# The category of Lie algebra representations

A representation of a Lie algebra $\mathfrak{g}$ on a vector space $V$ is a homomorphism of Lie algebras $\mathfrak{g} \to \mathfrak{gl}(V)$. We define morphisms between representations as intertwining linear maps as usual. Then we have a category $\mathsf{Rep}(\mathfrak{g})$ of representations of $\mathfrak{g}$.

I am wondering what are the essential properties of this category:

-is it semi-abelian or even abelian?

-Can we describe it as a functor category (as is the case for group representations)?

-Do any of the aforementioned properties depend in some way on either the Lie algebras or the vector spaces being finite-dimensional? Do they depend on the choice of field (or commutative ring)?

## 1 Answer

The category of representations of $\mathfrak{g}$ is not just abelian but is isomorphic to the category of modules over a certain (associative) $k$-algebra (where $k$ is the base field). Indeed, define $U(\mathfrak{g})$ to be the free associative $k$-algebra on the underlying $k$-vector space $\mathfrak{g}$, modulo relations that say that for each $x,y\in \mathfrak{g}$, $xy-yx=[x,y]$ (here the left-hand side is computed using the multiplication of our associative algebra, and the right-hand side is the bracket in $\mathfrak{g}$). That is, we "freely" construct an associative algebra from $\mathfrak{g}$ in which the bracket becomes the commutator operation on elements of $\mathfrak{g}$. Then it is straightforward to verify that a $U(\mathfrak{g})$-module is the same thing as a $\mathfrak{g}$-representation, giving an isomorphism of categories. The algebra $U(\mathfrak{g})$ is known as the universal enveloping algebra of $\mathfrak{g}$.

For any ring $R$, you can form an $Ab$-enriched category $BR$ with one object whose endomorphisms are $R$ (with addition in $R$ being the $Ab$-enrichment and multiplication in $R$ being composition of maps). An $R$-module is then the same thing as a functor $BR\to Ab$ which preserves the $Ab$-enrichment. In particular, taking $R=U(\mathfrak{g})$, this gives a description of the representation category of $\mathfrak{g}$ as a certain functor category.

None of this depends on finite-dimensionality, or even on $k$ being a field ($k$ could be any commutative ring).

• (You do need some flatness assumption, no?) – Mariano Suárez-Álvarez Mar 7 '16 at 2:58
• I don't think so? Why would you need a flatness assumption? – Eric Wofsey Mar 7 '16 at 3:01
• One needs the map from $g$ to $Ug$ to be injective, and that depends on PBW which is usually done only for projective algebras. I've always imagined it breaks done in the general case. – Mariano Suárez-Álvarez Mar 7 '16 at 3:04
• The map from $\mathfrak{g}$ to $U(\mathfrak{g})$ doesn't need to be injective to have the universal property and thus give an equivalence between representations of $\mathfrak{g}$ and $U(\mathfrak{g})$-modules. The kernel of $\mathfrak{g}\to U(\mathfrak{g})$ will act trivially on any representation. – Eric Wofsey Mar 7 '16 at 3:05