# the action of a Lie algebra on itself

Currently I've read the first 5, almost 6 chapters of Serre's "Lie algebras and Lie groups".

Let $L$ be a Lie algebra, and suppose $L$ is given as a subalgebra of $End(V)$, where $V$ is some finite dimensional vector space over an alg. closed field $k$.

Now, as usual one can identify $End(V) = V\otimes V^*$, where the isomorphism is given on simple tensors by sending $v\otimes w$ to the endomorphism $\psi_{v\otimes w} : z\mapsto w(z)\cdot v$.

Thus, $L$ is a subalgebra of $V\otimes V^*$, and it acts on $V\otimes V^*$ through its action on $V$ and hence on $V^*$, and hence on $V\otimes V^*$. I was trying to convince myself that through this action, $L$ is stable under the action of itself.

A priori, there are a number of ways one might view $L\subset End(V)$ acting on $V^*$. For example, at first I thought the right action is just by "precomposition", ie given $f\in V^*$ and $\varphi\in End(V)$, $\varphi f = f\circ\varphi$. However, this turns out to be wrong. The right one is given by: $$\varphi.f := - f\circ\varphi$$

Secondly, given an action of $\varphi$ on $V$, and on $W$, there is a natural action of $\varphi$ on $V\otimes W$ given on simple tensors by $\varphi(v\otimes w) = \varphi(v)\otimes\varphi(w)$, but again this turns out to be the wrong action. The right one in our context is instead $$\varphi.(v\otimes w) = \varphi.v\otimes w + v\otimes\varphi.w$$ In the case $W = V^*$, if one views $v\otimes w$ as the endomorphism $\psi_{v\otimes w} : z\mapsto w(z)\cdot v$, then we get: $$\varphi.(v\otimes w) = \varphi\circ \psi_{v\otimes w} + \psi_{v\otimes\varphi.w} = \varphi\circ\psi_{v\otimes w} - \psi_{v\otimes w}\circ\varphi$$

Thus, at last, if $x,\varphi\in End(V)$, then we get $$\varphi.x = \varphi\circ x + x\circ\varphi = [\varphi,x]$$ Thus the action of $\varphi\in End(V)$ on $End(V)$ is just the adjoint action. In particular, $L$ is invariant under $L$.

I suppose my question is - "Why?" As a novice to Lie algebras, this all seems rather strange to me. Why "should" $L$ leave itself invariant, viewed as a subspace of $End(V) = V\otimes V^*$? Intuitively, if $L$ is the tangent space of a certain Lie group, what is the meaning of this action of $L$ on itself? In the case $L = End(V)$, what do the other "natural" actions of $End(V)$ on itself by left composition, right composition, or both, mean in the Lie group context?

I suppose I'd welcome some examples that might shed light on how to think of these computations.

I get the feeling that a lot of this will become clear once I get into the Lie group parts of Serre's book, so I almost considered not posting this question, but I'm posting it anyway, partially just as a reminder to myself to continue to think about these questions as I continue reading.

• The action of L on itself is just the derivative of the action of the group by conjugation on itself. – Mariano Suárez-Álvarez Sep 4 '17 at 22:08