Group cohomology of $PSU(n)$

Question

I am having a hard time to find an explicit description of $H^4(B\ PSU(n),\mathbb Z)$, the fourth cohomology group of the projective special unitary group of rank $n$, with integer coefficients. I have a very vague intuition that perhaps $H^4=\mathbb Z\oplus\mathbb Z_n$, where the free factor is generated by (the image of) the second Chern class, and the torsion by some $\mathbb Z_n$-generalisation of a Stiefel-Whitney class. Is this completely off? If so, what is the correct description of this group?

(I would also be happy to hear about more general results regarding this problem, e.g. a description of the full cohomology ring, or the result for other simple non-simply-connected compact groups. Perhaps this should go into a separate post.)

Answer 1

Suppose $G$ is a Lie group. In general, I think computing the cohomology ring of $BG$ is non-trivial, even when $G$ is simple compact Lie group. Of course, when $H^\ast(G;\mathbb{Z})$ is torsion free (e.g., $G = SU(n), Sp(n)$)), then it's not bad. But even for $G = Spin(n)$, it's non-trivial.

All that said, I think calculating $H^4$ is tractable.

Proposition: Suppose $G$ is a connected compact Lie group. Then $H^4(BG)$ is a free abelian group. Further, if $G$ is semi-simple, then the number of factors is precisely the number of simple factors of $G$.

Proof: First we'll show $H^4(BG)$ is free abelian. Since it's finitely generated, it's enough to show it's torsion free. From the universal coefficients theorem, the torsion in $H^4(BG)$ is isomorphic to the torsion in $H_3(BG)$.

We claim that $H_3(BG) = 0$. To that end, notice that $BG$ is simply connected since $G$ is connected, so by the Hurewicz theorem, the map $\pi_3(BG)\rightarrow H_3(BG)$ is surjective. However, $\pi_3(BG)\cong \pi_2(G) = 0$, as it is for every Lie group. Thus, we conclude $H^4(BG)$ is torsion free.

To count the number of factors, we use the rational Hurewicz theorem. The rational homotopy group $\pi_2(BG)\otimes \mathbb{Q} = 0$ because $\pi_2(BG)\cong \pi_1(G)$ is a finite abelian group (since $G$ is semi-simple). Since $\pi_3(BG) = 0$ as mentioned above, we see that $BG$ is rationally $3$-connected. The rational Hurewicz theorem then gives that the map $\pi_4(BG)\otimes \mathbb{Q}\rightarrow H_4(BG;\mathbb{Q})$ is an isomorphism. By universal coefficients, $H_4(BG;\mathbb{Q})\cong H^4(BG;\mathbb{Q})$ and by the homology universal coeficients, $H^4(BG;\mathbb{Q})\cong H^4(BG;\mathbb{Z})\otimes\mathbb{Q}$.

So, we conclude the dimension of $\pi_4(BG)\otimes \mathbb{Q}$ (as a rational vector space) is equal to the rank of $H^4(BG;\mathbb{Z})$.

Now, $\pi_4(BG)\cong \pi_3(G)\cong \pi_3(\tilde{G}))$ where $\tilde{G}$ is the universal cover of $G$. The universal cover splits into a product of simple simply connected Lie groups, and it is known that $\pi_3 \cong \mathbb{Z}$ for each of those. $\square$

Answer 2

Because this is a central extension we have a fiber sequence $$BSU(n) \to BPSU(n) \to B^2(\Bbb Z/n).$$ From the fiber sequence $$\Bbb{CP}^\infty \xrightarrow{\otimes n} \Bbb{CP}^\infty \to B^2(\Bbb Z/n),$$ where the first map is the one that induces $\times n$ on $\pi_2$, it is straightforward to see that $H^*(B^2(\Bbb Z/n), \Bbb Z)$ is $\Bbb Z$ in degree zero, $\Bbb Z/n$ in degree 3, zero in degree 4, and $H^5(B^2(\Bbb Z/n);\Bbb Z) = \Bbb Z/n$ when $n$ is odd and is $\Bbb Z/2n$ when $n$ is even. Write $A(n)$ for this group.

and otherwise is supported in degrees $6$ and higher.

Now it is well known that $H^*(BSU(n);\Bbb Z) = \Bbb Z[c_2, \cdots, c_n]$, where $|c_i| = 2i$.

Thus the beginning of the first spectral sequence is

$$\begin{matrix} \Bbb Z & 0 & 0 & \Bbb Z/n & 0 & A(n) &\cdots\\ 0 & 0 & 0 & 0 & 0 & 0 &\cdots \\ 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\ 0 & 0 & 0 & 0 & 0 & 0 &\cdots \\ \Bbb Z & 0 & 0 & \Bbb Z/n & 0 & A(n) & \cdots \end{matrix}$$

Whatever the differential $d_4$ is, its kernel $E^{0,4}_5$ is still isomorphic to $\Bbb Z$, and it is the only thing that lies on the $s+t=4$ line. Thus $H^4(BPSU(n);\Bbb Z) = \Bbb Z$, generated by a class which pulls back to a multiple $a(n) c_2$, where $a(n)$ is an integer that divides $n$ when $n$ is odd or divided $2n$ when $n$ is even.

In the special case $BSO(3) = BPSU(2)$, we in fact have $H^*(BSO(3);\Bbb Z) \cong \Bbb Z[e,p_1]/(2e)$, where $|e| = 3$ and $|p_1| = 4$, which agrees with the beginning of the displayed spectral sequence.