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I want to show that $[v,0]=0=[0,v], \quad\forall v\in \mathfrak{g}$.

Is this done using the Jacobi identity? I am not sure how to do this, I just put $0$'s in the Jacobi identity, but it didn't give me anything to work with.

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  • $\begingroup$ Surely this is super simple, and I am missing something obvious $\endgroup$ Commented Aug 13, 2015 at 5:20
  • $\begingroup$ Is $0$ the identity element? Do you define $[a,b]$ as $ab-ba$? $\endgroup$
    – Tucker
    Commented Aug 13, 2015 at 5:28
  • $\begingroup$ @Tucker No sorry, just an abstract Lie bracket. Looks like the current answer works though, so thanks anyway :). As for identity element, I don't think Lie brackets can have an identity element $\endgroup$ Commented Aug 13, 2015 at 5:34
  • $\begingroup$ I meant an identity element of the group. $\endgroup$
    – Tucker
    Commented Aug 13, 2015 at 5:44
  • $\begingroup$ @Tucker I think it is the element $0\in \mathfrak{g}$ since the Lie bracket is defined as $\mathfrak{g}\times\mathfrak{g}\to \mathfrak{g}$ $\endgroup$ Commented Aug 13, 2015 at 5:55

3 Answers 3

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By definition, the Lie bracket of a Lie algebra is bilinear (see https://en.wikipedia.org/wiki/Lie_algebra), which means when you fix $v \in \mathfrak{g}$ the map

$[-,v]: \mathfrak{g} \rightarrow \mathfrak{g}$

is a linear map. That is for any $w_1 , w_2 \in \mathfrak{g}$ and any $a \in \mathbb{C}$ we have the identity

$[w_1 + a w_2,v] = [w_1,v] + a[w_2,v]$.

So if $w_1 = 0$ and $w_2$ is any vector in $\mathfrak{g}$ and $a=0$ then

$[0,v]=0\cdot[w_2,v]=0$.

The same works if $v$ is in the first part of the bracket.

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  • $\begingroup$ The argument as written looks circular to me. I think the answer by chansey is better. (Of course it works for any (bi)linear map.) $\endgroup$ Commented Sep 30, 2023 at 23:36
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$[v,0]=[v,0+0]$

$[v,0+0]=[v,0]+[v,0]$ (bilinearity)

$[v,0]=[v,0]+[v,0]$

$[v,0]=0$

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I would phrase it as: The left multiplication $\operatorname{ad}(X)$ is a linear function, i.e. $0$ is in its kernel, i.e. $\operatorname{ad}(X)(0)=[X,0]=0.$

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