Timeline for Isomorphism from $\mathbb{Z}$ to third homotopy group of compact simple Lie group
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2017 at 0:52 | comment | added | YCor | @AndréHenriques thanks, that's clear indeed! | |
| Feb 16, 2017 at 23:38 | comment | added | André Henriques | @YCor: yes, if $G$ is a simple simply connected Lie group, then $\pi_3(G)=H_3(G,\mathbb Z)$ is canonically isomorphic to $\mathbb Z$. Similarly, $H^3(G,\mathbb Z)$ is canonically isomorphic to $\mathbb Z$. For Lie algebra cohomology, $H^3(\mathfrak g,\mathbb R)=H^3(G,\mathbb R)$ is canonically $\mathbb R$. At the level of Lie algebra cohomology, one can write down a formula that makes it obvious that the isomorphism is canonical: the canonical generator of $H^3(\mathfrak g,\mathbb R)$ sends $X,Y,Z$ to $\langle [X,Y],Z\rangle$ where $\langle\,,\,\rangle$ is the basic inner product. | |
| Feb 14, 2017 at 3:43 | answer | added | Francois Ziegler | timeline score: 11 | |
| Feb 14, 2017 at 0:57 | comment | added | YCor | If I'm correct, the outer automorphism of $\mathfrak{su}_n$ acts as the identity on $\pi_3$ (I use identification with $H_3$ and then identification of the latter to the (predual of) 1-dimensional space of invariant quadratic forms, working now in $\mathfrak{sl}_n$ as we can complexify). It is true that automorphisms of compact simple Lie groups always act as the identity on $\pi_3$? | |
| Feb 13, 2017 at 21:32 | history | edited | YCor |
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| Feb 13, 2017 at 20:56 | comment | added | Kevin Casto | @user104853 Well, any automorphism of $G$ acts by post-composition on $\pi_3(G) = \mathbb{Z}$, and therefore must take $\rho$ to either itself or $-\rho$ in $\pi_3$. | |
| Feb 13, 2017 at 20:52 | comment | added | Allen Knutson | I believe that Jacobson-Morozov applies to compact groups, i.e. maps $\phi$ up to conjugacy are indexed by conjugacy classes of nilpotents in the complex group. And then I think that the $\rho$s that you're interested in are the ones corresponding to short simple roots, so in particular, conjugate. | |
| Feb 13, 2017 at 19:44 | comment | added | user104853 | @მამუკაჯიბლაძე I'm not sure how we would get $\varphi_*=\rho_*$. The point about inner automorphisms is that they can all be continuously deformed to the identity since $G$ is connected. I don't know any outer automorphism with this property. | |
| Feb 13, 2017 at 18:58 | comment | added | მამუკა ჯიბლაძე | E.g. $G$ may have outer automorphisms. But you probably would like to exclude these too? | |
| Feb 13, 2017 at 18:00 | review | First posts | |||
| Feb 13, 2017 at 18:17 | |||||
| Feb 13, 2017 at 17:56 | history | asked | user104853 | CC BY-SA 3.0 |