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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 $w=s_{i(1)}s_{i(2)}\dots s_{i(l)}$ …
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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 …
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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 …
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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 …
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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 …
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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 …
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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 …
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