 1 + k 2 + k 3 + k 4 = ∏ i ≤ 4 ∣ 1 − ζ k i ∣ 2 1 − ζ k 5 . The cycle e e e e ^ ( k ) = ∏ i ≤ 4 ( 1 − ζ k i ) \hat{e}(\mathbf{k})=\prod_{i\leq4}(1-\zeta^{k_i}) e ^ ( k ) = ∏ i ≤ 4 ( 1 − ζ k i ) u u u
∣ Q ( u , u ˉ ) ∣ = 5 − 4 ∏ i = 1 5 ∣ 1 − ζ k i ∣ . (20) \left|Q(u,\bar{u})\right|=5^{-4}\prod_{i=1}^{5}\left|1-\zeta^{k_i}\right|. \tag*{(20)} ∣ Q ( u , u ˉ ) ∣ = 5 − 4 i = 1 ∏ 5 1 − ζ k i . (20) This is nonzero, so u ˉ ≠ 0 \bar{u}\neq0 u ˉ = 0 (18) .
Conclusion of the proof
Inserting (19) and (20) in (18) ,
g b = 5 − 4 ∏ i = 1 4 ∣ 1 − ζ k i ∣ Γ ( k i / 5 ) 2 ⋅ 1 ∣ 1 − ζ k 5 ∣ Γ ( 1 − k 5 / 5 ) 2 . g_{\mathbf{b}}=5^{-4}\prod_{i=1}^{4}\left|1-\zeta^{k_i}\right|\Gamma(k_i/5)^{2}\cdot\frac{1}{\left|1-\zeta^{k_5}\right|\Gamma(1-k_5/5)^{2}}. g b = 5 − 4 i = 1 ∏ 4 1 − ζ k i Γ ( k i /5 ) 2 ⋅ ∣ 1 − ζ k 5 ∣ Γ ( 1 − k 5 /5 ) 2 1 . With ∣ 1 − ζ k ∣ = 2 sin ( π k / 5 ) \left|1-\zeta^{k}\right|=2\sin(\pi k/5) 1 − ζ k = 2 sin ( π k /5 ) Γ ( t ) Γ ( 1 − t ) = π / sin π t \Gamma(t)\Gamma(1-t)=\pi/\sin\pi t Γ ( t ) Γ ( 1 − t ) = π / sin π t
2 sin ( π t ) Γ ( t ) 2 = 2 π γ ( t ) , 1 2 sin ( π t ) Γ ( 1 − t ) 2 = γ ( t ) 2 π , 2\sin(\pi t)\Gamma(t)^{2}=2\pi\gamma(t),\qquad\frac{1}{2\sin(\pi t)\Gamma(1-t)^{2}}=\frac{\gamma(t)}{2\pi}, 2 sin ( π t ) Γ ( t ) 2 = 2 π γ ( t ) , 2 sin ( π t ) Γ ( 1 − t ) 2 1 = 2 π γ ( t ) , so g b = 5 − 4 ( 2 π ) 3 ∏ i = 1 5 γ ( k i / 5 ) g_{\mathbf{b}}=5^{-4}(2\pi)^{3}\prod_{i=1}^{5}\gamma(k_i/5) g b = 5 − 4 ( 2 π ) 3 ∏ i = 1 5 γ ( k i /5 ) C = ( 2 π ) 3 / 5 4 C=(2\pi)^{3}/5^{4} C = ( 2 π ) 3 / 5 4 Ω \Omega Ω k = 1 \mathbf{k}=1 k = 1 ∣ ∫ e Ω ∣ = ∣ P ∣ \left|\int_e\Omega\right|=|P| ∫ e Ω = ∣ P ∣ Γ ( ∑ i ≤ 4 k i / 5 ) = Γ ( 4 / 5 ) = Γ ( 1 − k 5 / 5 ) \Gamma(\sum_{i\leq4}k_i/5)=\Gamma(4/5)=\Gamma(1-k_5/5) Γ ( ∑ i ≤ 4 k i /5 ) = Γ ( 4/5 ) = Γ ( 1 − k 5 /5 ) ∥ Ω ∥ 2 = C γ ( 1 / 5 ) 5 \|\Omega\|^{2}=C\gamma(1/5)^{5} ∥Ω ∥ 2 = C γ ( 1/5 ) 5 ∂ ψ Res ( Ω 0 / F ψ ) = Res ( 5 x 1 ⋯ x 5 Ω 0 / F ψ 2 ) = 5 ω 1 \partial_{\psi}\operatorname{Res}(\Omega_{0}/F_{\psi})=\operatorname{Res}(5x_{1}\cdots x_{5}\Omega_{0}/F_{\psi}^{2})=5\omega_{1} ∂ ψ Res ( Ω 0 / F ψ ) = Res ( 5 x 1 ⋯ x 5 Ω 0 / F ψ 2 ) = 5 ω 1 ω 1 \omega_{1} ω 1 ( 2 , 1 ) (2,1) ( 2 , 1 ) ψ = 0 \psi=0 ψ = 0 Ω \Omega Ω
G ψ ψ ˉ ( 0 ) = 25 g 1 ∥ Ω ∥ 2 = 25 γ ( 2 / 5 ) 5 γ ( 1 / 5 ) 5 = 25 Γ ( 2 / 5 ) 5 Γ ( 4 / 5 ) 5 Γ ( 1 / 5 ) 5 Γ ( 3 / 5 ) 5 = 0.19226. \begin{aligned}
G_{\psi\bar{\psi}}(0)=\frac{25g_{1}}{\|\Omega\|^{2}}=25\frac{\gamma(2/5)^{5}}{\gamma(1/5)^{5}} \\
&=25\frac{\Gamma(2/5)^{5}\Gamma(4/5)^{5}}{\Gamma(1/5)^{5}\Gamma(3/5)^{5}}=0.19226.
\end{aligned} G ψ ψ ˉ ( 0 ) = ∥Ω ∥ 2 25 g 1 = 25 γ ( 1/5 ) 5 γ ( 2/5 ) 5 = 25 Γ ( 1/5 ) 5 Γ ( 3/5 ) 5 Γ ( 2/5 ) 5 Γ ( 4/5 ) 5 = 0.19226. Numerical values
With Γ ( 1 / 5 ) = 4.59084 \Gamma(1/5)=4.59084 Γ ( 1/5 ) = 4.59084 Γ ( 2 / 5 ) = 2.21816 \Gamma(2/5)=2.21816 Γ ( 2/5 ) = 2.21816 Γ ( 3 / 5 ) = 1.48919 \Gamma(3/5)=1.48919 Γ ( 3/5 ) = 1.48919 Γ ( 4 / 5 ) = 1.16423 \Gamma(4/5)=1.16423 Γ ( 4/5 ) = 1.16423 γ ( 1 / 5 ) = 3.94325 \gamma(1/5)=3.94325 γ ( 1/5 ) = 3.94325 γ ( 2 / 5 ) = 1.48951 \gamma(2/5)=1.48951 γ ( 2/5 ) = 1.48951 γ ( 3 / 5 ) \gamma(3/5) γ ( 3/5 ) γ ( 4 / 5 ) \gamma(4/5) γ ( 4/5 ) Table 2 are their inverses.
Appendix D. Shared characters on the octic and the dectic
This appendix completes the proof of Proposition 8.4 . By Lemma 8.3 it suffices to find, for each shift s ≤ d / 2 s\leq d/2 s ≤ d /2 b \mathbf{b} b b i + s ≢ n i − 1 ( m o d n i ) b_i+s\not\equiv n_i-1\pmod{n_i} b i + s ≡ n i − 1 ( mod n i ) i i i ∑ i a i = s \sum_i a_i=s ∑ i a i = s b ′ \mathbf{b}' b ′ d − s d-s d − s n 5 = 2 n_5=2 n 5 = 2 b 5 = 0 b_5=0 b 5 = 0 x 5 x_5 x 5 a 5 = s / 2 a_5=s/2 a 5 = s /2
The octic. Here b i ≤ 6 b_i\leq6 b i ≤ 6 b 1 + b 2 + b 3 + b 4 = 8 b_1+b_2+b_3+b_4=8 b 1 + b 2 + b 3 + b 4 = 8 s = 2 s=2 s = 2 s = 4 s=4 s = 4 b i + s ≤ 10 < 16 b_i+s\leq10<16 b i + s ≤ 10 < 16 a i = 1 a_i=1 a i = 1 b i ≥ 8 − s b_i\geq8-s b i ≥ 8 − s b i = 7 − s b_i=7-s b i = 7 − s ∑ i a i = s \sum_i a_i=s ∑ i a i = s s / 2 s/2 s /2 b i ≥ 8 − s b_i\geq8-s b i ≥ 8 − s
At s = 2 s=2 s = 2 ( 6 , 2 , 0 , 0 ) (6,2,0,0) ( 6 , 2 , 0 , 0 ) ( 6 , 1 , 1 , 0 ) (6,1,1,0) ( 6 , 1 , 1 , 0 ) ( 0 , 4 , 2 , 2 ) (0,4,2,2) ( 0 , 4 , 2 , 2 ) ( 0 , 3 , 3 , 2 ) (0,3,3,2) ( 0 , 3 , 3 , 2 ) 12 + 12 12+12 12 + 12 s = 4 s=4 s = 4 b b b ( 4 , 4 , 0 , 0 ) (4,4,0,0) ( 4 , 4 , 0 , 0 ) ( 0 , 0 , 4 , 4 ) (0,0,4,4) ( 0 , 0 , 4 , 4 )
The labels of the three lists are distinct: the sources at s = 2 s = 2 s = 2 s = 4 s = 4 s = 4 149 − 27 = 122 149 - 27 = 122 149 − 27 = 122
The dectic. Here b i ≤ 8 b_i \le8 b i ≤ 8 i ≤ 3 i \le3 i ≤ 3 b 4 ≤ 3 b_4 \le3 b 4 ≤ 3 b 1 + b 2 + b 3 + 2 b 4 = 10 b_1 + b_2 + b_3 + 2b_4 = 10 b 1 + b 2 + b 3 + 2 b 4 = 10 s = 2 s = 2 s = 2 s = 4 s = 4 s = 4 i ≤ 3 i \le3 i ≤ 3 b i + s ≤ 12 < 20 b_i + s \le12 < 20 b i + s ≤ 12 < 20 a i = 1 a_i = 1 a i = 1 b i ≥ 10 − s b_i \ge10 - s b i ≥ 10 − s b i = 9 − s b_i = 9 - s b i = 9 − s x 4 x_4 x 4 a 4 = ⌊ ( b 4 + s ) / 5 ⌋ a_4 = \lfloor(b_4 + s)/5 \rfloor a 4 = ⌊( b 4 + s ) /5 ⌋ b 4 + s ≡ 4 ( m o d 5 ) b_4 + s \equiv4 \pmod{5} b 4 + s ≡ 4 ( mod 5 )
# { i ≤ 3 : b i ≥ 10 − s } + a 4 = s / 2. \#\{i \le3 : b_i \ge10 - s\} + a_4 = s/2. # { i ≤ 3 : b i ≥ 10 − s } + a 4 = s /2. At s = 2 s = 2 s = 2 b 4 = 2 b_4 = 2 b 4 = 2 a 4 = 1 a_4 = 1 a 4 = 1 b 4 = 3 b_4 = 3 b 4 = 3 b i b_i b i b i b_i b i b 4 ≤ 1 b_4 \le1 b 4 ≤ 1 ( 8 , 0 , 0 , 1 ) (8, 0, 0, 1) ( 8 , 0 , 0 , 1 ) ( 8 , 2 , 0 , 0 ) (8, 2, 0, 0) ( 8 , 2 , 0 , 0 ) ( 8 , 1 , 1 , 0 ) (8, 1, 1, 0) ( 8 , 1 , 1 , 0 ) ( 0 , 2 , 2 , 3 ) (0, 2, 2, 3) ( 0 , 2 , 2 , 3 ) ( 0 , 4 , 2 , 2 ) (0, 4, 2, 2) ( 0 , 4 , 2 , 2 ) ( 0 , 3 , 3 , 2 ) (0, 3, 3, 2) ( 0 , 3 , 3 , 2 ) b i b_i b i b 4 = 3 b_4 = 3 b 4 = 3 ( c 1 , c 2 , c 3 , 3 ) (c_1, c_2, c_3, 3) ( c 1 , c 2 , c 3 , 3 ) c 1 + c 2 + c 3 = 4 c_1 + c_2 + c_3 = 4 c 1 + c 2 + c 3 = 4 ( c 1 + 2 , c 2 + 2 , c 3 + 2 , 0 ) (c_1 + 2, c_2 + 2, c_3 + 2, 0) ( c 1 + 2 , c 2 + 2 , c 3 + 2 , 0 )
At s = 4 s = 4 s = 4 b 4 = 0 b_4 = 0 b 4 = 0 b 4 ≥ 1 b_4 \ge1 b 4 ≥ 1 a 4 = 1 a_4 = 1 a 4 = 1 b i b_i b i b j + b k + 2 b 4 = 10 − b i ≤ 4 b_j + b_k + 2b_4 = 10 - b_i \le4 b j + b k + 2 b 4 = 10 − b i ≤ 4 b 4 ≥ 1 b_4 \ge1 b 4 ≥ 1 ( 6 , 2 , 0 , 1 ) (6, 2, 0, 1) ( 6 , 2 , 0 , 1 ) ( 6 , 1 , 1 , 1 ) (6, 1, 1, 1) ( 6 , 1 , 1 , 1 ) ( 6 , 0 , 0 , 2 ) (6, 0, 0, 2) ( 6 , 0 , 0 , 2 ) ( 7 , 1 , 0 , 1 ) (7, 1, 0, 1) ( 7 , 1 , 0 , 1 ) ( 8 , 0 , 0 , 1 ) (8, 0, 0, 1) ( 8 , 0 , 0 , 1 ) ( 0 , 6 , 4 , 0 ) (0, 6, 4, 0) ( 0 , 6 , 4 , 0 ) ( 0 , 5 , 5 , 0 ) (0, 5, 5, 0) ( 0 , 5 , 5 , 0 ) ( 0 , 4 , 4 , 1 ) (0, 4, 4, 1) ( 0 , 4 , 4 , 1 ) ( 1 , 5 , 4 , 0 ) (1, 5, 4, 0) ( 1 , 5 , 4 , 0 ) ( 2 , 4 , 4 , 0 ) (2, 4, 4, 0) ( 2 , 4 , 4 , 0 )
Sorting the labels of these lists by b 4 b_4 b 4 b 4 = 3 b_4 = 3 b 4 = 3 ( c 1 , c 2 , c 3 , 3 ) (c_1, c_2, c_3, 3) ( c 1 , c 2 , c 3 , 3 ) ( 0 , 2 , 2 , 3 ) (0, 2, 2, 3) ( 0 , 2 , 2 , 3 ) c = ( 0 , 2 , 2 ) \mathbf{c} = (0, 2, 2) c = ( 0 , 2 , 2 ) b 4 = 2 b_4 = 2 b 4 = 2 ( 0 , 4 , 2 , 2 ) (0, 4, 2, 2) ( 0 , 4 , 2 , 2 ) ( 0 , 3 , 3 , 2 ) (0, 3, 3, 2) ( 0 , 3 , 3 , 2 ) ( 6 , 0 , 0 , 2 ) (6, 0, 0, 2) ( 6 , 0 , 0 , 2 ) b 4 = 1 b_4 = 1 b 4 = 1 ( 8 , 0 , 0 , 1 ) (8, 0, 0, 1) ( 8 , 0 , 0 , 1 ) ( 6 , 2 , 0 , 1 ) (6, 2, 0, 1) ( 6 , 2 , 0 , 1 ) ( 6 , 1 , 1 , 1 ) (6, 1, 1, 1) ( 6 , 1 , 1 , 1 ) ( 7 , 1 , 0 , 1 ) (7, 1, 0, 1) ( 7 , 1 , 0 , 1 ) ( 0 , 4 , 4 , 1 ) (0, 4, 4, 1) ( 0 , 4 , 4 , 1 ) b 4 = 0 b_4 = 0 b 4 = 0 ( 8 , 2 , 0 , 0 ) (8, 2, 0, 0) ( 8 , 2 , 0 , 0 ) ( 8 , 1 , 1 , 0 ) (8, 1, 1, 0) ( 8 , 1 , 1 , 0 ) ( c 1 + 2 , c 2 + 2 , c 3 + 2 , 0 ) (c_1 + 2, c_2 + 2, c_3 + 2, 0) ( c 1 + 2 , c 2 + 2 , c 3 + 2 , 0 ) ( 0 , 6 , 4 , 0 ) (0, 6, 4, 0) ( 0 , 6 , 4 , 0 ) ( 0 , 5 , 5 , 0 ) (0, 5, 5, 0) ( 0 , 5 , 5 , 0 ) ( 1 , 5 , 4 , 0 ) (1, 5, 4, 0) ( 1 , 5 , 4 , 0 ) ( 2 , 4 , 4 , 0 ) (2, 4, 4, 0) ( 2 , 4 , 4 , 0 ) c = ( 0 , 2 , 2 ) \mathbf{c} = (0, 2, 2) c = ( 0 , 2 , 2 )
( 8 , 0 , 0 , 1 ) → s = 2 ( 0 , 2 , 2 , 3 ) → s = 2 ( 2 , 4 , 4 , 0 ) ← s = 4 ( 8 , 0 , 0 , 1 ) (8, 0, 0, 1) \xrightarrow{s=2} (0, 2, 2, 3) \xrightarrow{s=2} (2, 4, 4, 0) \xleftarrow{s=4} (8, 0, 0, 1) ( 8 , 0 , 0 , 1 ) s = 2 ( 0 , 2 , 2 , 3 ) s = 2 ( 2 , 4 , 4 , 0 ) s = 4 ( 8 , 0 , 0 , 1 ) and its permutations, the three triples of Proposition 8.4 (c). The remaining pairs are the 12 with c \mathbf{c} c ( 2 , 2 , 0 ) (2, 2, 0) ( 2 , 2 , 0 ) 6 + 3 6 + 3 6 + 3 s = 2 s = 2 s = 2 6 + 3 + 3 + 6 6 + 3 + 3 + 6 6 + 3 + 3 + 6 s = 4 s = 4 s = 4 145 − 87 + 42 = 100 145 - 87 + 42 = 100 145 − 87 + 42 = 100
The blocks of the mass map
Table 11 gives the data behind Table 4 . For each row we take the representative with ( i , j , k , l ) = ( 1 , 2 , 3 , 4 ) (i, j, k, l) = (1, 2, 3, 4) ( i , j , k , l ) = ( 1 , 2 , 3 , 4 ) b ∈ B χ \mathbf{b} \in B_\chi b ∈ B χ c ∈ B − χ \mathbf{c} \in B_{-\chi} c ∈ B − χ Table 4 , the vector h + b + c \mathbf{h} + \mathbf{b} + \mathbf{c} h + b + c r r r k \mathbf{k} k ( 10 ) (10) ( 10 ) ψ r ∏ i w i k i \psi^r \prod_i w_i^{k_i} ψ r ∏ i w i k i Theorem 8.5 . The other blocks of each row follow by permuting the coordinates of weight one. For the blocks of size 2 × 2 2 \times2 2 × 2 Δ \Delta Δ Lemma 8.7 is 6 for the self-dual sextic row, 0 and 8 for the second and third octic rows, and 0, 0, 10 and 10 for the sixth to ninth dectic rows; for instance in the eighth dectic row Δ = 4 + 6 − 0 − 0 = 10 \Delta= 4 + 6 - 0 - 0 = 10 Δ = 4 + 6 − 0 − 0 = 10 50 ⋅ 250 ψ 10 − 1 = 12500 ψ 10 − 1 50 \cdot250\,\psi^{10} - 1 = 12500\,\psi^{10} - 1 50 ⋅ 250 ψ 10 − 1 = 12500 ψ 10 − 1
Table 11. The entries of the blocks of Table 4, for the representative with ( i , j , k , l ) = ( 1 , 2 , 3 , 4 ) (i,j,k,l)=(1,2,3,4) ( i , j , k , l ) = ( 1 , 2 , 3 , 4 ) r r r h + b + c h+b+c h + b + c k k k
Exact certificates for the fixed threefold
The following rational identities finish the smooth-seed proof. In each row A q + B q ′ = 1 Aq+Bq'=1 A q + B q ′ = 1
q = 2 s 3 + 15 s 2 + 6 s + 1 , A = ( 21 s + 101 ) / 37 , B = − 7 s 2 / 37 − 307 s / 222 − 32 / 111 ; q = s 3 + 6 s 2 + 15 s + 2 , A = − s / 74 − 4 / 37 , B = s 2 / 222 + 5 s / 111 + 3 / 37 ; q = s 4 + 8 s 3 + 30 s 2 + 8 s + 1 , A = s 2 / 6 + 47 s / 48 + 19 / 8 , B = − s 3 / 24 − 21 s 2 / 64 − 29 s / 24 − 11 / 64. \begin{aligned}
q&=2s^{3}+15s^{2}+6s+1, & A&=(21s+101)/37,\\
&& B&=-7s^{2}/37-307s/222-32/111;\\
q&=s^{3}+6s^{2}+15s+2, & A&=-s/74-4/37,\\
&& B&=s^{2}/222+5s/111+3/37;\\
q&=s^{4}+8s^{3}+30s^{2}+8s+1, & A&=s^{2}/6+47s/48+19/8,\\
&& B&=-s^{3}/24-21s^{2}/64-29s/24-11/64.
\end{aligned} q q q = 2 s 3 + 15 s 2 + 6 s + 1 , = s 3 + 6 s 2 + 15 s + 2 , = s 4 + 8 s 3 + 30 s 2 + 8 s + 1 , A B A B A B = ( 21 s + 101 ) /37 , = − 7 s 2 /37 − 307 s /222 − 32/111 ; = − s /74 − 4/37 , = s 2 /222 + 5 s /111 + 3/37 ; = s 2 /6 + 47 s /48 + 19/8 , = − s 3 /24 − 21 s 2 /64 − 29 s /24 − 11/64. For g = s 2 − 8 s − 21 g=s^{2}-8s-21 g = s 2 − 8 s − 21 h = s 4 − 16 s 3 − 66 s 2 − 16 s + 89 h=s^{4}-16s^{3}-66s^{2}-16s+89 h = s 4 − 16 s 3 − 66 s 2 − 16 s + 89
( − 3 s 3 22 + 217 s 2 88 + 85 s 22 − 47 8 ) g + ( 3 s 22 − 11 8 ) h = 1. \left(-\frac{3s^{3}}{22}+\frac{217s^{2}}{88}+\frac{85s}{22}-\frac{47}{8}\right)g+\left(\frac{3s}{22}-\frac{11}{8}\right)h=1. ( − 22 3 s 3 + 88 217 s 2 + 22 85 s − 8 47 ) g + ( 22 3 s − 8 11 ) h = 1. These are identities in Q [ s ] \mathbb{Q}[s] Q [ s ]
For completeness, a reproducible description fixes every entry of the matrix in (13) . Order all bases lexicographically. For L 1 L 2 L_{1}L_{2} L 1 L 2 H 2 ( A , O ( − 4 , − 5 , 0 , 1 ) ) H^{2}(A,\mathcal{O}(-4,-5,0,1)) H 2 ( A , O ( − 4 , − 5 , 0 , 1 )) H 2 ( A , O ( − 2 , − 3 , 2 , 3 ) ) H^{2}(A,\mathcal{O}(-2,-3,2,3)) H 2 ( A , O ( − 2 , − 3 , 2 , 3 )) v v v y y y Theorem 14.2 . The source vectors range over
{ − 3 , − 2 , − 1 } × { − 4 , − 3 , − 2 , − 1 } × { 0 } × { 0 , 1 } . \{-3,-2,-1\}\times\{-4,-3,-2,-1\}\times\{0\}\times\{0,1\}. { − 3 , − 2 , − 1 } × { − 4 , − 3 , − 2 , − 1 } × { 0 } × { 0 , 1 } . and the target vectors range over
{ − 1 } × { − 2 , − 1 } × { 0 , 1 , 2 } × { 0 , 1 , 2 , 3 } . \{-1\}\times\{-2,-1\}\times\{0,1,2\}\times\{0,1,2,3\}. { − 1 } × { − 2 , − 1 } × { 0 , 1 , 2 } × { 0 , 1 , 2 , 3 } . Expand F 1 F_{1} F 1 ∑ e A e ∏ i x i 2 − e i y i e i \sum_{e} A_{e}\prod_{i}x_{i}^{2-e_{i}}y_{i}^{e_{i}} ∑ e A e ∏ i x i 2 − e i y i e i (11) . The integer matrix is
M ( 1 ) w , v = A w − v , M(1)_{w,v}=A_{w-v}, M ( 1 ) w , v = A w − v , where an absent coefficient is zero. Terms outside the target ranges represent zero Čech classes. This specifies a 24 by 24 integer matrix without a numerical approximation to π \pi π b b b ( v , b v ) (v,bv) ( v , b v ) v + ε b ( − 1 ) q b v v+\varepsilon b(-1)^{q}bv v + ε b ( − 1 ) q b v q q q a a a
M + + = ( 5 8 9 2 3 4 2 2 5 5 0 1 0 5 5 0 2 2 11 16 16 11 2 15 2 9 12 5 3 4 1 5 2 2 0 1 ) , det M + + = 792. M_{++}=
\begin{pmatrix}
5&8&9&2&3&4\\
2&2&5&5&0&1\\
0&5&5&0&2&2\\
11&16&16&11&2&15\\
2&9&12&5&3&4\\
1&5&2&2&0&1
\end{pmatrix},
\qquad\det M_{++}=792. M ++ = 5 2 0 11 2 1 8 2 5 16 9 5 9 5 5 16 12 2 2 5 0 11 5 2 3 0 2 2 3 0 4 1 2 15 4 1 , det M ++ = 792. M + − = ( 5 6 − 5 − 2 3 − 2 2 − 2 5 − 5 0 1 0 5 − 5 0 2 − 2 11 − 6 6 − 11 2 11 2 5 − 2 − 5 3 − 2 1 − 5 2 − 2 0 1 ) , det M + − = − 792. M_{+-}=
\begin{pmatrix}
5&6&-5&-2&3&-2\\
2&-2&5&-5&0&1\\
0&5&-5&0&2&-2\\
11&-6&6&-11&2&11\\
2&5&-2&-5&3&-2\\
1&-5&2&-2&0&1
\end{pmatrix},
\qquad\det M_{+-}=-792. M +− = 5 2 0 11 2 1 6 − 2 5 − 6 5 − 5 − 5 5 − 5 6 − 2 2 − 2 − 5 0 − 11 − 5 − 2 3 0 2 2 3 0 − 2 1 − 2 11 − 2 1 , det M +− = − 792. M − + = ( 5 1 2 2 0 1 2 9 12 5 3 4 11 16 16 11 2 15 0 5 5 0 2 2 2 2 5 5 0 1 1 12 9 2 3 4 ) , det M − + = 792. M_{-+}=
\begin{pmatrix}
5 & 1 & 2 & 2 & 0 & 1 \\
2 & 9 & 12 & 5 & 3 & 4 \\
11 & 16 & 16 & 11 & 2 & 15 \\
0 & 5 & 5 & 0 & 2 & 2 \\
2 & 2 & 5 & 5 & 0 & 1 \\
1 & 12 & 9 & 2 & 3 & 4
\end{pmatrix}, \qquad\det M_{-+}=792. M −+ = 5 2 11 0 2 1 1 9 16 5 2 12 2 12 16 5 5 9 2 5 11 0 5 2 0 3 2 2 0 3 1 4 15 2 1 4 , det M −+ = 792. M − − = ( 5 − 1 2 − 2 0 1 2 5 − 2 − 5 3 − 2 11 − 6 6 − 11 2 11 0 5 − 5 0 2 − 2 2 − 2 5 − 5 0 1 1 2 − 5 − 2 3 − 2 ) , det M − − = − 792. M_{--}=
\begin{pmatrix}
5 & -1 & 2 & -2 & 0 & 1 \\
2 & 5 & -2 & -5 & 3 & -2 \\
11 & -6 & 6 & -11 & 2 & 11 \\
0 & 5 & -5 & 0 & 2 & -2 \\
2 & -2 & 5 & -5 & 0 & 1 \\
1 & 2 & -5 & -2 & 3 & -2
\end{pmatrix}, \qquad\det M_{--}=-792. M −− = 5 2 11 0 2 1 − 1 5 − 6 5 − 2 2 2 − 2 6 − 5 5 − 5 − 2 − 5 − 11 0 − 5 − 2 0 3 2 2 0 3 1 − 2 11 − 2 1 − 2 , det M −− = − 792. The matrix code uses exact rational elimination and checks equivariance by reconstructing every image from the target eigenbasis. The nonvanishing assertion is a finite integer certificate, combined with the analytic transcendence argument for the actual parameter. It is not a finite search over moduli and is not a claimed Lean formalization.
Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL), Bhubaneswar, Odisha 751030, India
Formerly, University of Calcutta, Senate House, 87/1 College Street, Kolkata 700073, West Bengal, India
Email address: d.bhattacharjee@erl-forschung.de
Email address: itsdeep@live.com
Email address: egspl@proton.me
Formerly, National Taiwan University, Taipei 106319, Taiwan
Email address: rai.bhattacharya.in@gmail.com
