Symplectic Geometry of Character Varieties and SU(2) Lattice Gauge Theory I

Abstract

Associated to any finite graph $\Lambda$ is a closed surface ${\textbf{S}}={\textbf{S}}_\Lambda$, the boundary of a regular neighbourhood of an embedding of $\Lambda$ in any three manifold. The surface retracts to the graph, mapping loops on the surface to loops on the graph. The (SU(2)) character variety ${{\mathcal {M}}}$ of ${\textbf{S}}$ has a symplectic structure and associated Liouville measure; on the other hand, the character variety ${\textbf{M}}$ of $\Lambda$ carries a natural measure inherited from the Haar measure. Loops on ${\textbf{S}}$ define functions on the character varieties, the Wilson loops. By the works of W. Goldman, L. Jeffrey and J. Weitsman, the formalism of Duistermaat-Heckman applies to the relevant integrals over ${{\mathcal {M}}}$. We develop a calculus for calculating correlations of Wilson loops on ${{\mathcal {M}}}$ w.r.to the normalised Liouville measure, and present evidence that they approximate—for large graphs—the corresponding integrals over ${\textbf{M}}$. Lattice field theory involves integrals over ${\textbf{M}}$; we present “symplectic” analogues of expressions for partition functions, Wilson loop expectations, etc., in two and three space-time dimensions.

Appendix A: The operator M

Proposition A.1

Given $u,v \in [0,1]$ the operator $N_uN_v$ is trace class and

$$\begin{aligned} M(s,t) \equiv Tr(N_sN_t)= {\left\{ \begin{array}{ll} 2t(1-s)&{} \textrm{if} \ s \ge t\\ 2s(1-t)&{} \textrm{if} \ t \ge s\\ \end{array}\right. } \end{aligned}$$

Proof

The trace is given by the integral of the kernel along the diagonal. But $\int _0^1 \int _0^1 {\textbf{1}}_{{{\mathcal {P}}}_s} (u,v) {\textbf{1}}_{{{\mathcal {P}}}_t}(v,u) du dv$ is the area of the intersection ${{\mathcal {P}}}_s \cap {{\mathcal {P}}}_t$:

Here is a more computational proof. Since the operators $N_s,N_t$ are simultaneously diagonalised,

$$\begin{aligned} \begin{aligned} Tr(N_sN_t)&=\sum _{n=0}^\infty \frac{4}{(n+1)^2 \pi ^2} \sin {(n+1)\pi s} \sin {(n+1)\pi t}\\&=2\sum _{n=0}^\infty \frac{1}{(n+1)^2 \pi ^2} \{ \cos {2(n+1)\pi (\frac{s-t}{2})}-\cos {2(n+1)\pi (\frac{s+t}{2})} \} \end{aligned} \end{aligned}$$

We have $0 \le \frac{s+t}{2} \le 1$; further, If $s \ge t$, we also have $0 \le \frac{s-t}{2} \le 1$. So the trigonometric sums can be identified with the Bernoulli polynomials, and we get

$$\begin{aligned} \begin{aligned} Tr(N_sN_t)&=2\{B_2(\frac{s-t}{2})-B_2(\frac{s+t}{2})\}\\&=2\big \{\frac{(s-t)^2}{4}-\frac{s-t}{2}-\frac{(s+t)^2}{4}+\frac{s+t}{2}\big \}\\&=2(t-st)=2t(1-s) \end{aligned} \end{aligned}$$

$\square$

Corollary A.2

We have

$$\begin{aligned} \int dt_3 \vec {u}_{n_3}(t_3) N_{t_3}=\frac{\sqrt{2}}{(n_3+1) \pi } P_{n_3} \end{aligned}$$

where $P_{n}$ is the projection to the span of $\vec {u}_{n}$.

Proof

This is a direct computation.

$$\begin{aligned} \begin{aligned} \{[\int \vec {u}_{n_3}(t_3) N_{t_3}\ dt_3] \vec {u}_{n_1}\}(t_2)&= \int \vec {u}_{n_3}(t_3) \{N_{t_3} \vec {u}_{n_1}\}(t_2) \ dt_3 \\&=\int \vec {u}_{n_3}(t_3) \frac{2}{(n_1+1)\pi }\sin {(n_1+1) \pi t_3}\ \vec {u}_{n_1}(t_2)\ dt_3 \\&=\frac{\sqrt{2}}{(n_1+1) \pi } (\vec {u}_{n_3},\vec {u}_{n_1})_{L^2([0,1])} \vec {u}_{n_1}(t_2)\\&=\delta _{n_3,n_1}\frac{\sqrt{2}}{(n_1+1) \pi } \vec {u}_{n_1}(t_2) \end{aligned} \end{aligned}$$

$\square$

Let M denote the integral operator with kernel M(s, t).

Proposition A.3

The (normalised) eigenfunctions and corresponding eigenvalues of M are

$$\begin{aligned} (\vec {u}_n(t) \equiv \sqrt{2} \sin {(n+1) \pi t},\ \frac{2}{(n+1)^2 \pi ^2} ),\quad n=0,1, \dots \end{aligned}$$

Proof

The first proof uses the definition:

$$\begin{aligned} \begin{aligned} M \vec {u}_n (s)&=\int dt M(s,t) \vec {u}_n(t)=\int dt \ Tr(N_sN_t)\vec {u}_n(t)\\&=Tr(N_s [\int dt\ \vec {u}_n(t)N_t]))=\frac{\sqrt{2}}{(n+1) \pi }Tr(N_s P_{n})\\&=\frac{\sqrt{2}}{(n+1) \pi } \frac{2 \sin {(n+1) \pi s}}{(n+1) \pi }=\frac{2}{(n+1)^2 \pi ^2} \vec {u}_n (s) \end{aligned} \end{aligned}$$

The second proof uses the expression for the kernel M(s, t):

$$\begin{aligned}{} & {} M(s,t) \equiv Tr(N_sN_t)= {\left\{ \begin{array}{ll} 2t(1-s)&{} \textrm{if} \ s \ge t\\ 2s(1-t)&{} \textrm{if} \ t \ge s\\ \end{array}\right. }\\{} & {} M \vec {u}_n (s)=\int M(s,t) \vec {u}_n(t)\ dt\\{} & {} \quad =\sqrt{2}\int _0^s \ 2t(1-s) \sin {(n+1)\pi t}+\sqrt{2} \int _s^1 dt\ 2s(1-t) \sin {(n+1)\pi t} \ dt\\{} & {} \quad =-2s\sqrt{2}\int _0^1 \ t \sin {(n+1)\pi t}+\sqrt{2}\int _0^s dt\ 2t \sin {(n+1)\pi t}\\{} & {} \qquad +\,\sqrt{2} \int _s^1 dt\ 2s \sin {(n+1)\pi t} \ dt\\{} & {} \quad =-2s\sqrt{2}\{-\frac{1}{(n+1) \pi }\cos {(n+1)\pi }\}+2\sqrt{2}\{\frac{1}{(n+1)^2\pi ^2}\sin {(n+1)\pi s}\\{} & {} \qquad -\frac{s}{n\pi }\cos {(n+1)\pi s}\}-\sqrt{2}\frac{2s}{(n+1)\pi } (\cos {(n+1)\pi }-\cos {(n+1)\pi s})\\{} & {} \quad =\frac{2}{(n+1)^2 \pi ^2}\vec {u}_n(s) \end{aligned}$$

$\square$

We have used the following:

Lemma A.4

$$\begin{aligned} \int _0^s t \sin {xt} \ dt=\frac{1}{x^2}\sin {xs}-\frac{s}{x}\cos {xs} \end{aligned}$$

Proof

Set $I(x,s) \equiv -\int _0^s \cos {xt}\ dt=-\frac{1}{x}[\sin {xt}]^s_0=-\frac{1}{x}\sin {xs}$. Then

$$\begin{aligned} \begin{aligned} \int _0^s t \sin {xt} \ dt=\frac{d}{dx}I(x,s)&=\frac{1}{x^2}\sin {xs}-\frac{s}{x}\cos {xs} \end{aligned} \end{aligned}$$

$\square$

Here is a computation of the volume of ${{\mathcal {M}}}_{{\textsf{b}}}$ that mimics the computation using the Verlinde formula, except that we use our continuous analogue. Note first that for any $l \ge 1$

$$\begin{aligned} M^l \vec {u}_n=(\frac{2}{(n+1)^2 \pi ^2})^l \vec {u}_n \end{aligned}$$

If we express ${\textbf{S}}$ as a cyclic union of ${{\textsf{b}}}-1$ tori, and take the obvious periodic pants decomposition, the volume of the corresponding polytope is

$$\begin{aligned} \begin{aligned} Tr(M^{{{\textsf{b}}}-1})&=\sum _{n=0}^\infty (\frac{2}{(n+1)^2\pi ^2})^{{{\textsf{b}}}-1}= (\frac{2}{\pi ^2})^{{{\textsf{b}}}-1} \sum _{n=0}^\infty \frac{1}{(n+1)^{2({{\textsf{b}}}-1)}}\\&=\frac{2^{({{\textsf{b}}}-1)}}{\pi ^{2({{\textsf{b}}}-1)}} \zeta (({{\textsf{b}}}-1))\\ \end{aligned} \end{aligned}$$