the action of a Lie algebra on itself

Question

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.

Answer 1

I decided to revive this question because I believe it deserves a better fate. Although the comment by Mariano Suárez-Álvarez perfectly answers it from the point of view of the relation between Lie algebras and Lie groups, I believe there's something else at play on the way it was asked by oxeimon.

I'll try to answer it partially and hope that someone better informed than me will fill in the gaps.

oxeimon refers to the discussion in Serre's 'Lie groups and Lie algebras', Chapter V, section 1, entitled 'Complements on $\mathfrak g$-modules'. Let's follow the author's steps.

1. Definition of a $\mathfrak g$-module structure on a vector space $V$.

2. Definition of a $\mathfrak g$-module structure on the tensor product $V_1\otimes V_2$ of vector spaces $V_1,V_2$ equipped with $\mathfrak g$-module structures.

3. Definition of a $\mathfrak g$-module structure on the vector space ${\rm Hom}(V_1,V_2)$ of linear maps between vector spaces $V_1,V_2$ equipped with $\mathfrak g$-module structures.

At a first sight, one might be inclided to put these steps in other words by saying:

1. Definition of a category of $\mathfrak g$-modules. Let's call it $\mathfrak g$-${\bf mod}$. (Cf. category)

2. Definition of a monoidal structure on $\mathfrak g$-${\bf mod}$. (Cf. monoidal category)

3. Definition of a $\mathfrak g$-${\bf mod}$-category structure on $\mathfrak g$-${\bf mod}$. (Cf. enriched category)

What are the problems with these translations?

1. The author defines a $\mathfrak g$-module structure on a vector space $V$, but he doesn't specify what the morphisms between two such objects would be. One might guess that these should be just morphisms of vector spaces, i.e., linear maps, with some extra compatibility condition related to the $\mathfrak g$-module structure. However, note that in step 3 a $\mathfrak g$-module structure is defined in the whole vector space ${\rm Hom}(V_1,V_2)$, without any extra compatibility conditions, so that guess probably isn't what the author had in mind.

2. The author defines a $\mathfrak g$-module structure on $V_1\otimes V_2$, but it isn't at all clear how this definition could be thought in terms of a bifunctor, as the tensor product of vectors spaces itself is. Indeed, the formula employed here simply do not have a counterpart in that level of generality. Anyway, since we already don't have a proper notion of morphism, any attempts on that direction would probably be in vain. But one might want to try, it's fun:

$$x(v_1\otimes v_2)=(xv_1)\otimes v_2+v_1\otimes(xv_2),\qquad xv_1\in W_1, xv_2\in W_2\,?!$$

3. More or less the same as in steps 1 and 2.

In his question, oxeimon adds to step 3 something that doesn't appear in Serre's book. Namely, he/she tries to 'factorize' the definition of a $\mathfrak g$-module structure on ${\rm Hom}(V_1,V_2)$ as the combination of such a definition on $V^*$ with the definition on $V_1\otimes V_2$. I strongly appreciate and encourage this way of thought. Indeed, like the tensor product, dualization is functorial on vector spaces (perhaps the most well-known such thing), so this factorization could possibly help to understand how the above definitions appear in some 'natural' way.

However, once again, in the categorical level it simply doesn't work. In order to properly produce such a factorization, one would have to define a new 'functor' on vector spaces, for example as follows:

$${}^{-*}:{\bf Vect}\rightarrow{\bf Vect}$$

$$V\mapsto V^*\qquad f\mapsto -f^*$$

The problem is: it's not a functor. Just try to test this definition on identities, or see what happens with composition. So it seems that something is very deeply 'wrong' here.

If I'm allowed sort of a 'mathematical joke', I would like to share an attempt I made to fix these things, at least a little bit. Imagine that, instead of ${\bf Vect}$, we are dealing with a '$2$-category' (Cf. 2-category) whose objects and $1$-morphisms are the same as in ${\bf Vect}$, but with a $S_2$-action on each ${\rm Hom}(V_1,V_2)$, namely $(\pm1,f)\mapsto \pm f$, where $\pm1\in S_2$. Let us call it $S_2$-${\bf Vect}$.

Then, with the above definition, ${}^{-*}$ becomes a pseudofunctor (Cf. pseudofunctor) defined on $S_2$-${\bf Vect}$. =]

Jokes aside, my purpose with this answer is to show that, in spite of having explicitly said that his question was 'Why?' Serre's construction works, maybe oxeimon's intention might be rephrased as 'Why everything works, despite the fact that nothing works?'

EPILOGUE

I believe the mathematical community became accustomed, in the past few centuries, to explain what works with the most accurate level of detail one could imagine, and to leave more or less unexplained what doesn't work. Recently, however, there have been some serious attempts by prestigious mathematicians to overturn this habit, due certainly to the unprecedented easiness of communication provided by technology, but possibly not only because of that.

One might argue that it's good for students to try and learn what doesn't work by themselves, and I'm inclined to agree with that to some extent. Though, couldn't the same be said about virtually all mathematical ideas that happen to work? Alright, in a rapture of immodesty I would probably concede that I might have rediscovered, say, the quadratic formula all alone; but I would never concede the same about the cubic and quartic ones... What about the reason why the quintic case do not work?