If $\pi$ is a representation of a lie algebra $\mathfrak{g}$, why is $\pi(x)$ not required to be invertible?

Question

The definition of a representation of a group $G$ is a homomorphism $\pi: G\to GL(V)$. So here $\pi(x)$ is an invertible linear map $V \to V$.

The definition of a representation of a lie algebra $\mathfrak{g}$ is a homomorphism $\pi: \mathfrak{g} \to End(V)$. So here $\pi(x)$ is just a linear map.

What is the motivation for not requiring invertibility? Is this something with category theory?

Answer 1

Then what would $\pi(0)$ be? Since $\pi$ is a Lie algebra homomorphism, in particular it is a linear map, and therefore $\pi(0)=0$. And the null map is not invertible.

Besides, if $X\in\mathfrak g$, then the map$$\begin{array}{rccc}\operatorname{ad}(X)\colon&\mathfrak g&\longrightarrow&\mathfrak g\\&Y&\mapsto&[X,Y]\end{array}$$is never invertible (since $\operatorname{ad}(X)(X)=0$). But $\operatorname{ad}$ is the adjoint action, which is the natural action of $\mathfrak g$ on itself.

Answer 2

If $\pi: G \rightarrow \operatorname{GL}(V)$ is a representation of a Lie group $G$ on a finite dimensional vector space $V$, then the differential $d \pi: \mathfrak g \rightarrow \operatorname{End}(V)$ is a Lie algebra representation of the Lie algebra $\mathfrak g$ into the space of all linear maps from $V$ to itself.

Originally, Lie algebra representations showed up as differentials (tangent space maps) from ordinary group representations, which are invertible. These differentials are, in particular, linear maps. They send $0$ to $0$, and therefore you are forced to deal with non-invertible elements of $\operatorname{End}(V)$.

Answer 3

For a representation of a group you don’t need to require $π(g)$ to be invertible: a representation of group $G$ is a pair $(V, π)$ consisting of a vector space $V$ and a homomorphism of monoids $π \colon G \to \operatorname{End}(V)$. More explicitly: the map $π$ needs to be multiplicative, so that $π(gh) = π(g)π(h)$ for any two elements $g$ and $h$ of $G$, and it needs to satisfy $π(1_G) = 1_{\operatorname{End}(V)}$.

However, it then turns out that $π(g)$ is actually invertible for every element $g$ of $G$. This then allows us to regard $π$ not only as a homomorphism of monoids from $G$ to $\operatorname{End}(V)$, but also as a homomorphism of groups from $G$ to $\operatorname{GL}(V)$. (Which is why most people don’t bother with $\operatorname{End}(V)$ and just use $\operatorname{GL}(V)$ to begin with.)