# Stiefel-Whitney class of $O(2)$ and $SO(2)$

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

• When you write "the $G$-bundle", what do you mean? Nov 6, 2018 at 20:46
• @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory. Nov 6, 2018 at 20:53
• 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? Nov 6, 2018 at 21:12
• @JasonDeVito Thanks for the clarification. Yes, this is what I mean. Nov 6, 2018 at 21:15
• 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". Nov 6, 2018 at 21:21