Low dimensional integral cohomology of $BPSO(4n)$

Question

Toda has calculated the $\mathbb{Z}/2$‐cohomology ring of $BPSO(4n+2)$, and also gave the simple exceptional calculation of the $\mathbb{Z}/2$‐cohomology of $BPSO(4)$, in

• Hiroshi Toda, Cohomology of Classifying Spaces, Advanced Studies in Pure Mathematics 9, (1986) Homotopy Theory and Related Topics pp. 75-108, doi:10.2969/aspm/00910075

(This is also done in a different way in Kono–Mimura, referenced below, and by Baum–Browder) However for the general case of $BPSO(4n)$, we get a stability result for low-dimensional cohomology once $n\geq 2$, in particular $H^4(BPSO(8),\mathbb{Z}/2)$ gives all the higher-rank cases, too (this is in Kono and Mimura, On the Cohomology of the Classifying Spaces of $PSU(4n+2)$ and $PO(4n+2)$, doi:10.2977/prims/1195191887). Now I'm interested in the integral cohomology, rather than at the prime 2, mostly because amongst all compact simple Lie groups, $PSO(4n)$ is the only exceptional case where $H^3(PSO(4n),\mathbb{Z})$ is not $\mathbb{Z}$, but is $\mathbb{Z}\oplus \mathbb{Z}/2$, assuming $n\geq 2$.

I presume the stability result still holds for integral cohomology (correct me if I'm wrong!), and so aside from the exceptional case of $BPSO(4)$, I presume that knowing $H^4(BPSO(8),\mathbb{Z})$ means we would then know all the groups $H^4(BPSO(4n),\mathbb{Z})$.

Now I know that the torsion class in $H^3(PSO(4n),\mathbb{Z})$ is even equivariant for the conjugation action of $PSO(4n)$ on itself, that is, it comes from equivariant cohomology $H^3_{PSO(4n)}(PSO(4n),\mathbb{Z})$. But what I don't know is if this class comes from $H^4(BPSO(4n),\mathbb{Z})$.

• What's a reference that gives the fourth integral cohomology group of $BPO(4n)$?

Answer 1

Gawedzki and I have investigated this question for all compact simple Lie groups using the descent of multiplicative bundle gerbes from simply connected covers to quotients by subgroups of the center:

Gawedzki, Krzysztof, and Konrad Waldorf. "Polyakov-Wiegmann formula and multiplicative gerbes." Journal of High Energy Physics 2009.09 (2009): 073.
https://iopscience.iop.org/article/10.1088/1126-6708/2009/09/073/pdf

In your case, our results show that the pullback map $$ H^4(BPSO(4n),Z) \to H^4(BSpin(4n),Z)=\mathbb{Z} $$ is indeed injective, and that the image consists of all even $k\in \mathbb{Z}$ such that $4$ divides $kn$.

EDIT: I extracted this from our paper as follows. $H^4(BG,\mathbb{Z})$ classifies multiplicative bundle gerbes over $G$, and - that was our motivation there - Chern-Simons theories with gauge group $G$. In the paper we discuss the question when multiplicative bundle gerbes descent along the quotient map $\tilde G \to G:=\tilde G/Z$, where $\tilde G$ is a compact, connected, simply, simply-connected Lie group, and $Z$ is a subgroup of its center. For this purpose, we classified equivariant structures on multiplicative bundle gerbes over $\tilde G$, and obtained two results:

• Equivariant structures, if they exist, are in all cases unique. This shows injectivity of above map, or, in general, injectivity of the map $$H^4(BG,\mathbb{Z}) \to H^4(B\tilde G,\mathbb{Z}).$$

• The existence of equivariant structures is obstructed by the non-triviality of the so-called Felder-Gawedzki-Kupiainen-cocycle, which was known before for all $G$ and $Z$ as above. In the paper, we have a table on page 26 telling for all cases the obstructions ("called CS-constraints"), depending on $\tilde G$, $Z$, and the level $k\in \mathbb{Z}=H^4(B\tilde G,\mathbb{Z})$. In the relevant case of $\tilde G=Spin(4r)$ and $Z=\mathbb{Z}_2\times\mathbb{Z}_2$, this gives what I wrote above.

Answer 2

There is something special about $H^4$ that doesn't work in general. This way I can rely on a result proved by Henriques in

• The classification of chiral WZW models by $H^4_+(BG,\mathbb{Z})$, arXiv:1602.02968, doi:10.1090/conm/695

and discussed at $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group, rather than looking for a calculation of the cohomology of $BPSO(4n)$ in general. Namely that for any compact, connected Lie group $G$, $H^4(BG,\mathbb{Z}) = H^4(BT,\mathbb{Z})^W$, for $T\subset G$ a maximal torus and $W$ its Weyl group. Thus it is torsion free, and it's nothing to do with my special case. In particular the torsion class in $H^3(PSO(4n),\mathbb{Z})$ doesn't come from $BPSO(4n)$.