Recently, prompted by considerations in conformal field theory, I was lead to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.


By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.


However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a *meaning*. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:



> 
> **Question:** Can someone come up with a *non-computational* proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?
> 
> 
> 


[Added later]:
My paper on [WZW models and $H^4(BG,\mathbb Z)$](http://arxiv.org/abs/1602.02968) has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$.




---


For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].  
 Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi\_1(G)$. Then there is a Puppe sequence
$$
\pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2)
$$
It is a well known fact that $\pi\_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H\_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).


Now, here comes the computation:  

$H^\*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$  

from which it follows that for any finite abelian group $A$  

$H^\*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$  

from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$  

$H^\*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$


The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows:
$$
\begin{matrix}
\vdots & \vdots\\
H^4(B\tilde G) & 0 & \vdots & \vdots & & \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots\\
\mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\
\end{matrix}
$$
and the $d\_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$
cannot create torsion in degree four. QED
  
  

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d\_5$ differential is.



$H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
I try to give an argument without spectral sequences, not sure if this can be considered non-computational though. At least, there is a non-computational syllabus: *torsion classes in $H^4(BG,\mathbb{Z})$ would be characteristic classes of torsion bundles over $S^3$ but the latter have to be trivial.* 


Now for a slightly more detailed argument: first, the coefficient formula tells us that torsion in $H^4(BG,\mathbb{Z})$ comes from torsion in $H\_3(BG,\mathbb{Z})$ because torsion in $H\_4(BG,\mathbb{Z})$ would not survive the dualization. Since $G$ is connected, the Hurewicz theorem provides a surjection $\pi\_3(BG)\to H\_3(BG,\mathbb{Z})$, so any torsion class in $H\_3(BG,\mathbb{Z})$ comes from a map $S^3\to BG$ classifying a $G$-bundle over $S^3$. The corresponding clutching map $S^2\to G$ is homotopic to the constant map because $\pi\_2(G)$ is trivial. So we find that any $G$-bundle on $S^3$ is trivial, hence $H\_3$ and therefore $H^4$ of $BG$ do not have non-trivial torsion. 


