Complete Solution of the LSZ Model via Topological Recursion

Abstract

We prove that the Langmann–Szabo–Zarembo (LSZ) model with quartic potential, a toy model for a quantum field theory on noncommutative spaces grasped as a complex matrix model, obeys topological recursion of Chekhov, Eynard and Orantin. By introducing two families of correlation functions, one corresponding to the meromorphic differentials $\omega _{g,n}$ of topological recursion, we obtain Dyson–Schwinger equations that eventually lead to the abstract loop equations being, together with their pole structure, the necessary condition for topological recursion. This strategy to show the exact solvability of the LSZ model establishes another approach towards the exceptional property of integrability in some quantum field theories. We compare differences in the loop equations for the LSZ model (with complex fields) and the Grosse–Wulkenhaar model (with hermitian fields) and their consequences for the resulting particular type of topological recursion that governs the models.

Appendices

Appendix A. Solution of the 2-Point Function

Throughout the appendix, we will need the Lagrange interpolation formula that we want recall:

Lemma A.1

Let f be a polynomial of degree $d\ge 0$ and $x_1,\ldots ,x_{d+1}$ be pairwise distinct complex numbers. Then, for all $x\in \mathbb {C}$,

$$\begin{aligned} f(x)=L(x)\sum _{j=1}^{d+1}\frac{f_j}{(x-x_j)L'(x_j)},\qquad \text {where } L(x)=\prod _{j=1}^d(x-x_j) \text { and } f_j=f(x_j). \end{aligned}$$

This section is dedicated to the proof of the spectral curve, encoded in the solution of the analytically continued 2-point function.

1.1 Proof of Theorem 4.3

Most methods build on the solution strategy in the QKM, [SW19].⁶ We start with the complexified Dyson–Schwinger equation for the 2-point function. We assume that there are undetermined variable transforms x(z) and y(z) that can be uniquely specified by the upper ansatz. After transformation, we have for Eq. (3.1), $g=0$:

$$\begin{aligned}&\bigg (x(z)+y(w)+ \frac{\lambda }{N} \sum _{k=1}^{{\tilde{d}}} \tilde{r}_k \mathcal {G}^{(0)}(z,\tilde{\varepsilon }_k) \bigg )\mathcal {G}^{(0)}(z,w) \end{aligned}$$

(6.9)

$$\begin{aligned}&= 1+ \frac{\lambda }{N} \sum _{k=1}^d r_k\frac{\mathcal {G}^{(0)}(\varepsilon _k,w)-\mathcal {G}^{(0)}(z,w)}{x(\varepsilon _k)-x(z)}. \end{aligned}$$

(A.1)

The ansatz of Theorem 4.3 turns this equation into

$$\begin{aligned} (y(w) -y(z))\mathcal {G}^{(0)}(z,w) = 1+ \frac{\lambda }{N} \sum _{k=1}^d r_k \frac{\mathcal {G}^{(0)}(\varepsilon _k,w)}{x(\varepsilon _k)-x(z)}. \end{aligned}$$

(A.2)

We set $z= \hat{\tilde{w}}^l$ to get rid off the prefactor on the lhs under the finiteness assumption from Theorem 4.3 of $\mathcal {G}^{(0)}(z= \hat{\tilde{w}}^l,w)$. This leads to d equations since y was assumed to be of rational degree d

$$\begin{aligned} \frac{\lambda }{N} \sum _{k=1}^d r_k \frac{\mathcal {G}^{(0)}(\varepsilon _k,w)}{x(\hat{\tilde{w}}^l)-x(\varepsilon _k)}=1. \end{aligned}$$

Define the fundamental Lagrange interpolation polynomials $A(v) = \prod _{j=1}^d (v-x(\hat{\tilde{w}}^j))$ and $B(v) =\prod _{j=1}^d (v-x(\varepsilon _j))$. Then we can read off

$$\begin{aligned} \frac{\lambda r_k}{N} \mathcal {G}^{(0)}(\varepsilon _k,w) =- \frac{\prod _{j=1}^d (x(\varepsilon _k)-x(\hat{\tilde{w}}^j)}{\prod _{j \ne k}^d (x(\varepsilon _k)-x(\varepsilon _j))} \end{aligned}$$

(A.3)

since the following sum over all residues at $x(\varepsilon _j)$ for the poles in B(v) reads 1:

$$\begin{aligned} \sum _{j=1}^d {{\,\textrm{Res}\,}}_{v \rightarrow x(\varepsilon _k)} \frac{A(v)}{(x(\hat{\tilde{w}}^l)-v)B(v)} = \sum _{j=1}^d \frac{A(x(\varepsilon _k))}{(x(\hat{\tilde{w}}^l)-x(\varepsilon _j))B'(x(\varepsilon _k))}=1. \end{aligned}$$

With the same trick, we allow for $d+1$ factors in the interpolation formula and include x(z). This gives the possibility to write

$$\begin{aligned} \frac{\prod _j (x(z)-x(\hat{\tilde{w}}^j))}{\prod _j (x(z)-x(\varepsilon _j))} + \sum _{k=1}^d \frac{\prod _j(x(\varepsilon _k)-x(\hat{\tilde{w}}^j))}{(x(\varepsilon _k)-x(z))\prod _{j \ne k}(x(\varepsilon _k)-x(\varepsilon _j))}=1. \end{aligned}$$

Inserting Eq. (A.4) into Eq. (A.3) gives

$$\begin{aligned} \mathcal {G}^{(0)}(z,w) = \frac{1}{y(w)-y(z)} \prod _{j=1}^{d} \frac{x(z)-x\big (\widehat{\tilde{w}}^j\big )}{x(z)-x(\varepsilon _j)} \end{aligned}$$

(A.4)

and thus, we principally finished the proof of the first representation. However, it is still unclear why y(z) is necessarily given by $y(z)=-z+\frac{\lambda }{N}\sum _{n=1}^{d}\frac{r_n}{x'(\varepsilon _n)(z-\varepsilon _n)}+C$ where the constant C must vanish in the end, and the same for x(z). This can be shown using Liouville’s theorem. Inserting $z=\varepsilon _k$ into Eq. (A.5) and comparing with Eq. (A.4), y(z) must have d poles at $z= \varepsilon _k$. Moreover, we have a simple pole at $z=\infty$. Remembering the degree $d+1$ of the assumption, we already have the complete set of poles. Thus, $-z+\frac{\lambda }{N}\sum _{n=1}^{d}\frac{r_n}{x'(\varepsilon _n)(z-\varepsilon _n)}+C-y(z)$ is a bounded holomorphic function and by Liouville a constant. Analogously to [SW19], one can show with some technical effort, that the initial ansatz guarantees that $C=0$.

Now we exchange the role of x and y, where we have to take the DSE of Remark 3.2. Carrying out the analytic continuation of Definition 4.4, we get a second complexified DSE for the 2-point function

$$\begin{aligned} (x(z) -x(w))\mathcal {G}^{(0)}(z,w) = 1+ \frac{\lambda }{N} \sum _{k=1}^{\tilde{d}} \tilde{r}_k \frac{\mathcal {G}^{(0)}(z,\tilde{\varepsilon }_k)}{y(\tilde{\varepsilon }_k)-y(w)}. \end{aligned}$$

(A.5)

Applying exactly the same steps yields the second representation, where x(z) is necessarily given by $z-\frac{\lambda }{N}\sum _{k=1}^{\tilde{d}}\frac{\tilde{r}_k}{y'(\tilde{\varepsilon }_k)(z-\tilde{\varepsilon }_k)}$. $\square$

We remark that the form of the spectral cuarve is not surprising—it is the typical interwoven structure between x and y—as implicitly defined equations—that usually appears when solving a matrix model with external field (for instance the matrix model for simple Hurwitz numbers [BEMS11] or the Kontsevich model [EO07, Sec. 10.4]).

Appendix B. Proof of $\Omega ^{(0)}_2$

It remains the proof for the second part of the initial data, namely the Bergman kernel. By rational parametrisation of $\omega _{0,1}$, the spectral curve is a planar algebraic curve and thus, the shape of the fundamental form of the second kind is automatically determined, which we will show to coincide with $\Omega _{0,2}$.

Proposition B.1

The cylinder amplitude of the LSZ model reads

$$\begin{aligned} \Omega _{0,2}(z_1,z_2)=\frac{1}{x'(z_1)x'(z_2)(z_1-z_2)^2}. \end{aligned}$$

(A.6)

The proof is split into two lemmata that follow from techniques of [BHW22]. Taking Corollary 4.5 for $(g,n)=(0,2)$, setting $v=y(z)$ (the first term vanishes since H is rational in v) and inserting the DSE of the 2-point function with exchanged x and y (Eq. A.6), we deduce that $\Omega ^{(0)}_{2}$ satisfies the DSE

$$\begin{aligned}&x'(z)\mathfrak {G}_0(z)\Omega ^{(0)}_{2}(u,z) -\frac{\lambda }{N^2}\sum _{k,n=1}^{d,\tilde{d}} \frac{ r_k \tilde{r}_n \mathcal {T}^{(0)}(u\Vert \varepsilon _k,\tilde{\varepsilon }_n|)}{ (x(\varepsilon _k)-x(z))(y(\tilde{\varepsilon }_n)-y(z))}\\ =&-\frac{\partial }{\partial x(u)} \mathcal {G}^{(0)}(u,z) \;.\nonumber \end{aligned}$$

(B.1)

where $\mathfrak {G}_0(z)=-H_{0,1}(y(z);z)$ and thus $\mathfrak {G}_0(z)={{\,\textrm{Res}\,}}_{u\rightarrow z} \mathcal {G}^{(0)}(z,u)du$ by Theorem 4.3. To specify the pole structure of the 2-point function, we show the following decomposition:

Lemma B.2

The planar 2-point function can be rewritten in the following decomposition of its poles:

$$\begin{aligned}&\mathcal {G}^{(0)}(z,u) = \frac{1}{u-z} \bigg (1+ \frac{\lambda ^2}{N^2} \sum _{k,n=1}^d \sum _{l,m=1}^{\tilde{d}} \frac{C_{k,l}^{m,n}}{\Big (z-\widehat{\varepsilon }_k^{\,m}\Big )\Big (u-\widehat{\tilde{\varepsilon _l}}^n\Big )} \biggl ) \quad , \end{aligned}$$

(B.2)

where

$$\begin{aligned}&C_{k,l}^{m,n} = \frac{\Big (\widehat{\tilde{\varepsilon _l}}^n - \widehat{\varepsilon }_k^{\,m}\Big ) r_k \tilde{r}_l \mathcal {G}^{(0)}(\varepsilon _k,\tilde{\varepsilon }_l)}{x'\big (\widehat{\varepsilon }_k^{\,m} \big ) y' \big ( \widehat{\tilde{\varepsilon _l}}^n\Big ) \Big ( y\big (\widehat{\varepsilon }_k^{\,m} \big ) - y (\tilde{\varepsilon }_l) \Big ) \Big ( x\big (\widehat{\tilde{\varepsilon _l}}^n \big ) - x( \varepsilon _k) \Big ) }. \end{aligned}$$

Proof

To figure out all possible poles, we have to look at the two DSE’s (A.3) and (A.6). After dividing by $y(w)-y(z)$ (or $x(z)-x(w)$) and the regularity assumption of Theorem 4.3, possible poles are located at $z=u$ as well as at $z=\hat{\varepsilon }_k^m$ and $u=\widehat{\tilde{\varepsilon _l}}^n$. Then, we look at the function $(u-z) \mathcal {G}^{(0)}(u,z)$, which approaches 1 for $z,u \rightarrow \infty$. Again, the basic relations Eqs. (A.3) and (A.6) give the chance to directly read off the residues at $z=\widehat{\varepsilon _k}^m$ and $u=\widehat{\tilde{\varepsilon _l}}^n$ as stated. For the regularity discussion at $u={\hat{z}}^k$ and $u=\widehat{\tilde{z}^k}$, one can look at the finite limit of the product representation of $\mathcal {G}(z,u)$ given in Theorem 4.3. $\square$

Liouville’s theorem now proves:

Lemma B.3

Assume that (for generic u) the function $\Omega ^{(0)}_2(u,z)$ is regular at any zero z of $\mathfrak {G}_0(z)$. Then

$$\begin{aligned} x'(z)\Omega _{0,2}(z,u)-\frac{1}{x'(u)(z-u)^2}=0. \end{aligned}$$

(B.3)

Proof

We deduce the representation for $\mathfrak {G}_0(z)={{\,\textrm{Res}\,}}_{u\rightarrow z} \mathcal {G}^{(0)}(z,u)du$:

$$\begin{aligned}&\mathfrak {G}_0(z)= 1- \frac{\lambda ^2}{N^2} \sum _{k,n}^d \sum _{l,m}^{\tilde{d}} \frac{C_{k,l}^{m,n}}{\Big (z-\widehat{\varepsilon }_k^{\,m}\Big )\Big (z-\widehat{\tilde{\varepsilon _l}}^n\Big )} \end{aligned}$$

and therefore the partical fraction decomposition

$$\begin{aligned}&\mathcal {G}^{(0)}(u,z) = \frac{\mathfrak {G}_0(z)}{u-z} + \frac{\lambda ^2}{N^2} \sum _{k,n}^d \sum _{l,m}^{\tilde{d}} \frac{C_{k,l}^{m,n}}{\Big (z-\widehat{\varepsilon }_k^{\,m}\Big )\Big ( z-\widehat{\tilde{\varepsilon _l}}^n\Big )\Big (u-\widehat{\tilde{\varepsilon _l}}^n\Big )}. \end{aligned}$$

Comparing with Eq. (B.2), the right hand side of Eq. (B.4) reads with the upper partial fraction decomposition

$$\begin{aligned}&\frac{1}{\mathfrak {G}_0(z)}\bigg [ \frac{\lambda }{N^2}\sum _{k=1}^d \sum _{n=1}^{\tilde{d}} r_k \tilde{r}_n \frac{\mathcal {T}^{(0)}(u\Vert \varepsilon _k,\tilde{\varepsilon }_n|)}{ (x(\varepsilon _k)-x(z))(y(\tilde{\varepsilon }_n)-y(z))} \\&\quad + \frac{\lambda ^2}{N^2 x'(u)} \sum _{k,n}^d \sum _{l,m}^{\tilde{d}} \frac{C_{k,l}^{m,n}}{\Big (z-\widehat{\varepsilon }_k^{\,m}\Big )\Big (z-\widehat{\tilde{\varepsilon _l}}^n\Big )\Big (u-\widehat{\tilde{\varepsilon _l}}^n\Big )^2} \bigg ]. \end{aligned}$$

The simple poles at $z= \widehat{\varepsilon }_k^{\,m},\widehat{\tilde{\varepsilon _l}}^n$ cancel due to the prefactor $\frac{1}{\mathfrak {G}_0(z)}$. By assumption, we can exclude further poles on the rhs at any zero z of $\mathfrak {G}_0(z)$. Thus, the lhs and rhs of Eq. (B.4) must be a constant, and sending $z \rightarrow \infty$, the constant has to be zero as claimed. $\square$

Appendix C. Proofs of Sect. 5

Section 5 is completely free of any proofs. This has two reasons: First, the proofs are very technical without greater learning effects. Second, the proofs are carried out in quite similar manner to established results for hermitian fields. Proposition 5.1 and Theorem 5.8 show up in a slightly more general form for hermitian fields in [Hoc20]. The other two results are generalisations of results found in [BHW22, Chapter 4].

Proof of Proposition 5.1

Define the differential operator

$$\begin{aligned} {\hat{D}} = \frac{\partial ^{2b-2}}{\partial J_{p^2q^2}\partial J^\dagger _{q^2p^2}\ldots \partial J_{p^bq^b}\partial J^\dagger _{q^bp^b}}. \end{aligned}$$

The same steps as for Proposition 3.1, namely using Proposition 2.8, give:

$$\begin{aligned} G_{|pq|\mathcal {J}}&=N^{b-2} {\hat{D}} \frac{\partial ^2}{\partial J_{pq} \partial J^\dagger _{qp}} \log [\mathcal {Z}(J,J^\dagger )]_{J,J^\dagger =0}\\&=- \frac{\lambda N^{b-4} {\hat{D}}}{E_p+\tilde{E}_q} \frac{\partial }{\partial J_{qp}^\dagger } \frac{1}{ \mathcal {Z}(J,J^\dagger )}\sum _{m,n=1}^N\frac{\partial ^3}{\partial J_{pm}\partial J_{mn}^\dagger \partial J_{nq}} \mathcal {Z}(J,J^\dagger )_{J,J^\dagger =0}\\&= - \frac{\lambda N^{b-4} {\hat{D}}}{E_p+\tilde{E}_q} \frac{\partial }{\partial J_{qp}^\dagger } \frac{1}{\mathcal {Z}(J,J^\dagger )} \biggl [ \frac{\partial }{\partial J_{pq} } W_p[J,J^\dagger ] \mathcal {Z}(J,J^\dagger )\\&\quad + \sum _{m,n=1}^N \frac{N}{E_n-E_p} \frac{\partial }{\partial J_{nq}} \bigg ( J^\dagger _{mp} \frac{\partial }{\partial J^\dagger _{mn}}-J_{nm} \frac{\partial }{\partial J_{pm}} \bigg ) \mathcal {Z}(J,J^\dagger )\bigg ] _{J,J^\dagger =0}. \end{aligned}$$

Note that we generalised the steps in Example 2.6 to the general operator ${\hat{D}}$. For the second line, write

$$\begin{aligned} \frac{1}{\mathcal {Z}(J,J^\dagger )} \frac{\partial }{\partial J_{pq} } \big ( W_p[J,J^\dagger ] \mathcal {Z}(J,J^\dagger ) \big ) = \frac{\partial }{\partial J_{pq} } W_p[J,J^\dagger ] + W_p[J,J^\dagger ] \frac{\partial }{\partial J_{pq} } \log [\mathcal {Z}(J,J^\dagger )]\;, \end{aligned}$$

with $W_p[J,J^\dagger ]$ given by Proposition 2.8. Applying all derivatives on $W_p$ gives by definition at $J=J^\dagger =0$:

$$\begin{aligned} N^{b-2} {\hat{D}} \frac{\partial ^2 }{\partial J_{pq}\partial J_{qp}^\dagger } W_p[J,J^\dagger ] = \frac{1}{N^2} T_{p\Vert pq|\mathcal {J}|} . \end{aligned}$$

The second term, however, gives rise to a partitioned sum of two subsets, dependent on which subset of indices in ${\hat{D}}$ hit $\log [\mathcal {Z}(J,J^\dagger )]$ itself:

$$\begin{aligned} N^{b-4} {\hat{D}} \bigg ( W_p[J,J^\dagger ] \frac{\partial ^2 }{\partial J_{pq}\partial J_{qp}^\dagger } \log [\mathcal {Z}(J,J^\dagger )]\bigg )\bigg \vert _{J=0}= \sum _{\mathcal {J}'\uplus \mathcal {J}''=\mathcal {J}} T_{p\Vert \mathcal {J}'|} G_{|pq|\mathcal {J}''|}. \end{aligned}$$

Demanding $\mathcal {J}'\ne \emptyset$ gives $\Omega _pG_{|pq|\mathcal {J}}$ in Proposition 5.1. The other terms arise from the part of the Ward identity. Excluding the term $l=p$, we can write $\frac{1}{N}\sum _{ l=1, l\ne p}^N \frac{G_{|lq|\mathcal {J}|}-G_{|pq|\mathcal {J}|}}{E_l-E_p}$ with the same argument as in Proposition 3.1. The excluded term can be included in $\frac{1}{N^2} T_{p\Vert pq|\mathcal {J}|}$ to write it as $\frac{1}{N}\frac{\partial G_{|pq|\mathcal {J}|}}{\partial E_p}$. The other realisation of fixing the indices n, m by $p^s$ and $q^s$ (when $\frac{\partial }{\partial J_{p^sq^s}} \in \hat{D}$ hits $J_{mn}$ and vice versa) gives the very last term in Proposition 5.1 where two boundaries merge. $\square$

Proof of Proposition 5.5

The second term of the lhs of the DSE of Corollary 5.6 is rewritten into polynomial $f(.;w|I|\mathcal {J})$ of degree $d-1$ and a denominator also appearing in $\mathcal {G}^{(0)}(z,w)$.

$$\begin{aligned} -\frac{\lambda }{N}\sum _{k=1}^d r_k \frac{\mathcal {T}^{(g)}(I\Vert \varepsilon _k,w|\mathcal {J}|)}{x(\varepsilon _k)-x(z)}&=\frac{\frac{\lambda }{N}\sum _{k=1}^d r_k \mathcal {T}^{(g)}(I\Vert \varepsilon _k,w|\mathcal {J}|) \prod _{i\ne k}^d (x(z)-x(\varepsilon _i))}{\prod _{j=1}^d(x(z)-x(\varepsilon _j))}\\&=:\frac{f(x(z);w|I|\mathcal {J})}{\prod _{j=1}^d (x(z)-x(\varepsilon _j))}, \end{aligned}$$

with one of the product representations of the 2-point function, Theorem 4.3, application of the interpolation formula (see (A.1)) with $L_w(x(z)):=\prod _{j=1}^d (x(z)-x(\hat{\tilde{w}}^j)$, yields:

$$\begin{aligned} \frac{f(x(z);w|I|\mathcal {J})}{\prod _{j=1}^d (x(z)-x(\varepsilon _j))}&=\frac{ L_w(x(z))}{\prod _{j=1}^d (x(z)-x(\varepsilon _j))}\sum _{j=1}^d\frac{f(x(\hat{\tilde{w}}^j);w|I|\mathcal {J}|)}{(x(z)-x(\hat{\tilde{w}}^j) L_w'(x(\hat{\tilde{w}}^j)}\\&=-\lambda (y(w){-}y(z))\mathcal {G}^{(0)}(z,w) \sum _{j=1}^d{{\,\textrm{Res}\,}}_{t\rightarrow \hat{\tilde{w}}^j} \frac{f(x(t);w|I|\mathcal {J}|)x'(t)dt}{(x(z)-x(t)) L_w(x(t))}, \end{aligned}$$

where the analyticity of $f(x(z);w|I|\mathcal {J})$ at $z=\hat{\tilde{w}}^j$ was used. Next, insert the rhs of Corollary 5.6 for $z\mapsto t$ near $t=\hat{\tilde{w}}^j$ at which the first term of the lhs vanishes (here it is important that the integrand has only a simple pole at $t=\hat{\tilde{w}}^j$). Inserting it for $f(x(t);w|I|\mathcal {J})$ leads to

$$\begin{aligned}&-\frac{\lambda }{N}\sum _{k=1}^dr_k \frac{\mathcal {T}^{(g)}(I\Vert \varepsilon _k,w|)}{x(\varepsilon _k)-x(z)} \\&=-\lambda (y(w){-}y(z))\mathcal {G}^{(0)}(z,w) \sum _{j=1}^d {{\,\textrm{Res}\,}}_{t\rightarrow \hat{\tilde{w}}^j} \frac{x'(t)dt}{(x(z){-}x(t))(y(w){-}y(t))\mathcal {G}^{(0)}(t,w)} \nonumber \\&\quad \times \bigg [ \sum _{\begin{array}{c} I_1\uplus I_2=I\\ g_1+g_2=g,~(g_1,I_1)\ne (0,\emptyset ) \end{array}} \Omega ^{(g_1)}_{|I_1|+1}(I_1,t) \mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J}|) \nonumber \\&\quad +\mathcal {T}^{(g-1)}(I,t\Vert t,w|\mathcal {J}|) + \sum _{\begin{array}{c} I_1\uplus I_2=I\\ \mathcal {J}_1\uplus \mathcal {J}_2=\mathcal {J},~ \mathcal {J}_1\ne \emptyset \\ g_1+g_2=g \end{array}} \mathcal {T}^{(g_1)}(I_1,t\Vert \mathcal {J}_1|) \mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J}_2|) \nonumber \\&\quad + \sum _{i=1}^m \frac{\partial }{\partial x(u_i)} \Big (\frac{\mathcal {T}^{(g)}(I{\setminus } u_i\Vert u_i,w|\mathcal {J}|)}{ x(u_i)-x(t)}\Big ) \nonumber \\&\quad - \sum _{s=2}^b\frac{ \mathcal {T}^{(g)}(I\Vert z^s,w^s,z^s,w|\mathcal {J}{\setminus } J^s|) - \mathcal {T}^{(g)}(I\Vert t,w^s,z^s,w|\mathcal {J}{\setminus } J^s|) }{x(z^s)-x(t)}. \nonumber \end{aligned}$$

(B.4)

Next, compute for the same integrand the residue at $t= z$ (for arbitrary z):

$$\begin{aligned}&\lambda (y(w)-y(z))\mathcal {G}^{(0)}(z,w){{\,\textrm{Res}\,}}_{t\rightarrow z}\frac{x'(t)dt}{(x(z)-x(t))(y(w)-y(t))\mathcal {G}^{(0)}(t,w)} \\&\quad \times \bigg [ \sum _{\begin{array}{c} I_1\uplus I_2=I\\ g_1+g_2=g,~(g_1,I_1)\ne (0,\emptyset ) \end{array}} \Omega ^{(g_1)}_{|I_1|+1}(I_1,t) \mathcal {T}^{(g_2)}(I_2\Vert z,w|\mathcal {J}|) \\&\quad +\mathcal {T}^{(g-1)}(I,t\Vert t,w|\mathcal {J}|) + \sum _{\begin{array}{c} I_1\uplus I_2=I\\ \mathcal {J}_1\uplus \mathcal {J}_2=\mathcal {J},~ \mathcal {J}_1\ne \emptyset \\ g_1+g_2=g \end{array}} \mathcal {T}^{(g_1)}(I_1,t\Vert \mathcal {J}_1|) \mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J}_2|)\\&\quad + \sum _{i=1}^m \frac{\partial }{\partial x(u_i)} \Big (\frac{\mathcal {T}^{(g)}(I{\setminus } u_i\Vert u_i,w|\mathcal {J}|)}{ x(u_i)-x(t)}\Big )\\&\quad - \sum _{s=2}^b\frac{ \mathcal {T}^{(g)}(I\Vert z^s,w^s,z^s,w|\mathcal {J}{\setminus } J^s|) - \mathcal {T}^{(g)}(I\Vert t,w^s,z^s,w|\mathcal {J}{\setminus } J^s|) }{x(z^s)-x(t)}\\&=(y(w)-y(z))\mathcal {T}^{(g)}(I\Vert z,w|\mathcal {J}|) -\frac{\lambda }{N}\sum _{k=1}^d r_k \frac{\mathcal {T}^{(g)}(I\Vert \varepsilon _k,w|\mathcal {J}|)}{x(\varepsilon _k)-x(z)}\;. \end{aligned}$$

Summing both expressions finishes the proof. $\square$

Proof of Corollary 5.6

First of all, we need to proof analyticity at certain points: $\square$

Lemma C.1

Let $\mathcal {J}=\{J^2,\ldots ,J^b\}$ for $J^s=[z^s,w^s]$ and $I=\{u_1,\ldots ,u_m\}$. The generalised correlation functions $\mathcal {T}^{(g)}(I\Vert z,w|\mathcal {J})$ are analytic at $z=u_i$ and $z=z^s$.

Proof

In principal, this is clear when returning to the matrix base with a finite limit of coinciding indices. The result can be shown inductively in the Euler characteristic in DSE (5.4), with analogous considerations to [BHW22, Lemma 4.1]. $\square$

The strategy of the proof consists of rewriting a term in Proposition 5.5 in terms of all the others - but taking the residues at different points:

$$\begin{aligned}&\frac{\partial }{\partial x(u_i)}\sum _{j=1}^d {{\,\textrm{Res}\,}}_{t\rightarrow z,\hat{\tilde{w}}^j} \frac{x'(t)dt\prod _{k=1}^d(x(t)-x(\varepsilon _k)) }{(x(z)-x(t)) L_w(x(t))} \frac{\mathcal {T}^{(g)}(I\backslash u_i\Vert u_i,w|\mathcal {J})}{x(u_i)-x(t)}\\&=\frac{\partial }{\partial x(u_i)} \sum _{j=1}^d {{\,\textrm{Res}\,}}_{v\rightarrow x(z),x(\hat{\tilde{w}}^j)} \frac{dv\prod _{k=1}^d(v-x(\varepsilon _k)) }{(x(z)-v) L_w(v)} \frac{\mathcal {T}^{(g)}(I\backslash u_i\Vert u_i,w|\mathcal {J})}{x(u_i)-v}\\&=-\frac{\partial }{\partial x(u_i)}{{\,\textrm{Res}\,}}_{v\rightarrow x(u_i)} \frac{dv\prod _{k=1}^d(v-x(\varepsilon _k)) }{(x(z)-v) L_w(v)} \frac{\mathcal {T}^{(g)}(I\backslash u_i\Vert u_i,w|\mathcal {J})}{x(u_i)-v}\\&=\frac{1}{x'(u_i)}\frac{\partial }{\partial u_i} \frac{\prod _{k=1}^d(x(u_i)-x(\varepsilon _k)) }{(x(z)-x(u_i)) L_w(x(u_i))}\mathcal {T}^{(g)}(I\backslash u_i\Vert u_i,w|\mathcal {J})\\&={{\,\textrm{Res}\,}}_{t\rightarrow u_i}\frac{x'(t)dt \prod _{k=1}^d(x(t)-x(\varepsilon _k)) }{(x(z)-x(t)) L_w(x(t))}\Big \{\frac{1}{x'(u_i)x'(t)(t-u_i)^2 } \mathcal {T}^{(g)}(I\backslash u_i\Vert t,w|\mathcal {J})\Big \}\;, \end{aligned}$$

where we substituted $t\mapsto v=x(t)$, then moved the integration contour and finally represented the result in form of a residue formula. $\Omega ^{(0)}_2(z,w)$ is the only correlation function divergent on the diagonal so that the terms in $\{~ \}$ extend to $\sum _{I_1,I_2,g_1,g_2} \Omega ^{(g_1)}_{|I_1|+1}(I_1;t)\mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J})$ and finally to

$$\begin{aligned}&\sum _{\begin{array}{c} I_1\uplus I_2=I\\ g_1+g_2=g,~(g_1,I_1)\ne (0,\emptyset ) \end{array}} \Omega ^{(g_1)}_{|I_1|+1}(I_1,t) \mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J}|) \nonumber \\&\quad +\mathcal {T}^{(g-1)}(I,t\Vert t,w|\mathcal {J}|) + \sum _{\begin{array}{c} I_1\uplus I_2=I\\ \mathcal {J}_1\uplus \mathcal {J}_2=\mathcal {J},~ \mathcal {J}_1\ne \emptyset \\ g_1+g_2=g \end{array}} \mathcal {T}^{(g_1)}(I_1,t\Vert \mathcal {J}_1|) \mathcal {T}^{(g_2)}(I_2\Vert t,w|\mathcal {J}_2|) \nonumber \\&\quad + \sum _{s=2}^b\frac{ \mathcal {T}^{(g)}(I\Vert t,w^s,z^s,w|\mathcal {J}{\setminus } J^s|) }{x(z^s)-x(t)}, \end{aligned}$$

(C.1)

where the regularity at $z=u_i$ was exploited—under the residue operation, we added an effective zero. In the same manner, we rewrite the term

$$\begin{aligned}&\sum _{j=1}^d {{\,\textrm{Res}\,}}_{t\rightarrow z,\hat{\tilde{w}}^j} \frac{x'(t)dt\prod _{k=1}^d(x(t)-x(\varepsilon _k)) }{(x(z)-x(t)) L_w(x(t))} \frac{\mathcal {T}^{(g)}(I\Vert z^s,w^s,z^s,w|\mathcal {J}{\setminus } J^s|)}{x(z^s)-x(t)} \end{aligned}$$

for $s \in \{2,\ldots ,b\}$ into a residue formula where we take the residue at $t=z^s$ by deforming the contour. Since we have just a pole at order one at $t=z^s$, we set an in the integrand the function

$$\begin{aligned} \mathcal {T}^{(g)}(I\Vert z^s,w^s,z^s,w|\mathcal {J}{\setminus } J^s|)\quad \rightarrow \quad \mathcal {T}^{(g)}(I\Vert t,w^s,z^s,w|\mathcal {J}{\setminus } J^s|). \end{aligned}$$

Again, the regularity argument of the all the other terms for the integrand in Eq. (C.2) at $t=z^s$ allows for adding another zero. $\square$

Proof of Theorem 5.8

Assume $p^i_j$ and $q^i_j$ such that all $E_{p^i_j}$ and $\tilde{E}_{q^i_j}$ are pairwise different. Set $a=p^1_1$, $d=q^1_1$ and $c=q^1_{N_1}$ to have a clear distinction between these and the remaining $p^i_j$ and $q^i_j$. Again, we define a differential operator ${\hat{D}}_{dc}$:

$$\begin{aligned} \hat{D}_{dc}=\frac{\partial ^{2N_1+..+2N_b-2}}{\partial J^\dagger _{dp^1_2}\partial J_{p^1_2q^1_2}.. \partial J_{p^1_{N_1}c} \partial J_{p^2_1q^2_1}\partial J^\dagger _{q^2_1p^2_2}.. \partial J^\dagger _{q^2_{N_2}p^2_1} ..\partial J_{p^b_1q^b_1}.. \partial J^\dagger _{q^b_{N_b}p^b_1}}. \end{aligned}$$

Bringing the global denominator of the theorem to the other side yields by definition of the correlation function

$$\begin{aligned}&(\tilde{E}_{d}-\tilde{E}_{c})G_{|p_1^1 q^1_1..q^1_{N_1}|\mathcal {J}|}\\ =&(\tilde{E}_{d}-\tilde{E}_{c})N^{b-2} \hat{D}_{dc}\frac{\partial ^2}{\partial J^\dagger _{ca}\partial J_{ad}} \log \mathcal {Z}(J,J^\dagger )\big \vert _{J,J^\dagger =0}\\ =&N^{b-2} \hat{D}_{dc}\frac{\partial ^2}{\partial J^\dagger _{ca}\partial J_{ad}}[E_a+\tilde{E}_{d}-(E_a+\tilde{E}_{c})]\log \mathcal {Z}(J,J^\dagger )\big \vert _{J,J^\dagger =0}. \end{aligned}$$

We generalise the steps in Example 2.6 to the general operator ${\hat{D}}_{dc}$:

$$\begin{aligned}&C \hat{D}_{dc}N^{b-1}\bigg (\frac{\partial }{\partial J^\dagger _{ca}} \frac{\exp (-N S_{int}(\frac{1}{N}\partial _J,\frac{1}{N}\partial _{J^\dagger }))J_{da}}{\mathcal {Z}(J,J^\dagger )}\\&\quad - \frac{\partial }{\partial J_{ad}} \frac{\exp (-N S_{int}(\frac{1}{N}\partial _J,\frac{1}{N}\partial _{J^\dagger }))J^\dagger _{ac}}{\mathcal {Z}(J,J^\dagger )}\bigg )\mathcal {Z}_{free}(J,J^\dagger )\big \vert _{J,J^\dagger =0}\\&=-N^{b-4}\lambda \hat{D}_{dc}\sum _{n,m=1}^N\left( \frac{\partial }{\partial J^\dagger _{ca}} \frac{\frac{\partial ^3}{\partial J_{an}\partial J^\dagger _{nm}\partial J_{md}}}{\mathcal {Z}(J,J^\dagger )}-\frac{\partial }{\partial J_{ad}} \frac{ \frac{\partial ^3}{\partial J^\dagger _{cm}\partial J_{mn}\partial J^\dagger _{na}}}{\mathcal {Z}(J,J^\dagger )}\right) \mathcal {Z}(J,J^\dagger )\big \vert _{J,J^\dagger =0}. \end{aligned}$$

For $E_m=E_a$, the bracket vanishes for regular and non-regular terms. This comes from the fact that $\frac{\partial }{\partial J^\dagger _{ca}}$ and $\frac{\partial }{\partial J_{ad}}$ do not act on $\frac{1}{\mathcal {Z}(J,J^\dagger )}$ because it gives 0 after taking $J=0$ (no cycle in a), then setting $m=a$ yields the same four derivatives.

Therefore, we can assume $E_m\ne E_a$ and apply the simpler Ward identity of Proposition 2.7.

$$\begin{aligned}&=-\lambda \hat{D}_{dc}N^{b-3}\sum _{\begin{array}{c} n,m \\ m\ne a \end{array} }\bigg (\frac{\partial }{\partial J^\dagger _{ca}} \frac{\frac{\partial }{\partial J_{md}}}{\mathcal {Z}(J,J^\dagger )(E_m-E_a)} \big (J_{mn}\frac{\partial }{\partial J_{an}}- J^\dagger _{na}\frac{\partial }{\partial J^\dagger _{nm}}\big )\\&\quad +\frac{\partial }{\partial J_{ad}}\frac{ \frac{\partial }{\partial J^\dagger _{cm}}}{\mathcal {Z}(J,J^\dagger )(E_m-E_a)}\big (J_{an}\frac{\partial }{\partial J_{mn}}- J^\dagger _{nm}\frac{\partial }{\partial J^\dagger _{na}}\big )\bigg )\mathcal {Z}(J,J^\dagger )\big \vert _{J,J^\dagger =0}. \end{aligned}$$

The following derivatives lead to cancellations:

• in the first line $\frac{\partial }{\partial J^\dagger _{ca}}$ on $J^\dagger _{na}$ is cancelled by $\frac{\partial }{\partial J_{ad}}$ on $J_{an}$ in the second line

• if $\frac{\partial }{\partial J_{md}}$ acts on $J_{mn}$ is cancelled by $\frac{\partial }{\partial J^\dagger _{cm}}$ on $J^\dagger _{nm}$ in the second line

We end up with the surviving terms

$$\begin{aligned}&-\lambda \hat{D}_{dc}N^{b-3}\sum _{\begin{array}{c} n,m \\ m\ne a \end{array} }\frac{1}{E_m-E_a} \biggl (\frac{\partial }{\partial J^\dagger _{ca}}J_{mn}\frac{ \frac{\partial ^2}{\partial J_{an}\partial J_{md}}}{\mathcal {Z}(J,J^\dagger )}\\&\quad -\frac{\partial }{\partial J_{ad}}J^\dagger _{nm}\frac{ \frac{\partial ^2}{\partial J^\dagger _{cm}\partial J^\dagger _{na}}}{\mathcal {Z}(J,J^\dagger )}\biggl )\mathcal {Z}(J,J^\dagger )\big \vert _{J,J^\dagger =0}. \end{aligned}$$

We manipulate this equation using the general identity

$$\begin{aligned} \frac{\partial _x \partial _yf(x,y)}{f(x,y)}=\partial _x\partial _y\log [f(x,y)]+\partial _x\log [f(x,y)] \partial _y\log [f(x,y)] \end{aligned}$$

yields

$$\begin{aligned}&-\lambda \hat{D}_{dc}N^{b-3}\sum _{\begin{array}{c} n,m \\ m\ne a \end{array} }\frac{1}{E_m-E_a} \bigg \{\frac{\partial }{\partial J^\dagger _{ca}}J_{mn} \frac{\partial ^2}{\partial J_{an}\partial J_{md}}\log [\mathcal {Z}(J,J^\dagger )]\\&\quad -\frac{\partial }{\partial J_{ad}}J^\dagger _{nm} \frac{\partial ^2}{\partial J^\dagger _{cm}\partial J^\dagger _{na}}\log [\mathcal {Z}(J,J^\dagger )]\\&\quad + \frac{\partial }{\partial J^\dagger _{ca}}J_{mn} \bigg (\frac{\partial }{\partial J_{an}}\log [\mathcal {Z}(J,J^\dagger )]\bigg )\bigg (\frac{\partial }{\partial J_{md}}\log [\mathcal {Z}(J,J^\dagger )]\bigg )\\&\quad -\frac{\partial }{\partial J_{ad}}J^\dagger _{nm} \bigg (\frac{\partial }{\partial J^\dagger _{cm}}\log [\mathcal {Z}(J,J^\dagger )]\bigg ) \bigg (\frac{\partial }{\partial J^\dagger _{na}}\log [\mathcal {Z}(J,J^\dagger )]\bigg )\bigg \}\bigg \vert _{J,J^\dagger =0}. \end{aligned}$$

The n, m are fixed by a derivative acting on $J_{mn}$ (or $J^\dagger _{nm}$). We collect the following terms:

• either a derivative of the form $\frac{\partial }{\partial J^\dagger _{q^1_k p^1_{k}}}$ and $\frac{\partial }{\partial J_{p^1_k q^1_{k}}}$, respectively, fixes the n, m in the first two lines, which produces separated cycles

• or a derivative of the form $\frac{\partial }{\partial J^\dagger _{q^\beta _k p^\beta _{k}}}$ and $\frac{\partial }{\partial J_{p^\beta _k q^\beta _{k}}}$, respectively, with $\beta >1$ fixes the n, m, which merges the first cycle with the $\beta ^{\text {th}}$-cycle.

• in the last two lines it is only possible that a derivative of the form$\frac{\partial }{\partial J^\dagger _{q^1_k p^1_{k}}}$ and $\frac{\partial }{\partial J_{p^1_k q^1_{k}}}$, respectively, fixes the n, m, otherwise setting $J=0$ leads to vanishing contributions. Acting with the remaining derivatives of $\hat{D}_{dc}$ on the product of the logarithms by considering the Leibniz rule leads to the assertion for pairwise different $E_{p^i_j}$.

The expression stays true for coinciding $E_{p^i_j}$ since the lhs is regular which induces a well-defined limit of the rhs by continuation to differentiable functions. A genus expansion and application of the boundary creation $-N\frac{\partial }{\partial E_{p_i}}$ operator yields the assertion. $\square$