Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
1
vote
Accepted
Product of Bruhat Cells
Many thanks to Paul Garrett's comment above. Inspired by his comment I come up with a proof of the following equivalent statement.
Proposition. If $x\in BwB$ where $w=s_{i(1)}s_{i(2)}\dots s_{i(l)}$ …
1
vote
3
answers
514
views
The annihilator of a Borel subalgebra being its nilpotent radical
In the proof of Lemma 3.2.2 in Chriss and Ginzburg's Representation Theory and Complex Geometry, the final step states that "the annihilator $\mathfrak{b}^\perp\subset \mathfrak{g}^*$ gets identified …
1
vote
The annihilator of a Borel subalgebra being its nilpotent radical
This is a proof I came up myself. Since $\dim \mathfrak{b}+\dim\mathfrak{n}=\dim \mathfrak{g}$ and the Killing form $\kappa$ is non-degenerate, it suffices to prove that $\kappa(x,n)=0$ for any $x\in …
6
votes
2
answers
657
views
Product of Bruhat Cells
Fix a $(B,N)$ pair (Tits system) of a semisimple Lie group $G$. Let $u$ and $v$ be two Weyl group elements such that $l(uv)=l(u)+l(v)$. It is known that $BuvB=(BuB)(BvB)$ (see for example Humphreys's …
2
votes
2
answers
196
views
General linear group action on extensions of finite fields
Let $q$ be a prime power. Let $\mathbb{F}_q$ be the finite field with $q$ elements. Then $\mathbb{F}_{q^n}$ is a field extension of $\mathbb{F}_q$ of degree $n$ and can be considered as an $n$-dimensi …
6
votes
2
answers
511
views
The Analog of Borel Subgroup in a Compact Real Form
I recently learned that there is a one-to-one correspondence between isomorphism classes of complex reductive groups and isomorphism classes of compact connected real Lie groups given by taking a comp …
3
votes
Accepted
Computing affine Springer fibers
I came across this question while studying affine Springer fibers myself, and I hope this answer can help future learners.
Let us fix a Borel subgroup $B\subset G$ and a maximal torus $T\subset G$. Le …