Introduction
A closed string moving in ten dimensions has a tower of vibration modes, and the massless ones of the heterotic string on a product of four-dimensional space-time with a compact six-manifold X X X X X X E 6 E_6 E 6 X X X [1 ] and by Witten [2 ] , and the couplings of three charged fields were identified by Strominger [3 ] with an integral over X X X [4 ] computed on the quintic threefold. The question in the title is the one that this dictionary leaves open: a vibration mode is a cohomology class on X X X X X X 1
We answer it completely on one family, the Dwork pencil of quintics
X ψ = { x 1 5 + x 2 5 + x 3 5 + x 4 5 + x 5 5 − 5 ψ x 1 x 2 x 3 x 4 x 5 = 0 } ⊂ P 4 , (1) X_{\psi}=\{x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}+x_{5}^{5}-5\psi x_{1}x_{2}x_{3}x_{4}x_{5}=0\}\subset\mathbb{P}^{4}, \tag*{(1)} X ψ = { x 1 5 + x 2 5 + x 3 5 + x 4 5 + x 5 5 − 5 ψ x 1 x 2 x 3 x 4 x 5 = 0 } ⊂ P 4 , (1) smooth for ψ 5 ≠ 1 \psi^{5}\ne1 ψ 5 = 1 [5 ] and on which the Yukawa couplings were counted by Doran, Greene and Judes [6 ] . The group Γ ≅ ( Z / 5 ) 3 \Gamma\cong(\mathbb{Z}/5)^{3} Γ ≅ ( Z /5 ) 3 H 2 , 1 ( X ψ ) H^{2,1}(X_{\psi}) H 2 , 1 ( X ψ ) 27 27 27 x b x^{\mathbf{b}} x b b ∈ { 0 , 1 , 2 , 3 } 5 \mathbf{b}\in\{0,1,2,3\}^{5} b ∈ { 0 , 1 , 2 , 3 } 5
Theorem 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 2 (Couplings). Let ψ 5 ≠ 1 \psi^{5}\ne1 ψ 5 = 1 The cubic form of the pencil, normalised so that the monomial 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}\equiv(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 3 (Mass map). The invariant class h = x 1 x 2 x 3 x 4 x 5 h=x_{1}x_{2}x_{3}x_{4}x_{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 ) 1 1 1 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 27 27 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 g g
Theorem 2 is stronger than the count of Doran, Greene and Judes [6 ] , who found by a search that 1431 couplings are allowed by the symmetry; it shows that the symmetry is the only source of zeros, and gives every allowed coupling in closed form. The number 1431 is recovered by hand in § 5.1 . The theorems are proved in § 3 to § 6 after the dictionary between vibrations and particles is set out, with complete proofs, in § 2 . § 7 evaluates the Hodge norms at the Fermat point in terms of the Gamma function, which turns the mass formula of Theorem 3 into numbers there.
The fourth result concerns the choice of the vacuum. A conjecture of Bhattacharjee, Bhattacharya and Ghosh [33 ] , the Geometric Selection Conjecture, asks for a functional on the heterotic landscape, built from the Weil–Petersson metric, that is proper, has a unique minimum, takes that minimum at a phenomenologically viable vacuum and is invariant under mirror symmetry. The precise statement is recalled in § 9 .
Theorem 4 (Selection). The Geometric Selection Conjecture holds if and only if there is a viable vacuum, in the sense of conditions (C1) to (C9) of [33 ] , that is not moved by the mirror involution. On the Dwork pencil the mass condition (C8) holds in the precise form that the mass map of the invariant direction has full rank with entries 1 1 1 ψ \psi ψ ψ 2 \psi^{2} ψ 2 ψ ≠ 0 \psi\ne0 ψ = 0
The first part says that the functional carries no information beyond the vacuum it selects: once a viable vacuum is given, a functional with all the required properties is written down in § 9 from the metric alone, and conversely the minimum of any such functional is a viable vacuum. What remains of the conjecture is therefore the existence of one viable vacuum, which is where its content lies. The second part locates the Dwork pencil with respect to that question: its mass spectrum has the full rank and the hierarchy of scales that condition (C8) asks for, and the obstruction is the number of families, a hundred and one on the quintic and 100 / ∣ H ∣ 100/|H| 100/∣ H ∣ H H H
The same questions can be asked of the three other one-parameter 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 ) [24 ] , Klemm and Theisen [25 ] and Font [26 ] . 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 5 (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 \mathbf a,\mathbf b,\mathbf 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_ik_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.
So the two properties that make the Dwork pencil transparent are special to it. On the weighted pencils a shift of the exponents at the coordinates of weight one can be balanced by a shift at the coordinate of larger weight, the characters of Γ \Gamma Γ Theorem 4 therefore fails on all three pencils, and so does the three-family condition (C2). Theorem 5 is proved in § 8 , and 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 12500 ψ 10 = 1 12500\psi^{10}=1 12500 ψ 10 = 1
The count of families reaches beyond the four pencils. Of the fourteen one-parameter families of Calabi–Yau threefolds with hypergeometric Picard–Fuchs equations, thirteen are complete intersections in weighted projective space, and Proposition 27 shows that a finite group acting freely on a smooth member has order divisible only by the primes that divide the greatest common divisor g X g_X g X χ ( X , O X ( k ) ) \chi(X,\mathcal O_X(k)) χ ( X , O X ( k )) k ≥ 1 k\ge1 k ≥ 1
The constructions behind the four theorems were organised on the Anaxia engine of the BLLM programme [36 , 37 ] .2
History. The dictionary between the vibrations of the string and the fields in four dimensions on a Calabi–Yau background was written down in [1 ] ; the decomposition of the adjoint of E 8 E_8 E 8 E 6 × S U ( 3 ) E_6 \times SU(3) E 6 × S U ( 3 ) [2 ] discussed the breaking of E 6 E_6 E 6 27 27 27 Theorem 3 . Strominger [3 ] identified the cubic couplings of the 27 27 27 H 1 ( X , T X ) H^1(X,T_X) H 1 ( X , T X ) [8 ] and by Candelas and de la Ossa [9 ] . Candelas [4 ] computed the couplings on the Fermat quintic through the Jacobian ring, with the Gamma-function normalisation of the Hodge norms at that point; Greene and Plesser [7 ] introduced the quotient of the Dwork pencil by Γ \Gamma Γ [5 ] computed the coupling of h h h 5 / ( 1 − ψ 5 ) 5/(1-\psi^5) 5/ ( 1 − ψ 5 ) a = b = c = h a=b=c=h a = b = c = h Theorem 2 together with Remark 8. Doran, Greene and Judes [6 ] counted by computer the couplings not forced to vanish by Γ \Gamma Γ H 2 , 1 H^{2,1} H 2 , 1 [10 ] , and the perfectness of the residue pairing with the Hessian in the socle is Macaulay’s theorem in the form given by Carlson and Griffiths [11 ] . On the physical side, the holomorphic Yukawa couplings of heterotic line-bundle models were computed by Blesneag, Buchbinder, Candelas and Lukas [14 ] , and the physical, metric-normalised couplings of the quintic at the Fermat point by Butbaia, Mayorga Peña, Tan, Berglund, Hübsch, Jejjala and Mishra [15 ] with numerical Ricci-flat metrics, whose Gamma-function value for the invariant direction agrees with § 7 . Heterotic standard models with three families on other threefolds were constructed by Braun, He, Ovrut and Pantev [20 ] , Bouchard and Donagi [21 ] and Anderson, Gray, Lukas and Palti [22 , 23 ] ; the question of which classes carry which families is posed there in a form that depends on the choice of bundle, and the present paper answers it in the one case where every piece of the data is explicit. Distler and Greene [16 ] computed the normalised couplings of the Gepner model at the Fermat point exactly, and § 7 recovers their constant from the Hodge norms. The monodromy and intersection numbers used there go back to Pham [17 ] , Sebastiani and Thom [18 ] and Sakamoto [19 ] . Two further aspects of the same compactifications, the suppression of the electroweak scale by singularities of the threefold and fibrations of Calabi–Yau manifolds by circles, are treated in [34 , 35 ] . The weighted pencils of § 8 are the one-parameter models whose Picard–Fuchs equations, periods and mirror maps were found by Morrison [24 ] , Klemm and Theisen [25 ] and Font [26 ] , and the Hodge theory of quasi-smooth hypersurfaces in weighted projective space used there is due to Steenbrink [27 ] and Dolgachev [28 ] .
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 T_X 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 [12 ] ). 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,\bullet}\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 6 (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 ) copies of 27 ⊕ h 1 , 1 ( X ) copies of 27 ‾ , h^{2,1}(X)\text{ copies of }27\oplus h^{1,1}(X)\text{ copies of }\overline{27}, h 2 , 1 ( X ) copies of 27 ⊕ h 1 , 1 ( X ) copies of 27 , with the 27 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{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 T_X 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)\cong 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 ⟩ [8 , 9 ] .
Mass terms
A vacuum value v v v 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_{1}F,\ldots,\partial_{5}F) 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 [10 ] 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}\,\mathrm{d}x_{1}\wedge\cdots\wedge\widehat{\mathrm{d}x_{i}}\wedge\cdots\wedge\mathrm{d}x_{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 [11 ] , says that the Hessian Hess F = det ( ∂ i ∂ j F ) \operatorname{Hess}F=\det(\partial_{i}\partial_{j}F) Hess 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 6 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 [13 ] 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}=\{\mathbf{b}\in(\mathbb{Z}/5)^5:\sum b_i=0\}/\langle\mathbf{1}\rangle,\qquad|\widehat{\Gamma}|=125, Γ = { b ∈ ( Z /5 ) 5 : ∑ b i = 0 } / ⟨ 1 ⟩ , ∣ Γ ∣ = 125 , with b \mathbf{b} b a ↦ ζ a ⋅ b a\mapsto\zeta^{a\cdot b} a ↦ ζ a ⋅ b x b x^{\mathbf{b}} x b a ∈ K a\in K a ∈ K ζ a ⋅ b \zeta^{a\cdot b} ζ a ⋅ b b \mathbf{b} b
Let B B B b ∈ { 0 , 1 , 2 , 3 } 5 \mathbf{b}\in\{0,1,2,3\}^5 b ∈ { 0 , 1 , 2 , 3 } 5 ∣ b ∣ = ∑ b i = 5 |\mathbf{b}|=\sum b_i=5 ∣ b ∣ = ∑ b i = 5 x b x^{\mathbf{b}} x b b ∈ B \mathbf{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. Two elements b , b ′ ∈ B \mathbf{b},\mathbf{b}'\in B b , b ′ ∈ B b ′ ≡ b + k 1 ( m o d 5 ) \mathbf{b}'\equiv\mathbf{b}+k\mathbf{1}\pmod5 b ′ ≡ b + k 1 ( mod 5 ) k k k b ′ \mathbf{b}' b ′ { 0 , … , 4 } \{0,\ldots,4\} { 0 , … , 4 } b ′ \mathbf{b}' b ′ b + k 1 \mathbf{b}+k\mathbf{1} 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 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\le1 b i ≤ 1 b i − 2 b_i-2 b i − 2 b i ≥ 2 b_i\ge2 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 ′′ ( 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 1 \mathbf{1} 1 ( 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 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 7 (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 ) = 3200000 ( 1 − ψ 5 ) \lambda(\operatorname{Hess}F_\psi)=3200000(1-\psi^5) λ ( Hess F ψ ) = 3200000 ( 1 − ψ 5 )
Proof. The relations 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 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}+\mathbf{1}-5\mathbf{e}_i, \tag*{(6)} x e = ψ x e ′ , e ′ = e + 1 − 5 e i , (6) where e i 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 r m 0 \psi^r m_0 ψ r 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 3
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\mathbf{1} 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}}\equiv0 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 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 a a a 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 § 13 . 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 e ≡ ψ r m 0 x^e \equiv\psi^r m_0 x e ≡ ψ r m 0
H ( ψ ) = 3200000 − ( 2000000 + 1000000 + 375000 + 75000 − 187500 − 62500 ) ψ 5 = 3200000 ( 1 − ψ 5 ) . \begin{aligned}
H(\psi)=3200000-(2000000+1000000+375000+75000-187500-62500)\psi^5 \\
&=3200000(1-\psi^5).
\end{aligned} H ( ψ ) = 3200000 − ( 2000000 + 1000000 + 375000 + 75000 − 187500 − 62500 ) ψ 5 = 3200000 ( 1 − ψ 5 ) . Remark 8 (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 / ( 3200000 ( 1 − ψ 5 ) ) \lambda_{F_\psi}=\lambda/(3200000(1-\psi^5)) λ F ψ = λ / ( 3200000 ( 1 − ψ 5 )) a = b = c = h a=b=c=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 2 is stated for λ \lambda λ
Couplings
Proof of Theorem 2 . By (5) and Theorem 7 , Y ( x a , x b , x c ) = λ ( x a + b + c ) Y(x^a,x^b,x^c)=\lambda(x^{a+b+c}) Y ( x a , x b , x c ) = λ ( x a + b + c ) a + b + c a+b+c a + b + c ψ r \psi^r ψ r a + b + c ≡ ( 3 − r ) 1 a+b+c\equiv(3-r)\mathbf{1} a + b + c ≡ ( 3 − r ) 1 ψ ≠ 0 \psi\ne0 ψ = 0 § 5.1 . □
The count
A triple { a , b , c } \{a,b,c\} { a , b , c } B B B a + b + c ≡ c 1 ( m o d 5 ) a+b+c\equiv c\mathbf{1}\pmod5 a + b + c ≡ c 1 ( mod 5 ) c c c c ≡ 3 − r c\equiv3-r c ≡ 3 − r r r r c c c a + b a+b a + b Γ ^ B \widehat{\Gamma}_B Γ B B B B Theorem 1 ), so for each ordered pair ( a , b ) (a,b) ( a , b ) c c c − ( a + b ) -(a+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 a+b a + b
Excluded pairs. The entries of a + b a+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 a+b a + b ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) k k k k k k k = 0 , … , 4 k=0,\ldots,4 k = 0 , … , 4 p k p_k p k a 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 |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 ) . (7) (1+t)(1+t+t^2)(1+t+t^2+t^3)(t+t^2+t^3). \tag*{(7)} ( 1 + t ) ( 1 + t + t 2 ) ( 1 + t + t 2 + t 3 ) ( t + t 2 + t 3 ) . (7) which is 16 16 16 4 10 3 − 1920 = 8281 10^3-1920=8281 1 0 3 − 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\mathbf{c}_1 3 a ≡ c 1 a \mathbf{a} a { 0 , 1 , 2 , 3 } \{0,1,2,3\} { 0 , 1 , 2 , 3 } a = 1 \mathbf{a}=1 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}=1 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 7 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^ay^b P e ( x , y ) = ∑ x a y b 0 ≤ a , b ≤ 3 0\le a,b\le3 0 ≤ a , b ≤ 3 0 ≤ e − a − b ≤ 3 0\le e-a-b\le3 0 ≤ e − a − b ≤ 3
O ( 3 , 3 , 3 , 3 , 3 ) = 3391 , O ( 7 , 2 , 2 , 2 , 2 ) = 480 , O(3,3,3,3,3)=3391,\qquad O(7,2,2,2,2)=480, 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 , O(6,6,1,1,1)=162,\qquad O(5,5,5,0,0)=87, O ( 6 , 6 , 1 , 1 , 1 ) = 162 , O ( 5 , 5 , 5 , 0 , 0 ) = 87 , so the ordered counts by sector are 3391, 2400, 1620 and 870, with total 8281. The hundred unordered triples { a , a , c } \{\mathbf{a},\mathbf{a},\mathbf{c}\} { a , a , c } c ≠ a \mathbf{c}\ne\mathbf{a} c = a 2 a + c 2\mathbf{a}+\mathbf{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 , D r , T r O_r,D_r,T_r O r , D r , T r r r r
The mass map
Proof of Theorem 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)1 b + c + 1 ≡ ( 3 − r ) 1 c ≡ ( 2 − r ) 1 − b ( m o d 5 ) \mathbf{c}\equiv(2-r)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)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\le2 b i ≤ 2 r = 0 r=0 r = 0 c = 21 − b ∈ { 0 , 1 , 2 } 5 \mathbf{c}=21-\mathbf{b}\in\{0,1,2\}^5 c = 21 − b ∈ { 0 , 1 , 2 } 5 r = 1 r=1 r = 1 1 − b 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}=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 -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}=1 b = 1 21 − b 21-\mathbf{b} 21 − 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})=21-\mathbf{b} σ ( b ) = 21 − b b = 1 \mathbf{b}=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 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
Figure 3. The mass spectrum of Theorem 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 \le|\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
The Γ \Gamma Γ H 2 , 1 H^{2,1} H 2 , 1 Theorem 1 the components are the lines spanned by the x b x^b x b g b g_b g b (4) is then v v v ψ r ( b ) ( g b g σ b ) − 1 / 2 \psi^{r(b)}(g_b g_{\sigma b})^{-1/2} ψ r ( b ) ( g b g σ b ) − 1/2 ( b , σ b ) (b,\sigma b) ( b , σ b )
Remark 9 (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 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 1 / ( 1 − ψ 5 ) 1/(1-\psi^5) 1/ ( 1 − ψ 5 ) Ω \Omega Ω h h h [5 ] .
The Fermat point
At ψ = 0 \psi= 0 ψ = 0
Table 2. Hodge norms and masses at the Fermat point, from (8) and Theorem 3 . Norms are in units of C C C ν / C \nu/C ν / C ψ = 0 \psi=0 ψ = 0
Proposition 10 (Hodge norms at the Fermat point). Let X 0 X_0 X 0 ω b \omega_{\mathbf b} ω b x b x^{\mathbf b} x b b ∈ B \mathbf 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 \mathbf b b
g b = ⟨ ω b , ω b ‾ ⟩ = C ∏ i = 1 5 γ ( k i / 5 ) . (8) g_{\mathbf b}=\langle\omega_{\mathbf b},\overline{\omega_{\mathbf 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_{\mathbf b} ω b k = ( k i ) \mathbf k=(k_i) k = ( k i ) 1 ≤ k i ≤ 4 1\le k_i\le4 1 ≤ k i ≤ 4 ∑ k i = 10 \sum k_i=10 ∑ k i = 10 g b g_{\mathbf b} g b ω b \omega_{\mathbf 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_{\mathbf b}=|\int_P\omega_{\mathbf b}|^2/|P_{\mathbf k}\cdot P_{-\mathbf k}| 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 (8) . This is the computation of Candelas [4 ] on the Fermat quintic and, in the language of the Gepner model, of Distler and Greene [16 ] ; the steps are written out in Appendix § 15 . □
The proposition fixes the ratios of the Hodge norms inside H 2 , 1 H^{2,1} H 2 , 1 Theorem 3 need. Since 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_{\mathbf 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_{\mathbf 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 [16 ] expressed the normalised couplings of the Gepner model (see [15, eq. (4.3)]). The norm of $\Omega$, the class of $1$ with $\mathbf k=(1,1,1,1,1)$, has the same shape with a different constant, and the Weil–Petersson metric of the pencil at the Fermat point, computed from the periods by Candelas, de la Ossa, Green and Parkes [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 Bubeaia and collaborators [15 ] reproduce with numerical Ricci–flat metrics.5
Inserting (8) into Theorem 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_{\mathbf b}=g_{\sigma\mathbf b} g b = g σ b ν / g b \nu/g_{\mathbf b} ν / g b ν / g 1 \nu/g_1 ν / g 1 ∣ ψ ∣ |\psi| ∣ ψ ∣ ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2
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_i w_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
Weighted Fermat pencils
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 ) 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 ) (2) . We write
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 6
Lemma 11. 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 , χ = ( σ 3 ( w ) − d σ 2 ( w ) ) L 3 , (9) \begin{aligned}
L^3 &= \frac{d}{\prod_i w_i}, \qquad c_2 \cdot L = \sigma_2(w)L^3, \\
\chi&= \left(\sigma_3(w)-d\,\sigma_2(w)\right)L^3, \tag*{(9)}
\end{aligned} L 3 χ = ∏ i w i d , c 2 ⋅ L = σ 2 ( w ) L 3 , = ( σ 3 ( w ) − d σ 2 ( w ) ) L 3 , (9) 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_i x_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 − ∑ w i ) = O X K_{X}=\mathcal{O}_{X}(d-\sum w_{i})=\mathcal{O}_{X} K X = O X ( d − ∑ w i ) = O X [28 ] 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_{i}H) ∏ 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 + d L ) − 1 c(T_{X})=\prod_{i}(1+w_{i}L)(1+dL)^{-1} c ( T X ) = ∏ i ( 1 + w i L ) ( 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_{2}L^{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 ) 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_{1}F_{\psi},\ldots,\partial_{5}F_{\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 [27 ] and Dolgachev [28 ] : with
Ω 0 = ∑ i ( − 1 ) i w i x i d x 1 ∧ ⋯ ∧ d x i ^ ∧ ⋯ ∧ d x 5 , \Omega_{0}=\sum_{i}(-1)^{i}w_{i}x_{i}\,dx_{1}\wedge\cdots\wedge\widehat{dx_{i}}\wedge\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 [11 ] 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_{i}w_{i}b_{i}=d ∑ i w i b i = d
Lemma 12. Let ψ d δ ≠ 1 \psi^{d}\delta\ne1 ψ d δ = 1 x e x^{e} x e d d d x e ≡ c ψ r x b x^{e}\equiv c\psi^{r}x^{b} x e ≡ c ψ r x b R R R b ∈ B b\in B b ∈ B c c c r ≤ 1 r\le1 r ≤ 1 r ≤ 2 r\le2 r ≤ 2 x b x^{b} x b b ∈ B 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_{i}F_{\psi}/n_{i}=0 ∂ i F ψ / n i = 0
x i n i − 1 ≡ w i ψ ∏ j ≠ i x j . (10) x_{i}^{n_{i}-1}\equiv w_{i}\psi\prod_{j\ne i}x_{j}. \tag*{(10)} x i n i − 1 ≡ w i ψ j = i ∏ x j . (10) 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^{e} x e w i ψ x e ′ w_{i}\psi 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 i = n i e_{i}=n_{i} e i = n i e i ′ = 1 e'_{i}=1 e 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 − 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 i + w j ≥ d w_{i}+w_{j}\ge d w i + 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 \mathbf{1}\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 ′ = 1 e'=\mathbf{1} e ′ = 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_{i}x_{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 1 1 1 μ d \mu_d μ d ∑ i w i b i ≡ 0 ( m o d d ) \sum_iw_ib_i\equiv0\pmod d ∑ i w i b i ≡ 0 ( mod d ) w i n i = d w_in_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_iw_ib_i\equiv0\pmod d\right\}/\langle1\rangle. Γ = { b ∈ i ⨁ Z / n i : i ∑ w i b i ≡ 0 ( mod d ) } / ⟨ 1 ⟩ . A monomial x e x^{\mathbf e} x e d d d [ e ] [\mathbf e] [ e ] Ω 0 \Omega_0 Ω 0 ∏ i ε i = 1 \prod_i\varepsilon_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 ^ \widehat{\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 \varepsilon_i^{-1}\varepsilon_j^{-1} ε i − 1 ε j − 1 ∏ i ε i − 2 = 1 \prod_i\varepsilon_i^{-2}=1 ∏ i ε i − 2 = 1 R 3 d R_{3d} R 3 d F ψ F_\psi F ψ
Lemma 13 (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_ia_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}\quad\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_iw_ib_i=d\equiv0 ∑ i w i b i = d ≡ 0 b ′ ≡ b + s 1 \mathbf b'\equiv\mathbf b+s\mathbf1 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_iw_ib_i'=\sum_iw_ib_i+sd-d\sum_i a_i ∑ i w i b i ′ = ∑ i w i b i + s d − d ∑ i a i d d d ∑ i a i = s \sum_i a_i=s ∑ i a i = s
Proposition 14. Let ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1 The classes of labels that carry the same character are exactly the following, and every other label carries a character of its own.
(a) Sextic: the 4 pairs { x i 4 x 5 , x i 2 x j 2 x 5 } \{x_i^4x_5,x_i^2x_j^2x_5\} { x i 4 x 5 , x i 2 x j 2 x 5 } { 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 , x i 4 x j 2 } \{x_i^6x_j,x_i^4x_j^2\} { x i 6 x j , x i 4 x j 2 } { x i 6 x j x k , x j 3 x k 2 } \{x_i^6x_jx_k,x_j^3x_k^2\} { x i 6 x j x k , x j 3 x k 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 3 triples { x i 8 x 4 , x j 2 x k 2 x 4 4 , x i 2 x j 4 x k 4 } \{x_i^8x_4,x_j^2x_k^2x_4^4,x_i^2x_j^4x_k^4\} { x i 8 x 4 , x j 2 x k 2 x 4 4 , x i 2 x j 4 x k 4 } { x 1 c 1 x 2 c 2 x 3 c 3 x 4 , x 1 c 1 + 2 x 2 c 2 + 2 x 3 c 3 + 2 } \{x_1^{c_1}x_2^{c_2}x_3^{c_3}x_4,x_1^{c_1+2}x_2^{c_2+2}x_3^{c_3+2}\} { x 1 c 1 x 2 c 2 x 3 c 3 x 4 , x 1 c 1 + 2 x 2 c 2 + 2 x 3 c 3 + 2 } c 1 + c 2 + c 3 = 4 c_1+c_2+c_3=4 c 1 + c 2 + c 3 = 4 c c c ( 2 , 2 , 0 ) (2,2,0) ( 2 , 2 , 0 ) { x i 8 x j , x i 4 x j 6 } \{x_i^8x_j,x_i^4x_j^6\} { x i 8 x j , x i 4 x j 6 } { x i 8 x j x k , x j 3 x k 4 } \{x_i^8x_jx_k,x_j^3x_k^4\} { x i 8 x j x k , x j 3 x k 4 } { x i 6 x j 4 , x j 6 x k 4 } \{x_i^6x_j^4,x_j^6x_k^4\} { x i 6 x j 4 , x j 6 x k 4 } { x i 6 x j x k x 4 , x j 5 x k 5 } \{x_i^6x_jx_kx_4,x_j^5x_k^5\} { x i 6 x j x k x 4 , x j 5 x k 5 } { x i 7 x j x 4 , x i x j 4 x k } \{x_i^7x_jx_4,x_ix_j^4x_k\} { x i 7 x j x 4 , x i x j 4 x k } { x i 6 x 4 2 , x j 6 x k x 4 } \{x_i^6x_4^2,x_j^6x_kx_4\} { x i 6 x 4 2 , x j 6 x k 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 13 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 , 0 , 1 ) \mathbf b=(4,0,0,0,1) b = ( 4 , 0 , 0 , 0 , 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 s\mathbf1 s 1 ( π ( 0 , 1 , 3 , 4 ) , 2 ) (\pi(0,1,3,4),2) ( π ( 0 , 1 , 3 , 4 ) , 2 ) s ≠ 0 ( m o d 3 ) s\ne0\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 s_{1} s 1 ( π ( 0 , 2 , 4 , 6 ) , 1 ) (\pi(0,2,4,6),1) ( π ( 0 , 2 , 4 , 6 ) , 1 ) 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 ) − ( 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.
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^{\mathbf{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 ; (11) e_{i}=n_{i}-2-r+n_{i}k_{i}; \tag*{(11)} e i = n i − 2 − r + n i k i ; (11) 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 15 (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^{\mathbf{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^{\mathbf{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\mathbf{1} 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 \ge1 r ≥ 1 i i i n i k i ≥ r + 1 n_i k_i \ge r+1 n i k i ≥ r + 1 k i ≤ r / n i = r w i / d k_i \le 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 \le 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 \le r < d 1 ≤ r < d i i i e i ≥ n i − 1 e_i \ge n_i - 1 e i ≥ n i − 1 (10) gives 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 (11) 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 0 0 0 k i k_i k i e = n − 21 \mathbf{e} = \mathbf{n} - 2\mathbf{1} 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 16. 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 Theorem 5 (ii).
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\mathbf{1} 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 1 ≡ w 5 ψ x h P=x_1^1\equiv w_5\psi x^h P = x 1 1 ≡ w 5 ψ x h (10) at x 5 x_5 x 5 n 5 − 1 n_5-1 n 5 − 1 Corollary 16 the mass map of the invariant direction in the basis of labels is
M b c = λ ( x h + b + c ) , b , c ∈ B . M_{bc}=\lambda\left(x^{h+b+c}\right), \qquad\mathbf{b},\mathbf{c}\in B. M b c = λ ( x h + b + c ) , b , c ∈ B . Since h h h M b c ≠ 0 M_{bc}\ne0 M b c = 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 17. 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_{bc}M_{b'c'}-M_{bc'}M_{b'c}
=M_{bc'}M_{b'c}\left((\psi^d\delta)^{\Delta/d}-1\right), M b c 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 15 , 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 ) (11) , with k , k ′ \mathbf{k},\mathbf{k}' k , k ′ κ , κ ′ \boldsymbol{\kappa},\boldsymbol{\kappa}' κ , κ ′ n i ( k i + k i ′ − κ i − κ i ′ ) = Δ n_i(k_i+k_i'-\kappa_i-\kappa_i')=\Delta n i ( k i + k i ′ − κ i − κ i ′ ) = Δ i i i n i n_i n i Δ \Delta Δ d d d Δ \Delta Δ ∣ 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 18. 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 member 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\mathbf{1}-h-\mathbf{b} b ↦ n − 2 1 − h − b b ≤ n − 21 − h \mathbf{b}\le\mathbf{n}-2\mathbf{1}-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_{\mathrm{bc}} M bc b ∈ B χ \mathbf{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 14 and their negatives give the rows of Table 4 : for one representative of each orbit the character − χ -\chi − χ Theorem 15 ; Appendix E lists the sectors and the vectors k \mathbf{k} k i = 1 i=1 i = 1 h + ( 3 , 1 , 1 , 0 ) + ( 4 , 0 , 0 , 1 ) = ( 8 , 2 , 2 , 2 , 2 ) h+(3,1,1,0)+(4,0,0,1)=(8,2,2,2,2) h + ( 3 , 1 , 1 , 0 ) + ( 4 , 0 , 0 , 1 ) = ( 8 , 2 , 2 , 2 , 2 ) k = ( 1 , 0 , 0 , 0 , 1 ) \mathbf{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)=\mathbf{n}-2\mathbf{1} h + ( 3 , 1 , 1 , 1 , 0 ) + ( 0 , 2 , 2 , 2 , 0 ) = ( 4 , 4 , 4 , 4 , 1 ) = n − 2 1 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 ) 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 Lemma 17 . Adding the last column of 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_{\mathrm{bc}}\ne0 M bc = 0 h + b + c = n − 21 h+b+c=n-21 h + b + c = n − 21 c = n − 21 − h − b \mathbf{c}=\mathbf{n}-21-\mathbf{h}-\mathbf{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 \mathbf{b}\le\mathbf{n}-21-\mathbf{h} b ≤ n − 21 − h n − 21 − h = ( 3 , 3 , 3 , 3 , 0 ) \mathbf{n}-21-\mathbf{h}=(3,3,3,3,0) n − 21 − h = ( 3 , 3 , 3 , 3 , 0 ) ( 4 , 4 , 4 , 4 , 0 ) (4,4,4,4,0) ( 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 19. 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 6 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 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 11 ), 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 ∣ 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_0^3 L^3}{|H|}, \qquad c_2(Y)\mathbin{\cdot}D = \frac{m_0 c_2\mathbin{\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)\mathbin{\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) \equiv0 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)\mathbin{\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)\mathbin{\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 5. Smoothness and h 2 , 1 h^{2,1} h 2 , 1 Lemma 11 , part (i) is Proposition 14, part (ii) is Corollary 16 , part (iii) is Proposition 18 , and part (iv) is Proposition 19 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 20 (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 Theorem 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 1 × 1 1\times1 1 × 1 Δ = 0 \Delta=0 Δ = 0
The selection conjecture
The landscape and its metric
We use the setting of [33 ] . A heterotic vacuum is a pair ( X , V ) (X,V) ( X , V ) X X X V V V c 1 ( V ) = 0 c_1(V)=0 c 1 ( V ) = 0 c 2 ( T X ) − c 2 ( V ) c_2(TX)-c_2(V) c 2 ( T X ) − c 2 ( V ) V = T X V=T_X V = T X L \mathcal{L} L L τ \mathcal{L}_\tau L τ τ \tau τ g = G W P ⊕ G V g=G^{\mathrm{WP}}\oplus G^V g = G WP ⊕ G V L 2 L^2 L 2 7 μ \mu μ L μ \mathcal{L}^{\mu} L μ g g g [33 ] records for the Weil–Petersson metric. A vacuum is viable when it satisfies the nine conditions (C1)–(C9) of the phenomenological closure theorem of [33 ] : anomaly cancellation and stability, three net families, a Wilson line breaking to the Standard Model group, three families without chiral exotics, masses for the vectorlike exotics, doublet–triplet splitting, stabilised moduli, an acceptable rank and hierarchy of the Yukawa masses, and consistent supersymmetry breaking.
Conjecture 21 (Geometric Selection Conjecture [33 ] ). There is a functional S : L → [ 0 , ∞ ) \mathcal{S}:\mathcal{L}\to[0,\infty) S : L → [ 0 , ∞ ) g g g
(G1) S \mathcal{S} S L τ \mathcal{L}_\tau L τ
(G2) S \mathcal{S} S Y ∗ ∈ L \mathcal{Y}^*\in\mathcal{L} Y ∗ ∈ L
(G3) Y ∗ \mathcal{Y}^* Y ∗
(G4) S ∘ μ = S \mathcal{S}\circ\mu=\mathcal{S} S ∘ μ = S L μ \mathcal{L}^{\mu} L μ
The functional adds nothing.
Theorem 22. Conjecture 21 holds if and only if there is a viable vacuum V ⋆ \mathcal{V}^\star V ⋆ μ ( V ⋆ ) = V ⋆ \mu(\mathcal{V}^\star)=\mathcal{V}^\star μ ( V ⋆ ) = V ⋆ V ⋆ ∈ L μ \mathcal{V}^\star\in\mathcal{L}^{\mu} V ⋆ ∈ L μ
Proof. Suppose S \mathcal{S} S V ⋆ \mathcal{V}^\star V ⋆ V ⋆ ∈ L μ \mathcal{V}^\star\in\mathcal{L}^{\mu} V ⋆ ∈ L μ S ( μ ( V ⋆ ) ) = S ( V ⋆ ) \mathcal{S}(\mu(\mathcal{V}^\star))=\mathcal{S}(\mathcal{V}^\star) S ( μ ( V ⋆ )) = S ( V ⋆ ) μ V ⋆ \mu\mathcal{V}^\star μ V ⋆ μ V ⋆ = V ⋆ \mu\mathcal{V}^\star=\mathcal{V}^\star μ V ⋆ = V ⋆
Conversely, let V ⋆ \mathcal{V}^\star V ⋆ μ \mu μ μ \mu μ L τ ⋆ \mathcal{L}_{\tau^\star} L τ ⋆
A proper function on every component. A second countable manifold M M M ρ : M → [ 0 , ∞ ) \rho:M\to[0,\infty) ρ : M → [ 0 , ∞ ) K 1 ⊂ K 2 ⊂ ⋯ K_1\subset K_2\subset\cdots K 1 ⊂ K 2 ⊂ ⋯ K j K_j K j K j + 1 K_{j+1} K j + 1 M M M ( ϕ j ) (\phi_j) ( ϕ j ) ϕ j \phi_j ϕ j K j + 1 K_{j+1} K j + 1 K j − 1 K_{j-1} K j − 1 ρ = ∑ j j ϕ j \rho=\sum_j j\phi_j ρ = ∑ j j ϕ j ρ ≥ j \rho\geq j ρ ≥ j K j K_j K j ρ τ \rho_\tau ρ τ μ \mu μ L μ \mathcal{L}^{\mu} L μ ρ \rho ρ ρ + ρ ∘ μ \rho+\rho\circ\mu ρ + ρ ∘ μ ρ \rho ρ μ \mu μ ρ \rho ρ
A quadratic well at V ⋆ \mathcal{V}^\star V ⋆ Choose r > 0 r>0 r > 0 g g g V ⋆ \mathcal{V}^\star V ⋆ B ‾ \overline{B} B r r r V ⋆ \mathcal{V}^\star V ⋆ B B B f = d g ( V ⋆ , ⋅ ) 2 f=d_g(\mathcal{V}^\star,\cdot)^2 f = d g ( V ⋆ , ⋅ ) 2 V ⋆ ∈ L μ \mathcal{V}^\star\in\mathcal{L}^{\mu} V ⋆ ∈ L μ μ \mu μ V ⋆ \mathcal{V}^\star V ⋆ B B B f ∘ μ = f f\circ\mu=f f ∘ μ = f
Gluing. Let χ : [ 0 , ∞ ) → [ 0 , 1 ] \chi:[0,\infty)\to[0,1] χ : [ 0 , ∞ ) → [ 0 , 1 ] [ 0 , r 2 / 4 ] [0,r^2/4] [ 0 , r 2 /4 ] [ r 2 / 2 , ∞ ) [r^2/2,\infty) [ r 2 /2 , ∞ ) χ ∘ f \chi\circ f χ ∘ f B B B χ ∘ f \chi\circ f χ ∘ f B B B r / 2 r/\sqrt{2} r / 2 V ⋆ \mathcal{V}^\star V ⋆ c = r 2 / 4 c=r^2/4 c = r 2 /4
S = ( χ ∘ f ) f + ( 1 − χ ∘ f ) ( c + ρ ) \mathcal{S}=(\chi\circ f)f+(1-\chi\circ f)(c+\rho) S = ( χ ∘ f ) f + ( 1 − χ ∘ f ) ( c + ρ ) on L τ ⋆ \mathcal{L}_{\tau^\star} L τ ⋆ S = c + ρ \mathcal{S}=c+\rho S = c + ρ S \mathcal{S} S μ \mu μ V ⋆ \mathcal{V}^\star V ⋆ χ ∘ f = 1 \chi\circ f=1 χ ∘ f = 1 f f f V ⋆ \mathcal{V}^\star V ⋆ 0 < χ ∘ f < 1 0<\chi\circ f<1 0 < χ ∘ f < 1 f ≥ r 2 / 4 f\geq r^2/4 f ≥ r 2 /4 c + ρ ≥ c c+\rho\geq c c + ρ ≥ c χ ∘ f = 0 \chi\circ f=0 χ ∘ f = 0 c + ρ ≥ c c+\rho\geq c c + ρ ≥ c V ⋆ \mathcal{V}^\star V ⋆ { S ≤ t } \{\mathcal{S}\leq t\} { S ≤ t } L τ ⋆ \mathcal{L}_{\tau^\star} L τ ⋆ B ‾ ∪ { ρ ≤ t − c } \overline{B}\cup\{\rho\leq t-c\} B ∪ { ρ ≤ t − c } c + ρ c+\rho c + ρ S \mathcal{S} S
Figure 5 draws the functional of the proof on one component.
Remark 23 (Constructed from the metric). The phrase “constructed from the metric” in Conjecture 21 is not a mathematical condition, and the strongest condition it can be read as is invariance under a group of isometries of ( L , g ) (\mathcal{L},g) ( L , g ) G G G μ \mu μ ρ \rho ρ G G G G G G ∫ ρ ( h x ) d h ≤ t \int\rho(hx)\,dh\leq t ∫ ρ ( h x ) d h ≤ t ρ ( h x ) ≤ t \rho(hx)\leq t ρ ( h x ) ≤ t h h h x x x G ⋅ { ρ ≤ t } G\cdot\{\rho\leq t\} G ⋅ { ρ ≤ t } G G G G G G G G G
Remark 24 (Mirror pairs). A viable vacuum outside the standard embedding lies outside L μ \mathcal{L}^{\mu} L μ μ \mu μ ( 2 , 2 ) (2,2) ( 2 , 2 ) μ \mu μ L \mathcal{L} L
The Dwork pencil.
Proposition 25. (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 3 are ∣ ψ ∣ r ( b ) |\psi|^{r(b)} ∣ ψ ∣ r ( b ) ( g b g σ b ) − 1 / 2 (g_b g_{\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 )
Figure 5. The functional S \mathcal{S} S Theorem 22 on a two-dimensional component L ψ ∗ \mathcal{L}_{\psi^*} L ψ ∗ μ \mu μ r / 2 r/2 r /2 V ∗ \mathcal{V}^* V ∗ f f f r / 2 r/\sqrt{2} r / 2 c + ρ c+\rho c + ρ ρ \rho ρ μ \mu μ S \mathcal{S} S V ∗ \mathcal{V}^* V ∗
(iii) On the sextic, octic and dectic pencils of § 8 , with ψ ≠ 0 \psi\ne0 ψ = 0 ψ d δ ≠ 1 \psi^d\delta\ne1 ψ d δ = 1
Proof. Part (i) is Theorem 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 6 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
The proposition shows where the Dwork pencil stands with respect to the conjecture. The condition on masses is met in the strongest form one can ask of a single modulus: the mass matrix has full rank and three scales 1 1 1 ∣ ψ ∣ |\psi| ∣ ψ ∣ ∣ ψ ∣ 2 |\psi|^2 ∣ ψ ∣ 2 ψ → 0 \psi\to0 ψ → 0 [20 ] , of Bouchard and Donagi [21 ] , and the line-bundle models of Anderson, Gray, Lukas and Palti [22 , 23 ] , which also address (C5) and (C6). For none of them are (C7), (C8) and (C9) established, and the words “controlled”, “acceptable” and “consistent” in those conditions are fixed by comparison with measurement, not by a definition. By Theorem 22 this is the whole of the open part of the Geometric Selection Conjecture. The weighted pencils fail on both counts: they too never carry three families, and their mass map leaves generations massless at every member.
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 [31 ] . 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 25 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 + kc_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 26 (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 \ge1 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 \le j < p 1 ≤ j < p g j g^j g j [29 ] gives ∑ q ( − 1 ) q tr ( g j ∣ H q ( X , A m ) ) = 0 \sum_q (-1)^q \operatorname{tr}(g^j \mid H^q(X,A^m)) = 0 ∑ q ( − 1 ) q tr ( g j ∣ H q ( X , A m )) = 0 tr ( g j ∣ H 0 ( X , A m ) ) = 0 \operatorname{tr}(g^j \mid H^0(X,A^m)) = 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 □ \square □
Proposition 27. 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 11 , X X X K X ≅ O X ( ∑ j d j − ∑ i w i ) = O X K_X \cong\mathcal{O}_X(\sum_j d_j - \sum_i w_i) = \mathcal{O}_X K X ≅ O X ( ∑ j d j − ∑ i w i ) = O X [28 ] 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 [30 ] 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 Lemma 26 with 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 \ge1 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 + kc_2 \cdot L/12 χ ( X , k L ) = k 3 L 3 /6 + k c 2 ⋅ L /12 k = 0 k = 0 k = 0 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 \ge1 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 Lemma 11 , from 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
By Theorem 6 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 5 (iv) and Proposition 25 (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 [32 ] with a computer search; the proposition uses only the holomorphic Lefschetz formula and Riemann–Roch.8
Conclusion
On the Dwork pencil the question of the title has a complete answer. The hundred and one generations are labelled by characters of the symmetry group Γ \Gamma Γ ψ 70 \psi^{70} ψ 70 (6) and the grading of the Jacobian ring by characters, together with the general dictionary of § 2 .
Two questions are not settled by the geometry of the pencil. The breaking of E 6 E_6 E 6 100 / ∣ H ∣ 100/|H| 100/∣ H ∣ Proposition 25 ), and the same holds for every free quotient of the thirteen one-parameter complete intersections (Proposition 27 ). The selection of the vacuum is reduced by Theorem 22 to the existence of one viable vacuum, and that existence depends on conditions (C7) to (C9), whose thresholds are set by measurement. The same two facts drive § 8 , where the three other one-parameter families of Fermat type are treated. There the characters no longer separate the generations, the invariant deformation leaves 4, 24 and 36 of them massless at every member, and no member or free quotient carries three families. Among the four pencils of Fermat type, the quintic is the only one on which the symmetry labels every generation and the invariant deformation gives every generation a mass.
Data and code availability
The programs that repeat the finite computations of this paper are in its archive, release 1.2.0, https://github.com/creelie/strings/releases/tag/v1.2.0[38 ] , which also holds the figures and the record of the engine runs. They are written in C (the label and coupling counts), C++ (the expansion of the Hessian by permutations), Python (all counts, the intersection numbers of Appendix § 15 , the Gamma values, and the labels, characters, couplings and mass maps of the weighted pencils, and the intersection numbers, values χ ( X , k L ) \chi(X,kL) χ ( X , k L ) g X \mathfrak{g}_X g X Table 5 ), Julia (the Gamma values and the intersection numbers), Macaulay2 (the Jacobian ring of the quintic at two rational values of ψ \psi ψ ψ \psi ψ 3 d 3d 3 d [36 ] , described in [37 ] ; its record for the five targets of this paper is part of the archive. The programs and the engine runs were prepared by Deep Bhattacharjee with AI assistance from Claude (Anthropic).
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Funding. None. Competing interests. None.
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 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 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 = 3200000 ; 1^5:\ 1\ \text{permutation},\ m=0,\ +20^5=3200000; 1 5 : 1 permutation , m = 0 , + 2 0 5 = 3200000 ; 2 1 3 : 10 , odd , m = 2 : 10 ⋅ ( − 1 ) ⋅ 20 3 ⋅ 25 = − 2000000 ; 2\,1^3:\ 10,\ \text{odd},\ m=2:\ 10\cdot(-1)\cdot20^3\cdot25=-2000000; 2 1 3 : 10 , odd , m = 2 : 10 ⋅ ( − 1 ) ⋅ 2 0 3 ⋅ 25 = − 2000000 ; 3 1 2 : 20 , even , m = 3 : 20 ⋅ 20 2 ⋅ ( − 125 ) = − 1000000 ; 3\,1^2:\ 20,\ \text{even},\ m=3:\ 20\cdot20^2\cdot(-125)=-1000000; 3 1 2 : 20 , even , m = 3 : 20 ⋅ 2 0 2 ⋅ ( − 125 ) = − 1000000 ; 4 1 : 30 , odd , m = 4 : 30 ⋅ ( − 1 ) ⋅ 20 ⋅ 625 = − 375000 ; 4\,1:\ 30,\ \text{odd},\ m=4:\ 30\cdot(-1)\cdot20\cdot625=-375000; 4 1 : 30 , odd , m = 4 : 30 ⋅ ( − 1 ) ⋅ 20 ⋅ 625 = − 375000 ; 2 2 1 : 15 , even , m = 4 : 15 ⋅ 20 ⋅ 625 = 187500 ; 2^2\,1:\ 15,\ \text{even},\ m=4:\ 15\cdot20\cdot625=187500; 2 2 1 : 15 , even , m = 4 : 15 ⋅ 20 ⋅ 625 = 187500 ; 5 : 24 , even , m = 5 : 24 ⋅ ( − 3125 ) = − 75000 ; 5:\ 24,\ \text{even},\ m=5:\ 24\cdot(-3125)=-75000; 5 : 24 , even , m = 5 : 24 ⋅ ( − 3125 ) = − 75000 ; 3 2 : 20 , odd , m = 5 : 20 ⋅ ( − 1 ) ⋅ ( − 3125 ) = 62500 , 3\,2:\ 20,\ \text{odd},\ m=5:\ 20\cdot(-1)\cdot(-3125)=62500, 3 2 : 20 , odd , m = 5 : 20 ⋅ ( − 1 ) ⋅ ( − 3125 ) = 62500 , 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 − 3200000 -3200000 − 3200000 H ( ψ ) = 3200000 ( 1 − ψ 5 ) H(\psi)=3200000(1-\psi^5) H ( ψ ) = 3200000 ( 1 − ψ 5 )
Appendix B. The coupling count
Ordered triples by sector
For e e e O ( e ) O(e) O ( e ) ( a , b , c ) ∈ B 3 (\mathbf{a},\mathbf{b},\mathbf{c}) \in 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 B c ∈ B c ≤ e \mathbf{c}\leq\mathbf{e} c ≤ e s = e − c \mathbf{s}=\mathbf{e}-\mathbf{c} s = e − c a ∈ B \mathbf{a}\in B a ∈ B s − a ∈ B \mathbf{s}-\mathbf{a}\in 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)\leq a_i\leq\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. 25).
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 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 ) Since a 1 + b 1 ≤ 6 a_1+b_1\leq6 a 1 + b 1 ≤ 6 c 1 ≥ 1 c_1\geq1 c 1 ≥ 1 c i ≤ 2 c_i\leq2 c i ≤ 2 i ≥ 2 i\geq2 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\geq4-c_1 a 1 ≥ 4 − c 1 s i = 2 − c i ≤ 2 s_i=2-c_i\leq2 s i = 2 − c i ≤ 2 a i ≤ s i a_i\leq 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 )
Table 2 (PDF p. 25).
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\leq(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 ) Here c 3 , c 4 , c 5 ≤ 1 c_3,c_4,c_5\leq1 c 3 , c 4 , c 5 ≤ 1
Table 3 (PDF p. 25).
and O ( 6 , 6 , 1 , 1 , 1 ) = 162 O(6,6,1,1,1)=162 O ( 6 , 6 , 1 , 1 , 1 ) = 162 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\leq3 a i ≤ 3 1 ≤ a 3 ≤ 3 1\leq a_3\leq3 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
B.2. 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 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 20 + 30 = 50 20+30=50 20 + 30 = 50 D r = 0 , 50 , 20 , 30 D_r=0,50,20,30 D r = 0 , 50 , 20 , 30 § 5.1 .
Hodge norms at the Fermat point
This appendix proves Proposition 10 . Write G = μ 5 / μ 5 G=\mu_5/\mu_5 G = μ 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^{a\cdot\mathbf{k}} χ k ( a ) = ζ a ⋅ k
Eigenlines
C.1. 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 = 0 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=0}^{4}\zeta^{jk}\right)^5=\frac{1}{5}(4^5-4)=204=b_3(X_0) 5 1 ∑ j = 0 4 ( ∑ k = 0 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
C.2. 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 ˉ \bar{\chi} χ ˉ e e e u u u L ′ L' L ′ L L L u ˉ \bar{u} u ˉ e e e D ω b = α u ˉ D\omega_{\mathbf{b}}=\alpha\bar{u} D ω b = α u ˉ u ˉ ≠ 0 \bar{u}\ne0 u ˉ = 0
∫ e ω b = Q ( e , α u ˉ ) = α Q ( u , u ˉ ) , ∫ X 0 ω b ∧ ω b ‾ = ∣ α ∣ 2 Q ( u , u ˉ ) , \int_e\omega_{\mathbf{b}}=Q(e,\alpha\bar{u})=\alpha Q(u,\bar{u}),\qquad\int_{X_0}\omega_{\mathbf{b}}\wedge\overline{\omega_{\mathbf{b}}}=|\alpha|^2Q(u,\bar{u}), ∫ e ω b = Q ( e , α u ˉ ) = α Q ( u , u ˉ ) , ∫ X 0 ω b ∧ ω b = ∣ α ∣ 2 Q ( u , u ˉ ) , and therefore
g b = ∣ ∫ X 0 ω b ∧ ω b ‾ ∣ = ∣ ∫ e ω b ∣ 2 ∣ Q ( u , u ˉ ) ∣ . (12) 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}{\left|Q(u,\bar{u})\right|}. \tag*{(12)} g b = ∫ X 0 ω b ∧ ω b = ∣ Q ( u , u ˉ ) ∣ ∫ e ω b 2 . (12) It remains to compute both sides for one cycle.
The cycle of Pham
C.3. 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=-(k_1+k_2+k_3+k_4) k 5 = − ( k 1 + k 2 + k 3 + k 4 ) [17 ] showed that U U U − 1 -1 − 1 H 3 ( U , Z ) H_3(U,\mathbb{Z}) H 3 ( U , Z ) Z [ μ 5 ] \mathbb{Z}[\mu_5] Z [ μ 5 ]
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 , ∑ v i 5 = 1 } \{y_i=\varepsilon v_i,\ v_i\geq0,\ \sum v_i^5=1\} { y i = ε v i , v i ≥ 0 , ∑ v i 5 = 1 } ε 5 = − 1 \varepsilon^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 \mathbf{y}^{\mathbf{b}'}\,d\mathbf{y}/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 ) \mathbf{y}^{\mathbf{b}'}d\mathbf{y}/f_s^2=-\partial_s(\mathbf{y}^{\mathbf{b}'}d\mathbf{y}/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^{b'}\frac{dy}{f_s}=\int_{e_s}y^{b'}dy/df_s 2 π i 1 ∫ τ ( e ) y b ′ f s d y = ∫ 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\leq4}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_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^{b'}=\varepsilon^{|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=\varepsilon^{|b'|+4}5^{-4}\prod_{i\leq4}\Gamma(k_i/5)/\Gamma(\sum_{i\leq4}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\leq4}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 ) . (13) \left|\int_e\omega_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*{(13)} ∫ e ω b = 5 − 4 i = 1 ∏ 4 ∣1 − ζ k i ∣ Γ ( 1 − k 5 /5 ) ∏ i = 1 4 Γ ( k i /5 ) . (13) 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 ) \hat{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{\hat{x}(k)\hat{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 [18 ] , in the form of Sakamoto [19 ] 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}(k)\hat{z}(-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{|1-\zeta^{k_1+k_2+k_3+k_4}|}{\prod_{i\leq4}|1-\zeta^{k_i}|^2}
=\frac{|1-\zeta^{k_5}|}{\prod_{i\leq4}|1-\zeta^{k_i}|^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}(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 ∣ . (14) |Q(u,\bar{u})|=5^{-4}\prod_{i=1}^{5}|1-\zeta^{k_i}|. \tag*{(14)} ∣ Q ( u , u ˉ ) ∣ = 5 − 4 i = 1 ∏ 5 ∣1 − ζ k i ∣. (14) This is nonzero, so u ˉ ≠ 0 \bar{u}\ne0 u ˉ = 0 (12) .
Conclusion of the proof
Inserting (13) and (14) in (12) ,
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_b=5^{-4}\prod_{i=1}^{4}|1-\zeta^{k_i}|\Gamma(k_i/5)^2\cdot\frac{1}{|1-\zeta^{k_5}|\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 ) |1-\zeta^k|=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_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 k=1 k = 1 ∣ ∫ e Ω ∣ = ∣ P ∣ |\int_e\Omega|=|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 )
Appendix D. Shared characters on the octic and the dectic
This appendix completes the proof of Proposition 14. By Lemma 13 it suffices to find, for each shift s ≤ d / 2 s \le d/2 s ≤ d /2 b \mathbf{b} b b i + s ≠ n i − 1 ( m o d n i ) b_i+s \ne 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 \le6 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 \le10<16 b i + s ≤ 10 < 16 a i = 1 a_i=1 a i = 1 b i ≥ 8 − s b_i \ge8-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 \ge8-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 \mathbf{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 \pmod5 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 14(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
Appendix E. The blocks of the mass map
Table 6 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 (11) and the entry ψ r ∏ i w i k i \psi^r\prod_i w_i^{k_i} ψ r ∏ i w i k i Theorem 15 . 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 17 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 6. 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+\mathbf{b}+\mathbf{c} h + b + c k k k (11) .
References
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: itsdeep@live.com URL: https://orcid.org/0000-0003-0466-750X
Formerly, National Taiwan University, Taipei 106319, Taiwan Email address: rai.bhattacharya.in@gmail.com URL: https://orcid.org/0009-0002-2254-3914
