Lie Algebra $[v,0]=0=[0,v], \quad\forall v\in \mathfrak{g}$.

Question

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.

Answer 1

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.

Answer 2

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

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

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

$[v,0]=0$

Answer 3

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