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 $H^4(BG,\mathbb Z)$ 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.

  • 5
    $\begingroup$ I am well out of my depth here, but is there a stupid reason why this is not true for finite groups? $\endgroup$ Sep 5, 2014 at 16:33
  • 9
    $\begingroup$ @DanielBarter: in some sense yes. Cohomology of a finite groups is always annihilated by the order of the group, by a transfer argument. An easy explicit example of torsion in $H^4$ for finite groups appears e.g. for cyclic and dihedral groups. So the connectedness assumption is relevant. $\endgroup$ Sep 5, 2014 at 18:03
  • 6
    $\begingroup$ If $G$ is simply connected then $\pi_i(G)$ is torsion-free for $i<4$ since $\pi_2$ of a Lie group is zero and $\pi_3$ is always torsion-free. Thus $\pi_i(BG)$ is torsion-free for $i\le 4$ and the result follows from mod C Hurewicz. $\endgroup$ Sep 5, 2014 at 21:50
  • 2
    $\begingroup$ Quick thought, probably flawed: $Tors(H^4(BG;Z)) = H^3(BG;\mathbb{R}/Z) = H^3(BT;\mathbb{R}/Z)^W$ where $T\subset G$ is maximal torus and $W$ is Weyl group, and I think $H^*(BT;\mathbb{R}/Z)$ has only even-dimensional generators? The first "=" is taken from the Dijkgraaf-Witten paper, I don't immediately see why it's true, but I do see that $H^k(BG;\mathbb{R}/Z)\hookrightarrow H^{k+1}(BG;Z)$ when $k$ is odd, with image given by kernel of $H^{k+1}(BG;Z)\to H^{k+1}(BG;\mathbb{R})$ which at least contains the torsion elements. Perhaps then $Tors(H^\text{even}(BG;\mathbb{Z})) = 0$? Doubtful. $\endgroup$ Sep 8, 2014 at 9:15
  • 3
    $\begingroup$ @Ben: $G=SO(3)$ is an easier example. $\endgroup$ Sep 8, 2014 at 18:08

2 Answers 2


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.

  • $\begingroup$ I like this argument. $\endgroup$ Oct 1, 2014 at 18:52
  • 2
    $\begingroup$ Note that this argument proves the slightly stronger statement $H_3(BG,\mathbb Z)=0$. $\endgroup$ Oct 1, 2014 at 19:48
  • 1
    $\begingroup$ Don't you need $G$ to be simply connected to apply the Hurewicz theorem? $\endgroup$ Oct 1, 2014 at 21:24
  • 4
    $\begingroup$ @DanPetersen: I don't think so. The group $G$ is connected, so $BG$ is simply connected. In this situation, the Hurewicz theorem applied to $BG$ gives a bijection $\pi_2(BG)\to H_2(BG)$ and a surjection $\pi_3(BG)\to H_3(BG)$. The latter is the thing I use to represent any homology class by some bundle. $\endgroup$ Oct 2, 2014 at 4:15
  • 1
    $\begingroup$ @AntonFetisov: not quite. For homology, $H_3(K(\pi,2),\mathbb{Z})=0$ by the Hurewicz theorem. André's calculation is about cohomology $H^3(BG,\mathbb{Z})$ whose torsion, by the universal coefficient formula, comes from the torsion in $H_2(BG,\mathbb{Z})\cong\pi_1(G)$. I think my argument applies not just to simply connected Lie groups. $\endgroup$ Oct 2, 2014 at 4:17

Here is another spectral sequence argument. It is mainly of interest because it is different, but I think that it is slightly easier and will detail how. The spectral sequence is used by Deligne in the paper cited in the comments by Guest, but only for the simply connected case. The advantage is that André's argument required knowing that $H^4(K(\pi,2);\mathbb Z)$ is torsion-free, while this argument does not. Also, it does not seem to require knowing that $\pi_2$ of a Lie group vanishes or that $\pi_3$ is torsion-free. Indeed, it appears to prove those facts, but they are probably closely related to the required input. What it does require knowing is that $K/T$, the quotient of a compact group by its maximal torus, has cohomology that is torsion-free and concentrated in even degrees. This is usually shown by the Schubert (or Bruhat) decomposition into even dimensional cells. The spectral sequence is the Serre spectral sequence for the fiber sequence $K/T\to BT\to BK$, namely $H^*(BK; H^*(K/T))\Rightarrow H^*(BT)$. Since $BK$ is simply connected, the cohomology is untwisted; and since the coefficients are torsion-free (another advantage over the other spectral sequence), the $E_2$ becomes a tensor product: $H^*(BK)\otimes H^*(K/T)\Rightarrow H^*(BT)$ and takes this form: $$\begin{matrix} \vdots\\ * & \vdots \\ 0 & 0 \\ * & 0 & * \\ 0 & 0 & 0 & 0 \\ \mathbb Z & 0 & * & * & H^4(BK) & \cdots \\ \end{matrix}$$

There is no room for a differential to come from or hit $H^4(BK)$, so it injects into $H^4(BT)$, which is torsion-free. Injecting into a torsion-free group is slightly nicer than in the other spectral sequence, where it is an extension of two torsion-free groups. In the simply connected case, $H^4(BK)=H^4(BT)^W$, but that is not true in general (cf the spin characteristic class $\frac{p_1}2$). and that's also true in the non-simply connected case (cf comments below).

  • $\begingroup$ Ben: I'm not sure what you mean by "cf the characteristic class $\frac{p_1}2$". Could you please elaborate? $\endgroup$ Sep 9, 2014 at 14:13
  • 1
    $\begingroup$ I was confused. Now I am so confused that not only do I not have an example that doesn't fill up the invariants, but my analyses reach contradictions. $\endgroup$ Sep 9, 2014 at 20:07
  • $\begingroup$ I think that even in the non-simply connected case $H^4(BK)=H^4(BT)^W$. The argument goes roughly as follows. Consider a finite cover $\tilde K$ of $K$ which is a product of some simply connected group and some torus. Let $A$ be the kernel of the projection map from $\tilde K$ to $K$. Then compare the Serre spectral sequences for $B\tilde K\to BK\to K(A,2)$ and $B\tilde T\to BT\to K(A,2)$. Key fact: the map of SS is injective in all the relevant bi-degrees, namely (0,2) (0,4) (3,0) (3,2) (5,0). The $E_\infty$ term for $H^4(BK)$ is therefore given by $H^4(BT)\cap H^4(B\tilde K)$. $\endgroup$ Sep 9, 2014 at 22:15
  • 1
    $\begingroup$ That makes sense, though I am still very confused. That answers your question, right? It identifies the group with a group that is obviously torsion-free. $\endgroup$ Sep 10, 2014 at 5:51
  • $\begingroup$ I don't know that it exactly answers the question, as the argument is still quite computational, but it certainly provides a very valuable piece of information. $\endgroup$ Sep 10, 2014 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.