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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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3 Answers 3

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Let $\mathfrak g$ have Cartan subalgebra $\mathfrak h$, let $X_\alpha\in\mathfrak g_\alpha$ be a non-zero vector. Then you are looking at nilpotent elements of the form $\sum_{\alpha\in \Theta}X_\alpha$ for some subset $\Theta\subset\Delta$.

Note that each subset $\Theta\subset \Delta$ we can associate a parabolic subalgebra $\mathfrak p_\Theta\subset \mathfrak g$ generated by $\mathfrak h$ and $\mathfrak g_\alpha$ where $\alpha\in\Delta\cup(-\Theta)$, with Levi subalgebra $\mathfrak l_\Theta$ generated by $\mathfrak h$ and $\mathfrak g_\alpha$ for $\alpha\in\Theta\cup(-\Theta)$. Now from e.g. Section 4.1 of Collingwood-McGovern we see that the orbit of $\sum_{\alpha\in\Theta}X_\alpha$ is the principal nilpotent orbit of $\mathfrak l_\Theta$.

Thus, the nilpotent orbits you obtain are precisely those whose intersection with some Levi subalgebra of $\mathfrak g$ is principal.

Many nilpotent orbits do not satisfy this. For example, in type $C_n$ nilpotent orbits are parametrized by partitions of $2n$ with odd parts occuring with even multiplicity. However, the Levi subalgebras of $C_n$ are of the form $A_{i_1}\times\cdots \times A_{i_m}\times C_k$ so the nilpotent orbits obtained from them correspond to partitions where all but one part appears with even multiplicity (so the partition $[4,2]$ in type $C_3$ does not arise from your construction).

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  • $\begingroup$ @kenta_suzuki Thank you! $\endgroup$
    – Alexander
    Commented Apr 1 at 12:22
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Actually I think this has to fail for $G_2$. Using fundamental coweights, we can construct a semisimple element that scales the simple root vectors $u_1, u_2$ by arbitrary nonzero constants $c_1, c_2$. Hence there can be at most $4$ orbits that appear in this way. But $G_2$ has $5$ nilpotent orbits.

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No, not every nilpotent element lies in the span of simple root vectors. In general, the Lie algebra of a reductive group (G) decomposes into a direct sum of its semisimple and nilpotent parts under the adjoint action of (G). While the simple root vectors span the semisimple part, there's no guarantee that they span the nilpotent part. However, for semisimple Lie algebras, every nilpotent element can be written as a linear combination of positive root vectors (with respect to a chosen Cartan subalgebra), and similarly, as a linear combination of negative root vectors.

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