Timeline for Bruhat cell of a Coxeter element
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2019 at 17:12 | comment | added | Ami | @user44191 oops yes you are of course right, I meant that every element of $J(\mathbb Z)$ is conjugate to an element of $BwB$ in $\operatorname{SL}_n(\mathbb Z)$ | |
| Oct 8, 2019 at 16:23 | comment | added | user44191 | @Ami For $n \geq 3$, it is not true that $BwB \supset J$. For example, for $n = 3$, the minor gotten by removing the 1st row and 3rd column of any element of $BwB$ is $0$, if I haven't made a mistake, which is not true for every element of $J$. | |
| Oct 8, 2019 at 15:53 | comment | added | Ami | @user44191 what about when $G=\operatorname{SL}_{n}(\mathbb C)$ and $w$ is the permutation matrix of $(123..n)$ when $B$ is the upper triangular matrices and $T$ the diagonal matrices. Then $J=\{A\in \operatorname{SL}_{n}(\mathbb C)|\operatorname{disc}(A)\neq 0\}$ is Zariski open and contained in $BwB$? | |
| Oct 8, 2019 at 15:23 | history | undeleted | Ami | ||
| Oct 6, 2019 at 19:54 | history | deleted | Ami | via Vote | |
| Aug 31, 2019 at 19:57 | comment | added | Jim Humphreys | @Ami: By the way, it seems misleading to refer to 'Bruhat-Coxeter cells' here. Coxeter had nothing to do with this idea. The tag 'topological-groups' is also misleading, since an affine algebraic group (with the Zariski topology) usually fails to be a topological group: the Zariski topology is typically not Hausdorff. | |
| Aug 28, 2019 at 19:24 | comment | added | Jim Humphreys | @Ami: I don't have a suitable reference in mind. But in any case, I don't see why a Coxeter element of $W$ should plaly a special role in the Bruhat decomposition here, or why other elements of $W$ should automatically be involved in a Chevalley group over $\mathbb{Z}$. (I did study arithmetic groups at an early stage but got more attracted to modular representation theory of algebraic groups.) | |
| Aug 27, 2019 at 20:33 | comment | added | Ami | @JimHumphreys is it also true that for any algebraic extension of $\mathbb Z$, $\overline {\mathbb Z}$, that $G(\overline {\mathbb Z})$ has a Tits system, with the same presentation for the Weyl group in $G(\mathbb C)$? Is there a good source for that? | |
| Aug 23, 2019 at 23:23 | history | edited | Ami | CC BY-SA 4.0 |
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| Aug 23, 2019 at 23:22 | comment | added | Ami | I meant the more general version by Steinberg, so you are saying that the Tits system of $G(\mathbb Z)$ is the Tits system of $G(\mathbb C)$ intersected with $G(\mathbb Z)$ and have the same presentation for the Weyl group in $G(\mathbb C)$? so $G(\mathbb Z)$ contains nontrivial elements from every cell, but I don't see how it shows $H$ contain a conjugation to an element in a Bruhat-Coexeter cell? Also, I forgot to mention that $H$ has a finite index. | |
| Aug 23, 2019 at 19:55 | comment | added | Jim Humphreys | What exactly do you mean by "Chevalley group"? This could be the adjoint-type group, as in Chevalley's 1955 paper, based on the adjoint representation, or could be the more general version treated in Steinberg's 1967-68 Yale lectures, based on a more general faithful representation for groups such as SL$_n(\mathbb{C})$. In either setting, the group over $\mathbb{Z}$ has a Tits system (BN-pair), with "Weyl group" the full $W$ involved. This ensures nonrtrivial intersection with each Bruhat cell. | |
| Aug 23, 2019 at 1:28 | comment | added | user44191 | "I know that the Bruhat cells that are not the big cell are closed"? Only the small cell (corresponding to $e \in W$) is closed, unless you mean the closed Bruhat cells (in which case, the big cell is the entire variety). The rest are the intersection of a closed set (the closed Bruhat cell) with an open set (the complement of the union of Bruhat cells with smaller $w$ in the Bruhat order). The statement after is still true (they contain no open sets), though. | |
| Aug 23, 2019 at 0:59 | history | edited | Ami | CC BY-SA 4.0 |
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| Aug 22, 2019 at 20:37 | history | asked | Ami | CC BY-SA 4.0 |