In http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1104202513, 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.

  • 5
    $\begingroup$ @Blake: what is your question? $\endgroup$ – André Henriques Nov 2 '11 at 16:12
  • $\begingroup$ Wild guess: the question is "Does Witten's claim hold water?". $\endgroup$ – S. Carnahan Nov 3 '11 at 8:30
  • 1
    $\begingroup$ 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)? $\endgroup$ – Blake Nov 3 '11 at 14:39

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.