In Link, Witten claims (p. 54) that to quantize the moduli space of flat $SL(2,\mathbb{C})$ connections on a torus, one can simply quantize the cotangent bundle of a real torus and take the part invariant by the Weyl group $W$.
I agree that $T^*(T\times T)/W$ is the moduli space of flat connections on the torus. However, it seems to me that the correct symplectic form should be the following: if we parametrize $T\times T$ with coordinates $(r_1, \theta _1, r_2, \theta _2)$ (quotienting the $\theta$ coordinates appropriately) then the symplectic form $\omega$ should be induced by $dr_1\wedge dr_2+d\theta _1 \wedge d\theta _2$. This restricts to the correct symplectic form on the real torus $T\times T$ obtained by taking the $r_1 = r_2 = 1$ subspace (which appears to be the $SU(2)$ character variety with its correct symplectic structure). But this is not the symplectic form induced by the cotangent structure, which would be $\omega'=d\theta _1 \wedge dr_1 + d\theta _2 \wedge dr_2$. On the other hand $\omega'$ vanishes on the real torus corresponding to $SU(2)$ representations, and does not seem to be the form induced by the character variety, which comes equipped with a natural symplectic form.
This makes a substantial difference in quantization because $\omega'$ is exact and hence we can take the trivial line bundle for prequantization, and then in a real polarization obtain a Hilbert space $L^2(T\times T)$ (which is what Witten claims to be the quantization of the moduli space). However, if we take $\omega$ as the symplectic form, $\omega$ is non-zero in cohomology, and we will end up with a more complicated quantization.
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5$\begingroup$ @Blake: what is your question? $\endgroup$– André HenriquesCommented Nov 2, 2011 at 16:12
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1$\begingroup$ Wild guess: the question is "Does Witten's claim hold water?". $\endgroup$– S. Carnahan ♦Commented Nov 3, 2011 at 8:30
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2$\begingroup$ Yeah, I guess my question is - is Witten correct? If so, why is $\omega'$ the correct symplectic form to use? If one uses $\omega$ (which appears to me to be the correct symplectic form to use) then the space of Bohr-Sommerfeld leaves in a real polarization will be a finite set of real lines in $\mathbb{C}^*$ each of constant $\theta$, and the space of states will be the Weyl-invariant part of $L^2(\mathbb{R}^n)$, which is obviously quite different than Witten's claim. Which is correct (if either)? $\endgroup$– BlakeCommented Nov 3, 2011 at 14:39
1 Answer
Let me call your $\omega$ as $\omega_I$. The symplectic form you get from the Chern-Simons action is $k\omega_I+s\omega_K$, where $\omega_K$ is one of the Kähler forms on the Hitchin space, which, in particular, is exact. If you choose a real polarization as Witten does, the Hilbert space is $\Gamma(Bun_GX,Det^{\otimes k})$, where $Det$ is the determinant bundle whose first Chern class $[\omega_I]$. One should note that the polarization is not the naive vertical polarization on $T^*Bun_GX$ since the fibers are not Lagrangian for $k\neq 0$.
Narasimhan-Seshadri identifies $Bun_GX$ with the character variety for the compact group, which in genus 1 is $T\times T\ /W$. So, the Hilbert space is $\Gamma(T\times T\ /W, Det^{\otimes k})$, precisely what Witten claims after eq. (5.11).
