Introduction
A self-avoiding path at its critical weight can make a long excursion without paying an exponential cost. A basic quantitative question is how much of its weight reaches the opposite side of a strip. On the regular honeycomb lattice we prove that this crossing mass has power − 1 / 4 -1/4 − 1/4
The model and the result
We use the honeycomb lattice dual to the equilateral triangular tiling of side length one, with one triangular edge direction horizontal. The separation of consecutive horizontal band boundaries is d 0 = 3 / 2 d_0=\sqrt{3}/2 d 0 = 3 /2 port is the midpoint of an edge on the boundary of a union of triangles. A path between distinct ports follows dual edges, including the two terminal half-edges, visits each dual vertex at most once, and meets the boundary only at its endpoints. Its length ∣ γ ∣ |\gamma| ∣ γ ∣ ρ ∣ γ ∣ \rho^{|\gamma|} ρ ∣ γ ∣
ρ = 1 2 + 2 , c = cos ( 3 π / 8 ) . (1) \rho=\frac{1}{\sqrt{2+\sqrt{2}}}, \qquad c=\cos(3\pi/8). \tag*{(1)} ρ = 2 + 2 1 , c = cos ( 3 π /8 ) . (1) This vertex convention agrees with the usual edge convention for the connective constant up to the fixed endpoint factor. Duminil-Copin and Smirnov proved that the reciprocal of ρ \rho ρ [9 ] .
Let S N S_N S N N N N N ≥ 1 N\ge1 N ≥ 1 Z \mathbb{Z} Z 0 0 0 Z D ( a , b ) Z_D(a,b) Z D ( a , b ) a a a b b b D D D arch ; a path ending at the top is called a bridge . In both cases all other points of the path are in the interior of the strip. Define
K N ( k ) = Z S N ( 0 , k ) , A N = ∑ k ≠ 0 K N ( k ) , B N = ∑ b on the top Z S N ( 0 , b ) , m N = ∑ k ≥ 1 k K N ( k ) . (2) \begin{aligned}
K_N(k)&=Z_{S_N}(0,k),\\
A_N&=\sum_{k\ne0}K_N(k), \qquad B_N=\sum_{\text{$b$ on the top}} Z_{S_N}(0,b),\\
m_N&=\sum_{k\ge1} kK_N(k). \tag*{(2)}
\end{aligned} K N ( k ) A N m N = Z S N ( 0 , k ) , = k = 0 ∑ K N ( k ) , B N = b on the top ∑ Z S N ( 0 , b ) , = k ≥ 1 ∑ k K N ( k ) . (2) The sums run over actual admissible boundary ports. Adjacent bottom ports have horizontal separation one, so m N m_N m N N / 2 + Z N/2+\mathbb{Z} N /2 + Z B N B_N B N Figure 1 shows the convention.
Figure 1. Port and length conventions. The path passes through six adjacent triangles of the triangular tiling, so it visits six vertices of the honeycomb dual and has port weight ρ 6 \rho^6 ρ 6
Theorem 1.1 (Critical strip mass). For every integer N ≥ 1 N \ge1 N ≥ 1 (2) are finite. With constants independent of N N N
c A N + B N = 1 , m N + 1 − m N ≍ B N , m N ≍ N 3 / 4 , B N ≍ N − 1 / 4 . (3) cA_N+B_N=1,\qquad m_{N+1}-m_N\asymp B_N,\qquad m_N\asymp N^{3/4},\qquad B_N\asymp N^{-1/4}. \tag*{(3)} c A N + B N = 1 , m N + 1 − m N ≍ B N , m N ≍ N 3/4 , B N ≍ N − 1/4 . (3) Moreover, B N B_N B N N N N
Here f N ≍ g N f_N\asymp g_N f N ≍ g N B 0 = 1 B_0=1 B 0 = 1 B N ≍ ( 1 + N ) − 1 / 4 B_N\asymp(1+N)^{-1/4} B N ≍ ( 1 + N ) − 1/4 N ≥ 0 N\ge0 N ≥ 0
Earlier work
The two-dimensional polymer exponents arose from the dilute O ( n ) O(n) O ( n ) n = 0 n=0 n = 0 [22 ] . Lawler, Schramm and Werner developed the conjectural conformally invariant scaling picture for planar self-avoiding walk [21 ] , Sections 3.3.1, 3.4.3 and 4.1. Their boundary exponent 5 / 8 5/8 5/8 − 5 / 4 -5/4 − 5/4 − 1 / 4 -1/4 − 1/4 [9 ] , Section 4.
Duminil-Copin and Smirnov proved the critical-point prediction using a parafermionic observable whose local cancellation gives a positive boundary identity [9 ] , Lemmas 1 and 2. Their strip argument gives bounds of order 1 / N 1/N 1/ N 1 1 1
The decay of this mass was subsequently proved by Beaton, Bousquet-Mélou, de Gier, Duminil-Copin and Guttmann in their study of surface adsorption [2 ] , Theorem 10. Glazman and Manolescu obtained a shorter proof, a logarithmic bound along a sequence of heights, and invariance of the half-plane boundary two-point function under columnwise rhombic deformations with angles in [ π / 3 , 2 π / 3 ] [\pi/3, 2\pi/3] [ π /3 , 2 π /3 ] [16 ] . Krachun and Panagiotis proved a polynomial upper bound B N ≤ 100 N − 10 − 10 B_N \le100N^{-10^{-10}} B N ≤ 100 N − 1 0 − 10 [20 ] . Theorem 1.1 establishes the predicted strip power with uniform multiplicative constants in the port convention above. The exponent 3 / 4 3/4 3/4 m N m_N m N
Proof strategy
The boundary identity alone does not determine this power. Our route is to calculate m N m_N m N B N B_N B N
In Section 2, cancellation of closed excursions gives the boundary identity and controls finite-height transfer matrices. We introduce rhombic weights with one spectral parameter in each row. A cut between columns records vacant positions and occupied positions paired by path pieces on one side of the cut. The stationary vacuum vector records the weighted sums of these partial pairings with no exterior source; a source vector has one additional strand connected to a specified boundary source. Their local relations belong to the dilute-loop integrable structure; the relation with discrete parafermions was developed by Ikhlef and Cardy, and the weighted-walk normalization and corrected general formulas were treated by Glazman [18 , 15 ] . We prove the exact local identities used here in Appendix A.
Section 3 constructs a scalar Pfaffian p N p_N p N [8 , 13 , 14 ] . Here the degree and stationarity arguments are proved for zero loop weight. Section 4 cuts an arch at every column between its endpoints. This converts m N m_N m N p N + 1 / p N p_{N+1}/p_N p N + 1 / p N m N + 1 − m N ≍ B N m_{N+1}-m_N \asymp B_N m N + 1 − m N ≍ B N
Section 5 converts the homogeneous polynomial problem into a positive integral over 0 < y 1 < ⋯ < y N < 1 0<y_1<\cdots<y_N<1 0 < y 1 < ⋯ < y N < 1
∏ i = 1 N ω ( y i ) ∏ i < j ( y j − y i ) 2 y i + y j , ω ( y ) = 1 2 ( 1 + y 2 y − 1 ) . \prod_{i=1}^{N}\omega(y_i)\prod_{i<j}\frac{(y_j-y_i)^2}{y_i+y_j}, \qquad\omega(y)=\frac{1}{2}\left(\sqrt{\frac{1+y}{2y}}-1\right). i = 1 ∏ N ω ( y i ) i < j ∏ y i + y j ( y j − y i ) 2 , ω ( y ) = 2 1 ( 2 y 1 + y − 1 ) . Positivity of a Cauchy-moment determinant makes the coincident-parameter limit nonsingular. The pair interaction in the resulting integral is the one appearing in generalized Bures ensembles [12 ] . The derivation from the strip polynomial fixes the normalization and identifies the integral with the arch moment. Finally, Section 6 compares this measure with Bures–Lagüerre laws. Schur’s Pfaffian identity and de Bruijn’s integration formula give their normalizing constants; positive association gives the comparison. The smallest-coordinate estimates yield m N ≍ N 3 / 4 m_N \asymp N^{3/4} m N ≍ N 3/4
Positive flux and finite strip transfers
We write
t = 3 8 , λ = π 8 , d = cos ( 2 λ ) = sin ( 2 λ ) , C = cos λ . (4) t=\frac{3}{8}, \qquad\lambda=\frac{\pi}{8}, \qquad d=\cos(2\lambda)=\sin(2\lambda), \qquad C=\cos\lambda. \tag*{(4)} t = 8 3 , λ = 8 π , d = cos ( 2 λ ) = sin ( 2 λ ) , C = cos λ . (4) Then c = cos ( 3 λ ) c=\cos(3\lambda) c = cos ( 3 λ ) ρ = C − c = 1 / ( 2 C ) \rho=C-c=1/(2C) ρ = C − c = 1/ ( 2 C ) (1) .
The boundary identity controls positive path sums before any spectral parameter is introduced. We use it first to make the fixed-height transfer construction convergent. The local identities then allow the stationary vector to be recovered by polynomial interpolation.
Lemma 2.1 (Boundary identity). Let D D D a a a W ( γ ) W(\gamma) W ( γ )
∑ b ∈ ∂ D ∑ γ : a → b ρ ∣ γ ∣ e i t W ( γ ) = 1. (5) \sum_{b\in\partial D}\ \sum_{\gamma:a\to b}\rho^{|\gamma|}e^{itW(\gamma)}=1. \tag*{(5)} b ∈ ∂ D ∑ γ : a → b ∑ ρ ∣ γ ∣ e i t W ( γ ) = 1. (5) If D D D 1 / c 1/c 1/ c
Proof. Send mass one along the inward half-edge at a a a ρ e i λ \rho e^{i\lambda} ρ e iλ ρ e − i λ \rho e^{-i\lambda} ρ e − iλ 2 ρ cos λ = 1 2\rho\cos\lambda=1 2 ρ cos λ = 1 D D D D D D
For a stopped collision, the newly closed cycle is simple and has a disjoint incoming stem. The stem lies outside the cycle: it starts on ∂ D \partial D ∂ D π / 3 \pi/3 π /3 − π / 3 -\pi/3 − π /3 4 π / 3 4\pi/3 4 π /3 − 4 π / 3 -4\pi/3 − 4 π /3 e i t ( 4 π / 3 ) + e − i t ( 4 π / 3 ) = 0 e^{it(4\pi/3)}+e^{-it(4\pi/3)}=0 e i t ( 4 π /3 ) + e − i t ( 4 π /3 ) = 0 (5) .
For a path ending at b b b b b b a a a π / 2 \pi/2 π /2
W ( γ ) = π − turn ∂ D ( b → a ) . W(\gamma)=\pi-\operatorname{turn}_{\partial D}(b\to a). W ( γ ) = π − turn ∂ D ( b → a ) . For a convex boundary the latter turn lies in [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] ∣ W ( γ ) ∣ ≤ π |W(\gamma)|\le\pi ∣ W ( γ ) ∣ ≤ π cos ( t W ( γ ) ) ≥ cos ( 3 π / 8 ) = c \cos(tW(\gamma))\ge\cos(3\pi/8)=c cos ( t W ( γ )) ≥ cos ( 3 π /8 ) = c □ \square □
We next introduce auxiliary spectral parameters. A column has N N N i i i u i / t u_i/t u i / t u i u_i u i
a ( u ) = cos ( 2 u − 3 λ ) , L ( u ) = sin ( 2 λ + u ) sin ( 3 λ + u ) = C + a ( u ) 2 , A ( u ) = d sin ( 3 λ − u ) L ( u ) , U ( u ) = d sin u L ( u ) , B ( u ) = sin u sin ( 3 λ − u ) L ( u ) = a ( u ) − c a ( u ) + C , E ( u ) = sin ( 3 λ − u ) sin ( 2 λ − u ) L ( u ) , F ( u ) = sin ( u − λ ) sin u L ( u ) . (6) \begin{aligned}
a(u)&=\cos(2u-3\lambda),&
L(u)&=\sin(2\lambda+u)\sin(3\lambda+u)=\frac{C+a(u)}{2},\\
A(u)&=\frac{d\sin(3\lambda-u)}{L(u)},&
U(u)&=\frac{d\sin u}{L(u)},\\
B(u)&=\frac{\sin u\sin(3\lambda-u)}{L(u)}
=\frac{a(u)-c}{a(u)+C},\\
E(u)&=\frac{\sin(3\lambda-u)\sin(2\lambda-u)}{L(u)},&
F(u)&=\frac{\sin(u-\lambda)\sin u}{L(u)}. \tag*{(6)}
\end{aligned} a ( u ) A ( u ) B ( u ) E ( u ) = cos ( 2 u − 3 λ ) , = L ( u ) d sin ( 3 λ − u ) , = L ( u ) sin u sin ( 3 λ − u ) = a ( u ) + C a ( u ) − c , = L ( u ) sin ( 3 λ − u ) sin ( 2 λ − u ) , L ( u ) U ( u ) F ( u ) = sin ( 2 λ + u ) sin ( 3 λ + u ) = 2 C + a ( u ) , = L ( u ) d sin u , = L ( u ) sin ( u − λ ) sin u . (6) These weights come from Nienhuis’s integrable O ( n ) O(n) O ( n ) [23 ] , in the self-avoiding-walk normalization displayed in [16 ] ; the general parafermionic and integrable local formulas are discussed in [18 , 15 ] . The empty tile has weight one. A single connection between opposite ports has weight B B B A A A U U U E E E F F F u = λ u = \lambda u = λ 2 λ 2\lambda 2 λ ρ \rho ρ u = 0 u = 0 u = 0 3 λ 3\lambda 3 λ
The dilute diagram algebra and its Yang–Baxter operators have an algebraic construction in Grimm and Pearce [17 ] . Their distinction between gauge-invariant periodic partition functions and other gauge-dependent quantities is relevant here: we specify the caps, scalar factors and port ordering, and prove the identities for these choices.
We use the planar diagram algebra whose inputs and outputs are ordered slots, each vacant or occupied. Composition glues matching slots and annihilates a mismatch or a closed loop. Tensor product means juxtaposition; transpose reflects a diagram, interchanging inputs and outputs. The two-slot operator R ( v ) R(v) R ( v ) 1 1 1 A ( v ) A(v) A ( v ) B ( v ) B(v) B ( v ) U ( v ) U(v) U ( v ) E ( v ) E(v) E ( v ) F ( v ) F(v) F ( v ) Figure 2 .
Figure 2. The nine local diagrams of R ( v ) R(v) R ( v ) v v v
Number the rows from bottom to top. A column swept from left to right uses R ( 3 λ − u i ) R(3\lambda-u_i) R ( 3 λ − u i ) i i i i i i i + 1 i+1 i + 1
A ( 3 λ − u ) = U ( u ) , U ( 3 λ − u ) = A ( u ) , B ( 3 λ − u ) = B ( u ) , E ( 3 λ − u ) = F ( u ) . A(3\lambda-u)=U(u), \qquad U(3\lambda-u)=A(u), \qquad B(3\lambda-u)=B(u), \qquad E(3\lambda-u)=F(u). A ( 3 λ − u ) = U ( u ) , U ( 3 λ − u ) = A ( u ) , B ( 3 λ − u ) = B ( u ) , E ( 3 λ − u ) = F ( u ) . The last relation also gives F ( 3 λ − u ) = E ( u ) F(3\lambda-u)=E(u) F ( 3 λ − u ) = E ( u ) Figure 3 records the port order.
Figure 3. From a geometric rhombus to a column operator. South and west are the ordered inputs; east and north are the ordered outputs. The operator is drawn with its inputs below and outputs above, so the rhombic parameter u u u 3 λ − u 3\lambda-u 3 λ − u
Define S 2 S_2 S 2 ρ \rho ρ ρ \rho ρ S 3 S_3 S 3
Lemma 2.2 (Local diagram identities). Writing R i R_i R i i , i + 1 i, i+1 i , i + 1
R 2 ( y ) R 1 ( x + y ) R 2 ( x ) = R 1 ( x ) R 2 ( x + y ) R 1 ( y ) , (7) R_2(y)R_1(x+y)R_2(x)=R_1(x)R_2(x+y)R_1(y), \tag*{(7)} R 2 ( y ) R 1 ( x + y ) R 2 ( x ) = R 1 ( x ) R 2 ( x + y ) R 1 ( y ) , (7) R ( s λ ) = S s S s t , s = 2 , 3 , (8) R(s\lambda)=S_sS_s^{\mathsf{t}}, \qquad s=2,3, \tag*{(8)} R ( s λ ) = S s S s t , s = 2 , 3 , (8) R 2 ( x ) R 1 ( x + s λ ) ( I ⊗ S s ) = ( S s ⊗ I ) R ^ s ( x ) , (9) R_2(x)R_1(x+s\lambda)(I\otimes S_s)=(S_s\otimes I)\widehat{R}_s(x), \tag*{(9)} R 2 ( x ) R 1 ( x + s λ ) ( I ⊗ S s ) = ( S s ⊗ I ) R s ( x ) , (9) R ( x ) R ( − x ) = I . (10) R(x)R(-x)=I. \tag*{(10)} R ( x ) R ( − x ) = I . (10) Here R ^ 2 ( x ) = R ( x + λ ) \widehat{R}_2(x)=R(x+\lambda) R 2 ( x ) = R ( x + λ ) R ^ 3 ( x ) \widehat{R}_3(x) R 3 ( x )
The complete coefficient proof is in Appendix A. For a cut between two columns of N N N L N 0 \mathcal{L}_N^0 L N 0 L N 1 \mathcal{L}_N^1 L N 1 Figure 4 illustrates.
Figure 4. A single arch may cross a cut several times. In this schematic example the left half has a source strand at slot 1 and pairs (2, 3), (5, 6); the right half has pairs (1, 2), (3, 5) and a source strand at slot 6. Slot 4 is vacant on both sides. Each source strand is exposed, meaning that no pair encloses its slot. Gluing the two states recovers one path from a a a b b b
Before fixing its two auxiliary ends, the complete column is the map
M X = R N ( v N ) ⋯ R 1 ( v 1 ) , v i = 3 λ − u i , X = ( u 1 , … , u N ) . (11) M_X=R_N(v_N)\cdots R_1(v_1),\qquad v_i=3\lambda-u_i,\qquad X=(u_1,\ldots,u_N). \tag*{(11)} M X = R N ( v N ) ⋯ R 1 ( v 1 ) , v i = 3 λ − u i , X = ( u 1 , … , u N ) . (11) Its ordered inputs are (aux, 1 , … , N 1,\ldots,N 1 , … , N ( 1 , … , N , a u x ) (1,\ldots,N,\mathrm{aux}) ( 1 , … , N , aux ) T N ( ϵ ) : L N 0 → L N ϵ T_N^{(\epsilon)}:\mathcal{L}_N^0\to\mathcal{L}_N^\epsilon T N ( ϵ ) : L N 0 → L N ϵ ϵ = 0 , 1 \epsilon=0,1 ϵ = 0 , 1 T N = T N ( 0 ) T_N=T_N^{(0)} T N = T N ( 0 ) b N : L N 0 → L N 1 b_N:\mathcal{L}_N^0\to\mathcal{L}_N^1 b N : L N 0 → L N 1 b N n o r t h b_N^{\mathrm{north}} b N north − - − X X X X ∗ = ( 3 λ − u 1 , … , 3 λ − u N ) X^\ast=(3\lambda-u_1,\ldots,3\lambda-u_N) X ∗ = ( 3 λ − u 1 , … , 3 λ − u N ) M X − = R N ( u N ) ⋯ R 1 ( u 1 ) M_{X}^{-}=R_N(u_N)\cdots R_1(u_1) M X − = R N ( u N ) ⋯ R 1 ( u 1 ) T N v T_Nv T N v v v v
Lemma 2.3 (Decay and rational continuation). If every u i u_i u i { 0 , λ , 2 λ , 3 λ } \{0,\lambda,2\lambda,3\lambda\} { 0 , λ , 2 λ , 3 λ } T N T_N T N T N ( 1 ) T_N^{(1)} T N ( 1 ) P N P_N P N e i u i e^{\mathrm{i}u_i} e i u i
Proof. First take a homogeneous triangular parallelogram of height H > 0 H>0 H > 0 M M M M → ∞ M\to\infty M → ∞ H H H H H H M < L M<L M < L L L L M M M q H ( M ) q_H(M) q H ( M ) C H C_H C H
ρ C H ∑ M < L q H ( M ) ≤ 1 / c . \rho^{C_H}\sum_{M<L}q_H(M)\leq1/c. ρ C H M < L ∑ q H ( M ) ≤ 1/ c . Thus q H ( M ) → 0 q_H(M)\to0 q H ( M ) → 0
For a mixed list of the four allowed parameters, collapse the flat rows. The row vectors belong to { 1 , e i π / 3 , e 2 i π / 3 , − 1 } \{1,e^{\mathrm{i}\pi/3},e^{2\mathrm{i}\pi/3},-1\} { 1 , e i π /3 , e 2 i π /3 , − 1 } O ( N ) O(N) O ( N ) H H H M − O ( N ) M-O(N) M − O ( N ) N N N N N N
Every nonzero diagram contributing to either restricted transfer power has a component connecting the left and right cuts. In the vacuum sector, otherwise the nonempty input pairing would close into a loop; in the source sector, the source must reach the output. Cut all components at the two ends of the homogeneous core. There are only boundedly many endpoint patterns, depending on N N N 1 / c 1/c 1/ c H > 0 H>0 H > 0 M − κ X M-\kappa_X M − κ X 0 ≤ κ X = O ( N ) 0\leq\kappa_X=O(N) 0 ≤ κ X = O ( N ) Q N Q_N Q N ∥ ⋅ ∥ max \lVert\cdot\rVert_{\max} ∥ ⋅ ∥ m a x
∥ Q N M ∥ max + ∥ ( T N ( 1 ) ) M ∥ max ≤ C N ∑ a ∈ E H q H , a ( M − κ X ) ⟶ 0. \lVert Q_N^M\rVert_{\max}+\lVert(T_N^{(1)})^M\rVert_{\max}\leq C_N\sum_{a\in\mathcal{E}_H}q_{H,a}(M-\kappa_X)\longrightarrow0. ∥ Q N M ∥ m a x + ∥( T N ( 1 ) ) M ∥ m a x ≤ C N a ∈ E H ∑ q H , a ( M − κ X ) ⟶ 0. Here E H \mathcal{E}_H E H q H , a q_{H,a} q H , a a a a C N C_N C N N N N H = 0 H=0 H = 0 M M M
At the specified parameters all tile weights are nonnegative. The matrices obtained by taking absolute tile weights vary continuously, so their restricted spectral radii remain less than one in a small complex neighborhood. This also proves the asserted absolute convergence. The reflected argument is identical.
Writing the vacuum matrix in the empty/nonempty decomposition gives
T N = ( 1 0 v Q ) . T_N=\begin{pmatrix}1&0\\ \mathbf{v}&Q\end{pmatrix}. T N = ( 1 v 0 Q ) . Its normalized stationary vector is ( 1 , ( I − Q ) − 1 v ) (1,(I-Q)^{-1}\mathbf{v}) ( 1 , ( I − Q ) − 1 v ) P N P_N P N
We now pass from local identities to column identities. This also records the source insertions needed for the arch moment. Let X ′ X' X ′ X X X i , i + 1 i,i+1 i , i + 1 δ = u i + 1 − u i \delta=u_{i+1}-u_i δ = u i + 1 − u i
M X ( I a u x ⊗ R i ( δ ) ) = ( R i ( δ ) ⊗ I a u x ) M X ′ . (12) M_X(I_{\mathrm{aux}}\otimes R_i(\delta))=(R_i(\delta)\otimes I_{\mathrm{aux}})M_{X'}. \tag*{(12)} M X ( I aux ⊗ R i ( δ )) = ( R i ( δ ) ⊗ I aux ) M X ′ . (12) The tensor positions reflect the different input and output orders in (11) . If u i + 1 = u i + s λ u_{i+1}=u_i+s\lambda u i + 1 = u i + s λ s = 2 , 3 s=2,3 s = 2 , 3 X r e d X_{\mathrm{red}} X red u i + λ u_i+\lambda u i + λ s = 2 s=2 s = 2 s = 3 s=3 s = 3 S s S_s S s L N − r s ϵ \mathcal{L}^{\epsilon}_{N-r_s} L N − r s ϵ L N ϵ \mathcal{L}^{\epsilon}_N L N ϵ r 2 = 1 r_2=1 r 2 = 1 r 3 = 2 r_3=2 r 3 = 2
M X ( I a u x ⊗ S s ) = ( S s ⊗ I a u x ) M X r e d . (13) M_X(I_{\mathrm{aux}}\otimes S_s)=(S_s\otimes I_{\mathrm{aux}})M_{X_{\mathrm{red}}}. \tag*{(13)} M X ( I aux ⊗ S s ) = ( S s ⊗ I aux ) M X red . (13) Indeed its argument is x = v i + 1 x=v_{i+1} x = v i + 1 s = 2 s=2 s = 2 x + λ = 3 λ − ( u i + λ ) x+\lambda=3\lambda-(u_i+\lambda) x + λ = 3 λ − ( u i + λ )
( S s t ⊗ I a u x ) M X − = M X r e d − ( I a u x ⊗ S s t ) . (14) (S_s^{\mathsf{t}}\otimes I_{\mathrm{aux}})M_X^{-}=M_{X_{\mathrm{red}}}^{-}(I_{\mathrm{aux}}\otimes S_s^{\mathsf{t}}). \tag*{(14)} ( S s t ⊗ I aux ) M X − = M X red − ( I aux ⊗ S s t ) . (14) All three equations leave the auxiliary endpoints untouched. We can therefore impose either vacant ends or either source-insertion condition. With X X X ϵ = 0 , 1 \epsilon=0,1 ϵ = 0 , 1
T X ( ϵ ) R i ( δ ) = R i ( δ ) T X ′ ( ϵ ) , b X R i ( δ ) = R i ( δ ) b X ′ , T_X^{(\epsilon)}R_i(\delta)=R_i(\delta)T_{X'}^{(\epsilon)},\qquad\mathbf{b}_XR_i(\delta)=R_i(\delta)\mathbf{b}_{X'}, T X ( ϵ ) R i ( δ ) = R i ( δ ) T X ′ ( ϵ ) , b X R i ( δ ) = R i ( δ ) b X ′ , T X ( ϵ ) S s = S s T X r e d ( ϵ ) , b X S s = S s b X r e d , T_X^{(\epsilon)}S_s=S_sT_{X_{\mathrm{red}}}^{(\epsilon)},\qquad\mathbf{b}_XS_s=S_s\mathbf{b}_{X_{\mathrm{red}}}, T X ( ϵ ) S s = S s T X red ( ϵ ) , b X S s = S s b X red , S s t T X ( ϵ ) , − = T X r e d ( ϵ ) , − S s t , S s t b X − = b X r e d − S s t . (15) S_s^{\mathsf{t}}T_X^{(\epsilon),-}=T_{X_{\mathrm{red}}}^{(\epsilon),-}S_s^{\mathsf{t}},\qquad S_s^{\mathsf{t}}\mathbf{b}_X^{-}=\mathbf{b}_{X_{\mathrm{red}}}^{-}S_s^{\mathsf{t}}. \tag*{(15)} S s t T X ( ϵ ) , − = T X red ( ϵ ) , − S s t , S s t b X − = b X red − S s t . (15) The same equations hold with b n o r t h \mathbf{b}^{\mathrm{north}} b north b \mathbf{b} b
The normalized stationary vector is now determined under exchange and reduction:
P N ( … , u i , u i + 1 , … ) = R i ( u i + 1 − u i ) P N ( … , u i + 1 , u i , … ) , (16) P_N(\ldots,u_i,u_{i+1},\ldots)=R_i(u_{i+1}-u_i)P_N(\ldots,u_{i+1},u_i,\ldots), \tag*{(16)} P N ( … , u i , u i + 1 , … ) = R i ( u i + 1 − u i ) P N ( … , u i + 1 , u i , … ) , (16) P N = S s P r e d when u i + 1 = u i + s λ . (17) P_N=S_sP_{\mathrm{red}}\qquad\text{when }u_{i+1}=u_i+s\lambda. \tag*{(17)} P N = S s P red when u i + 1 = u i + s λ . (17) The column equations prove stationarity of the right sides. Their empty coordinate is one: a cap producing an empty output from a nonempty pairing would close a zero-weight loop. Uniqueness in Lemma 2.3 proves exchange near the all-zero parameter list, and reduction near a mixed list with u i = 0 u_i=0 u i = 0
At u 1 = 3 λ u_1=3\lambda u 1 = 3 λ u 1 = 0 u_1=0 u 1 = 0 P N − 1 P_{N-1} P N − 1 P N − 1 P_{N-1} P N − 1 2 , … , N 2,\ldots,N 2 , … , N R ( 0 ) = I R(0)=I R ( 0 ) = I R ( 3 λ ) = S 3 S 3 t R(3\lambda) = S_3S_3^{\mathsf{t}} R ( 3 λ ) = S 3 S 3 t u k u_k u k π \pi π T N T_N T N D k D_k D k k k k D k D_k D k P N ( X + π e k ) = D k P N ( X ) P_N(X + \pi e_k) = D_kP_N(X) P N ( X + π e k ) = D k P N ( X ) e k e_k e k k k k
The polynomial vacuum
The stationary vacuum is initially only a rational vector. To extract a physical observable at coincident parameters, we need a polynomial numerator with a controlled degree. We first construct its scalar normalization and then prove the vector degree bound by interpolation and stationarity. Exchange equations, degree bounds and reductions at special rapidities have earlier uses in the dense O ( 1 ) O(1) O ( 1 ) [8 ] and in the open dilute O ( 1 ) O(1) O ( 1 ) [13 , 14 ] . The latter comparison concerns the fusion construction of the polynomial normalization. A related verification of a Pfaffian normalization by deletion identities and degree bounds appears for the crossing O ( 1 ) O(1) O ( 1 ) [7 ] , Section 2.6. We construct the present zero-loop-value scalar and vector with their own degree and stationarity proofs.
Write X = ( u 1 , … , u N ) X = (u_1,\ldots,u_N) X = ( u 1 , … , u N ) a i = a ( u i ) a_i = a(u_i) a i = a ( u i )
D ( a , b ) = a 2 + b 2 + 2 d a b − d 2 , D(a,b) = a^2 + b^2 + 2dab - d^2, D ( a , b ) = a 2 + b 2 + 2 d ab − d 2 , p N ( a 1 , … , a N ) = ∏ i < j D ( a i , a j ) a i − a j Pf [ ( a i − a j ) ( a i + a j + ρ ) D ( a i , a j ) ] . (18) p_N(a_1,\ldots,a_N) = \prod_{i<j}\frac{D(a_i,a_j)}{a_i-a_j}\operatorname{Pf}\left[\frac{(a_i-a_j)(a_i+a_j+\rho)}{D(a_i,a_j)}\right]. \tag*{(18)} p N ( a 1 , … , a N ) = i < j ∏ a i − a j D ( a i , a j ) Pf [ D ( a i , a j ) ( a i − a j ) ( a i + a j + ρ ) ] . (18) For odd N N N p 0 = 1 p_0 = 1 p 0 = 1
Lemma 3.1 (Scalar reduction identities). The function p N p_N p N N − 1 N-1 N − 1 p N − 1 p_{N-1} p N − 1
p N ( Y , c ) = p N − 1 ( Y ) ∏ z ∈ Y ( z + C ) . (19) p_N(Y,c) = p_{N-1}(Y)\prod_{z\in Y}(z+C). \tag*{(19)} p N ( Y , c ) = p N − 1 ( Y ) z ∈ Y ∏ ( z + C ) . (19) For f j = cos ( w + j λ ) f_j = \cos(w+j\lambda) f j = cos ( w + j λ )
p ( Y , f 0 , f 6 ) = p ( Y ) ( f 0 + f 6 + ρ ) ∏ z ∈ Y ( z − f − 6 ) ( z − f 12 ) , (20) p(Y,f_0,f_6) = p(Y)(f_0+f_6+\rho)\prod_{z\in Y}(z-f_{-6})(z-f_{12}), \tag*{(20)} p ( Y , f 0 , f 6 ) = p ( Y ) ( f 0 + f 6 + ρ ) z ∈ Y ∏ ( z − f − 6 ) ( z − f 12 ) , (20) p ( Y , f − 2 , f 2 ) = p ( Y , f 0 ) ( f − 2 + f 2 + ρ ) ∏ z ∈ Y ( z + f 0 ) . (21) p(Y,f_{-2},f_2) = p(Y,f_0)(f_{-2}+f_2+\rho)\prod_{z\in Y}(z+f_0). \tag*{(21)} p ( Y , f − 2 , f 2 ) = p ( Y , f 0 ) ( f − 2 + f 2 + ρ ) z ∈ Y ∏ ( z + f 0 ) . (21) Here the subscript of p p p
Proof. After multiplication by all D D D (18) . Counting degrees before and after division gives degree at most N − 1 N-1 N − 1 − 1 -1 − 1 D ( z , c ) = ( z − c ) ( z + C ) D(z,c)=(z-c)(z+C) D ( z , c ) = ( z − c ) ( z + C ) (19) .
At D ( f 0 , f 6 ) = 0 D(f_0,f_6)=0 D ( f 0 , f 6 ) = 0 D ( f 0 , z ) = ( z − f − 6 ) ( z − f 6 ) D(f_0,z)=(z-f_{-6})(z-f_6) D ( f 0 , z ) = ( z − f − 6 ) ( z − f 6 ) (20) .
Prove the merge identity by induction on r = ∣ Y ∣ r=|Y| r = ∣ Y ∣ r = 0 r=0 r = 0 p 2 ( a , b ) = a + b + ρ p_2(a,b)=a+b+\rho p 2 ( a , b ) = a + b + ρ r ≥ 1 r\geq1 r ≥ 1 z z z r + 1 r+1 r + 1 z z z z = − f 0 z=-f_0 z = − f 0 f − 2 f_{-2} f − 2 f 2 f_2 f 2 z = c z=c z = c
( C + f − 2 ) ( C + f 2 ) = ( c + f 0 ) ( C + f 0 ) . (C+f_{-2})(C+f_2)=(c+f_0)(C+f_0). ( C + f − 2 ) ( C + f 2 ) = ( c + f 0 ) ( C + f 0 ) . They also agree at each deletion value of z z z g 0 , g 6 g_0,g_6 g 0 , g 6 g j = cos ( v + j λ ) g_j=\cos(v+j\lambda) g j = cos ( v + j λ )
∏ ℓ = ± 2 ( f ℓ − g − 6 ) ( f ℓ − g 12 ) = ( f 0 + g 0 ) ( f 0 + g 6 ) ( f 0 − g − 6 ) ( f 0 − g 12 ) . \prod_{\ell=\pm2}(f_\ell-g_{-6})(f_\ell-g_{12})=(f_0+g_0)(f_0+g_6)(f_0-g_{-6})(f_0-g_{12}). ℓ = ± 2 ∏ ( f ℓ − g − 6 ) ( f ℓ − g 12 ) = ( f 0 + g 0 ) ( f 0 + g 6 ) ( f 0 − g − 6 ) ( f 0 − g 12 ) . This is the product-to-sum identity obtained from shifts by ± 2 λ \pm2\lambda ± 2 λ 2 ( r − 1 ) 2(r-1) 2 ( r − 1 ) − f 0 -f_0 − f 0 c c c r + 1 r+1 r + 1 r r r
Lemma 3.2 (Vacuum degree bound). The vector W N = p N P N W_N=p_NP_N W N = p N P N e i u 1 , … , e i u N e^{iu_1},\ldots,e^{iu_N} e i u 1 , … , e i u N − ( 2 N − 2 ) -(2N-2) − ( 2 N − 2 ) 2 N − 2 2N-2 2 N − 2 y = u N y=u_N y = u N b = a ( y ) b=a(y) b = a ( y ) b b b N − 1 N-1 N − 1 W N − 1 W_{N-1} W N − 1 sin y \sin y sin y b b b N − 2 N-2 N − 2 W 0 = 1 W_0=1 W 0 = 1
Proof. The assertion for N = 1 N=1 N = 1 W W W N N N W j = p j P j W_j=p_jP_j W j = p j P j
Deleting a vacant last slot identifies that sector with L N − 1 0 \mathcal{L}_{N-1}^{0} L N − 1 0 L N − 1 1 \mathcal{L}_{N-1}^{1} L N − 1 1 v v v o o o
Construction and regularity. For each earlier row i < N i<N i < N
y i = u i + 3 λ , β i = a ( y i ) , γ i = a ( u i − 3 λ ) , σ i = a ( u i + 6 λ ) . y_i=u_i+3\lambda,\qquad\beta_i=a(y_i),\qquad\gamma_i=a(u_i-3\lambda),\qquad\sigma_i=a(u_i+6\lambda). y i = u i + 3 λ , β i = a ( y i ) , γ i = a ( u i − 3 λ ) , σ i = a ( u i + 6 λ ) . Let R i \mathcal{R}_i R i y i y_i y i i i i y i − u j y_i-u_j y i − u j i < j < N i<j<N i < j < N [16 ] , [17 ] , and (20), the value required at y = y i y=y_i y = y i
Y i = ( a i + β i + ρ ) ∏ j < N j ≠ i ( a j − γ i ) ( a j − σ i ) R i S 3 W N − 2 ( X ∖ { u i , u N } ) . (22) Y_i=(a_i+\beta_i+\rho)\prod_{\substack{j<N\\j\ne i}}(a_j-\gamma_i)(a_j-\sigma_i)\mathcal{R}_iS_3W_{N-2}(X\setminus\{u_i,u_N\}). \tag*{(22)} Y i = ( a i + β i + ρ ) j < N j = i ∏ ( a j − γ i ) ( a j − σ i ) R i S 3 W N − 2 ( X ∖ { u i , u N }) . (22) These data agree with the genuine rational vector p N P N p_NP_N p N P N u i = 0 u_i=0 u i = 0
For distinct nodes put
ℓ i ( b ) = ∏ j < N j ≠ i b − β j β i − β j . \ell_i(b)=\prod_{\substack{j<N\\j\ne i}}\frac{b-\beta_j}{\beta_i-\beta_j}. ℓ i ( b ) = j < N j = i ∏ β i − β j b − β j . Define the candidate by the two interpolation formulas
W v ( b ) = W N − 1 ∏ i < N ( b − β i ) + ∑ i < N ( Y i ) v ℓ i ( b ) , W_v(b) = W_{N-1} \prod_{i<N}(b-\beta_i) + \sum_{i<N}(Y_i)_v\ell_i(b), W v ( b ) = W N − 1 i < N ∏ ( b − β i ) + i < N ∑ ( Y i ) v ℓ i ( b ) , W o ( y ) = sin y ∑ i < N ( Y i ) o sin y i ℓ i ( b ) . (23) W_o(y) = \sin y \sum_{i<N}\frac{(Y_i)_o}{\sin y_i}\ell_i(b). \tag*{(23)} W o ( y ) = sin y i < N ∑ sin y i ( Y i ) o ℓ i ( b ) . (23) The vacant part has degree at most N − 1 N-1 N − 1 W N − 1 W_{N-1} W N − 1 sin y \sin y sin y N − 2 N-2 N − 2 W W W p N p_N p N
We check carefully that this construction has no poles in earlier spectral variables. Each moving box has two sine factors in its denominator; they divide the factors ( a j − γ i ) ( a j − σ i ) (a_j-\gamma_i)(a_j-\sigma_i) ( a j − γ i ) ( a j − σ i ) (22) . The possible divisor sin y i \sin y_i sin y i x = u i x=u_i x = u i v = u j v=u_j v = u j
( a ( v ) − a ( x − 3 λ ) ) ( a ( v ) − a ( x + 6 λ ) ) L ( x + 3 λ − v ) = − 4 sin ( x + v − 6 λ ) sin ( x + v + 3 λ ) , \frac{(a(v)-a(x-3\lambda))(a(v)-a(x+6\lambda))}{L(x+3\lambda-v)}
=-4\sin(x+v-6\lambda)\sin(x+v+3\lambda), L ( x + 3 λ − v ) ( a ( v ) − a ( x − 3 λ )) ( a ( v ) − a ( x + 6 λ )) = − 4 sin ( x + v − 6 λ ) sin ( x + v + 3 λ ) , a ( x ) + a ( x + 3 λ ) + ρ = 4 c sin ( x + 3 λ ) sin ( 3 λ − x ) . a(x)+a(x+3\lambda)+\rho=4c\sin(x+3\lambda)\sin(3\lambda-x). a ( x ) + a ( x + 3 λ ) + ρ = 4 c sin ( x + 3 λ ) sin ( 3 λ − x ) . The remaining possible divisors come from colliding interpolation nodes. Since
β i − β j = − 2 sin ( u i + u j + 3 λ ) sin ( u i − u j ) , \beta_i-\beta_j=-2\sin(u_i+u_j+3\lambda)\sin(u_i-u_j), β i − β j = − 2 sin ( u i + u j + 3 λ ) sin ( u i − u j ) , the two components of β i = β j \beta_i=\beta_j β i = β j
u i = u j ( m o d π ) , u i + u j = − 3 λ ( m o d π ) . u_i=u_j\pmod{\pi},\qquad u_i+u_j=-3\lambda\pmod{\pi}. u i = u j ( mod π ) , u i + u j = − 3 λ ( mod π ) . On the first component the data, including the divided occupied data, agree by their identities with the true vector and its occupancy signs. This can be verified near the regular mixture u i = u j = 0 u_i=u_j=0 u i = u j = 0 y i = y j = 3 λ y_i=y_j=3\lambda y i = y j = 3 λ σ i , σ j \sigma_i,\sigma_j σ i , σ j W W W
For its degree, send an earlier u k u_k u k (6) . Data with i ≠ k i\ne k i = k a k N − 1 a_k^{N-1} a k N − 1 W N − 2 W_{N-2} W N − 2 ℓ i ( b ) \ell_i(b) ℓ i ( b ) i = k i=k i = k a k 1 + 2 ( N − 2 ) a_k^{1+2(N-2)} a k 1 + 2 ( N − 2 ) O ( a k − ( N − 2 ) ) O(a_k^{-(N-2)}) O ( a k − ( N − 2 ) ) W N − 1 ∏ i < N ( b − β i ) W_{N-1}\prod_{i<N}(b-\beta_i) W N − 1 ∏ i < N ( b − β i ) − ( 2 N − 2 ) -(2N-2) − ( 2 N − 2 ) 2 N − 2 2N-2 2 N − 2 u k ↦ u k + π u_k\mapsto u_k+\pi u k ↦ u k + π
Relations in the earlier rows. Earlier-row braid covariance follows from the same last-variable interpolation: the exchange box is independent of the last variable, the leading coefficients agree by induction, and every node datum agrees by the rational vacuum exchange identity. Uniqueness of the interpolant then gives covariance.
We have established the degree bounds and prescribed node values of W W W
If adjacent earlier rows satisfy u r + 1 = u r + s λ u_{r+1}=u_r+s\lambda u r + 1 = u r + s λ s = 2 , 3 s=2,3 s = 2 , 3 (20) or (21) . The quotient has degree one in b b b i ≠ r + 1 i\ne r+1 i = r + 1 u r = 0 u_r=0 u r = 0 u i = 0 u_i=0 u i = 0 y i = 3 λ y_i=3\lambda y i = 3 λ i = r + 1 i=r+1 i = r + 1 b − σ r b-\sigma_r b − σ r b + a ( u r + λ ) b+a(u_r+\lambda) b + a ( u r + λ ) a r − γ i a_r-\gamma_i a r − γ i a r − σ i a_r-\sigma_i a r − σ i r r r
At u 1 = 0 u_1=0 u 1 = 0 3 λ 3\lambda 3 λ (17) , replacing their smaller vacuum vector by W N − 1 W_{N-1} W N − 1 ∏ j = 2 N ( a j + C ) \prod_{j=2}^{N}(a_j+C) ∏ j = 2 N ( a j + C ) 1 1 1 A ( y ) A(y) A ( y ) U ( y ) U(y) U ( y ) B ( y ) B(y) B ( y ) b + C b+C b + C 1 1 1 u i = 0 u_i=0 u i = 0 i > 1 i>1 i > 1 u 1 = 3 λ u_1=3\lambda u 1 = 3 λ i = 1 i=1 i = 1 y i = 6 λ y_i=6\lambda y i = 6 λ a ( y i ) + C a(y_i)+C a ( y i ) + C a 1 + β 1 + ρ a_1+\beta_1+\rho a 1 + β 1 + ρ
On each specialization the nodes are generically distinct and their dividing sines nonzero; all other cases follow by continuation.
Stationarity. Consider the polynomial stationarity residual
V = ( ∏ j = 1 N L ( u j ) ) ( T N − I ) W . V=\left(\prod_{j=1}^{N}L(u_j)\right)(T_N-I)W. V = ( j = 1 ∏ N L ( u j ) ) ( T N − I ) W . At each last-variable node it vanishes, because the data came from the genuine stationary vector. To determine its degree, put n = N − 1 n=N-1 n = N − 1 q = W o / sin y q=W_o/\sin y q = W o / sin y Λ n = ∏ j < N L ( u j ) \Lambda_n=\prod_{j<N}L(u_j) Λ n = ∏ j < N L ( u j ) b n n o r t h b_n^{\mathrm{north}} b n north C n n o r t h : L n 1 → L n 0 C_n^{\mathrm{north}}:\mathcal{L}_n^1\to\mathcal{L}_n^0 C n north : L n 1 → L n 0 n n n y y y
The last box contributes 1 1 1 A ( y ) A(y) A ( y ) U ( y ) U(y) U ( y ) B ( y ) B(y) B ( y ) L ( y ) = ( b + C ) / 2 L(y)=(b+C)/2 L ( y ) = ( b + C ) /2 sin y sin ( 3 λ − y ) = ( b − c ) / 2 \sin y\sin(3\lambda-y)=(b-c)/2 sin y sin ( 3 λ − y ) = ( b − c ) /2
V v = Λ n 2 ( ( b + C ) ( T n − I ) W v + d ( b − c ) C n n o r t h q ) , (24) V_v=\frac{\Lambda_n}{2}\left((b+C)(T_n-I)W_v+d(b-c)C_n^{\mathrm{north}}q\right), \tag*{(24)} V v = 2 Λ n ( ( b + C ) ( T n − I ) W v + d ( b − c ) C n north q ) , (24) V o sin y = Λ n 2 ( ( b − c ) T n ( 1 ) q − ( b + C ) q + 2 d b n n o r t h W v ) . (25) \frac{V_o}{\sin y}=\frac{\Lambda_n}{2}\left((b-c)T_n^{(1)}q-(b+C)q+2d b_n^{\mathrm{north}}W_v\right). \tag*{(25)} sin y V o = 2 Λ n ( ( b − c ) T n ( 1 ) q − ( b + C ) q + 2 d b n north W v ) . (25) The second block has degree at most N − 1 N-1 N − 1 N N N
[ b N ] V v = Λ n 2 ( T n − I ) W n = 0 [b^N]V_v=\frac{\Lambda_n}{2}(T_n-I)W_n=0 [ b N ] V v = 2 Λ n ( T n − I ) W n = 0 by the inductive stationarity. Thus V v V_v V v N − 1 N-1 N − 1 V V V ∏ i < N ( b − β i ) \prod_{i<N}(b-\beta_i) ∏ i < N ( b − β i ) u k u_k u k 2 N 2N 2 N V V V 0 0 0 3 λ 3\lambda 3 λ u j = u i + s λ u_j = u_i + s\lambda u j = u i + s λ i < j < N i < j < N i < j < N s = 2 , 3 s = 2, 3 s = 2 , 3 (10) .
For fixed k < N k < N k < N
( b − β k ) sin u k sin ( u k − 3 λ ) ∏ j < k s ∈ { 2 , 3 } sin ( u k − u j − s λ ) ∏ k < j < N s ∈ { 2 , 3 } sin ( u j − u k − s λ ) . (b-\beta_k)\sin u_k\sin(u_k-3\lambda)
\prod_{\substack{j<k\\s\in\{2,3\}}}\sin(u_k-u_j-s\lambda)
\prod_{\substack{k<j<N\\s\in\{2,3\}}}\sin(u_j-u_k-s\lambda). ( b − β k ) sin u k sin ( u k − 3 λ ) j < k s ∈ { 2 , 3 } ∏ sin ( u k − u j − s λ ) k < j < N s ∈ { 2 , 3 } ∏ sin ( u j − u k − s λ ) . For generic other variables this divisor has Laurent extremes ± 2 N \pm2N ± 2 N b − β k b-\beta_k b − β k 2 ( N − 2 ) 2(N-2) 2 ( N − 2 ) 2 N − 4 2N-4 2 N − 4 u k ↦ u k + π u_k\mapsto u_k+\pi u k ↦ u k + π e i u k e^{\mathrm{i}u_k} e i u k k k k V V V V = 0 V=0 V = 0 W W W p N p_N p N W = p N P N W=p_NP_N W = p N P N □ \square □
We now reconnect the polynomial vacuum with a positive path observable. Cutting a rightward arch at every column between its endpoints counts it exactly as many times as its horizontal displacement. We use this pairing of source states to identify the arch moment with a polynomial coefficient. A stepped-boundary flux calculation then compares its successive increments with the bridge mass.
For a repeated rhombic column X = ( u 1 , … , u N ) X=(u_1,\ldots,u_N) X = ( u 1 , … , u N ) K X ( k ) K_X(k) K X ( k ) k k k m ( X ) = ∑ k ≥ 1 k K X ( k ) m(X)=\sum_{k\geq1}kK_X(k) m ( X ) = ∑ k ≥ 1 k K X ( k ) Figure 4 . Using the south-source column b N b_N b N
H N = ( I − T N ( 1 ) ) − 1 b N P N . (26) H_N=(I-T_N^{(1)})^{-1}b_NP_N. \tag*{(26)} H N = ( I − T N ( 1 ) ) − 1 b N P N . (26) and define H N − H_N^{-} H N − P N P_N P N
m ( X ) = Glue ( H N − , H N ) . (27) m(X)=\operatorname{Glue}(H_N^{-},H_N). \tag*{(27)} m ( X ) = Glue ( H N − , H N ) . (27) Indeed expanding the two resolvents places one source on either side of the cut. For endpoint separation k k k k k k m m m
Related finite-strip current calculations use bilinear link-state pairings, fusion reductions and interpolation [6 , 11 ] . Those O ( 1 ) O(1) O ( 1 )
Proposition 4.1 (Arch formula). Let η = 2 d − 1 \eta=2d-1 η = 2 d − 1
g X ( z ) = p N + 1 ( a 1 , … , a N , z ) p N ( a 1 , … , a N ) . (28) g_X(z)=\frac{p_{N+1}(a_1,\ldots,a_N,z)}{p_N(a_1,\ldots,a_N)}. \tag*{(28)} g X ( z ) = p N ( a 1 , … , a N ) p N + 1 ( a 1 , … , a N , z ) . (28) Then g X g_X g X N N N X X X
m ( X ) = ∑ i = 1 N a i − η [ z N − 1 ] g X ( z ) . (29) m(X)=\sum_{i=1}^{N}a_i-\eta[z^{N-1}]g_X(z). \tag*{(29)} m ( X ) = i = 1 ∑ N a i − η [ z N − 1 ] g X ( z ) . (29) The right side is understood by continuation at removable singularities.
Proof. We identify m m m 2 N 2N 2 N p N 2 m p_N^2m p N 2 m a i a_i a i (29) .
Symmetries and reductions. Moving an exchange box through both halves of (27) , using (15) , proves symmetry in the rows. At an ascending merge or deletion specialization, the same identities for the vacuum and south-source columns give
P N = S s P r e d , H N = S s H r e d . P_N=S_sP_{\mathrm{red}},\qquad H_N=S_sH_{\mathrm{red}}. P N = S s P red , H N = S s H red . The reflected column identity (14) gives
S s t P N − = P r e d − , S s t H N − = H r e d − . S_s^{\mathsf t}P_N^{-}=P_{\mathrm{red}}^{-},\qquad S_s^{\mathsf t}H_N^{-}=H_{\mathrm{red}}^{-}. S s t P N − = P red − , S s t H N − = H red − . For the first reflected equality, the projected vector is stationary with empty coordinate one. For the second, the projected source equation is
( I − T r e d ( 1 ) , − ) S s t H N − = b r e d − S s t P N − = b r e d − P r e d − . (30) (I-T_{\mathrm{red}}^{(1),-})S_s^{\mathsf t}H_N^{-}=b_{\mathrm{red}}^{-}S_s^{\mathsf t}P_N^{-}=b_{\mathrm{red}}^{-}P_{\mathrm{red}}^{-}. \tag*{(30)} ( I − T red ( 1 ) , − ) S s t H N − = b red − S s t P N − = b red − P red − . (30) Uniqueness near a mixed specialization and rational continuation therefore identify the projected vector with H r e d − H_{\mathrm{red}}^{-} H red − m m m S s t S s = I S_s^{\mathsf t}S_s=I S s t S s = I π \pi π
There is also a separate reflection symmetry u j ↔ 3 λ − u j u_j\leftrightarrow3\lambda-u_j u j ↔ 3 λ − u j ℓ ≥ 1 \ell\geq1 ℓ ≥ 1 A U B ℓ − 1 AU B^{\ell-1} A U B ℓ − 1 a j = c a_j=c a j = c A U = 0 AU=0 A U = 0
A polynomial bound from the source sector. To obtain the degree bound, first use a north source. Add a last row y y y
( P N + 1 ) o = B ( y ) T N ( 1 ) ( P N + 1 ) o + U ( y ) b N n o r t h ( P N + 1 ) v . (31) (P_{N+1})_o=B(y)T_N^{(1)}(P_{N+1})_o+U(y)b_N^{\mathrm{north}}(P_{N+1})_v. \tag*{(31)} ( P N + 1 ) o = B ( y ) T N ( 1 ) ( P N + 1 ) o + U ( y ) b N north ( P N + 1 ) v . (31) As b = a ( y ) → ∞ b=a(y)\to\infty b = a ( y ) → ∞ ( P N + 1 ) v → P N (P_{N+1})_v\to P_N ( P N + 1 ) v → P N B ( y ) → 1 B(y)\to1 B ( y ) → 1 ( P N + 1 ) o / U ( y ) (P_{N+1})_o/U(y) ( P N + 1 ) o / U ( y ) U ( y ) = 2 d sin y / ( b + C ) U(y)=2d\sin y/(b+C) U ( y ) = 2 d sin y / ( b + C ) p N p_N p N
1 2 d [ b N − 1 ] ( W N + 1 ) o sin y . \frac{1}{2d^{[b^{N-1}]}}\frac{(W_{N+1})_o}{\sin y}. 2 d [ b N − 1 ] 1 sin y ( W N + 1 ) o . The lemma bounds its Laurent exponents in each earlier variable by 2 N 2N 2 N u i u_i u i 3 λ − u i 3\lambda-u_i 3 λ − u i p N p_N p N
Hence both p N H N p_N H_N p N H N p N H N ‾ p_N H_{\overline{N}} p N H N (27) , p N 2 m p_N^2m p N 2 m 4 N 4N 4 N π \pi π a i = cos ( 2 u i − 3 λ ) a_i=\cos(2u_i-3\lambda) a i = cos ( 2 u i − 3 λ ) 2 N 2N 2 N π \pi π z = e 2 i u i z=e^{2iu_i} z = e 2 i u i z ↦ e 6 i λ / z z\mapsto e^{6i\lambda}/z z ↦ e 6 iλ / z a i a_i a i
Interpolation. Let the right side of (29) be m ~ \widetilde{m} m
g m e r g e ( z ) = ( z + f 0 ) g r e d ( z ) , g d e l e t e ( z ) = ( z − f − 6 ) ( z − f 12 ) g r e d ( z ) , g f l a t ( z ) = ( z + C ) g r e d ( z ) . g_{\mathrm{merge}}(z)=(z+f_0)g_{\mathrm{red}}(z),\qquad g_{\mathrm{delete}}(z)=(z-f_{-6})(z-f_{12})g_{\mathrm{red}}(z),\qquad g_{\mathrm{flat}}(z)=(z+C)g_{\mathrm{red}}(z). g merge ( z ) = ( z + f 0 ) g red ( z ) , g delete ( z ) = ( z − f − 6 ) ( z − f 12 ) g red ( z ) , g flat ( z ) = ( z + C ) g red ( z ) . The elementary identities
f − 2 + f 2 − f 0 = η f 0 , f 0 + f 6 = − η ( f − 6 + f 12 ) , c = η C f_{-2}+f_2-f_0=\eta f_0,\qquad f_0+f_6=-\eta(f_{-6}+f_{12}),\qquad c=\eta C f − 2 + f 2 − f 0 = η f 0 , f 0 + f 6 = − η ( f − 6 + f 12 ) , c = η C show that m ~ \widetilde{m} m m m m p N 2 m p_N^2m p N 2 m p N 2 m ~ p_N^2\widetilde{m} p N 2 m 2 N 2N 2 N a i a_i a i N N N a j = cos θ j a_j=\cos\theta_j a j = cos θ j cos ( θ j ± 4 λ ) \cos(\theta_j\pm4\lambda) cos ( θ j ± 4 λ ) cos ( θ j ± 6 λ ) \cos(\theta_j\pm6\lambda) cos ( θ j ± 6 λ ) c c c 4 N − 3 4N-3 4 N − 3 2 N 2N 2 N N ≥ 2 N\ge2 N ≥ 2 m ( ∅ ) = 0 m(\varnothing)=0 m ( ∅ ) = 0
m ( u ) = A ( u ) U ( u ) ( 1 − B ( u ) ) 2 = a ( u ) − c 1 + d = a ( u ) − η ( a ( u ) + ρ ) . m(u)=\frac{A(u)U(u)}{(1-B(u))^2}=\frac{a(u)-c}{1+d}=a(u)-\eta(a(u)+\rho). m ( u ) = ( 1 − B ( u ) ) 2 A ( u ) U ( u ) = 1 + d a ( u ) − c = a ( u ) − η ( a ( u ) + ρ ) . In the size-zero quotient convention use g 0 = 1 g_0=1 g 0 = 1 [ z − 1 ] g 0 = 0 [z^{-1}]g_0=0 [ z − 1 ] g 0 = 0
At X = ( λ , … , λ ) X=(\lambda,\ldots,\lambda) X = ( λ , … , λ ) ρ \rho ρ a i = C a_i=C a i = C m ( X ) = m N m(X)=m_N m ( X ) = m N
m N = N C − η lim a 1 , … , a N → C [ z N − 1 ] g X ( z ) . (32) m_N=NC-\eta\lim_{a_1,\ldots,a_N\to C}[z^{N-1}]g_X(z). \tag*{(32)} m N = N C − η a 1 , … , a N → C lim [ z N − 1 ] g X ( z ) . (32) This coefficient has a removable limit because m m m g X g_X g X
Lemma 4.2 (Moment increments and bridge mass). For homogeneous honeycomb strips, c A N + B N = 1 cA_N+B_N=1 c A N + B N = 1 m N + 1 − m N ≍ B N m_{N+1}-m_N\asymp B_N m N + 1 − m N ≍ B N N ≥ 1 N\ge1 N ≥ 1
Proof. Apply Lemma 2.1 to parallelogram truncations of the strip. The left and right boundary terms tend to zero by Lemma 2.3: after the source column, propagate in the one-source sector to the distant cut, where the exit is the only occupied boundary port. The remaining vacuum states are bounded. The other sums increase to their infinite-strip limits. Bottom exits have winding ± π \pm\pi ± π c A N + B N = 1 cA_N+B_N=1 c A N + B N = 1 B N B_N B N
Add one triangular row above the right half of S N S_N S N 1 , 2 , … 1,2,\ldots 1 , 2 , … p p p K ‾ ( i , j ) \overline{K}(i,j) K ( i , j ) Figure 5 .
Figure 5. The stepped strip used to compare m N + 1 m_{N+1} m N + 1 m N m_N m N
From a bottom source i i i
sin ( 3 λ ) ( ∑ j < i K ‾ ( i , j ) − ∑ j > i K ‾ ( i , j ) ) + sin λ Z ( i , p ) = 0. (33) \sin(3\lambda)\left(\sum_{j<i}\overline{K}(i,j)-\sum_{j>i}\overline{K}(i,j)\right)+\sin\lambda\,Z(i,p)=0. \tag*{(33)} sin ( 3 λ ) ( j < i ∑ K ( i , j ) − j > i ∑ K ( i , j ) ) + sin λ Z ( i , p ) = 0. (33) Indeed horizontal top exits have winding zero, and closing along the leftward boundary gives winding π / 3 \pi/3 π /3 p p p N + 1 N+1 N + 1
Sum (33) over − L ≤ i ≤ L -L \le i \le L − L ≤ i ≤ L N N N N + 1 N+1 N + 1 N + 1 N+1 N + 1
m N + 1 − m N = sin λ sin ( 3 λ ) ∑ i Z ( p , i ) . (34) m_{N+1}-m_N=\frac{\sin\lambda}{\sin(3\lambda)}\sum_i Z(p,i). \tag*{(34)} m N + 1 − m N = sin ( 3 λ ) sin λ i ∑ Z ( p , i ) . (34) Turning down immediately from p p p ∑ i Z ( p , i ) ≥ ρ B N \sum_i Z(p,i)\ge\rho B_N ∑ i Z ( p , i ) ≥ ρ B N
For the upper bound, order the path’s visits to the interface between the base strip and the new row from left to right. In the new row, whose top is vacant, disjoint intervals pair p p p
At u = λ u=\lambda u = λ A / ( 1 − B ) A/(1-B) A / ( 1 − B ) 1 / ( 2 c ) 1/(2c) 1/ ( 2 c ) A U / ( 1 − B ) = 2 c B AU/(1-B)=2cB A U / ( 1 − B ) = 2 c B B N B_N B N
∑ i Z ( p , i ) ≤ A 1 − B ∑ r ≥ 0 ( 1 2 c 2 c B ) r B N = A ( 1 − B ) 2 B N . \sum_i Z(p,i)\le\frac{A}{1-B}\sum_{r\ge0}\left(\frac{1}{2c}2cB\right)^r B_N=\frac{A}{(1-B)^2}B_N. i ∑ Z ( p , i ) ≤ 1 − B A r ≥ 0 ∑ ( 2 c 1 2 c B ) r B N = ( 1 − B ) 2 A B N . Here B = B ( λ ) = ρ 2 < 1 B=B(\lambda)=\rho^2<1 B = B ( λ ) = ρ 2 < 1 N N N
A positive integral for the homogeneous moment
The arch formula (29) is exact, but its Pfaffian quotient does not directly show the size of m N m_N m N X X X b = cos w b=\cos w b = cos w
Q X ( b ) = ( b − C ) g X ( b ) ∏ i = 1 N ( b − a i ) ( cos ( w − 2 λ ) − a i ) ( cos ( w + 2 λ ) − a i ) , (35) Q_X(b)=(b-C)g_X(b)\prod_{i=1}^{N}(b-a_i)(\cos(w-2\lambda)-a_i)(\cos(w+2\lambda)-a_i), \tag*{(35)} Q X ( b ) = ( b − C ) g X ( b ) i = 1 ∏ N ( b − a i ) ( cos ( w − 2 λ ) − a i ) ( cos ( w + 2 λ ) − a i ) , (35) Φ X ( w ) = sin w Q X ( cos w ) . \Phi_X(w)=\sin w\,Q_X(\cos w). Φ X ( w ) = sin w Q X ( cos w ) . The paired shifted factors have product b 2 − 2 d a i b + a i 2 − d 2 b^2-2da_i b+a_i^2-d^2 b 2 − 2 d a i b + a i 2 − d 2 Q X Q_X Q X 4 N + 1 4N+1 4 N + 1
Lemma 5.1 (Missing harmonics). Every sine harmonic of Φ X \Phi_X Φ X h = b 2 ( 1 − b 2 ) = ( 1 − cos 4 w ) / 8 h=b^2(1-b^2)=(1-\cos4w)/8 h = b 2 ( 1 − b 2 ) = ( 1 − cos 4 w ) /8
Q X ( b ) = P ( h ) + b J ( h ) + b 2 E ( h ) , deg P , deg J ≤ N , deg E ≤ N − 1. (36) Q_X(b)=P(h)+bJ(h)+b^2\mathcal{E}(h),\qquad\deg P,\deg J\le N,\qquad\deg\mathcal{E}\le N-1. \tag*{(36)} Q X ( b ) = P ( h ) + b J ( h ) + b 2 E ( h ) , deg P , deg J ≤ N , deg E ≤ N − 1. (36) Proof. Fix i i i f j = cos ( v + j λ ) f_j=\cos(v+j\lambda) f j = cos ( v + j λ ) f 0 = a i f_0=a_i f 0 = a i (35) gives Φ X ( v ± 2 λ ) = 0 \Phi_X(v\pm2\lambda)=0 Φ X ( v ± 2 λ ) = 0 Φ X ( v + 6 λ ) + Φ X ( v − 6 λ ) = 0 \Phi_X(v+6\lambda)+\Phi_X(v-6\lambda)=0 Φ X ( v + 6 λ ) + Φ X ( v − 6 λ ) = 0 a j a_j a j f 6 , f − 6 , f 4 , f − 4 , − f 0 f_6,f_{-6},f_4,f_{-4},-f_0 f 6 , f − 6 , f 4 , f − 4 , − f 0
R ± = sin ( v ± 6 λ ) ( f ± 6 − C ) ( f 0 + f ± 6 + ρ ) ( f ± 6 − f 0 ) ( f ± 4 − f 0 ) . R_\pm=\sin(v\pm6\lambda)(f_{\pm6}-C)(f_0+f_{\pm6}+\rho)(f_{\pm6}-f_0)(f_{\pm4}-f_0). R ± = sin ( v ± 6 λ ) ( f ± 6 − C ) ( f 0 + f ± 6 + ρ ) ( f ± 6 − f 0 ) ( f ± 4 − f 0 ) . Product-to-sum and ρ = 2 c d \rho=2cd ρ = 2 c d
R ± = cos ( 2 v ) ( f ± 6 − C ) sin ( v ± 3 λ ) ( cos ( v ± 3 λ ) + d ) . R_\pm=\cos(2v)(f_{\pm6}-C)\sin(v\pm3\lambda)(\cos(v\pm3\lambda)+d). R ± = cos ( 2 v ) ( f ± 6 − C ) sin ( v ± 3 λ ) ( cos ( v ± 3 λ ) + d ) . The last two expressions after removing cos ( 2 v ) \cos(2v) cos ( 2 v ) v = ± 3 λ , ± 5 λ , ± 7 λ v=\pm3\lambda,\pm5\lambda,\pm7\lambda v = ± 3 λ , ± 5 λ , ± 7 λ 2 π 2\pi 2 π
Consider the sum of the four shifts of Φ X \Phi_X Φ X ± 2 λ , ± 6 λ \pm2\lambda,\pm6\lambda ± 2 λ , ± 6 λ k k k 4 cos ( 4 k λ ) cos ( 2 k λ ) 4\cos(4k\lambda)\cos(2k\lambda) 4 cos ( 4 k λ ) cos ( 2 k λ ) k k k 4 N + 2 4N+2 4 N + 2 sin 4 w \sin4w sin 4 w cos 4 w \cos4w cos 4 w N − 1 N-1 N − 1 N N N a i a_i a i
Every polynomial in b b b P 0 ( h ) + b P 1 ( h ) + b 2 P 2 ( h ) + b 3 P 3 ( h ) P_0(h)+bP_1(h)+b^2P_2(h)+b^3P_3(h) P 0 ( h ) + b P 1 ( h ) + b 2 P 2 ( h ) + b 3 P 3 ( h ) b 4 = b 2 − h b^4=b^2-h b 4 = b 2 − h sin w \sin w sin w
sin w b 3 = 1 4 sin 2 w + 1 8 sin 4 w . \sin w\,b^3=\frac{1}{4}\sin2w+\frac{1}{8}\sin4w. sin w b 3 = 4 1 sin 2 w + 8 1 sin 4 w . Its projection onto multiples of four is 1 8 sin 4 w P 3 ( ( 1 − cos 4 w ) / 8 ) \frac{1}{8}\sin4w\,P_3((1-\cos4w)/8) 8 1 sin 4 w P 3 (( 1 − cos 4 w ) /8 ) P 3 = 0 P_3=0 P 3 = 0 deg Q X = 4 N + 1 \deg Q_X=4N+1 deg Q X = 4 N + 1
At the homogeneous parameters a i = C a_i=C a i = C (35) become ( b − C ) ( b − c ) (b-C)(b-c) ( b − C ) ( b − c )
Q ( b ) = ( b − C ) 2 N + 1 ( b − c ) N g ( b ) = P ( h ) + b J ( h ) + b 2 E ( h ) , deg g = N . (37) Q(b)=(b-C)^{2N+1}(b-c)^N g(b)=P(h)+bJ(h)+b^2\mathcal{E}(h),\qquad\deg g=N. \tag*{(37)} Q ( b ) = ( b − C ) 2 N + 1 ( b − c ) N g ( b ) = P ( h ) + b J ( h ) + b 2 E ( h ) , deg g = N . (37) Lemma 5.2 (Nonsingularity and integral representation). The monic problem (37) has a unique solution, which is the limit of g X g_X g X as all a i → C a_i \to C a i → C
ω ( x ) = 1 2 1 + x 2 x − 1 , 0 < x < 1 , (38) \omega(x)=\frac{1}{2}\sqrt{\sqrt{\frac{1+x}{2x}}-1},\qquad0<x<1, \tag*{(38)} ω ( x ) = 2 1 2 x 1 + x − 1 , 0 < x < 1 , (38) and let ( y 1 , … , y N ) (y_1,\ldots,y_N) ( y 1 , … , y N )
∏ i = 1 N ω ( y i ) ∏ i < j ( y j − y i ) 2 y i + y j , 0 < y 1 < ⋯ < y N < 1. (39) \prod_{i=1}^{N}\omega(y_i)\prod_{i<j}\frac{(y_j-y_i)^2}{y_i+y_j},\qquad0<y_1<\cdots<y_N<1. \tag*{(39)} i = 1 ∏ N ω ( y i ) i < j ∏ y i + y j ( y j − y i ) 2 , 0 < y 1 < ⋯ < y N < 1. (39) Then
m N = η π ∫ 0 1 ( 1 − E ∏ i = 1 N y i − x y i + x ) ω ( x ) x d x . (40) m_N=\frac{\eta}{\pi}\int_0^1\left(1-\mathbb{E}\prod_{i=1}^{N}\frac{y_i-x}{y_i+x}\right)\frac{\omega(x)}{x}\,dx. \tag*{(40)} m N = π η ∫ 0 1 ( 1 − E i = 1 ∏ N y i + x y i − x ) x ω ( x ) d x . (40) Proof. We first turn the zero-multiplicity conditions into Cauchy-transform moment equations. A positive determinant will make those equations uniquely solvable and control the coalescing limit. We then identify the resulting coefficient with the expectation in (40) . The proposed probability law is well defined: ω ( x ) = O ( x − 1 / 4 ) \omega(x)=O(x^{-1/4}) ω ( x ) = O ( x − 1/4 ) y j ≤ 1 y_j\le1 y j ≤ 1
Cauchy moment equations. The two prescribed zeros satisfy h ( C ) = h ( c ) = 1 / 8 h(C)=h(c)=1/8 h ( C ) = h ( c ) = 1/8 z = ( 8 h − 1 ) − 1 z=(8h-1)^{-1} z = ( 8 h − 1 ) − 1
b 2 ( 1 − b 2 ) = 1 + 1 / z 8 . b^2(1-b^2)=\frac{1+1/z}{8}. b 2 ( 1 − b 2 ) = 8 1 + 1/ z . We use the two branches near infinity tending to C C C c c c z − 1 z^{-1} z − 1 h ′ ( C ) h'(C) h ′ ( C ) h ′ ( c ) h'(c) h ′ ( c ) (37) , multiply by z N z^N z N h = ( 1 + 1 / z ) / 8 h=(1+1/z)/8 h = ( 1 + 1/ z ) /8 P ~ \widetilde{P} P J ~ \widetilde{J} J E ~ \widetilde{E} E E ~ ( 0 ) = 0 \widetilde{E}(0)=0 E ( 0 ) = 0 deg E ≤ N − 1 \deg E\le N-1 deg E ≤ N − 1 C C C
f ( z ) = 1 + ( 1 − 1 / z ) / 2 2 , l ( z ) = 1 + ( 1 + 1 / z ) / 2 . f(z)=\sqrt{\frac{1+\sqrt{(1-1/z)/2}}{2}},\qquad l(z)=\sqrt{1+\sqrt{(1+1/z)/2}}. f ( z ) = 2 1 + ( 1 − 1/ z ) /2 , l ( z ) = 1 + ( 1 + 1/ z ) /2 . These formulas continue f f f [ 0 , 1 ] [0,1] [ 0 , 1 ] l l l [ − 1 , 0 ] [-1,0] [ − 1 , 0 ] C C C 2 N + 1 2N+1 2 N + 1 (37) becomes O ( z − 2 N − 1 ) O(z^{-2N-1}) O ( z − 2 N − 1 ) c c c N N N O ( z − N ) O(z^{-N}) O ( z − N ) z N z^N z N
F ( z ) = P ~ ( z ) + f ( z ) J ~ ( z ) + f ( z ) 2 E ~ ( z ) = O ( z − N − 1 ) , (41) \mathcal{F}(z)=\widetilde{P}(z)+f(z)\widetilde{J}(z)+f(z)^2\widetilde{E}(z)=O(z^{-N-1}), \tag*{(41)} F ( z ) = P ( z ) + f ( z ) J ( z ) + f ( z ) 2 E ( z ) = O ( z − N − 1 ) , (41) D ( z ) = J ~ ( z ) + l ( z ) E ~ ( z ) = O ( 1 ) . (42) \mathcal{D}(z)=\widetilde{J}(z)+l(z)\widetilde{E}(z)=O(1). \tag*{(42)} D ( z ) = J ( z ) + l ( z ) E ( z ) = O ( 1 ) . (42) For the second statement, evaluate the same polynomial expression on the branch tending to c c c z N z^N z N O ( 1 ) O(1) O ( 1 ) C C C D \mathcal{D} D
With subscripts + + + − - −
f + ( x ) − f − ( x ) = 2 i ω ( x ) , l + ( − x ) − l − ( − x ) = − 2 2 i ω ( x ) , f + ( x ) + f − ( x ) = l ( x ) . f_+(x)-f_-(x)=2i\omega(x),\qquad l_+(-x)-l_-(-x)=-2\sqrt{2}\,i\omega(x),\qquad f_+(x)+f_-(x)=l(x). f + ( x ) − f − ( x ) = 2 iω ( x ) , l + ( − x ) − l − ( − x ) = − 2 2 iω ( x ) , f + ( x ) + f − ( x ) = l ( x ) . At the outer endpoints these functions are bounded. At zero, F = O ( ∣ z ∣ − 1 / 4 ) \mathcal{F}=O(\lvert z\rvert^{-1/4}) F = O (∣ z ∣ − 1/4 ) D ( z ) → D 0 : = J ~ ( 0 ) \mathcal{D}(z)\to D_0:=\widetilde{\mathcal{J}}(0) D ( z ) → D 0 := J ( 0 ) E ~ ( 0 ) = 0 \widetilde{\mathcal{E}}(0)=0 E ( 0 ) = 0
F ( z ) = 1 π ∫ 0 1 D ( x ) ω ( x ) x − z d x , (43) \mathcal{F}(z)=\frac{1}{\pi}\int_0^1\frac{\mathcal{D}(x)\omega(x)}{x-z}\,dx, \tag*{(43)} F ( z ) = π 1 ∫ 0 1 x − z D ( x ) ω ( x ) d x , (43) ∫ 0 1 x j D ( x ) ω ( x ) d x = 0 , 0 ≤ j < N , (44) \int_0^1x^j\mathcal{D}(x)\omega(x)\,dx=0,\qquad0\leq j<N, \tag*{(44)} ∫ 0 1 x j D ( x ) ω ( x ) d x = 0 , 0 ≤ j < N , (44) D ( z ) = D 0 − 2 π z ∫ 0 1 E ~ ( − y ) / y z + y ω ( y ) d y . (45) \mathcal{D}(z)=D_0-\frac{\sqrt{2}}{\pi}z\int_0^1\frac{\widetilde{\mathcal{E}}(-y)/y}{z+y}\,\omega(y)\,dy. \tag*{(45)} D ( z ) = D 0 − π 2 z ∫ 0 1 z + y E ( − y ) / y ω ( y ) d y . (45) The moment equations are the first N N N (43) at infinity. To obtain (45) , subtract the bounded-at-infinity representation at zero. Its integrand is integrable there because E ~ ( − y ) / y \widetilde{\mathcal{E}}(-y)/y E ( − y ) / y
To control the endpoint at zero, use ω ( x ) = O ( x − 1 / 4 ) \omega(x)=O(x^{-1/4}) ω ( x ) = O ( x − 1/4 ) E ~ ( z ) = O ( z ) \widetilde{\mathcal{E}}(z)=O(z) E ( z ) = O ( z ) (45) , split at y = x y=x y = x
C ( ∫ 0 x y − 1 / 4 d y + x ∫ x 1 y − 5 / 4 d y ) = O ( x 3 / 4 ) . C\left(\int_0^x y^{-1/4}\,dy+x\int_x^1y^{-5/4}\,dy\right)=O(x^{3/4}). C ( ∫ 0 x y − 1/4 d y + x ∫ x 1 y − 5/4 d y ) = O ( x 3/4 ) . This estimate will also justify the singular moment integral below.
Nonsingularity and coalescence. The polynomial E ~ ( − y ) / y \widetilde{\mathcal{E}}(-y)/y E ( − y ) / y N − 1 N-1 N − 1 D 0 D_0 D 0
M j k = ∫ 0 1 ∫ 0 1 x j + 1 y k x + y ω ( x ) ω ( y ) d x d y , 0 ≤ j , k < N . (46) M_{jk}=\int_0^1\int_0^1\frac{x^{j+1}y^k}{x+y}\omega(x)\omega(y)\,dx\,dy,\qquad0\leq j,k<N. \tag*{(46)} M j k = ∫ 0 1 ∫ 0 1 x + y x j + 1 y k ω ( x ) ω ( y ) d x d y , 0 ≤ j , k < N . (46) Its determinant is strictly positive. This is the Cauchy-bimoment positivity argument used in [3 ] ; we give the finite determinant calculation for the present weights. Expand the determinant and antisymmetrize successively in the x x x y y y ∏ i x i \prod_i x_i ∏ i x i
det [ 1 x i + y j ] i , j = ∏ i < j ( x j − x i ) ( y j − y i ) ∏ i , j ( x i + y j ) > 0. \det\left[\frac{1}{x_i+y_j}\right]_{i,j}=\frac{\prod_{i<j}(x_j-x_i)(y_j-y_i)}{\prod_{i,j}(x_i+y_j)}>0. det [ x i + y j 1 ] i , j = ∏ i , j ( x i + y j ) ∏ i < j ( x j − x i ) ( y j − y i ) > 0. For completeness this identity follows by multiplying through by the denominator: the numerator is alternating in each family and has exactly the degree of the two Vandermondes. The remaining constant is one, recursively evaluating at x N = − y N x_N=-y_N x N = − y N
Consider now a solution of the homogeneous linear problem for g g g deg g ≤ N − 1 \deg g\leq N-1 deg g ≤ N − 1 b 4 N + 1 b^{4N+1} b 4 N + 1 Q Q Q
[ b 4 N + 1 ] Q = ( − 8 ) N D 0 , [b^{4N+1}]Q=(-8)^N D_0, [ b 4 N + 1 ] Q = ( − 8 ) N D 0 , we obtain D 0 = 0 D_0=0 D 0 = 0 (46) forces E ~ = 0 \widetilde{\mathcal{E}}=0 E = 0 (45) gives J ~ = 0 \widetilde{\mathcal{J}}=0 J = 0 P ~ = 0 \widetilde{\mathcal{P}}=0 P = 0 g = 0 g=0 g = 0 g ( b ) = b N + ∑ r = 0 N − 1 γ r b r g(b)=b^N+\sum_{r=0}^{N-1}\gamma_r b^r g ( b ) = b N + ∑ r = 0 N − 1 γ r b r b 3 P 3 ( h ) b^3P_3(h) b 3 P 3 ( h ) N N N N × N N\times N N × N ( γ 0 , … , γ N − 1 ) (\gamma_0,\ldots,\gamma_{N-1}) ( γ 0 , … , γ N − 1 ) g g g g X g_X g X a i a_i a i a i a_i a i C C C g X g_X g X g g g
The moment coefficient and its expectation. For this monic solution D 0 = ( − 8 ) − N D_0=(-8)^{-N} D 0 = ( − 8 ) − N G ( z ) = D ( z ) / D 0 G(z)=D(z)/D_0 G ( z ) = D ( z ) / D 0 [ b 4 N ] Q = ( − 8 ) N P ~ ( 0 ) [b^{4N}]Q=(-8)^N\widetilde{P}(0) [ b 4 N ] Q = ( − 8 ) N P ( 0 ) (37) gives the coefficient needed by the arch formula:
[ b N − 1 ] g ( b ) = P ~ ( 0 ) D 0 + ( 2 N + 1 ) C + N c . [b^{N-1}]g(b)=\frac{\widetilde{P}(0)}{D_0}+(2N+1)C+Nc. [ b N − 1 ] g ( b ) = D 0 P ( 0 ) + ( 2 N + 1 ) C + N c . Substitution into (32) , using C = η ( 2 C + c ) C=\eta(2C+c) C = η ( 2 C + c )
m N = − η ( C + P ~ ( 0 ) D 0 ) = η π ∫ 0 1 ( 1 − G ( x ) ) ω ( x ) x d x . (47) m_N=-\eta\left(C+\frac{\widetilde{P}(0)}{D_0}\right)=\frac{\eta}{\pi}\int_0^1(1-G(x))\frac{\omega(x)}{x}\,dx. \tag*{(47)} m N = − η ( C + D 0 P ( 0 ) ) = π η ∫ 0 1 ( 1 − G ( x )) x ω ( x ) d x . (47) For the second equality, use
f ( z ) = C + 1 π ∫ 0 1 ω ( x ) x − z d x f(z)=C+\frac{1}{\pi}\int_0^1\frac{\omega(x)}{x-z}\,dx f ( z ) = C + π 1 ∫ 0 1 x − z ω ( x ) d x and subtract D 0 f D_0f D 0 f (43) , then let z → 0 − z\to0^- z → 0 − (45) gives D ( x ) − D 0 = O ( x 3 / 4 ) D(x)-D_0=O(x^{3/4}) D ( x ) − D 0 = O ( x 3/4 ) O ( x − 1 / 2 ) O(x^{-1/2}) O ( x − 1/2 ) f ( z ) ( J ~ ( z ) − D 0 ) f(z)(\widetilde{\mathcal{J}}(z)-D_0) f ( z ) ( J ( z ) − D 0 ) f ( z ) 2 E ~ ( z ) f(z)^2\widetilde{\mathcal{E}}(z) f ( z ) 2 E ( z )
The moment equations now have a unique solution. It remains to exhibit their probabilistic solution, thereby identifying G G G (40) . Let W n W_n W n (39) with n n n ( 0 , 1 ) n (0,1)^n ( 0 , 1 ) n n = N n=N n = N
∏ i = 1 N y i − x y i + x = 1 − x ∑ i = 1 N α i x + y i , α i = 2 ∏ j ≠ i y j + y i y j − y i . \prod_{i=1}^{N}\frac{y_i-x}{y_i+x}=1-x\sum_{i=1}^{N}\frac{\alpha_i}{x+y_i},\qquad\alpha_i=2\prod_{j\ne i}\frac{y_j+y_i}{y_j-y_i}. i = 1 ∏ N y i + x y i − x = 1 − x i = 1 ∑ N x + y i α i , α i = 2 j = i ∏ y j − y i y j + y i . After multiplication by the density, the factors in α i \alpha_i α i ( 0 , 1 ) N (0,1)^N ( 0 , 1 ) N x / ( x + y i ) x/(x+y_i) x / ( x + y i ) 1 − x ∫ 0 1 P ( y ) ω ( y ) / ( x + y ) d y 1-x\int_0^1P(y)\omega(y)/(x+y)\,dy 1 − x ∫ 0 1 P ( y ) ω ( y ) / ( x + y ) d y deg P ≤ N − 1 \deg P\le N-1 deg P ≤ N − 1 x k ω ( x ) x^k\omega(x) x k ω ( x ) k < N k<N k < N x x x ( N + 1 ) (N+1) ( N + 1 )
x k ω ( x ) W N ( y 1 , … , y N ) ∏ i = 1 N y i − x y i + x = x k W N + 1 ( x , y 1 , … , y N ) ∏ i = 1 N ( y i − x ) . x^k\omega(x)W_N(y_1,\ldots,y_N)\prod_{i=1}^{N}\frac{y_i-x}{y_i+x}
=
\frac{x^kW_{N+1}(x,y_1,\ldots,y_N)}{\prod_{i=1}^{N}(y_i-x)}. x k ω ( x ) W N ( y 1 , … , y N ) i = 1 ∏ N y i + x y i − x = ∏ i = 1 N ( y i − x ) x k W N + 1 ( x , y 1 , … , y N ) . The left side is absolutely integrable, since the product has modulus at most one. Symmetrizing the right side in all N + 1 N+1 N + 1
∑ r = 0 N z r k ∏ j ≠ r ( z j − z r ) = 0 , 0 ≤ k < N . \sum_{r=0}^{N}\frac{z_r^k}{\prod_{j\ne r}(z_j-z_r)}=0,\qquad0\le k<N. r = 0 ∑ N ∏ j = r ( z j − z r ) z r k = 0 , 0 ≤ k < N . Thus the expectation has the same polynomial Cauchy representation and vanishing moments as G G G (46) now identifies this expectation with G G G (47) . ∙იდ
The smallest coordinate and the strip exponent
The positive integral reduces the growth of the arch moment to the coordinates nearest zero. We prove the required bounds by comparing with an explicitly normalized Bures–Laguerre density. The comparison is at the level of positive measures, so it also controls the possible sign changes in the product appearing inside the integral.
The integral formula immediately gives
E y 1 − 1 / 4 ≲ m N ≲ E ( ∑ i = 1 N y i − 1 ) 1 / 4 . (48) \mathbb{E}y_1^{-1/4} \lesssim m_N \lesssim\mathbb{E}\left(\sum_{i=1}^{N}y_i^{-1}\right)^{1/4}. \tag*{(48)} E y 1 − 1/4 ≲ m N ≲ E ( i = 1 ∑ N y i − 1 ) 1/4 . (48) To verify both estimates pointwise, put R ( x ) = ∏ i ( y i − x ) / ( y i + x ) R(x)=\prod_i(y_i-x)/(y_i+x) R ( x ) = ∏ i ( y i − x ) / ( y i + x ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 1 − R ( x ) ≥ 0 1-R(x)\ge0 1 − R ( x ) ≥ 0
1 − R ( x ) ≤ 2 min { 1 , x ∑ i y i − 1 } . 1-R(x)\le2\min\left\{1,x\sum_i y_i^{-1}\right\}. 1 − R ( x ) ≤ 2 min { 1 , x i ∑ y i − 1 } . For y 1 / 3 < x < y 1 / 2 y_1/3<x<y_1/2 y 1 /3 < x < y 1 /2 1 / 2 1/2 1/2 1 − R ( x ) ≥ 1 / 2 1-R(x)\ge1/2 1 − R ( x ) ≥ 1/2 ω ( x ) ≲ x − 1 / 4 \omega(x)\lesssim x^{-1/4} ω ( x ) ≲ x − 1/4 ( 0 , 1 ) (0,1) ( 0 , 1 ) ω ( x ) ≍ x − 1 / 4 \omega(x)\asymp x^{-1/4} ω ( x ) ≍ x − 1/4 ( 0 , 1 / 2 ) (0,1/2) ( 0 , 1/2 ) x = ( ∑ i y i − 1 ) − 1 x=\left(\sum_i y_i^{-1}\right)^{-1} x = ( ∑ i y i − 1 ) − 1 (48) .
The reference inverse-moment estimates below will contribute a factor N 1 / 2 N^{1/2} N 1/2 N − 1 N^{-1} N − 1 N 1 / 4 N^{1/4} N 1/4 N 3 / 4 N^{3/4} N 3/4
For comparison we use the generalized Bures–Laguerre family [12 ] , in ordered coordinates and with parameter a > 0 a>0 a > 0
1 Z n ( a ) ∏ i = 1 n r i a − 1 e − r i ∏ i < j ( r j − r i ) 2 r i + r j , 0 < r 1 < ⋯ < r n < ∞ . (49) \frac{1}{Z_n(a)}\prod_{i=1}^{n}r_i^{a-1}e^{-r_i}\prod_{i<j}\frac{(r_j-r_i)^2}{r_i+r_j},\qquad0<r_1<\cdots<r_n<\infty. \tag*{(49)} Z n ( a ) 1 i = 1 ∏ n r i a − 1 e − r i i < j ∏ r i + r j ( r j − r i ) 2 , 0 < r 1 < ⋯ < r n < ∞. (49) Expectations and probabilities for this law will carry subscripts n , a n,a n , a a − 1 a-1 a − 1 a a a [12 ] , equation (1.5) plus one. Our ordered normalizer is the corresponding unordered integral divided by n ! n! n ! 2 _2 2 [4 ] ; see also [10 ] . We give a direct conditioning proof for the ordered pair kernel used here.
Lemma 6.1 (Association and monotone comparison). Let I ⊂ ( 0 , ∞ ) I\subset(0,\infty) I ⊂ ( 0 , ∞ ) n ≥ 1 n\ge1 n ≥ 1 w 1 , … , w n w_1,\ldots,w_n w 1 , … , w n I I I
∏ i = 1 n w i ( x i ) ∏ i < j K ( x i , x j ) , K ( x , y ) = ( y − x ) 2 x + y , x 1 < ⋯ < x n , x i ∈ I , \prod_{i=1}^{n}w_i(x_i)\prod_{i<j}K(x_i,x_j),\qquad K(x,y)=\frac{(y-x)^2}{x+y},\qquad x_1<\cdots<x_n,\quad x_i\in I, i = 1 ∏ n w i ( x i ) i < j ∏ K ( x i , x j ) , K ( x , y ) = x + y ( y − x ) 2 , x 1 < ⋯ < x n , x i ∈ I , has finite positive normalizing integral. Its probability law is positively associated: increasing integrable functions have nonnegative covariance whenever the covariance is defined. The assertion also holds after truncation to a shorter common interval. Consequently, a coordinatewise increasing likelihood ratio shifts the coordinates upwards in stochastic order, and a decreasing one shifts them downwards.
Proof. The mixed derivative of each pair contribution is
∂ 2 ∂ x ∂ y log K ( x , y ) = 2 ( y − x ) 2 + 1 ( x + y ) 2 > 0. \frac{\partial^2}{\partial x\partial y}\log K(x,y)=\frac{2}{(y-x)^2}+\frac{1}{(x+y)^2}>0. ∂ x ∂ y ∂ 2 log K ( x , y ) = ( y − x ) 2 2 + ( x + y ) 2 1 > 0. We prove association by induction on the number of coordinates, initially for bounded increasing functions. Conditional on the largest coordinate t t t s < t s<t s < t s s s t t t s s s t t t t t t
The law of total covariance now has two nonnegative terms: conditional covariance by induction, and covariance of the two conditional expectations by the one-dimensional inequality
Cov ( F ( T ) , G ( T ) ) = 1 2 E [ ( F ( T ) − F ( T ′ ) ) ( G ( T ) − G ( T ′ ) ) ] ≥ 0. \operatorname{Cov}(F(T),G(T))=\frac{1}{2}\mathbb{E}[(F(T)-F(T'))(G(T)-G(T'))]\ge0. Cov ( F ( T ) , G ( T )) = 2 1 E [( F ( T ) − F ( T ′ )) ( G ( T ) − G ( T ′ ))] ≥ 0. Here T ′ T' T ′ T T T L L L E [ F L ] / E L ≥ E F \mathbb{E}[FL]/\mathbb{E}L\ge\mathbb{E}F E [ F L ] / E L ≥ E F F F F L L L
The upper comparison will use a = 1 / 2 a=1/2 a = 1/2 a = 1 a=1 a = 1 a = 1 a=1 a = 1 a = 2 a=2 a = 2
Lemma 6.2 (Laguerre normalization and coordinate bounds). For fixed a > 0 a>0 a > 0 n ≥ 1 n\ge1 n ≥ 1
Z n ( a ) = ∏ i = 0 n − 1 Γ ( a + i ) ∏ 0 ≤ i < j < n j − i 2 a + i + j . (50) Z_n(a)=\prod_{i=0}^{n-1}\Gamma(a+i)\prod_{0\le i<j<n}\frac{j-i}{2a+i+j}. \tag*{(50)} Z n ( a ) = i = 0 ∏ n − 1 Γ ( a + i ) 0 ≤ i < j < n ∏ 2 a + i + j j − i . (50) *There exist constants K a , C a K_a,C_a K a , C a a a a P n , a ( r n ≤ K a n ) \mathbb{P}_{n,a}(r_n\le K_an) P n , a ( r n ≤ K a n ) n n n D k = a k + k ( k − 1 ) / 2 , D_k=ak+k(k-1)/2, D k = ak + k ( k − 1 ) /2 ,
P n , a ( r k < h ) ≤ min { 1 , ( C a h n 2 k 3 ) D k } , h > 0 , 1 ≤ k ≤ n . (51) \mathbb{P}_{n,a}(r_k<h)\le\min\left\{1,\left(\frac{C_ahn^2}{k^3}\right)^{D_k}\right\},\qquad h>0,\quad1\le k\le n. \tag*{(51)} P n , a ( r k < h ) ≤ min { 1 , ( k 3 C a h n 2 ) D k } , h > 0 , 1 ≤ k ≤ n . (51) In particular,
E n , 1 / 2 ( ∑ i r i − 1 ) 1 / 4 ≲ n 1 / 2 , (52) \mathbb{E}_{n,1/2}\left(\sum_i r_i^{-1}\right)^{1/4}\lesssim n^{1/2}, \tag*{(52)} E n , 1/2 ( i ∑ r i − 1 ) 1/4 ≲ n 1/2 , (52) E n , 2 ∑ i r i − 1 ≲ n 2 , (53) \mathbb{E}_{n,2}\sum_i r_i^{-1}\lesssim n^2, \tag*{(53)} E n , 2 i ∑ r i − 1 ≲ n 2 , (53) E n , 1 r 1 − 1 / 4 ≳ n 1 / 2 . (54) \mathbb{E}_{n,1}r_1^{-1/4}\gtrsim n^{1/2}. \tag*{(54)} E n , 1 r 1 − 1/4 ≳ n 1/2 . (54) The constants in the last three bounds are absolute.
Proof. Normalization. Write the pair product as a Vandermonde determinant times the Schur Pfaffian with entries ( s − r ) / ( s + r ) (s-r)/(s+r) ( s − r ) / ( s + r ) [19 ] , equation (1.2), is
Pf [ x j − x i x j + x i ] i < j = ∏ i < j x j − x i x j + x i . \operatorname{Pf}\left[\frac{x_j-x_i}{x_j+x_i}\right]_{i<j}=\prod_{i<j}\frac{x_j-x_i}{x_j+x_i}. Pf [ x j + x i x j − x i ] i < j = i < j ∏ x j + x i x j − x i . For even size, clearing denominators leaves an alternating polynomial of Vandermonde degree, so only its constant factor needs checking. Pairing two opposite arguments gives that constant recursively as one. For odd size, add an argument tending to infinity; the resulting augmenting column consists of ones.
Expand the determinant and the Pfaffian and integrate over the unordered region, divided by n ! n! n ! [5 ] ; the general antisymmetric-kernel form, including its odd-size border, is stated in [1 ] . The result is the Pfaffian of double moments, with the single moments as augmenting entries in odd size. Absolute integrability follows from ∣ ( s − r ) / ( s + r ) ∣ ≤ 1 \lvert(s-r)/(s+r)\rvert\le1 ∣( s − r ) / ( s + r )∣ ≤ 1 a > 0 a>0 a > 0 i , j i,j i , j
∫ 0 ∞ ∫ 0 ∞ r a + i − 1 s a + j − 1 e − r − s s − r s + r d r d s = Γ ( a + i ) Γ ( a + j ) j − i 2 a + i + j . \int_0^\infty\int_0^\infty r^{a+i-1}s^{a+j-1}e^{-r-s}\frac{s-r}{s+r}\,dr\,ds=\Gamma(a+i)\Gamma(a+j)\frac{j-i}{2a+i+j}. ∫ 0 ∞ ∫ 0 ∞ r a + i − 1 s a + j − 1 e − r − s s + r s − r d r d s = Γ ( a + i ) Γ ( a + j ) 2 a + i + j j − i . Use r + s r+s r + s r / ( r + s ) r/(r+s) r / ( r + s ) ( a + i ) / ( 2 a + i + j ) (a+i)/(2a+i+j) ( a + i ) / ( 2 a + i + j ) a + i a+i a + i (50) .
Coordinate tails and inverse moments. For the maximum, condition on its value s s s s s s K ( r , s ) ≤ s K(r,s)\le s K ( r , s ) ≤ s 0 < r < s 0<r<s 0 < r < s
P n , a ( r n > K n ) ≤ Z n − 1 ( a ) Z n ( a ) ∫ K n ∞ s a + n − 2 e − s d s . \mathbb{P}_{n,a}(r_n>K n)\le\frac{Z_{n-1}(a)}{Z_n(a)}\int_{K n}^{\infty}s^{a+n-2}e^{-s}\,ds. P n , a ( r n > K n ) ≤ Z n ( a ) Z n − 1 ( a ) ∫ K n ∞ s a + n − 2 e − s d s . The explicit normalizer gives
Z n − 1 ( a ) Z n ( a ) = 1 Γ ( a + n − 1 ) ∏ j = 1 n − 1 2 a + 2 n − 2 − j j ≤ e a ( 2 e ) n − 1 Γ ( a + n − 1 ) . \frac{Z_{n-1}(a)}{Z_n(a)}=\frac{1}{\Gamma(a+n-1)}\prod_{j=1}^{n-1}\frac{2a+2n-2-j}{j}\le\frac{e^a(2e)^{n-1}}{\Gamma(a+n-1)}. Z n ( a ) Z n − 1 ( a ) = Γ ( a + n − 1 ) 1 j = 1 ∏ n − 1 j 2 a + 2 n − 2 − j ≤ Γ ( a + n − 1 ) e a ( 2 e ) n − 1 . For n ≥ 2 n\ge2 n ≥ 2 2 ( a + n − 1 ) 2(a+n-1) 2 ( a + n − 1 ) ( n − 1 ) ! ≥ ( ( n − 1 ) / e ) n − 1 (n-1)!\ge((n-1)/e)^{n-1} ( n − 1 )! ≥ (( n − 1 ) / e ) n − 1 ( 1 + a / ( n − 1 ) ) n − 1 ≤ e a (1+a/(n-1))^{n-1}\le e^a ( 1 + a / ( n − 1 ) ) n − 1 ≤ e a n = 1 n=1 n = 1
1 Γ ( a + n − 1 ) ∫ K n ∞ s a + n − 2 e − s d s ≤ e − K n / 2 2 a + n − 1 . \frac{1}{\Gamma(a+n-1)}\int_{K n}^{\infty}s^{a+n-2}e^{-s}\,ds\le e^{-Kn/2}2^{a+n-1}. Γ ( a + n − 1 ) 1 ∫ K n ∞ s a + n − 2 e − s d s ≤ e − K n /2 2 a + n − 1 . Choosing K K K n n n
For the small-coordinate estimate, integrate the first k k k h h h h h h
P n , a ( r k < h ) ≤ h D k a k k ! Z n − k ( a + k ) Z n ( a ) . (55) \mathbb{P}_{n,a}(r_k<h)\le\frac{h^{D_k}}{a^k k!}\frac{Z_{n-k}(a+k)}{Z_n(a)}. \tag*{(55)} P n , a ( r k < h ) ≤ a k k ! h D k Z n ( a ) Z n − k ( a + k ) . (55) with Z 0 = 1 Z_0=1 Z 0 = 1 (50) ,
Z n − k ( a + k ) Z n ( a ) = ∏ i = 0 k − 1 Γ ( a + i ) − 1 ∏ 0 ≤ i < j < k ( 1 + 2 a + 2 i j − i ) . \frac{Z_{n-k}(a+k)}{Z_n(a)}
=\prod_{i=0}^{k-1}\Gamma(a+i)^{-1}\prod_{0\le i<j<k}\left(1+\frac{2a+2i}{j-i}\right). Z n ( a ) Z n − k ( a + k ) = i = 0 ∏ k − 1 Γ ( a + i ) − 1 0 ≤ i < j < k ∏ ( 1 + j − i 2 a + 2 i ) . For gaps j − i ≤ k j-i\le k j − i ≤ k 2 a + 2 i ≤ C a k 2a+2i\le C_a k 2 a + 2 i ≤ C a k i < k i<k i < k
∑ j = 1 k log ( 1 + C a k / j ) ≤ k log ( 1 + C a ) + k log k − log ( k ! ) = O a ( k ) . \sum_{j=1}^{k}\log(1+C_a k/j)\le k\log(1+C_a)+k\log k-\log(k!)=O_a(k). j = 1 ∑ k log ( 1 + C a k / j ) ≤ k log ( 1 + C a ) + k log k − log ( k !) = O a ( k ) . Thus the short gaps contribute O a ( k 2 ) = O a ( D k ) O_a(k^2)=O_a(D_k) O a ( k 2 ) = O a ( D k ) log ( 1 + x ) ≤ x \log(1+x)\le x log ( 1 + x ) ≤ x
∑ i < k ( 2 a + 2 i ) ∑ j = k + 1 n − 1 − i 1 j ≤ 2 D k log ( n / k ) . \sum_{i<k}(2a+2i)\sum_{j=k+1}^{n-1-i}\frac{1}{j}\le2D_k\log(n/k). i < k ∑ ( 2 a + 2 i ) j = k + 1 ∑ n − 1 − i j 1 ≤ 2 D k log ( n / k ) . This includes k = n k=n k = n
∏ i = 0 k − 1 Γ ( a + i ) ≥ k D k e − O a ( D k ) . (56) \prod_{i=0}^{k-1}\Gamma(a+i)\ge k^{D_k}e^{-O_a(D_k)}. \tag*{(56)} i = 0 ∏ k − 1 Γ ( a + i ) ≥ k D k e − O a ( D k ) . (56) Here is a direct estimate. For q = a + i q=a+i q = a + i [ q , q + 1 ] [q,q+1] [ q , q + 1 ] log Γ ( q ) ≥ ( q − 1 ) log q − q − O a ( 1 ) \log\Gamma(q)\ge(q-1)\log q-q-O_a(1) log Γ ( q ) ≥ ( q − 1 ) log q − q − O a ( 1 ) log k \log k log k ∑ i < k ( a + i ) log ( ( a + i ) / k ) ≥ − k 2 / e \sum_{i<k}(a+i)\log((a+i)/k)\ge-k^2/e ∑ i < k ( a + i ) log (( a + i ) / k ) ≥ − k 2 / e D k log k − O a ( D k ) D_k\log k-O_a(D_k) D k log k − O a ( D k ) k log k k\log k k log k O a ( D k ) O_a(D_k) O a ( D k ) k k k a − k / k ! a^{-k}/k! a − k / k ! C a D k C_a^{D_k} C a D k
For any 0 < p < D k 0<p<D_k 0 < p < D k
E n , a r k − p ≤ D k D k − p ( C a n 2 k 3 ) p . (57) \mathbb{E}_{n,a}r_k^{-p}\le\frac{D_k}{D_k-p}\left(\frac{C_a n^2}{k^3}\right)^p. \tag*{(57)} E n , a r k − p ≤ D k − p D k ( k 3 C a n 2 ) p . (57) At a = 1 / 2 a=1/2 a = 1/2 ( k , p ) = ( 1 , 1 / 4 ) (k,p)=(1,1/4) ( k , p ) = ( 1 , 1/4 ) p = 1 p=1 p = 1 k ≥ 2 k\ge2 k ≥ 2 E r 1 − 1 / 4 ≲ n 1 / 2 \mathbb{E}r_1^{-1/4}\lesssim n^{1/2} E r 1 − 1/4 ≲ n 1/2 E ∑ k ≥ 2 r k − 1 ≲ n 2 \mathbb{E}\sum_{k\ge2}r_k^{-1}\lesssim n^2 E ∑ k ≥ 2 r k − 1 ≲ n 2 a = 2 a=2 a = 2 p = 1 p=1 p = 1 k k k
The minimum at a = 1 a=1 a = 1 The density of the smallest coordinate at s > 0 s>0 s > 0
e − s Z n − 1 ( 2 ) Z n ( 1 ) E n − 1 , 2 [ 1 { r i > s ∀ i } ∏ i ( 1 − s / r i ) 2 1 + s / r i ] . e^{-s}\frac{Z_{n-1}(2)}{Z_n(1)}\mathbb{E}_{n-1,2}\left[\mathbf{1}_{\{r_i>s\ \forall i\}}\prod_i\frac{(1-s/r_i)^2}{1+s/r_i}\right]. e − s Z n ( 1 ) Z n − 1 ( 2 ) E n − 1 , 2 [ 1 { r i > s ∀ i } i ∏ 1 + s / r i ( 1 − s / r i ) 2 ] . The normalizer ratio is exactly n ( n + 1 ) / 2 n(n+1)/2 n ( n + 1 ) /2 q s ( r ) = 1 { r > s } ( 1 − s / r ) 2 / ( 1 + s / r ) q_s(r)=\mathbf{1}_{\{r>s\}}(1-s/r)^2/(1+s/r) q s ( r ) = 1 { r > s } ( 1 − s / r ) 2 / ( 1 + s / r ) r > 0 r>0 r > 0 r ≤ s r\le s r ≤ s 0 ≤ q s ( r ) ≤ 1 0\le q_s(r)\le1 0 ≤ q s ( r ) ≤ 1 1 − q s ( r ) ≤ 3 s / r 1-q_s(r)\le3s/r 1 − q s ( r ) ≤ 3 s / r
∏ i q s ( r i ) ≥ 1 − 3 s ∑ i r i − 1 . \prod_i q_s(r_i)\ge1-3s\sum_i r_i^{-1}. i ∏ q s ( r i ) ≥ 1 − 3 s i ∑ r i − 1 . By (53), the expectation is at least 1 / 2 1/2 1/2 0 < s < ε n − 2 0<s<\varepsilon n^{-2} 0 < s < ε n − 2 ε > 0 \varepsilon>0 ε > 0 n 2 n^2 n 2 n − 2 n^{-2} n − 2 r 1 − 1 / 4 ≳ n 1 / 2 r_1^{-1/4}\gtrsim n^{1/2} r 1 − 1/4 ≳ n 1/2 n = 1 n=1 n = 1
Completion of Theorem 1.1. We compare (39) with the reference laws to estimate the two sides of (48). For the upper estimate, first condition the target law on y N < 1 / 2 y_N<1/2 y N < 1/2 y i = r i / ( b N ) y_i=r_i/(bN) y i = r i / ( b N ) a = 1 / 2 a=1/2 a = 1/2 r N < b N / 2 r_N<bN/2 r N < b N /2
ω ( y ) y 1 / 2 e b N y . \omega(y)y^{1/2}e^{bNy}. ω ( y ) y 1/2 e b N y . Direct differentiation gives
d d y log ( ω ( y ) y 1 / 2 ) = 1 − 2 y − 2 y ( 1 + y ) 4 y ( 1 − y ) ≥ − 4 , 0 < y ≤ 1 2 . \frac{d}{dy}\log\left(\omega(y)y^{1/2}\right)=\frac{1-2y-\sqrt{2y(1+y)}}{4y(1-y)}\ge-4,\qquad0<y\le\frac{1}{2}. d y d log ( ω ( y ) y 1/2 ) = 4 y ( 1 − y ) 1 − 2 y − 2 y ( 1 + y ) ≥ − 4 , 0 < y ≤ 2 1 . Indeed the numerator is positive for y ≤ 1 / 8 y\le1/8 y ≤ 1/8 − 1 -1 − 1 1 / 4 1/4 1/4 N ≥ 1 N\ge1 N ≥ 1 b ≥ 4 b\ge4 b ≥ 4 b / 2 ≥ K 1 / 2 b/2\ge K_{1/2} b /2 ≥ K 1/2 N N N
E ( ∑ i y i − 1 ) 1 / 4 ≲ ( b N ) 1 / 4 E N , 1 / 2 ( ∑ i r i − 1 ) 1 / 4 ≲ N 3 / 4 . \mathbb{E}\left(\sum_i y_i^{-1}\right)^{1/4}\lesssim(bN)^{1/4}\mathbb{E}_{N,1/2}\left(\sum_i r_i^{-1}\right)^{1/4}\lesssim N^{3/4}. E ( i ∑ y i − 1 ) 1/4 ≲ ( b N ) 1/4 E N , 1/2 ( i ∑ r i − 1 ) 1/4 ≲ N 3/4 . For the lower estimate, set y i = 2 z i / ( 1 + z i ) y_i=2z_i/(1+z_i) y i = 2 z i / ( 1 + z i ) 0 < z 1 < ⋯ < z N < 1 0<z_1<\cdots<z_N<1 0 < z 1 < ⋯ < z N < 1
K ( 2 z i 1 + z i , 2 z j 1 + z j ) = 2 K ( z i , z j ) ( 1 + z i ) ( 1 + z j ) ( 1 + 2 z i z j / ( z i + z j ) ) . K\left(\frac{2z_i}{1+z_i},\frac{2z_j}{1+z_j}\right)=\frac{2K(z_i,z_j)}{(1+z_i)(1+z_j)(1+2z_i z_j/(z_i+z_j))}. K ( 1 + z i 2 z i , 1 + z j 2 z j ) = ( 1 + z i ) ( 1 + z j ) ( 1 + 2 z i z j / ( z i + z j )) 2 K ( z i , z j ) . Together with the Jacobian this gives one-variable factors ω ( 2 z / ( 1 + z ) ) ( 1 + z ) − ( N + 1 ) \omega(2z/(1+z))(1+z)^{-(N+1)} ω ( 2 z / ( 1 + z )) ( 1 + z ) − ( N + 1 ) ( 1 + 2 z i z j / ( z i + z j ) ) − 1 (1+2z_i z_j/(z_i+z_j))^{-1} ( 1 + 2 z i z j / ( z i + z j ) ) − 1 z i = r i / ( b N ) z_i=r_i/(bN) z i = r i / ( b N ) a = 1 a=1 a = 1 0 < b < 1 / 2 0<b<1/2 0 < b < 1/2 r N < b N r_N<bN r N < b N
ω ( 2 z 1 + z ) ( 1 + z ) − ( N + 1 ) e b N z \omega\left(\frac{2z}{1+z}\right)(1+z)^{-(N+1)}e^{bNz} ω ( 1 + z 2 z ) ( 1 + z ) − ( N + 1 ) e b N z and the additional pair factors just displayed. Each is decreasing in every coordinate: ω \omega ω − ( N + 1 ) / ( 1 + z ) + b N < 0 -(N+1)/(1+z)+bN<0 − ( N + 1 ) / ( 1 + z ) + b N < 0 0 < z < 1 0<z<1 0 < z < 1 z z z y 1 ≤ 2 z 1 y_1\le2z_1 y 1 ≤ 2 z 1
E y 1 − 1 / 4 ≥ 2 − 1 / 4 ( b N ) 1 / 4 E N , 1 r 1 − 1 / 4 ≳ N 3 / 4 . \mathbb{E}y_1^{-1/4}\ge2^{-1/4}(bN)^{1/4}\mathbb{E}_{N,1}r_1^{-1/4}\gtrsim N^{3/4}. E y 1 − 1/4 ≥ 2 − 1/4 ( b N ) 1/4 E N , 1 r 1 − 1/4 ≳ N 3/4 . (48) now proves m N ≍ N 3 / 4 m_N\asymp N^{3/4} m N ≍ N 3/4
It remains to extract the bridge estimate from Lemma 4.2. By monotonicity, summing the lower increment bound over 1 ≤ j < N 1\le j<N 1 ≤ j < N ( N − 1 ) B N ≲ m N (N-1)B_N\lesssim m_N ( N − 1 ) B N ≲ m N B N ≲ N − 1 / 4 B_N\lesssim N^{-1/4} B N ≲ N − 1/4 N N N L L L m L N m_{LN} m L N m N m_N m N
N 3 / 4 ≲ m L N − m N ≲ ∑ j = N L N − 1 B j ≤ ( L − 1 ) N B N . N^{3/4}\lesssim m_{LN}-m_N\lesssim\sum_{j=N}^{LN-1}B_j\le(L-1)NB_N. N 3/4 ≲ m L N − m N ≲ j = N ∑ L N − 1 B j ≤ ( L − 1 ) N B N . This proves B N ≳ N − 1 / 4 B_N\gtrsim N^{-1/4} B N ≳ N − 1/4
Local coefficient identities
This appendix proves Lemma 2.2 in the diagram convention of Section 2. The finite tables also specify every nonzero local coefficient used by the interpolation argument.
Proof. Equation (8) follows by direct substitution. For completeness we record a coefficient check of the splitting identity. Subscripts 1, 2 mean arguments x + s λ x+s\lambda x + s λ x x x q q q k k k q = k = ρ q=k=\rho q = k = ρ s = 2 s=2 s = 2 q = 1 q=1 q = 1 k = 0 k=0 k = 0 s = 3 s=3 s = 3 R ^ s \hat{R}_s R ^ s B = 1 B=1 B = 1
Table 1.
For s = 2 s=2 s = 2
Table 2.
Here are the substitutions verifying these equalities. Put s j = sin ( x + j λ ) s_j=\sin(x+j\lambda) s j = sin ( x + j λ ) x + j λ x+j\lambda x + j λ s j + 2 s j + 3 s_{j+2}s_{j+3} s j + 2 s j + 3
Table 3.
Substitution uses the following product-to-sum identities:
s j + 8 = − s j , s j s j + 1 + s j + 4 s j + 5 = C , s 2 s 6 = d 2 − s 0 2 , s 1 s 5 = d 2 − s 7 2 , s_{j+8}=-s_j,\qquad s_js_{j+1}+s_{j+4}s_{j+5}=C,\qquad s_2s_6=d^2-s_0^2,\qquad s_1s_5=d^2-s_7^2, s j + 8 = − s j , s j s j + 1 + s j + 4 s j + 5 = C , s 2 s 6 = d 2 − s 0 2 , s 1 s 5 = d 2 − s 7 2 , ρ ( s 1 − s 7 ) = s 0 , ρ ( s 6 − s 0 ) = s 7 , ρ ( s 1 s 2 − s 6 s 7 ) = s 0 s 4 , ρ ( s 5 s 6 − s 0 s 1 ) = s 7 s 3 , \rho(s_1-s_7)=s_0,\qquad\rho(s_6-s_0)=s_7,\qquad\rho(s_1s_2-s_6s_7)=s_0s_4,\qquad\rho(s_5s_6-s_0s_1)=s_7s_3, ρ ( s 1 − s 7 ) = s 0 , ρ ( s 6 − s 0 ) = s 7 , ρ ( s 1 s 2 − s 6 s 7 ) = s 0 s 4 , ρ ( s 5 s 6 − s 0 s 1 ) = s 7 s 3 , ρ ( s 7 s 0 + s 4 s 5 ) = s 6 s 2 , ρ ( s 2 s 3 + s 7 s 0 ) = s 1 s 5 , s 1 s 6 − s 7 s 0 = ρ d 2 , \rho(s_7s_0+s_4s_5)=s_6s_2,\qquad\rho(s_2s_3+s_7s_0)=s_1s_5,\qquad s_1s_6-s_7s_0=\rho d^2, ρ ( s 7 s 0 + s 4 s 5 ) = s 6 s 2 , ρ ( s 2 s 3 + s 7 s 0 ) = s 1 s 5 , s 1 s 6 − s 7 s 0 = ρ d 2 , s 1 s 6 s 5 = s 7 s 2 s 3 + d 2 s 0 , s 1 s 6 s 2 = s 0 s 4 s 5 + d 2 s 7 s_1s_6s_5=s_7s_2s_3+d^2s_0,\qquad s_1s_6s_2=s_0s_4s_5+d^2s_7 s 1 s 6 s 5 = s 7 s 2 s 3 + d 2 s 0 , s 1 s 6 s 2 = s 0 s 4 s 5 + d 2 s 7 The last two trigonometric identities also follow from the preceding formulas for s 1 s 5 s_1s_5 s 1 s 5 s 6 s 2 s_6s_2 s 6 s 2
For (10), the sector with one strand reduces to multiplication of two matrices with diagonal entries A A A B B B
E ( x ) E ( − x ) = 1 , U ( x ) + E ( x ) U ( − x ) = 0 , E(x)E(-x)=1,\qquad U(x)+E(x)U(-x)=0, E ( x ) E ( − x ) = 1 , U ( x ) + E ( x ) U ( − x ) = 0 , E ( x ) F ( − x ) + F ( x ) E ( − x ) + U ( x ) U ( − x ) = 0. E(x)F(-x)+F(x)E(-x)+U(x)U(-x)=0. E ( x ) F ( − x ) + F ( x ) E ( − x ) + U ( x ) U ( − x ) = 0. A cap composed with a cup creates a loop and therefore vanishes. These identities again follow from (6).
To prove (7), first set y = s λ y=s\lambda y = s λ s = 2 , 3 s=2,3 s = 2 , 3 ( I ⊗ S s ) R ^ s b ( x ) ( S s ⊗ I ) (I\otimes S_s)\widehat{R}^{\,b}_s(x)(S^s\otimes I) ( I ⊗ S s ) R s b ( x ) ( S s ⊗ I ) x + y = s λ x+y=s\lambda x + y = s λ ( − x , x + y ) (-x,x+y) ( − x , x + y ) y = 0 y=0 y = 0 R ( 0 ) = I R(0)=I R ( 0 ) = I
Multiply the difference of the two sides by their three scalar denominators, and call the resulting diagram sum F ( x , y ) \mathcal{F}(x,y) F ( x , y ) e i y e^{iy} e i y − 4 -4 − 4 4 4 4 D j D_j D j − 1 -1 − 1 j j j
R i ( v + π ) = D i + 1 R i ( v ) D i . R_i(v+\pi)=D_{i+1}R_i(v)D_i. R i ( v + π ) = D i + 1 R i ( v ) D i . On either side of the braid relation the two internal D 2 D_2 D 2 π \pi π
F ( x , y + π ) = D 3 F ( x , y ) D 1 . \mathcal{F}(x,y+\pi)=D_3\mathcal{F}(x,y)D_1. F ( x , y + π ) = D 3 F ( x , y ) D 1 . Every boundary coefficient has fixed parity. After multiplication by a Laurent monomial it is a polynomial of degree at most four in e 2 i y e^{2iy} e 2 i y y = 0 , 2 λ , 3 λ , 2 λ − x , 3 λ − x y=0,2\lambda,3\lambda,2\lambda-x,3\lambda-x y = 0 , 2 λ , 3 λ , 2 λ − x , 3 λ − x π \pi π x x x
