Matrix group integrals, surfaces, and mapping class groups II: $\textrm{O}\left( n\right)$ and $\textrm{Sp}\left( n\right)$

Abstract

Let w be a word in the free group on r generators. The expected value of the trace of the word in r independent Haar elements of $\textrm{O}(n)$ gives a function ${\mathcal {T}}r_{w} ^{\textrm{O}}(n)$ of n. We show that ${\mathcal {T}}r_{w} ^{\textrm{O}}(n)$ has a convergent Laurent expansion at $n=\infty$ involving maps on surfaces and $L^{2}$-Euler characteristics of mapping class groups associated to these maps. This can be compared to known, by now classical, results for the GUE and GOE ensembles, and is similar to previous results concerning ${\textrm{U}}\left( n\right)$, yet with some surprising twists. A priori to our result, ${\mathcal {T}}r_{w} ^{\textrm{O}}(n)$ does not change if w is replaced with $\alpha (w)$ where $\alpha$ is an automorphism of the free group. One main feature of the Laurent expansion we obtain is that its coefficients respect this symmetry under $\textrm{Aut}({\textbf{F}} _{r})$. As corollaries of our main theorem, we obtain a quantitative estimate on the rate of decay of ${\mathcal {T}}r _{w}^{\textrm{O}}(n)$ as $n\rightarrow \infty$, we generalize a formula of Frobenius and Schur, and we obtain a universality result on random orthogonal matrices sampled according to words in free groups, generalizing a theorem of Diaconis and Shahshahani. Our results are obtained more generally for a tuple of words $w_{1},\ldots ,w_{\ell }$, leading to functions ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{O}}$. We also obtain all the analogous results for the compact symplectic groups $\textrm{Sp}(n)$ through a rather mysterious duality formula.

A Proof of Theorem 1.2: relationship between \(\textrm{O}\) and \(\textrm{Sp}\)

Here we prove Theorem 1.2. Throughout this appendix, fix n with $2n\ge N\left( w_{1},\ldots ,w_{\ell }\right)$, the latter defined in (3.2). For $i\in \left[ 2n\right]$ denote

$$\begin{aligned} \hat{i}{\mathop {=}\limits ^{{\textrm{def}}}}{\left\{ \begin{array}{ll} i+n &{} \textrm{if}~1\le i\le n,\\ i-n &{} \textrm{if}~n+1\le i\le 2n, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \xi \left( i\right) {\mathop {=}\limits ^{{\textrm{def}}}}\textrm{sign}\left( n+\frac{1}{2}-i\right) ={\left\{ \begin{array}{ll} 1 &{} \textrm{if}~1\le i\le n,\\ -1 &{} \textrm{if}~n+1\le i\le 2n. \end{array}\right. } \end{aligned}$$

Recall that we think of $\textrm{Sp}\left( n\right)$ as a subgroup of $\textrm{GL}_{2n}\left( {\textbf{C}}\right)$, and that the matrix J was defined in (1.2). The following lemma follows easily from (1.3) and the fact that $A^{-1}=J^{T}A^{T}J$ for $A\in \textrm{Sp}\left( n\right)$.

Lemma A.1

If $A\in \textrm{Sp}\left( n\right)$ and $i,j\in \left[ 2n\right]$, then

$$\begin{aligned} \left( A^{-1}\right) _{i,j}=\xi \left( i\right) \xi \left( j\right) A_{\hat{j},\hat{i}}. \end{aligned}$$

(A.1)

Our first goal is to obtain an analog of Theorem 3.4 for $\textrm{Sp}\left( n\right)$, namely, to obtain a formula for ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right)$ as a finite sum over systems of matchings, only with an additional sign associated with every such system; see Proposition A.5 for the precise statement.

We recall some of the notation we use here. Let $2L=2\sum _{x\in B}L_{x}=\sum _{j=1}^{\ell }\left| w_{j}\right|$ denote the total number of letters in $w_{1},\ldots ,w_{\ell }$. The jth boundary component of every surface in $\textsf{Surfaces} ^{*}\left( w_{1},\ldots ,w_{\ell }\right)$ is subdivided to $\left| w_{j}\right|$ intervals corresponding to the letters of $w_{j}$, and we denoted by $\mathcal {I},\mathcal {I}^{+},\mathcal {I}^{-}$ the sets of all 2L intervals, the subset of intervals corresponding to positive letters and its complement, respectively. Likewise, we denote by $\mathcal {I}_{x},\mathcal {I}_{x}^{+},\mathcal {I}_{x}^{-}$ the analogous sets of intervals corresponding to the instances of $x\in B$. We again identify $\mathcal {I}_{x}$ with the set $[2L_{x}]$, for each $x\in B$, in the same way as in Section 2.1.4. Similarly to the notation from Section 2.1.4, we denote by $\mathcal {A}=\mathcal {A}(w_{1},\ldots ,w_{\ell })$ the set of index assignments

$$\begin{aligned} \textbf{a}:\left\{ p_{I}\left( k\right) \,\big |\,I\in \mathcal {I},k\in \left\{ 0,1\right\} \right\} \rightarrow \left[ 2n\right] , \end{aligned}$$

where for every two immediately adjacent marked points p, q in $\cup _{j=1}^{\ell }C\left( w_{j}\right)$ that belong to different intervals in $\mathcal {I}$ we have $\textbf{a}\left( p\right) =\textbf{a}\left( q\right)$. (Note the range here is $\left[ 2n\right]$ and not $\left[ n\right]$ as in Section 2.1.4). Given $\textbf{a}\in \mathcal {A}$, let $\hat{\textbf{a}}$ be the assignment obtained after applying (A.1), namely,

$$\begin{aligned} \hat{\textbf{a}}(p_{I}(i))={\left\{ \begin{array}{ll} \textbf{a}(p_{I}(i)) &{} \text {if} I\in \mathcal {I}^{+}\\ \widehat{\textbf{a}(p_{I}(i))} &{} \text {if} I\in \mathcal {I}^{-}. \end{array}\right. } \end{aligned}$$

As we shall use Theorem 3.1 for evaluating ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right)$, we need the following expression which gathers the total sign contribution for a given system of matchings $\textbf{m}=\{(m_{x,0},m_{x,1})\}_{x\in B}\in \textrm{MATCH}^{\kappa \equiv 1}$ and an assignment $\textbf{a}$. Recall the notation $\delta _{\textbf{i},m}^{\textrm{Sp}}$ from Theorem 3.1. We let

$$\begin{aligned} \Delta \left( \textbf{a},\textbf{m}\right){} & {} {\mathop {=}\limits ^{{\textrm{def}}}} \left[ \prod _{I\in \mathcal {I}^{-}}\xi \left( \textbf{a}(p_{I}(0))\right) \xi \left( \textbf{a}(p_{I}(1))\right) \right] \cdot \prod _{\begin{array}{c} {\scriptscriptstyle x\in B}\\ k=0,1 \end{array} }\delta _{\hat{\textbf{a}}|_{\left\{ p_{I}\left( k\right) \,\big |\,I\in \mathcal {I}_{x}\right\} },m_{x,k}}^{\textrm{Sp}}\\{} & {} = \left[ \prod _{I\in \mathcal {I}^{-}}\xi \left( \textbf{a}(p_{I}(0))\right) \xi \left( \textbf{a}(p_{I}(1))\right) \right] \cdot \prod _{\begin{array}{c} {\scriptscriptstyle x\in B}\\ k=0,1 \end{array} }\prod _{\begin{array}{c} {\scriptstyle {\scriptstyle \left( p_{I}(k),p_{J}(k)\right) }}\\ {\scriptstyle \textrm{matched}~\textrm{by}~m_{x,k}} \end{array} }\left\langle e_{\hat{\textbf{a}}(p_{I}(k))},e_{\hat{\textbf{a}}(p_{J}(k))}\right\rangle _{\textrm{Sp}}, \end{aligned}$$

where in the innermost product, each matched pair appears once and is given its predetermined order. Note that for $i,j\in \left[ 2n\right]$, we have

$$\begin{aligned} \left\langle e_{i},e_{j}\right\rangle _{\textrm{Sp}}=e_{i}^{T}Je_{j}=\delta _{\hat{i},j}\xi \left( i\right) , \end{aligned}$$

(A.2)

where here $\delta$ is the Kronecker delta. Also notice that $\Delta \left( \textbf{a},\textbf{m}\right) \in \left\{ -1,0,1\right\}$. We say $\textbf{a}\vdash ^{*}\textbf{m}$ if $\Delta (\textbf{a},\textbf{m})\ne 0$. Therefore,

$$\begin{aligned} \Delta \left( \textbf{a},\textbf{m}\right) =\textbf{1}_{\textbf{a}\vdash ^{*}\textbf{m}}\cdot \left[ \prod _{I\in \mathcal {I}^{-}}\xi \left( \textbf{a}(p_{I}(0))\right) \xi \left( \textbf{a}(p_{I}(1))\right) \right] \cdot \prod _{\begin{array}{c} {\scriptstyle x\in B}\\ {\scriptstyle k=0,1} \end{array} }\prod _{\begin{array}{c} {\scriptstyle {\scriptstyle \left( p_{I}(k),p_{J}(k)\right) }}\\ {\scriptstyle \textrm{matched}~\textrm{by}~m_{x,k}} \end{array} }\xi \left( \hat{\textbf{a}}\left( p_{I}\left( k\right) \right) \right) . \end{aligned}$$

(A.3)

Definition A.2

Let $\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}$. Call a matching arc of $\textbf{m}$ orientable if it pairs an interval in $\mathcal {I}^{\pm }$ with an interval in $\mathcal {I}^{\mp }$, and non-orientable otherwise. Let m be one of the matchings in $\textbf{m}$. In every pair $\left( m_{(2t-1)},m_{(2t)}\right)$ we think of the corresponding matching arc in $\Sigma _{\textbf{m}}$ as directed from its origin – the interval corresponding to $m_{(2t-1)}$, to its terminus – the interval associated with $m_{(2t)}$. Let D be a type-o disc of $\Sigma _{\textbf{m}}$. Every interval in $\mathcal {I}$ that meets $\partial D$ has an orientation coming from the given orientation of $\partial \Sigma _{\textbf{m}}$. We say that two intervals that meet $\partial D$ are co-oriented (relative to D) if their orientation induces the same orientation on $\partial D$, and counter-oriented otherwise. Note that a matching arc is orientable if and only if it matches two co-oriented intervals meeting $\delta D$.

In the computation of $\Delta \left( \textbf{a},\textbf{m}\right)$, we attribute every sign that appears in (A.3) to one of the type-o discs of $\Sigma _{\textbf{m}}$. Indeed, every matching arc is at the boundary of exactly one type-o disc, and every $p_{I}(k)$ also belongs to exactly one type-o disc.

Lemma A.3

(Computation of $\Delta (\textbf{a},\textbf{m})$) Assume $\textbf{a},\textbf{a}_{1},\textbf{a}_{2}\in \mathcal {A}$ and $\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}$, and let D be a type-o disc in $\Sigma _{\textbf{m}}$.

• 1. The number of non-orientable matching arcs along $\partial D$ is even.

• 2. If $\textbf{a}\vdash ^{*}\textbf{m}$, the total sign contribution of D to $\Delta (\textbf{a},\textbf{m})$ is the product of:

  $\left( i\right)$ the sign of the index⁸ given by $\textbf{a}$ at the origin of every matching arc with origin in $\mathcal {I}^{+}$,

  $\left( ii\right)$ the sign of the index given by $\textbf{a}$ at the terminus of every matching arc with terminus in $\mathcal {I}^{-}$, and

  $\left( iii\right)$ $\left( -1\right)$ for every matching arc with origin in $\mathcal {I}^{-}$.

• 3. If $\textbf{a}\vdash ^{*}\textbf{m}$ and $p_{I}(k),p_{J}(k)$ are matched by any $m_{x,k}$ then $\textbf{a}(p_{I}(k))\equiv \textbf{a}(p_{J}(k))\bmod n$, and moreover, $\textbf{a}(p_{I}(k))=\textbf{a}(p_{J}(k))$ if and only if $m_{x,k}$ corresponds to an orientable matching arc.

• 4. For fixed $\textbf{m}$, the number of $\textbf{a}$ with $\textbf{a}\vdash ^{*}\textbf{m}$ is $\left( 2n\right) ^{\#\{\text {type}-o\,\text { discs of}\,{ \Sigma _{\textbf{m}}\}}}$.

• 5. If $\textbf{a}_{1},\textbf{a}_{2}\vdash ^{*}\textbf{m}$, then $\Delta (\textbf{a}_{1},\textbf{m})=\Delta (\textbf{a}_{2},\textbf{m})$.

The final statement allows us to define:

Definition A.4

For every $\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}$ we let $\Delta (\textbf{m}){\mathop {=}\limits ^{{\textrm{def}}}}\Delta (\textbf{a},\textbf{m})$, defined by any $\textbf{a}\vdash ^{*}\textbf{m}$.

Proof of Lemma A.3

We prove part by part.

Part 1. The first point is due to the fact that the boundary components of $\Sigma$ have built-in orientation, and along the boundary of D, the orientation of intervals meeting $\partial \Sigma$ is preserved when going along an orientable matching arc, and flipped along a non-orientable matching are. But $\partial D$ is a loop, so the number of orientation flips must be even.

Part 2. This follows from (A.3) by checking case by case over all possibilities.

Part 3. We have $\textbf{a}\vdash ^{*}\textbf{m}$ if and only if for all ordered matched pairs $p_{I}(k),p_{J}(k)$ of any $m_{x,k}$

$$\begin{aligned} \textbf{a}\left( p_{I}(k)\right) ={\left\{ \begin{array}{ll} \widehat{\textbf{a}\left( p_{J}(k)\right) } &{} \text {if}\, I \text {and} J \text {are in the same set} {\mathcal {I}^{\pm }},\\ \textbf{a}\left( p_{J}(k)\right) &{} \text {if} I \text {and} J \text {are in}\, \mathcal {I}^{\pm } \text {and} \mathcal {I}^{\mp }, \text {respectively.} \end{array}\right. } \end{aligned}$$

(A.4)

This means that when $\textbf{a}\in \mathcal {A}$ and $\textbf{a}\vdash ^{*}\textbf{m}$, there is a constraint on the values of $\textbf{a}$ at every pair of points that are adjacent on the boundary of some type-o disc of $\Sigma _{\textbf{m}}$. This is similar to the situation for the orthogonal group, but the constraints are more complicated now. The constraint implies that the values of $\textbf{a}$ on the points $p_{I}(k)$ in the boundary of a fixed type-o disc D of $\Sigma _{\textbf{m}}$ are determined by the value at any fixed point $p_{D}$ on the boundary of that disc. The values of $\textbf{a}$ are constant along segments of $\partial D$, except for segments that are matching arcs joining intervals in the same set $I_{x}^{\pm }$, across which the value of $\textbf{a}$ jumps by $n\bmod 2n$. These are the non-orientable matching arcs defined in Definition A.2.

Part 4. It now follows from Parts 1 and 3 that if for each type-o disc D of $\Sigma _{\textbf{m}}$, we choose $\textbf{a}(p)$ for some p in $\partial D$, then there exists a unique $\textbf{a}\in \mathcal {A}$ with these prescribed values and such that $\textbf{a}\vdash ^{*}\textbf{m}$. Hence, for any $\textbf{m}$, there are $(2n)^{\#\{\text {type}-o \text {discs of} {\Sigma _{\textbf{m}}\}}}$ elements of $\mathcal {A}$ with $\textbf{a}\vdash ^{*}\textbf{m}$.

Part 5. We need to show that for $\textbf{m}$ fixed, all the $\Delta (\textbf{a},\textbf{m})$ have the same sign. Indeed, we collect the contribution to the sign of every o-disc D separately, and show it does not depend on the particular assignment of indices along $\partial D$. There are two options for the signs of these indices, where one is a complete negation of the other. Recall that the sign of $\Delta (\textbf{a},\textbf{m})$ splits up into three types of contributions according to Part 2. The contribution from $\left( iii\right)$ clearly does not depend on $\textbf{a}$. Now consider the $\left( i\right)$- and $\left( ii\right)$-type contributions.

• If $\alpha$ is an orientable matching arc, its $\left( i\right)$- and $\left( ii\right)$-type contributions to $\Delta (\textbf{a}_{i},\textbf{m})$ are always 1 in total. This is surely the case if $\alpha$ is directed from $I\in I_{x}^{-}$ to $J\in I_{x}^{+}$. But it is also the case when $\alpha$ is directed the other way round, as the signs of both indices at its endpoints are identical. Hence the contributions of type (i) and type (ii) of orientable matching arcs to either $\Delta (\textbf{a}_{1},\textbf{m})$ or $\Delta (\textbf{a}_{2},\textbf{m})$ is equal to 1.

• Note from the discussion in the proof of Part 3, that $\textbf{a}_{1}$ and $\textbf{a}_{2}$ are related by a sequence of the following type of flip-moves: choose a type-o disc D of $\Sigma _{\textbf{m}}$, and modify $\textbf{a}_{1}$ by adding n to $\textbf{a}(p_{I}(k))$ modulo 2n, for every $p_{I}(k)$ that meets $\partial D$. Now for any given non-orientable matching arc $\alpha$, its $\left( i\right)$- and $\left( ii\right)$-type contribution is the sign of one of the endpoints. Hence the effect of a flip-move on $\textbf{a}_{1}$ at a disc D is to change the type (i) and (ii) contributions to $\Delta (\textbf{a}_{1},\textbf{m})$ by $(-1)^{\#\text {non-orientable matching arcs of}\, \textbf{m} \text {meeting}\, {D}}$. On the other hand, by Part 1, the total number of non-orientable matching arcs of $\textbf{m}$ meeting D is even.

This concludes the proof of Lemma A.3. $\square$

We can now prove the analog of Theorem 3.4 for ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right)$.

Proposition A.5

For $2n\ge N$

$$\begin{aligned} {\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right) =\sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\left( 2n\right) ^{\# \{\text {type-}o \text {discs of}\, {\Sigma _{\textbf{m}}\}}}\Delta \left( \textbf{m}\right) \prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{Sp}}(m_{x,0},m_{x,1};n), \end{aligned}$$

with $\Delta \left( \textbf{m}\right) \in \left\{ 1,-1\right\}$ as defined in Definition A.4.

Proof

Let g(I) be as in $\S$3.2. Assume $2n\ge N$. By the same arguments that led to (3.7), incorporating (A.1) and using Theorem 3.1, we have

$$\begin{aligned} {\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}(n)=\sum _{\textbf{a}\in \mathcal {A}(w_{1},\ldots ,w_{\ell })}\sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\Delta (\textbf{a},\textbf{m})\prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{Sp}}(m_{x,0},m_{x,1};n). \end{aligned}$$

This formula was the original motivation for introducing $\Delta (\textbf{a},\textbf{m})$. Now using Lemma A.3, Parts 4 and 5, and interchanging the sums over $\textbf{a}$ and $\textbf{m}$ gives

$$\begin{aligned} {\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}(n)&=\sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\left( \prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{Sp}}(m_{x,0},m_{x,1};n)\right) \left( \sum _{\textbf{a}\in \mathcal {A}(w)} {\Delta (\textbf{a},\textbf{m})}\right) \\&=\sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\left( 2n\right) ^{\#\{\text {type-}o \text {discs of} {\Sigma _{\textbf{m}}\}}}\Delta \left( \textbf{m}\right) \prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{Sp}}(m_{x,0},m_{x,1};n) \end{aligned}$$

as required. $\square$

We can now prove the main result of this subsection and show that ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right) =\left( -1\right) ^{\ell }\cdot {\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{O}}\left( -2n\right)$ for large n.

Proof of Theorem 1.2

It follows from Proposition A.5 and Lemma 3.2 that

$$\begin{aligned} {\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}\left( n\right)= & {} \sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\left( 2n\right) ^{\#\{o\text {-discs of}\, {\Sigma _{\textbf{m}}\}}}\Delta \left( \textbf{m}\right) \prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{Sp}}(m_{x,0},m_{x,1};n)\nonumber \\= & {} \sum _{\textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}\left( 2n\right) ^{\#\{o\text {-discs of}\, {\Sigma _{\textbf{m}}\}}}\nonumber \\{} & {} \cdot \prod _{x\in B}\left( -1\right) ^{L_{x}}\cdot \textrm{sign}\left( \sigma _{m_{x,0}}^{-1}\cdot \sigma _{m_{x,1}}\right) \cdot {\textrm{Wg}} _{L_{x}}^{\textrm{O}}(m_{x,0},m_{x,1};-2n)\cdot \Delta \left( \textbf{m}\right) \nonumber \\= & {} \sum _{{\scriptscriptstyle \textbf{m}\in \textrm{MATCH}^{\kappa \equiv 1}}}\left( -2n\right) ^{\#\{o\text {-discs of}\, {\Sigma _{\textbf{m}}\}}}\prod _{x\in B}{\textrm{Wg}} _{L_{x}}^{\textrm{O}}\left( m_{x,0},m_{x,1};-2n\right) \Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right) \nonumber \\ \end{aligned}$$

(A.5)

where

$$\begin{aligned} \Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right) {\mathop {=}\limits ^{{\textrm{def}}}}\left( -1\right) ^{\#\{ \text {type-}o \text {discs of}\, {\Sigma _{\textbf{m}}\}}}\cdot \left( -1\right) ^{L}\cdot \Delta \left( \textbf{m}\right) \cdot \prod _{x\in B}\textrm{sign}\left( \sigma _{m_{x,0}}^{-1}\sigma _{m_{x,1}}\right) . \end{aligned}$$

We now show that $\Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right)$ is independent of $\textbf{m}$ and equal to $\left( -1\right) ^{\ell }$. This will complete the proof by combining (A.5) with Theorem 3.4.

Our strategy for proving that $\Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right) \equiv \left( -1\right) ^{\ell }$ consists of three parts:

• 1. The fact there are $r=|B|$ different types of letters in $w_{1},\ldots ,w_{\ell }$ can be ignored, and all letters may be considered as identical.

• 2. If $\mathcal {I}=\mathcal {I}^{+}$, there is one particular set of matchings $\textbf{m}$ for which $\Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right) =\left( -1\right) ^{\ell }$.

• 3. The value of $\Xi \left( w_{1},\ldots ,w_{\ell };\textbf{m}\right)$ does not change if we make small local changes: $\left( a\right)$ flipping the direction of one matching arc, $\left( b\right)$ exchanging the termini of two matching arcs, or $\left( c\right)$ flipping the orientation of one of the letters in the word from positive to negative or vice versa.

During this proof we consider the matchings $m_{x,i}$ of $\textbf{m}$ as matchings of the letters of the words $w_{1},\ldots ,w_{\ell }$, this is possible since the letters are in one-to-one correspondence with the intervals $\mathcal {I}$.

Part I: Consider all letters as identical First, recall the definition from $\S$2.1.1 of the permutation $\sigma _{m}\in S_{2k}$ associated with the matching m belonging to $\textbf{m}$, and note that the order of the pairs in m does not affect the sign of $\sigma _{m}$, nor $\Delta \left( \textbf{m}\right)$, so we ignore it here. (In contrast, the order within each pair does affect these quantities.) As a result, we can treat all matchings $\left\{ m_{x,0}\right\} _{x\in B}$ as a single matching $m_{0}\in M_{L}$ of the whole collection of intervals $\mathcal {I}$, where we keep track of the order within each pair, namely, of which endpoint is the origin and which the terminus of every matching arc. Similarly, we replace $\left\{ m_{x,1}\right\} _{x\in B}$ with a single matching $m_{1}\in M_{L}$ of $\mathcal {I}$. The corresponding permutations $\sigma _{m_{0}}$ and $\sigma _{m_{1}}$ lie in $S_{2L}$. From every pair of matchings $m_{0},m_{1}\in M_{L}$, we can construct a corresponding surface $\Sigma _{m_{0},m_{1}}$ as in $\S$2.1. We define $\Delta \left( m_{0},m_{1}\right)$ accordingly. It is thus enough to show that for every $m_{0},m_{1}\in M_{L}$,

$$\begin{aligned}{} & {} \Xi \left( w_{1},\ldots ,w_{\ell };m_{0},m_{1}\right) {\mathop {=}\limits ^{{\textrm{def}}}}\left( -1\right) ^{\#\{ \text {type-}o \text {discs of}\,{ \Sigma _{m_{0},m_{1}}\}}}\cdot \left( -1\right) ^{L}\cdot \nonumber \\ {}{} & {} \Delta \left( m_{0},m_{1}\right) \cdot \textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right) =\left( -1\right) ^{\ell }. \end{aligned}$$

(A.6)

Part II: Particular matchings $m_{0},m_{1}$ Next, we show that the equality (A.6) holds for a particular pair $m_{0},m_{1}\in M_{L}$, when all letters in $w_{1},\ldots ,w_{\ell }$ are positive, namely, when $\mathcal {I}=\mathcal {I}^{+}$. The pair will satisfy $m_{0}=m_{1}$, and so $\textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right) =1$. We partition the words $w_{1},\ldots ,w_{\ell }$ into singletons of even-length words and pairs of odd-length words. The matchings $m_{0}$ and $m_{1}$ will only pair letters of words in the same block of this partition. It is enough to prove (A.6) for every connected component of $\Sigma _{m_{0},m_{1}}$ separately.

First consider the case of a single, even-length word w (which, by abuse of notation, has length 2L). Let each of $m_{0},m_{1}$ pair the first interval to the second, the third to the fourth, and so forth. It is easy to check that in this case, there is exactly one type-o disc, with 2L non-orientable matching arcs at its boundary, all directed, say, clockwise. In every compatible assignment of indices $\textbf{a}\vdash ^{*}\left( m_{0},m_{1}\right)$, the sign $\xi$ flips along every matching arc, and as all letters are positive, exactly half of the matching arcs contribute $\left( -1\right)$ (see Lemma A.3, Part 2), so $\Delta \left( m_{0},m_{1}\right) =\left( -1\right) ^{L}$ in this case. Hence the left hand side of (A.6) is $\left( -1\right) ^{1}\cdot \left( -1\right) ^{L}\cdot \left( -1\right) ^{L}\cdot 1=\left( -1\right)$, which is the desired outcome as $\ell =1$. See the left hand side of Fig. 4.

Fig. 4

On the left hand side, there is one word of even length (6 in this example) with all letters positive, and $m_{0}=m_{1}$ match $I_{1}\rightarrow I_{2}$, $I_{3}\rightarrow I_{4}$ and $I_{5}\rightarrow I_{6}$. An easy computation gives that $\Xi \left( w;m_{0},m_{1}\right) =-1$ in this case. On the right hand side, there are two words of odd length each (7 and 5 in this example) with all letters positive. Here, $m_{0}=m_{1}$ match $I_{1}\rightarrow I_{2}$, $I_{3}\rightarrow I_{4}$, $I_{5}\rightarrow I_{6}$, $I'_{1}\rightarrow I'_{2}$, $I'_{3}\rightarrow I'_{4}$ and $I_{7}\rightarrow I'_{5}$. An easy analysis gives that $\Xi \left( w_{1},w_{2};m_{0},m_{1}\right) =1$ in this case

Second, consider the case of a pair of odd-length words $w_{1},w_{2}$, of total length 2L. Let each of $m_{0},m_{1}$ pair the first interval of each word with the second one, the third with the fourth and so on, and pair the last interval of $C\left( w_{1}\right)$ with the last interval of $C\left( w_{2}\right)$. Again, it is easy to verify there is a single type-o disc in $\Sigma _{m_{0},m_{1}}$, with 2L non-orientable matching arcs at its boundary. At the boundary of the type-o disc there are $\left| w_{1}\right|$ successive o-points of $C\left( w_{1}\right)$, and then $\left| w_{2}\right|$ o-points of $C\left( w_{2}\right)$, where the matching arcs separating these two sequences are the two matchings arcs connecting the last interval of $w_{1}$ with the last interval of $w_{2}$. In every compatible assignment $\textbf{a}\vdash ^{*}\left( m_{0},m_{1}\right)$, the signs $\xi$ alternate, and so every pair of matching arcs connecting the same two intervals of the same word contributes $\left( -1\right)$ to $\Delta \left( m_{0},m_{1}\right)$. However, both matching arcs connecting the last intervals have the same sign at their origins, and so their contribution is 1. This shows that $\Delta \left( m_{0},m_{1}\right) =\left( -1\right) ^{L-1}$ in this case. Hence the left hand side of (A.6) is $\left( -1\right) ^{1}\cdot \left( -1\right) ^{L}\cdot \left( -1\right) ^{L-1}\cdot 1=1$, which is the desired outcome as $\ell =2$. See the right hand side of Fig. 4.

Part III: $\Xi$ is invariant under local modifications Finally, we show that the three local modifications we specified above do not alter the value of $\Xi \left( w_{1},\ldots ,w_{\ell };m_{0},m_{1}\right)$. As applying suitable steps of all three types leads from the instance described in part II of this proof to any given pair of matchings $m_{0},m_{1}$ and to any orientation of the 2L letters (positive/negative), this will complete the proof. Note that none of these changes affect the total number of letters, L, so we ought to show that they do not alter the product $\left( -1\right) ^{\#\{\text {type-}o \text {discs of}\, {\Sigma _{m_{0},m_{1}}\}}}\cdot \Delta \left( m_{0},m_{1}\right) \cdot \textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right)$.

We begin with flipping the direction of one matching arc. Obviously, this does not change the number of type-o discs. It does change the sign of one of $\sigma _{m_{0}}$ or $\sigma _{m_{1}}$, and therefore the sign of $\sigma _{m_{0}}^{-1}\sigma _{m_{1}}$, but it also changes the contribution of this matching arc to $\Delta \left( m_{0},m_{1}\right)$: this follows from a simple case-by-case analysis of whether the origin of the matching arc is in $\mathcal {I}^{+}$ or in $\mathcal {I}^{-}$, and likewise the terminus of the arc. The analysis is based on Lemma A.3 and (A.4).

Next, consider a switch between the termini of two matching arcs $\alpha _{1}$ and $\alpha _{2}$ of, say, $m_{0}$. This switch changes the sign of $\sigma _{m_{0}}$ and therefore of $\sigma _{m_{0}}^{-1}\sigma _{m_{1}}$. We distinguish between three cases and show that in each one of them, there is one more sign change that cancels with the change in $\textrm{sign}(\sigma _{m_{0}}^{-1}\sigma _{m_{1}})$:

• Assume that $\alpha _{1}$ and $\alpha _{2}$ both belong to the same type-o disc D and are directed along the same orientation of $\partial D$. Then switching the termini splits D into two discs, and so the sign of $\left( -1\right) ^{\#\{\text {type}-o \text {discs of}\, {\Sigma _{m_{0},m_{1}}\}}}$ flips. Any compatible assignment $\textbf{a}$ before the switch remains compatible after it, and the combined sign contribution of the two arcs (as in Lemma A.3, Part 2) remains unchanged. See Figure 5

• Assume that $\alpha _{1}$ and $\alpha _{2}$ both belong to the same type-o disc D and are directed along different orientations of $\partial D$. Of the two components of $\partial D\setminus \left( \alpha _{1}\cup \alpha _{2}\right)$, one, denoted $C_{o}$, has the origins of $\alpha _{1}$ and $\alpha _{2}$ as endpoints, and the other, denoted $C_{t}$, has the two termini as endpoints. Switching the termini corresponds to reflecting $C_{t}$ – see Figure 6. By the definition of compatible assignments, every piece of $\partial D\cap \partial \Sigma$ is assigned a well-defined index in $\left[ 2n\right]$, and by Lemma A.3, Part 3, two different pieces of $\partial D\cap \partial \Sigma$ are assigned the same index if and only if the corresponding orientations induced by $\partial \Sigma$ induce, in turn, the same orientation on $\partial D$. This means that if we preserve the assignment along $C_{o}$, the signs along $C_{t}$ must be flipped. The number of type-o discs is preserved. In the terminology of Lemma A.3, type-$\left( iii\right)$ contributions to $\Delta \left( m_{0},m_{1}\right)$ do not change. The sign contributions of the matching arcs along $C_{o}$ do not change. Also, the contribution of orientable arcs along $C_{t}$ does not change, nor does the type-$\left( i\right)$ contribution of $\alpha _{1}$ and $\alpha _{2}$. However, $\Delta \left( m_{0},m_{1}\right)$ does flip. To see this, denote by $\partial _{1},\partial _{2}$ the two connected components of $\partial D\cap \partial \Sigma$ which contain (as endpoints) the two termini $t\left( \alpha _{1}\right)$ and $t\left( \alpha _{2}\right)$, respectively ($\partial _{1}$ and $\partial _{2}$ may be equal). Let $i_{1}$ and $i_{2}$ be the indices corresponding to $\partial _{1}$ and $\partial _{2}$ in some compatible assignment (before the flip of $\alpha _{1}$ and $\alpha _{2}$).

  If the orientation of $\partial \Sigma$ along $\partial _{1}$ and $\partial _{2}$ induces the same orientation on $\partial D$, then $i_{1}=i_{2}$ and of the two intervals at the termini of $\alpha _{1}$ and $\alpha _{2}$, one is in $\mathcal {I}^{+}$ and the other in $\mathcal {I}^{-}$. Thus the total type-$\left( ii\right)$ contribution of $\alpha _{1}$ and $\alpha _{2}$ flips. As $C_{t}$ contains an even number of non-orientable matching arcs in this case, the total sign contribution of the non-orientable arcs along $C_{t}$ is preserved (as in the proof of Lemma A.3, Part 5).

  If the orientation of $\partial \Sigma$ along $\partial _{1}$ and $\partial _{2}$ induces different orientations on $\partial D$, then $i_{2}=\widehat{i_{1}}$ and the two letters at the termini are both positive or both negative. In this case, the total type-$\left( ii\right)$ contribution of $\alpha _{1}$ and $\alpha _{2}$ is unchanged, the total type-$\left( i\right)$ and type-$\left( ii\right)$ contribution of every orientable arc alongs $C_{t}$ is unchanged, but the same contribution of every non-orientable arc along $C_{t}$ is flipped, and the total number of non-orientable arcs along $C_{t}$ is odd (by our assumption about $\partial _{1}$ and $\partial _{2}$).

• The third and last case is the one where $\alpha _{1}$ and $\alpha _{2}$ belong to different type-o discs. Switching their termini then leads to merging the two discs into one. In the united type-o disc, the two arcs are “co-oriented”, so this case is the reverse of the first one, and $\Delta \left( m_{0},m_{1}\right)$ remains unchanged.

Fig. 5

The left part depicts a type-o disc D with two co-directed matching arcs $\alpha _{1}$ and $\alpha _{2}$ at its boundary. Switching their termini results in splitting D into two separate type-o discs: $D_{1}$ and $D_{2}$, as in the right hand side. This move flips both $\textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right)$ and $\left( -1\right) ^{\#\{\text {type-}o\,\text {discs of}\, {\Sigma _{m_{0},m_{1}}\}}}$, but leaves $\Delta \left( m_{0},m_{1}\right)$, and therefore also $\Xi \left( w_{1},\ldots ,w_{\ell };m_{0},m_{1}\right)$, unchanged

Fig. 6

The left part depicts a type-o disc D with two counter-directed matching arcs $\alpha _{1}$ and $\alpha _{2}$ at its boundary. The two connected components of $D\setminus \left( \alpha _{1}\cup \alpha _{2}\right)$ are denoted $C_{o}$ and $C_{t}$. Switching the termini of $\alpha _{1}$ and $\alpha _{2}$ results in reflecting $C_{t}$, as in the right hand side. This move flips both $\textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right)$ and $\Delta \left( m_{0},m_{1}\right)$, but leaves $\left( -1\right) ^{\#\{\text {type-}o\,\, \text {discs of}\, {\Sigma _{m_{0},m_{1}}\}}}$, and therefore also $\Xi \left( w_{1},\ldots ,w_{\ell };m_{0},m_{1}\right)$, unchanged

The final small change we consider is that of flipping some letter from being positive to negative, namely, of flipping an interval in some $C\left( w_{j}\right)$ from $\mathcal {I}^{\pm }$ to $\mathcal {I}^{\mp }$. Here, $\textrm{sign}\left( \sigma _{m_{0}}^{-1}\sigma _{m_{1}}\right)$ is unchanged. By the first local modification in this part of the proof, we may assume without loss of generality that this letter is at the termini of two matching arcs, $\alpha _{1}$ and $\alpha _{2}$. A similar argument as in the previous paragraph would show that:

• Assume that $\alpha _{1}$ and $\alpha _{2}$ belong to the same type-o disc D with the same orientation. The flip of the letter then splits D into two type-o discs. Denote by $\partial _{1}$ and $\partial _{2}$ the pieces of $\partial D\cap \partial \Sigma$ at the termini of $\alpha _{1}$ and $\alpha _{2}$. They must be counter-oriented. We may preserve the same assignment of indices as before the flip of the letter, but then the type-$\left( ii\right)$ contribution of both arcs flips when the letter is flipped. No other change in sign contributions occurs.

• Assume that $\alpha _{1}$ and $\alpha _{2}$ belong to the same type-o disc D with opposite orientations. The flip of the letter preserves the number of type-o discs and corresponds to reflecting $C_{t}$. Here $\partial _{1}$ and $\partial _{2}$ are co-oriented and the signs along $C_{t}$ must be flipped. There is no change to $\Delta \left( m_{0},m_{1}\right)$: the total type-$\left( ii\right)$ contribution of $\alpha _{1}$ and $\alpha _{2}$ is 1 before and after the flip, and the number of non-orientable arcs along $C_{t}$ is even.

• If $\alpha _{1}$ and $\alpha _{2}$ belong to different type-o disc, the flip is the reverse of the first case.

This completes the proof of Theorem 1.2. $\square$

We now have the analog of Corollary 3.5 for $G=\textrm{Sp}$: ‘

Corollary A.6

There is a rational function $\overline{{\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}}\in {{\textbf{Q}}}(n)$ such that for $2n\ge \max \{L_{x}\,:\,x\in B\}$, ${\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}(n)$ is given by evaluating $\overline{{\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}}$ at n.

Proof

Theorem 1.2 shows that we can obtain $\overline{{\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{Sp}}}$ by switching n with $-2n$ in $\overline{{\mathcal {T}}r_{w_{1},\ldots ,w_{\ell }} ^{\textrm{O}}}$ (the rational function from Corollary 3.5) and multiplying by $(-1)^{\ell }$. $\square$