Group cohomology of $PSU(n)$

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.

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. $\square$