Can someone tell me a reference which proves this? - as to how does the bulk partition function of Chern-Simons' theory get completely determined by the WZW theory (its conformal blocks) on its boundary..

  • I see that Toshitake Kohno's book seems to do a thorough study of this but its not beginner friendly at all! (...so I am kind of looking for something that will give a "simpler" proof and give me a stepping stone into that book...)

  • I am often told that Witten's QFT and Jones polynomial paper proves this but I am hard pressed to find it there. If anyone can spot the proof there, it would help still..

  • $\begingroup$ There are some mathoverflow answers to this question. I recommend you do a bit of searching there. $\endgroup$ Commented Aug 19, 2013 at 19:22
  • $\begingroup$ Here is another very nice review: pos.sissa.it/271/001 $\endgroup$ Commented May 17, 2021 at 11:03

3 Answers 3


There are several different relations between Chern-Simons/WZW models, and there are several way to show these. A nice paper doing this in a concrete way is Elitzur et al Nucl.Phys. B326 (1989) 108.

The Chern-Simons theory on a compact spatial manifold give rise to a finite dimensional Hilbert space (only global degrees of freedom) which turns out to be isomorphic to the space of conformal blocks of a WZW model (which is also finite dimensional, since there are finite number of WZW primaries under the associated affine Lie algebra).

If you however put the theory on a manifold with boundary, there will be local degrees of freedom near the boundary and the Hilbert space infinite dimensional (the dynamics of the boundary degrees of freedom are controlled by a WZW model). Let me go through a simple example of the latter type, you can fill out the detailed calculations.

The action is given by

$$ S[a] = \frac k{4\pi}\int_\mathcal M\text{tr}\left(a\wedge\text d a + \frac 23 a\wedge a\wedge a\right).$$ One can show that for $k\in\mathbb Z$ and the boundary condition $a_0\big|_{\partial\mathcal M} = 0$, $e^{iS[a]}$ is gauge invariant and the equations for motions well-defined. Next we need to fix the gauge appropriately, lets now assume our three-manifold has the following simple form $\mathcal M = \mathbb R\times\Sigma$. Make a temporal decomposition $\text d = \partial_0\text dx^0 + \tilde{\text d}$, where $\tilde{\text d} = \partial_i\text dx^i$, and $a = \tilde a_0 + \tilde a$, where $\tilde a_0 = a_0\text dx^0$ and $\tilde a = a_i\text dx^i$ $(i=1,2)$. With this decomposition we get the following action

$$S[a] = -\frac k{4\pi}\int_\mathcal M\text{tr}\left(\tilde a\wedge\partial_0\tilde a\right)\wedge\text dx^0 + \frac k{2\pi}\int_\mathcal M\text{tr}\left(\tilde a_0\wedge \tilde f\right),$$ where $\tilde f = \tilde{\text d}\tilde a +\tilde a\wedge\tilde a$. It is clear that $\tilde a_0$ is just a Lagrange multiplier and we fix the gauge as $a_0 = 0$ (everywhere, not just on the boundary). Alternatively, integrate out $a_0$ and we get $\delta(\tilde f)$ in the path-integral. We therefore have the following action and constraint

$$S[\tilde a, \tilde a_0=0] = -\frac k{4\pi}\int_\mathcal M\text{tr}\left(\tilde a\wedge\partial_0\tilde a\right)\wedge\text dx^0, \qquad \tilde f = \tilde{\text d}\tilde a +\tilde a\wedge\tilde a=0.$$ Thus the phase-space of the theory is the moduli space of flat connections on $\Sigma$. Whether the phase-space has finite or infinite volume, depends of whether $\Sigma$ has a boundary or not.

For simplicity, let us restrict to the simple manifold $\mathcal M = \mathbb R\times D^2$, where $\Sigma=D^2$ is the $2$-disc. Since $\pi_1(\mathcal M)=0$, there no non-trivial Wilson loops/holonomies (since the Wilson loop only depend on the homotopy class of a curve, for flat-connections) and there are thereby no topological degrees of freedom. In this case we can solve the flat connection constraint $\tilde f = 0$ by letting the gauge field be a pure-gauge

$$\tilde a = -\tilde dUU^{-1},$$ where $U:\mathcal M\rightarrow G$ is a single-valued, group-valued function. Here, $U$, parametrize the local degrees of freedom (of the Chern-Simons theory) modulo gauge redundancies. The action that determines the dynamics of $U$ is found by substituting $\tilde a$ in the above action. Using the coordinates $(t,r,\theta)$, we find

\begin{align*} S_{CWZW}&[U] = S\left[\tilde a=-\tilde{\text d} UU^{-1},\tilde a_0=0\right],\\ &= \frac k{4\pi}\int_{\partial\mathcal M}\text{tr}\left(\partial_\theta U^{-1}\partial_tU\right)\text d^2x + \frac k{12\pi}\int_{\mathcal M}\text{tr}\left([\text d UU^{-1}]^3\right),\\ &= \frac k{4\pi}\int_{\partial\mathcal M}\text{tr}\left(\partial_\theta U^{-1}\partial_tU\right)\text d^2x + \frac k{12\pi}\int_{\mathcal M}\text{tr}\left(\epsilon^{\mu\nu\rho}\partial_\mu UU^{-1}\partial_\nu UU^{-1}\partial_\rho UU^{-1}\right)\text d^3 x. \end{align*} Formally, one also has to check that the path-integral does not come with any Jacobian

$$\int\mathcal D\tilde a\delta(\tilde f) = \int\mathcal DU,$$ where $\mathcal DU$ comes from the Haar measure of $G$. This shows what you were looking for, that the partition function of the Chern-Simons theory is determined by a (chiral) WZW model on the boundary. For more general $\mathcal M$, one can do a similar calculation with a few extra elements. See the above reference for details.

There are of course other ways to show this relation. One can for example show that the Dirac bracket on the phase-space (moduli space of flat-connections) reduce to the affine Lie algebra $\hat{\mathfrak g}_k$ (which is the Chiral algebra of the WZW model). There are also approaches where functional equations for the wave functional are derived. One can also use canonical quantization as Witten does, exploding the fact that the moduli space of flat connections (modulo gauge transformation) is a Kähler manifold and the symplectic form represents the first Chern class of a holomorphic line bundle. This last approach is more abstract and less direct, than the one taken above.

  • 1
    $\begingroup$ I don't really know how to see the flatness of the connection drops out. I know it is the classical solution of the action, but I don't get why it is enforced even at the quantum level. You say if I integrate out $a_0$ I should get a delta-function in the path integral, forcing $f = 0$? Is there a simple way of seeing how that delta-function arises? $\endgroup$ Commented Feb 26, 2016 at 0:31
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    $\begingroup$ The delta function has a simple Fourier transform in terms of an integral. An analogous identity holds for the delta functional in terms of a path integral. This identity is being used here. $\endgroup$
    – Mtheorist
    Commented Oct 3, 2017 at 13:25
  • $\begingroup$ @Heidar Very nice answer! Could you give more details on how the path integral measures are equal to each other? $\endgroup$ Commented Jul 5, 2021 at 2:25

A decent review is around p. 30 of

For more pointers see on the nLab at AdS3-CFT2 and CS-WZW correspondence.


I recommend

M. Bos and V.P. Nair. Coherent State Quantization of Chern-Simons Theory. International Journal of Modern Physics A, A5:959, 1990.

Also, I wrote a short review of it: check the second section of my paper on Yang-Mills-Chern-Simons theory (http://arxiv.org/abs/1311.1853) .


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