Timeline for Bott-Samelson construction of a perfect Morse function on G/T
Current License: CC BY-SA 3.0
10 events
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| Sep 6, 2016 at 23:55 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 7, 2016 at 23:12 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 28, 2016 at 0:39 | comment | added | user21574 | $G/T\cong G^{\mathbb C}/P$, so see my question long times ago, mathoverflow.net/questions/156883/… , if it is ok, don't forget vote ;) | |
| Jan 28, 2016 at 0:37 | comment | added | user36931 | @HassanJolany These facts are very classical and I'm well aware of how to prove them... the varieties all have Schubert stratifications so certainly no facts by Debarre are needed. Also one has to check something about $\pi_2$ not $\pi_1$. My goal is to explain this to an undergraduate student with no background in algebriac geometry and not much rep theory... Cheers | |
| Jan 28, 2016 at 0:35 | comment | added | user36931 | @LiviuNicolaescu Thank you for the notes. This symplectic point of view is very nice and of course could be an entry point into other subjects. | |
| Jan 28, 2016 at 0:34 | comment | added | user21574 | @user36931 It is known fact from Olivier Debarre that any smooth, complex, rationally connected projective variety is simply connected. So, because $G/T$ is rationally coonected projective variety , so $\pi_1(G/T)=e$ | |
| Jan 28, 2016 at 0:26 | comment | added | user36931 | @HassanJolany As I mentioned in the question, the goal is to \emph{prove} $\pi_2(G)=0$. There are of course other ways to do this, but one standard way is via using some suitable Bruhat or Morse stratification to a priori prove that $\pi_2(G/T)$ is torsion-free. | |
| Jan 27, 2016 at 23:36 | comment | added | Liviu Nicolaescu | Maybe Sect. 3.4 of these notes www3.nd.edu/~lnicolae/Morse2nd.pdf might be more appealing to your student. $G/T$ is a coadjoint orbit and as such it is equipped with many perfect Morse functions. In these notes I also discuss special coadjoint orbits such as Grassmannians and flag manifolds. | |
| Jan 27, 2016 at 23:31 | comment | added | user21574 | Take $G$ be connected and simply-connected. From the principal $G_a$-bundle $G_a\rightarrow G\rightarrow G/G_x=\mathcal{O}_x$, where $\mathcal{O}_x$,is the coadjoint orbit we obtain a long-exact sequence $$\ldots\rightarrow\pi_i(G_x)\rightarrow\pi_i(G)\rightarrow\pi_i(\mathcal{O}_x) \rightarrow \pi_{i-1}(G_x)\rightarrow\ldots.$$ Since $\pi_1(G)=0$ and $\pi_2(G)=0$, exactness considerations imply that $\pi_1(G_x)=\pi_2(\mathcal{O}_x)$. | |
| Jan 27, 2016 at 23:13 | history | asked | user36931 | CC BY-SA 3.0 |