If $G$ leaves stable a subspace $W$ of $K^n$, prove that $\mathfrak{g} \subseteq \mathfrak{gl}(n,K)$ does likewise

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.

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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

1 Answer

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

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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