Timeline for on the open bruhat cell
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2013 at 20:43 | comment | added | Jim Humphreys | @prochet: With this version of the open cell, which contains $B$, Marguax's comment takes care of your first question (since every element of $G$ is conjugate to an element of $B$ by standard structure theory) and probably also the second. | |
| Oct 6, 2013 at 19:49 | comment | added | prochet | yes sorry, I corrected it. | |
| Oct 6, 2013 at 19:49 | history | edited | prochet | CC BY-SA 3.0 |
correct statement
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| Oct 6, 2013 at 19:41 | comment | added | Jim Humphreys | The open Bruhat cell is the double coset $Bw_0 B$ in the Bruhat decomposition of $G$ (whereas the Schubert cells live in the flag variety isomorphic to $G/B$). Aside from this, the identity element of $G$ lies in the double coset $B$ but not in any other double coset, so your union of conjugates can't be all of $G$. The question needs some reformulation, I think. | |
| Oct 6, 2013 at 19:15 | history | edited | prochet | CC BY-SA 3.0 |
fixed typos
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| Oct 6, 2013 at 19:12 | comment | added | Marguax | Is $G$ connected reductive (so "open schubert cell" makes sense)? If so then every $h \in G$ lies in a Borel, and every Borel is conjugate to a single Borel, so since an open cell contains a Borel we are done (for your first question). | |
| Oct 6, 2013 at 17:54 | history | asked | prochet | CC BY-SA 3.0 |