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












3














$O(2)$ is an extension of $mathbb{Z}_2$ by $SO(2)$,
$$1to SO(2) to O(2)to mathbb{Z}_2 to 1$$
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 $w_2(SO(2))$?











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  • When you write "the $G$-bundle", what do you mean?
    – Jason DeVito
    Nov 6 '18 at 20:46










  • @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
    – user34104
    Nov 6 '18 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 $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
    – Jason DeVito
    Nov 6 '18 at 21:12












  • @JasonDeVito Thanks for the clarification. Yes, this is what I mean.
    – user34104
    Nov 6 '18 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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
    – Jason DeVito
    Nov 6 '18 at 21:21


















3














$O(2)$ is an extension of $mathbb{Z}_2$ by $SO(2)$,
$$1to SO(2) to O(2)to mathbb{Z}_2 to 1$$
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 $w_2(SO(2))$?











share|cite|improve this question
























  • When you write "the $G$-bundle", what do you mean?
    – Jason DeVito
    Nov 6 '18 at 20:46










  • @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
    – user34104
    Nov 6 '18 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 $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
    – Jason DeVito
    Nov 6 '18 at 21:12












  • @JasonDeVito Thanks for the clarification. Yes, this is what I mean.
    – user34104
    Nov 6 '18 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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
    – Jason DeVito
    Nov 6 '18 at 21:21
















3












3








3


1





$O(2)$ is an extension of $mathbb{Z}_2$ by $SO(2)$,
$$1to SO(2) to O(2)to mathbb{Z}_2 to 1$$
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 $w_2(SO(2))$?











share|cite|improve this question















$O(2)$ is an extension of $mathbb{Z}_2$ by $SO(2)$,
$$1to SO(2) to O(2)to mathbb{Z}_2 to 1$$
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 $w_2(SO(2))$?








general-topology algebraic-topology differential-topology






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 5 at 22:07









wonderich

2,08631230




2,08631230










asked Nov 6 '18 at 20:41









user34104user34104

1478




1478












  • When you write "the $G$-bundle", what do you mean?
    – Jason DeVito
    Nov 6 '18 at 20:46










  • @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
    – user34104
    Nov 6 '18 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 $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
    – Jason DeVito
    Nov 6 '18 at 21:12












  • @JasonDeVito Thanks for the clarification. Yes, this is what I mean.
    – user34104
    Nov 6 '18 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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
    – Jason DeVito
    Nov 6 '18 at 21:21




















  • When you write "the $G$-bundle", what do you mean?
    – Jason DeVito
    Nov 6 '18 at 20:46










  • @JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
    – user34104
    Nov 6 '18 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 $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
    – Jason DeVito
    Nov 6 '18 at 21:12












  • @JasonDeVito Thanks for the clarification. Yes, this is what I mean.
    – user34104
    Nov 6 '18 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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
    – Jason DeVito
    Nov 6 '18 at 21:21


















When you write "the $G$-bundle", what do you mean?
– Jason DeVito
Nov 6 '18 at 20:46




When you write "the $G$-bundle", what do you mean?
– Jason DeVito
Nov 6 '18 at 20:46












@JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
– user34104
Nov 6 '18 at 20:53




@JasonDeVito I mean principle bundle with structure group $G$, the usual definition in the standard construction in gauge theory.
– user34104
Nov 6 '18 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 $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
– Jason DeVito
Nov 6 '18 at 21:12






I am not familiar with gauge theory. I've also never seen the notation $w_2(G)$ used for a principal $G$-bundle $Prightarrow B$, 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 $Qrightarrow B$ where $Q$ happens to have the form $Ptimes_{SO(2)} O(2)$ for some principal $SO(2)$-bundle $Prightarrow B$. And you're asking how $w_2(P),w_2(Q)in H^2(B;mathbb{Z}/2mathbb{Z})$ compare. Is this an accurate restatement of your first question?
– Jason DeVito
Nov 6 '18 at 21:12














@JasonDeVito Thanks for the clarification. Yes, this is what I mean.
– user34104
Nov 6 '18 at 21:15




@JasonDeVito Thanks for the clarification. Yes, this is what I mean.
– user34104
Nov 6 '18 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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
– Jason DeVito
Nov 6 '18 at 21:21






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 $phi_P:Brightarrow BSO(2)$ for the $P$ bundle above should be the lift of the classifying map $phi_Q:Brightarrow BO(2)$. It is known (though don't know of a reference right off hand) that the map $H^ast(BO(n);mathbb{Z}/2mathbb{Z})rightarrow H^ast(BSO(n);mathbb{Z}/2mathbb{Z})$ maps $w_1 in H^1(BO(n);mathbb{Z}/2mathbb{Z})$ to $0$, but otherwise is the "identity map".
– Jason DeVito
Nov 6 '18 at 21:21












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