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YCor
YCor
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Let $G$ be a reductive group (for simplicity, we can work over $\mathbb{C}$). Let $N \in \mathrm{Lie} G$$N \in \operatorname{Lie} G$ be nilpotent. Does there exist a Borel pair $(B,T)$ of $G$ such that $N$ lies in the span of the simple root vectors for $(B,T)$? This is true for $\mathrm{GL}_n$ by considering the Jordan form of a nilpotent matrix.

Let $G$ be a reductive group (for simplicity, we can work over $\mathbb{C}$). Let $N \in \mathrm{Lie} G$ be nilpotent. Does there exist a Borel pair $(B,T)$ of $G$ such that $N$ lies in the span of the simple root vectors for $(B,T)$? This is true for $\mathrm{GL}_n$ by considering the Jordan form of a nilpotent matrix.

Let $G$ be a reductive group (for simplicity, we can work over $\mathbb{C}$). Let $N \in \operatorname{Lie} G$ be nilpotent. Does there exist a Borel pair $(B,T)$ of $G$ such that $N$ lies in the span of the simple root vectors for $(B,T)$? This is true for $\mathrm{GL}_n$ by considering the Jordan form of a nilpotent matrix.

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Alexander
Alexander
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Does every nilpotent lie in the span of simple root vectors?

Let $G$ be a reductive group (for simplicity, we can work over $\mathbb{C}$). Let $N \in \mathrm{Lie} G$ be nilpotent. Does there exist a Borel pair $(B,T)$ of $G$ such that $N$ lies in the span of the simple root vectors for $(B,T)$? This is true for $\mathrm{GL}_n$ by considering the Jordan form of a nilpotent matrix.