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For any $x\neq0$ in a Lie algebra $L$, is there always a matrix representation $\rho:L\to\mathfrak{gl}(V)$ such that $\rho(x)^2\neq0$ ?

(Of course $\rho(x)^2$ means ordinary multiplication/composition, not the commutator.)

All the spaces involved are finite-dimensional, over $\mathbb R$ (or some field with characteristic $0$, or not $2$; but maybe this is irrelevant).

This question generalizes to $\rho(x)^k\neq0$ for various $k$, and further to $\det\!\big(\rho(x)\big)\neq0$.

mr_e_man
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1 Answers1

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Yes (for arbitrary finite-dimensional Lie algebras over a field with $2\neq 0$).

Indeed, first using Ado's theorem, it is enough to assume that $\mathfrak{g}=\mathfrak{gl}_n(K)$. For a nonzero $x$ therein, if $x^2\neq 0$ we are done. Otherwise, we can suppose that $x=E_{1n}$ ($xe_n=e_1$, $xe_i=0$ for $i<n$). Then in the second symmetric power $S^2(K^n)$, we have $x(e_n\odot e_n)=2(e_1\odot e_n)$ and $x(e_1\odot e_n)=e_1\odot e_1$, so the corresponding operator doesn't square to $0$.

For $\rho(x)^k\neq 0$ you can argue similarly (probably excluding positive characteristic $\le k$).

For $\det\neq 0$, it does not work: for $E_{12}\in\mathfrak{sl}_2(K)$ in characteristic $0$ is mapped to a nilpotent operator by every finite-dimensional representation. (I don't know about positive characteristic.)

YCor
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  • I think you mean $\mathfrak{g}\subseteq\mathfrak{gl}_n(K)$, or whatever the "subalgebra" symbol is. – mr_e_man Sep 16 '19 at 00:23
  • You made me learn that representations extend naturally to symmetric powers (and tensor powers)! That is, $\rho_{S^k}([x,y])=[\rho_{S^k}(x),\rho_{S^k}(y)]$ even though $\rho_{S^k}(xy)\neq\rho_{S^k}(x)\rho_{S^k}(y)$. Indeed, shouldn't tensor powers work as well as symmetric powers in your solution? – mr_e_man Sep 16 '19 at 00:26
  • Also, the precise form $E_{1n}$ is not necessary (and I don't see that it can always be reached); your solution works even if $xe_i\neq0$ for some $i<n$. Finally, do you have a proof for your last claim, that any $\rho(E_{12})$ is nilpotent? – mr_e_man Sep 16 '19 at 00:50
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    The last claim is a basic fact on representations of $\mathfrak{sl}_2$ in characteristic zero. Actually, it's even true if you take $x$ inside the $2$-dimensional Lie algebra with basis $(h,x)$ and bracket $[h,x]=2x$. See for instance Corollary 3.2 in these lectures of mine. – YCor Sep 16 '19 at 06:06
  • I haven't tried to understand $V_t$ (though I see the definition), but here's a different proof starting from $(H-2-t)X=X(H-t)$. Taking the determinant, if $\det X\neq0$ then the characteristic polynomial $p_H(t)=p_H(t+2)$ is periodic. But non-constant polynomials (over which fields?) are never periodic. (And $p_H$ is not constant, having a term $t^{\dim V}$.) Therefore $\det X=0$. – mr_e_man Sep 16 '19 at 19:24
  • Of course, my proof is weaker; vanishing determinant does not imply nilpotence. – mr_e_man Sep 16 '19 at 19:44