Questions from Chern-Simons theory

Question

I am currently reading an article about TFTs (DW - Group Cohomology and TFTs), and I have a few questions:

(1) Let $M$ be a 3-man., then we know there exists some 4-man. $B$ such that $\partial B = M$. Now, let $E$ be a $G$-bundle over $M$, when can we extend $E$ to a $G$-bundle over $B$ and a connection $A$ over $E$ to a connection $A'$ over $E'$?

(2) If we can construct such an $E'$ over $B$ then we can re-define the CS action as $CS(A') = \frac{k}{8\pi^2}\int_B Tr(F'\wedge F')$, where $F'$ is the curvature of $A'$. According to the paper, if $k$ is an integer then $CS(A')$ is independent, mod 1, of the choice of $B$ and of the extension $E$ and $A$. How do you see this?

(3) Later on they give a better def of the action, $CS = \frac{1}{n}\left(\int_B \frac{k}{8\pi^2}Tr(F'\wedge F') - \langle \gamma^*\omega,B\rangle\right)$, where $\gamma : B \rightarrow BG$ and $\omega$ is an integer-valued cocycle. They then go on to say that if $B$ is closed then $\int_B \frac{k}{8\pi^2}Tr(F'\wedge F') = \langle \gamma^*\omega,B\rangle$ and so $CS$ is independent of $B$ and the way we have continued the bundle and the connection. Why is this so?

Answer 1

(1) Obstruction theory: there is a sequence of obstructions to extending a bundle On $M$ to all of $B$. For homotopy theoretic purposes, a bundle is the same as a map to $BG$. The obstructions live in $H^{n}(B,M;\pi_{n-1}(BG))$. These groups are trivial for a $4$-manifold $B$ when $BG$ is $3$-connected, which happens as along as $G$ is connected and simply-connected.

(2) Chern-Weil theory: $\frac{1}{4 \pi^2}F \wedge F$ represents the first Pontrjagin class, which is integral. If you have two choices of nullbordism, glue them together to get a closed manifold. The difference (because of orientation) of the integrals over the two nullbordisms is the same as the integral over the closed manifold. The role of the $k$ is not so clear to me.

(3) The cocylce $\omega$ has to be $k$ times the Pontrjagin class, then this is true. The same argument as in (2) should give the result.

Answer 2

This is not an answer to your questions, but rather a way to avoid them. Additionally, the assumption that the group $G$ is connected or simply-connected becomes unnecessary.

For a general (compact) Lie group $G$, the level $k$ is not an integer but a class $k \in H^4(BG,\mathbb{Z})$. (If the group is simple, connected and simply-connected, then $H^4(BG,\mathbb{Z}) = \mathbb{Z}$; this reproduces the integer.)

Associated to the class $k$ and the connection $A$ on the bundle $E$ is a bundle 2-gerbe with connection over $M$, the Chern-Simons 2-gerbe $\mathbb{CS}(k,A)$. Connections on bundle 2-gerbes have $U(1)$-valued holonomies around closed oriented 3-manifolds, and so one defines the (exponentiated) Chern-Simons action $$ S_{CS}(k,A) := Hol_{\mathbb{CS}(k,A)}(M) \in U(1). $$

With this definition, it is not necessary to extend the manifold $M$ or the bundle $E$.

However, if extensions of $M$, $E$ and $A$ exist one has the following: there is a Stokes' Theorem for 2-gerbe holonomy, namely $$ Hol_{\mathbb{G}}(M) = \exp \left ( \int_B \mathrm{curv}(\mathbb{G}) \right ). $$ Now, the curvature of the Chern-Simons 2-gerbe $\mathbb{CS}(k,A)$ is $\frac{1}{8\pi^2} b_{k} ( F \wedge F )$, where the bilinear form $b_{k}: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ corresponds to $k$ under the Chern-Weil homomorphism. So we have $$ S_{CS}(k,A) = \exp \left ( \frac{1}{8\pi^2} \int_B b_{k}(F \wedge F) \right ) $$ which is the formula you have asked for in (2).