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If we have a real Lie algebra $(V, [,])$ then the Lie bracket is a bilinear map $$ [, ]: V \times V \to V $$ Assuming $V$ is finite dimensional, this map is bounded and so, for any choice of norm, there exists a constant $C$ such that $$ \|[v, w]\| \leq C\|v\|\|w\|. $$ If $V$ is a compact semisimple Lie algebra then there is a natural choice of norm given by the Killing form. If we pick this norm, can we say anything about the optimal values of $C$? For example, it could be the case that with the Killing form $$ \|[v, w]\| \leq \|v\|\|w\|, $$ but it may also be possible to construct examples where $C$ is arbitrarily large.

Holmes
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  • "[...] then there is a natural choice of norm given by the Killing form" -- how so? The Killing form is basically never positive definite (if the ground field is $\mathbb R$), cf. https://math.stackexchange.com/q/2546778/96384; and if the ground field is $\mathbb C$, it is not even sesquilinear (the Killing form is always bilinear). So it's not an inner product. – Torsten Schoeneberg Apr 17 '24 at 16:01
  • Ah, I'm unsure of the correct lie algebra terminology - I want to restrict to real Lie algebras where the killing form is negative definite (I should have made this more clear in the question). – Holmes Apr 18 '24 at 11:22
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    Then you need compact semisimple (real) Lie algebras. You can (and should) edit that in the question. – Torsten Schoeneberg Apr 18 '24 at 16:22
  • Thanks for editing. Now why don't you try the easiest example $\mathfrak{su}_2$ (a three-dimensional compact Lie algebra) and share the results? – Torsten Schoeneberg Apr 19 '24 at 23:16
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    $su(2) \cong \mathbb R^3$ with the Lie bracket given by the cross product. The negative of the killing form is 2 times the euclidean inner product. With the euclidean inner product, the formula for the cross product in terms of $\sin$ gives $| w \wedge v | < |v||w|$. Hence with respect the the killing form norm $|v\wedge v| \leq \frac{1}{\sqrt 2}|v||w|$. – Holmes Apr 20 '24 at 13:50

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