# Cohomology of finite quotients of Lie groups

Let $G$ denote the $Spin(n)$ group with $n>4$ and let $\Gamma$ be a cyclic subgroup $G$ of a prime order $p >2$. When does the projection $G \to G/\Gamma$ induce a surjection between cohomology groups $H^3$ with integral coefficients?

-
"Finite quotient" usually means that $G/\Gamma$ is finite not that $\Gamma$ is finite as in your question. – Jim Conant Jun 22 '11 at 11:41
No, I would not like to assume that $\Gamma$ is in the center. If my Lie group is $Spin$, as in the question, and the order of the group not divisible by $2$, it cannot be in the center anyway. – Alexander Lytchak Jun 22 '11 at 17:15
@Alexander: Right, I didn't notice "prime". Well, I would know how to use it: for central subgroups the question can be reformulated in terms of levels of orbifold WZW models, and these are all known. – Konrad Waldorf Jun 22 '11 at 17:45
For $G=SU(3)$ and $\Gamma=Z(G)$, the map $is$ surjective, for example. More general, for $SU(n)$ the map is surjective if and only if $n$ is odd. – Konrad Waldorf Jun 22 '11 at 17:48
@Konrad: This sounds intersting and would help me to understand the situation at least a little bit. Is it difficult to see? Or is there a reference for this claim? – Alexander Lytchak Jun 22 '11 at 17:57

## 2 Answers

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.

EDIT: For non-central groups one can still use the basic gerbe $\mathcal{G^1}$, but I don't know if the obstruction classes are then still accessible for calculations.

-
It sounds very interesting. Unfortunately, I do not know what gerbes are have to believe you without understanding the reason. Your answer shows that the problem is probably very difficult in general. – Alexander Lytchak Jun 22 '11 at 19:38

Here's another approach using the Leray-Serre spectral sequence. Using the fact that $\Gamma$ acts freely on $G$, the map $G \to G/\Gamma$ is a covering space, and so the cohomology of $G/\Gamma$ may be computed by the spectral sequence

$$E_2^{s, t} = H^s(\Gamma, H^t(G)) \implies H^{s+t}(G/\Gamma).$$

Here $H^s(\Gamma, M)$ is the s'th group cohomology of $\Gamma$ with coefficients in the $\Gamma$-representation $M$. We record three facts to start:

1. If $\Gamma \cong \mathbb{Z} / p$, then $H^s(\Gamma, \mathbb{Z})$ is 0 for $s$ odd, and $\mathbb{Z} / p$ for $s$ even.
2. For $G=Spin(n)$, $H^t(G) = \mathbb{Z}, 0, 0, \mathbb{Z}$, for $t=0,1,2,3$.
3. For $p$ odd, $\mathbb{Z} / p$ cannot act nontrivially on $\mathbb{Z}$, since $Aut(\mathbb{Z}) = \mathbb{Z} / 2$.

Let's compute part of the $E_2$-term of the spectral sequence. Fact 2 implies that the $t=1$ and $2$ rows vanish entirely. Fact 3 implies that $H^0(G) = \mathbb{Z}$ and $H^3(G) = \mathbb{Z}$ are trivial $\Gamma$-modules, so the $t=0$ and $3$ rows are the group cohomology described in Fact 1.

Thus the only possible term that can contribute to $H^3(G/\Gamma)$ is $E_2^{0, 3} = H^3(G) = \mathbb{Z}$. There is, however, a possiblity of a single differential in this region of the spectral sequence, namely

$$d_4: E_2^{0, 3} = \mathbb{Z} \longrightarrow E_2^{4, 0} = H^4(\Gamma, \mathbb{Z}) = \mathbb{Z} / p.$$

Therefore, $H^3(G/\Gamma)$ surjects onto $H^3(G)$ precisely when this differential $d_4= 0$. We note that if it's not 0, it is surjective; thus at worst $H^3(G/\Gamma)$ may be identified with an index $p$ subgroup of $H^3(G)$.

So, how do we compute $d_4$? I claim that it's given as:

$$H^3(G) \cong H^4(BG) \to H^4(B\Gamma) = H^4(\Gamma)$$

where the map is the restriction in cohomology, induced by the (inclusion) homomorphism $\Gamma \subseteq G$. This can be seen, for instance, by comparing this with the spectral sequence for the (rather dumb) fibration $G \to G/G=pt$.

So a long winded answer to your question is: The map is surjective if and only if none of the $H^4(BG) = \mathbb{Z}$ is supported on $H^4(\Gamma) = \mathbb{Z} /p$. I would imagine that determining when that is the case is highly dependent upon the subgroup in question.

-