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An (almost) duplicate of the question I want to ask is Adjoint map is Lie homomorphism.

I went to that question to find the answer to the question it asks: How do I prove that $ad_x$ is a Lie homomorphism?

I now am a little confused. I thought that I wanted to prove that $ad_x$ is a Lie homomorphism, but now I'm not sure that's actually true.

I now realise that perhaps I was confused (as that asker was) and that $ad$ is a Lie homomorphism, where $ad : g \rightarrow End(g)$ sends $x \in g$ to $ad_x$. $ad_x$ is potentially not a Lie homomorphism (I have no reason to think it is.

Question 1: Is $ad_x:g \rightarrow g, y \mapsto [x,y]$ a Lie homomorphism? I've tried to prove it (because I misunderstood something), and I can't, so I think not.

Question 2: Considering instead $ad:g \rightarrow End(g), x \mapsto ad_x$, we need to show that $ad([x,y]) = [ad(x),ad(y)]$

It suffices to show that these are identical on every $z \in g$, so we want that $ad([x,y])(z) = [ad(x),ad(y)](z)$

And I'm not sure how to proceed from here. Am I getting the right idea or should I be doing this differently?

Matt
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1 Answers1

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The Jacobi identity for a Lie algebra $L$ just says that $$ \operatorname{ad}([x,y])(z)=[[x,y],z]=[x,[y,z]]-[y,[x,z]]=(\operatorname{ad}(x)\operatorname{ad}(y))(z)-(\operatorname{ad}(y)\operatorname{ad}(x))(z)=[\operatorname{ad}(x),\operatorname{ad}(y)](z). $$ Here the last Lie bracket is for endomorphisms $A,B$, given by $[A,B]:=AB-BA$.

Dietrich Burde
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  • For future readers: I was quite confused for a while because I kept thinking, we're trying not to assume $[\cdot,\cdot]$ is simply a commutator, whereas in this solution, we must assume that it is. But ad is a map from $g$ to End$(g)$, and each space has its own Lie bracket. We don't need to assume the Lie bracket on $g$ is a commutator, but we do assume that the Lie bracket on End$(g)$ is the commutator of two elements. Indeed, this is a natural Lie product on End$(g)$, so this is the "right" homomorphism to consider. – WillG Oct 15 '21 at 18:43
  • Yes, the Lie bracket on $\mathfrak{gl}(V)$ is given by $[A,B]=AB-BA$, but in general the Lie bracket is only a symbol. Some people write $x\cdot y$ instead of $[x,y]$ then. – Dietrich Burde Oct 15 '21 at 18:47