# Is a principal $\mathbb{Z}_2\ltimes PSU(4)$-bundle over a 3-manifold $M$ equivalent to an element in $H^1(M,\mathbb{Z}_2)\times H^2(M,\mathbb{Z}_4)$?

Given a 3-manifold $$M$$ and a principal $$\mathbb{Z}_2\ltimes PSU(4)$$-bundle $$P$$ over $$M$$ whose isomorphism class is represented by the homotopy class of a map $$f:M\to B(\mathbb{Z}_2\ltimes PSU(4))$$ where $$\mathbb{Z}_2$$ acts on $$PSU(4)$$ by the outer automorphism.

Since there is a (right split) short exact sequence of groups: $$1\to PSU(4)\to \mathbb{Z}_2\ltimes PSU(4)\to\mathbb{Z}_2\to1,$$ we have a fiber sequence $$BPSU(4)\to B(\mathbb{Z}_2\ltimes PSU(4))\to B\mathbb{Z}_2.$$ Inspired by my previous question, there seems to be no differentials in the Serre spectral sequence $$H^p(B\mathbb{Z}_2,H^q(BPSU(4),\mathbb{Z}_2))\Rightarrow H^{p+q}(B(\mathbb{Z}_2\ltimes PSU(4)),\mathbb{Z}_2)$$ for $$p+q\le3$$.

So I think $$P$$ can be identified with a principal $$\mathbb{Z}_2\times PSU(4)$$-bundle over $$M$$ whose isomorphism class is represented by the homotopy class of a pair of maps:

$$g:M\to B\mathbb{Z}_2$$ and $$h:M\to BPSU(4).$$

By my previous another question, for 3-manifold $$M$$, $$h$$ is equivalent to a map $$h':M\to B^2\mathbb{Z}_4.$$

Now the homotopy class of $$g$$ is a cohomology class in $$H^1(M,\mathbb{Z}_2)$$, the homotopy class of $$h'$$ is a cohomology class in $$H^2(M,\mathbb{Z}_4)$$.

So my question reduces to the above steps. Can you help me prove or disprove them?

Thank you!

• The thing that I worry about is that I do not see a canonical way to "tensor a real line bundle in" so that $w_1(P \otimes \xi) = w_1(P) + w_1(\xi)$. If you could, this would be easy: tensor with the line bundle corresponding to $w_1(P)$ and now you are left with a $PSU(4)$-bundle, which you see the classification of. Instead, maybe what happens is more like the case of $O(2)$: there you still have a class $w \in H^1(M;\Bbb Z/2)$, but the second class lies in $H^2(M;\Bbb Z_w)$, the $w$-twisted cohomology, instead of $H^2(M;\Bbb Z) = [M, BSO(2)]$. – user98602 Dec 5 '18 at 22:17
• You have written equality when it is unclear what that means. Is this an abstract isomorphism of groups? There won't be any geometric meaning to it, and it's unclear whether you can link the discussion of characteristic classes between the two cases. In particular, I do not see how you cook up the $PSU(4)$ bundle over $M$ other than by appealing to an abstract calculation of these bordism groups (without geometric content). – user98602 Dec 8 '18 at 13:36
• @MikeMiller Sorry, my edit changed my original question, I'm just afraid that the answer to my original question is negative, maybe my original question is more meaningful. – Borromean Dec 8 '18 at 14:15
• I suspect the answer to your original question is indeed negative but I hope that it has an interesting answer nonetheless. For instance, perhaps analagous to my first comment, the second characteristic classes lies in twisted $\Bbb Z/4$-cohomology? This would be interesting (at least to me). – user98602 Dec 8 '18 at 14:17
• can you guys post an answer ??? -- the bounty ends soon. – annie heart Dec 14 '18 at 17:01