$O(2)$ is an extension of $\mathbb{Z}_2$ by $SO(2)$, $$1\to SO(2) \to O(2)\to \mathbb{Z}_2 \to 1$$ Suppose the generator of $SO(2)$ is $j$, and the generator of $\mathbb{Z}_2$ is $r$, then $O(2)$ is generated by the pair $(j, r)$ with $rj=j^{-1} r$ and $r^2=1$. Let $w_i(G)$ be the $i$-th stiefel whitney class of the $G$-bundle.

Then what is the relation between $w_2(O(2))$ and $w_2(SO(2))$? Are they the same?

If we instead of considering $O(2)$, but consider the group $Pin(2)^-$, which is generated by $(j, r)$ with $rj=j^{-1}r$, $r^2 j= j r^2$ and $r^4=1$,

then what is the relation between $w_2(Pin(2)^-)$ and $w_2(SO(2))$?

  • $\begingroup$ When you write "the $G$-bundle", what do you mean? $\endgroup$ Nov 6, 2018 at 20:46
  • $\begingroup$ @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory. $\endgroup$
    – user34104
    Nov 6, 2018 at 20:53
  • $\begingroup$ I am not familiar with gauge theory. I've also never seen the notation $w_2(G)$ used for a principal $G$-bundle $P\rightarrow B$, only $w_2(P)$. You also haven't mentioned how "the $SO(2)$-bundle" and "the $O(2)$-bundle" are related, so here is a guess: You have a principal $O(2)$ bundle $Q\rightarrow B$ where $Q$ happens to have the form $P\times_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $P\rightarrow B$. And you're asking how $w_2(P),w_2(Q)\in H^2(B;\mathbb{Z}/2\mathbb{Z})$ compare. Is this an accurate restatement of your first question? $\endgroup$ Nov 6, 2018 at 21:12
  • $\begingroup$ @JasonDeVito Thanks for the clarification. Yes, this is what I mean. $\endgroup$
    – user34104
    Nov 6, 2018 at 21:15
  • $\begingroup$ I don't have time to write a full answer now (and I don't know the answer in the $Pin$ case), but for the case I just described above, we should have $w_2(P) = w_2(Q)$. Shortly, the classifying map $\phi_P:B\rightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $\phi_Q:B\rightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^\ast(BO(n);\mathbb{Z}/2\mathbb{Z})\rightarrow H^\ast(BSO(n);\mathbb{Z}/2\mathbb{Z})$ maps $w_1 \in H^1(BO(n);\mathbb{Z}/2\mathbb{Z})$ to $0$, but otherwise is the "identity map". $\endgroup$ Nov 6, 2018 at 21:21


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