Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a closed subgroup (algebraic group) of $GL(n,K)$. If $G$ leaves stable a subspace $W$ of $K^n$, prove that $\mathfrak{g} \subseteq \mathfrak{gl}(n,K)$ does likewise. Converse?

Here, $K$ is an algebraically closed field. $\mathfrak{g}$, $\mathfrak{gl}(n,K)$ are the Lie algebras of $G$ and $GL(n,K)$ respectively.

I think $\mathfrak{g}$ acts on $K^n$ by matrix production. But I find it difficult to corresponds $G$ with $\mathfrak{g}$, especially in the matrix condition. Thanks.

share|improve this question
    
I don't know if this is correct: Suppose that with a given base $\{v_1,\cdots,v_m,v_{m+1},\cdots,v_n \}$ of $K^n$, $m\leq n$, the subspace $W$ is spanned by $v_1,\cdots,v_m$. Then any matrix of $G$ is of the form $\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}$, i.e., for any $(a_{ij})\in G$, $a_{ij}=0$ whenever $i>m$ and $j < m$. –  ShinyaSakai Oct 17 '11 at 12:46
    
Let $I$ denote the ideal in $K[GL(n,K)]$ vanishing on $G$, then $T_{ij}\in I$ if $i>m$ and $j<m$.When $i>m$ and $j<m$, for any $\mathrm{x}\in\mathfrak{g}$, $\mathrm{x}(T_{ij})=0$. So $\mathrm{x}=(\mathrm{x}(T_{ij}))$ is also of the form $\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}$, thus leaves $W$ stable. –  ShinyaSakai Oct 17 '11 at 12:46
add comment

1 Answer

up vote 2 down vote accepted

Given $X \in \mathfrak g, w \in W$, $\exp(tX) \cdot w \in W$. Now differentiate at $t=0$...

share|improve this answer
5  
That works greatly if you have an exponential map and, moreover, you can differentiate it :) –  Mariano Suárez-Alvarez Jul 18 '11 at 1:11
    
Ah ok...I guess I was assuming $K = \mathbb R$ or $\mathbb C$. –  Eric O. Korman Jul 21 '11 at 15:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.