3
$\begingroup$

$O(2)$ 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 $w_2(SO(2))$?

$\endgroup$
6
  • $\begingroup$ When you write "the $G$-bundle", what do you mean? $\endgroup$
    – Jason DeVito
    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? $\endgroup$
    – Jason DeVito
    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". $\endgroup$
    – Jason DeVito
    Nov 6, 2018 at 21:21

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.