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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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    Let $I$ denote the ideal in $K[GL(n,K)]$ vanishing on $G$, then $T_{ij}\in I$ if i>m and j.When i>m and j, 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.2011-10-17

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

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    Ah ok...I guess I was assuming $K = \mathbb R$ or $\mathbb C$.2011-07-21