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  Number of Physical States of a $U(1)$ Chern-SImons Theory on a Riemann Surface of Genus $g$

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In A Duality Web in 2+1 Dimensions and Condensed Matter Physics, the authors claimed in Appendix B that for a for a $U(1)_{k}$ Chern-Simons theory defined on a Riemann surface $\Sigma$ of genus $g$, the number of physical states is $k^{g}$.

Can anybody tell me how to calculate the number of physical states of an Abelian Chern-Simons theory on a Riemann surface? Is there any reference that I can follow to understand the above statement?

I also posted my question here.

asked Dec 22, 2018 in Theoretical Physics by Libertarian Feudalist Bot (270 points) [ no revision ]

1 Answer

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I give below one way to understand this result (there are probably many).

The problem is to show that the dimension of the Hilbert space of $U(1)$ level $k$ Chern-Simons theory on a genus $g$ surface is $k^g$.

The classical phase space of $U(1)$ Chern-Simons theory on a genus $g$ surface is the space $M$ of $U(1)$-flat connections on a genus $g$ surface: it is a torus of dimension $2g$ (indeed, a $U(1)$-flat connection is specified by the $2g$ monodromies around a basis of $2g$ 1-cycles on the surface).

The Hilbert space of quantum states is "the quantization" of the phase space of classical states. One way to "quantize" is holomorphic quantization (see Section 3.1 of https://projecteuclid.org/euclid.cmp/1104178138 ). If one picks a complex structure on the Riemann surface, then the phase space $M$ becomes a complex manifold (an abelian variety), the symplectic form becomes the curvature of an holomorphic line bundle $L$, and the Hilbert space is the space of holomorphic sections of $L$. More precisely, relevant holomorphic line bundles are of the form $L^{\otimes k}$, where $L$ is a specific line bundle (principal polarization) and $k$ is the level. The dimension of the space of holomorphic sections of $L^{\otimes k}$ can be computed using the Hirzebruch-Riemann-Roch theorem (higher cohomoloy groups vanish, Todd class is $1$ for an abelian variety, $M$ has complex dimension $g$):

$\int_M ch(L^{\otimes k})=k^{g} \int_M \frac{c_1(L)^g}{g!}$

One can show that $ \int_M \frac{c_1(L)^g}{g!}=1$ (it is almost the definition of $L$: it is the simplest possible non-trivial case, corresponding to $k=1$).

answered Dec 22, 2018 by 40227 (5,140 points) [ revision history ]

Thank you very much for answering my question. I also found a physical method from David Tong's lecture notes on quantum hall effect.

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