Obstruction to extending $G$-bundle to 4-dimensions in Chern-Simons theory I am reading Dijkgraaf and Witten's paper on Chern-Simons and finite gauge groups and something they have written about the obstruction to extending the bundle to the 4-manifold confuses me.
My understanding is the following. I have a $G$-bundle $E$ on smooth 3-manifold $M$, which I would like to extend to a 4-manifold $B$ so that it restricts to $E$ on the boundary. The obstruction to doing this is the cohomology group $H^4(M,\pi_3(G))$ - essentially because you know you can define $E$ on 3-cells $D^3$ but sections $s|_{\partial D^4}\to G$ might be non-trivial elements in $\pi_3(G)$. You can look in the classifying space as well - for classifying map $\gamma: M\to BG$ an element $\gamma(M)\in H^4(BG,\pi_3(G))$ should be the same obstruction.
What Dijkgraaf and Witten say is that the obstruction is measured by the image $\gamma[M]$ in the cohomology group $H^3(BG,\mathbb{Z})$. So this seems to be both the wrong cohomology group (3 not 4) and the wrong coefficients ($\mathbb{Z}$ not $\pi_3(G)$). Now, I know there is an isomorphism
$$H^k(BG,\mathbb{Z})\cong H^{k-1}(BG,\mathbb{R}/\mathbb{Z})$$
for finite groups and for many Lie groups the cohomology vanishes, but I don't see how the coefficients are changing. At this stage in the paper, they haven't specified the exact character of $G$ yet (I think it's connected only).
Can someone correct my understanding or match these two analysis together?
 A: Your self-answer has the right idea, but I'd like to make a few things clearer.
Let $M$ be a $3$-manifold and $E$ be a principal $G$-bundle over $M$ with classifying map $\gamma: M \longrightarrow BG$. We want to find a smooth $4$-manifold $B$ with $\partial B = M$ and and extension $\tilde{E}$ of $E$ to $B$, i.e. $\tilde{E}|_{\partial B} = E$.
To make things easier, Dijkgraaf and Witten let $M$ (respectively $B$) be a smooth singular $3$-chain (respectively $4$-chain). Given $B$, finding $\tilde{E}$ is equivalent to finding an extension $\tilde{\gamma}: B \longrightarrow BG$ of $\gamma$. If such a $\tilde{\gamma}$ exists, then $\tilde{B} = \tilde{\gamma}(B)$ is a smooth singular $4$-chain in $BG$ with boundary $\gamma(M)$.
Now $\gamma(M) \in C_3(BG; \Bbb Z)$. If our $\tilde{B}$ exists, then $\gamma(M) = \partial \tilde{B} \in \mathrm{Im}(\partial)$, so that
$$[\gamma(M)] = \gamma_\ast [M] = 0 \in H_3(BG; \Bbb Z).$$
Hence there is an extension of $E$ to $B$ if and only if $\gamma_\ast [M] = 0 \in H_3(BG; \Bbb Z)$. Note that it is not necessary for $H_3(BG; \Bbb Z)$ to be trivial (perhaps this is a typo in your self-answer); we only need $\gamma_\ast [M] = 0$.
Here the word "obstruction" isn't really referring to the general machinery of obstruction theory. The "Main Theorem of Obstruction Theory" tells us what the obstruction to extending a map defined on the $k$-skeleton of a relative CW-complex $(X, A)$ to the $(k+1)$-skeleton is. Note that if our map is defined on the $3$-manifold $\partial B = M$ and we want to extend it to $B$, this doesn't really fit into the framework of obstruction theory. If we take $X = (B, \varnothing)$, then $M$ contains a $3$-cell so we have $\gamma$ defined on part of the $3$-skeleton of $B$. But $B$ itself can have cells of dimensions $0$ through $3$, so $\gamma$ isn't actually defined on the entire $3$-skeleton, and this isn't simply a case of extending from $X^{(3)}$ to $X^{(4)}$. If we try $X = (B, M)$, then $\gamma$ is only defined on the $-1$-skeleton $X^{(-1)} = M$, and we want to extend all the way to the $4$-skeleton. Hence from this point of view we have multiple obstructions.
Hence we see that obstruction theory isn't really the way to go for this extension problem, and instead we just use the simpler concept of homology of singular chains.
A: Trying to answer my own question - misunderstanding Dijkgraaf and Witten! They say "The obstruction to the existence of such a 4-chain (which has a bundle that restricts to $E$ on the boundary) is measured by the image $\gamma_*(M)$ in the cohomology group $H_3(BG,\mathbb{Z})$." In fact, the bundle over the 4-chain can always be found by restriction of the universal bundle. In order for the 4-chain $\sigma$ itself to exist (at least uniquely) we must have every $\delta \in C_3(BG)$ be a boundary of some $\sigma\in C_4(BG)$ (in homology, not cohomology...). If $\delta = \partial \sigma$ then $\partial \delta =0~\forall \delta$ so $H_3=0$.
The phrase "obstruction theory" immediately lead me to think about how the bundle was extended, but we just need to make sure that such a 4-chain even exists in the first place! Can I get a comment or two to make sure this is correct?
