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Let $\mathfrak{g}\subset gl(V)$ be a Lie algebra. Is the set of nilpotents of $\mathfrak{g}$ a lie subalgebra? To be more precise, let $A$ and $B$ be nilpotent matrices. Then is $AB-BA$ also nilpotent?

  • It's true in finite-dimensional solvable Lie subalgebras of $\mathfrak{gl}(V)$ (assuming the field has characteristic zero). – YCor Apr 17 '17 at 18:47

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No. Take $A=\pmatrix{0&1\\0&0}$ and $B=\pmatrix{0&0\\1&0}$.

Angina Seng
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  • Thanks. A proof in my Lie algebra textbook says "consider the lie subalgebra consisting of all the nilpotent elements". Is it just the algebra generated by all the nilpotent elements? –  Apr 15 '17 at 16:42
  • @AyushKhaitan Which book? Which Lie algebra? – Angina Seng Apr 15 '17 at 17:52