# Chern Simons Theory over S^3 as integral - what is domain of integration?

I found these nice lecture notes Lectures on localization and matrix models in supersymmetric Chern-Simons-matter theories so I am hoping to understand some parts of the Chern Simons theory better.

As an exercise in stenography I wrote down the Chern Simons action:

$$S = \frac{k}{4\pi} \int_{M_3} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$$

where $A$ is a connection on $M_3$ over some Lie group $G$. For now, I do not even bother asking why $A \wedge A \wedge A \neq 0$ or why this is a top-level form (a 3-form).

In many cases the Chern Simons path integral simplify or "localizes" to a matrix integral, so that's whaty I am interested in today.

$$Z(S^3) = \int \prod_{i=1}^N d\mu_i \prod_{i < j} \sinh \left( \frac{\mu_i - \mu_j}{2} \right) \;\mathrm{exp}\left( - \frac{k}{8\pi}\sum \mu_i^2 \right)$$

what is the domain of integration here -- is it $\mathbb{R}^n$ ? I have no way of verifying this integral if I don't know the domain. Usually they just say the "Cartan" but I have on idea what that is.

This is very general. If instead of $G = U(N)$ we have any $G$:

$$Z(S^3) \propto \int \prod_{i=1}^N d\mu_i \prod_{i < j} 2\sinh \left( \frac{\alpha \cdot \mu }{2} \right) \;\mathrm{exp}\left( - \frac{k}{8\pi}\sum \mu_i^2 \right)$$

where $\alpha = e_i - e_j$ in the case $G = U(N)$ so this becomes the first formula.

At first I thought this integral was Haar measure over the Unitary group, but that is an integral over the maximal torus $[0, 2\pi]^n$ in the Lie group instead of the Lie algebra. So I am really confused.

Wilson loops can be evaluated exactly for this Chern Simons theory.

$$\langle W \rangle = \frac{e^{\frac{N\pi i}{k}}}{N} \frac{\sin \frac{N\pi}{k}}{\sin \frac{\pi}{k}}$$

Again I just found this in a paper and I am trying to understand this integral better. Apparently this is for fudamental representation of $U(N)$. In another paper, I found:

$$\langle W_{\wedge^m \square} \rangle = \dots = \left[ \begin{array}{c} N \\ m \end{array} \right]$$

This might be the same formula. I know it also deals with a 3-sphere. If $m = 1$ the right side is $[N] = \frac{q^N - q^{-N}}{q - q^{-1}}$ with $q = e^{2\pi i t}$ and the sine functions emerge.

I am nervous because that paper says $G = SU(N)$ instead of $G = U(N)$ (so determinant is always 1). Hopefully by now physicists agree on what the value of the Chern Simons path integral over the 3-sphere should be in this case... so it makes sense to ask for clarifications.

My apologies in advance for confusing all the terminology.

• $A^3$ is not zero because it's a $\mathfrak{g}$-valued form and the various components do not necessarily commute. It's top-valued because the connection is a one-form and wedging three one-forms gives a three-form, which is top-valued on $M^3$. Also, you are missing a $\mathrm{Tr}_\mathfrak{g}$ in your definition of the CS action. – Ryan Unger Jan 17 '16 at 21:52
• It's completely unclear what your actual question is. If your question is about the domain of integration, then there is a lost of irrelevant stuff here. " Usually they just say the "Cartan"" - this seems to indicate the integral is over the Cartan subalgebra/maximal torus. – ACuriousMind Jan 17 '16 at 22:11
• Dear OP, it seems like you are missing a lof of background to understand the relation between Chern-Simons and matrix models. I recommend you to study chapters 1 and 2 of "Chern–Simons theory, matrix models and topological strings" by Marcus Mariño (Oxford Univ. Press 2005). – Hans Moleman Jan 17 '16 at 22:13
• @ACuriousMind In the case of $G = SU(N)$ or $G = U(N)$ those integrals can be written as integrals over subsets of $\mathbb{R}^N$ and I am asking which ones... I am having difficulty computing the regions in each case. – john mangual Jan 17 '16 at 22:14
• @HansMoleman Would you care to supply some of that background? Without a domain of integration there is no way to verify those integrals are correct except to take physicist's word for it. Marino's book looks interesting though. – john mangual Jan 17 '16 at 22:16