What is the categorical significance of the trivial $\mathfrak{g}$-module in the category of $\mathfrak{g}$-mod?

Question

This question may be trivial for experts.

Let $\mathfrak{g}$ be a Lie algebra over a field $k$ and consider the category $\mathfrak{g}$-mod of $\mathfrak{g}$ modules. We can add suitable conditions, like finite dimension, semisimple, if necessary.

We call the 1-dimensional $k$-space with $\mathfrak{g}$ acts by $0$ the trivial $\mathfrak{g}$-module. This $\mathfrak{g}$ module is kind of special in the category $\mathfrak{g}$-mod. But it is not the initial or terminal object (the $0$-module is both initial and terminal).

$\textbf{My question}$ is: is there any categorical significance of this trivial $\mathfrak{g}$ module? Do we need to take the tensor structure into account?

Answer 1

It's the tensor unit: for any module $V$ we have $V \otimes k \cong k \otimes V \cong V$ as $\mathfrak{g}$-modules.

There are cases in which you must take the tensor structure into account to distinguish the trivial module from the other simples. One example is a semisimple complex Lie algebra, whose module category is semisimple, so equivalent (as an abelian category, not as a tensor category) to the category of representations of a certain quiver with no arrows. Here we clearly can't tell the difference between simples without some extra structure.

As another example (not quite the same thing you are talking about, but related) consider representations of the restricted Lie algebra $\mathfrak{sl}_2$ in characteristic $p>2$. The module category of any non-semisimple block is equivalent to the category of representations of the algebra $kC_2 \rtimes k[x,y]/(x^2, y^2)$ where $C_2=\langle g \rangle$ is cyclic of order two, and $g$ conjugates each of $x$ and $y$ to its additive inverse. This algebra has two simple modules, each one-dimensional with $g$ acting as $\pm 1$. The algebra has an automorphism sending $g$ to $-g$ and fixing $x,y$, which lifts to an automorphism of the module category exchanging the two simple modules. Thus you can't tell the difference between them if you only know the structure of the category.

As usual there is a lot of insight to be had from the case of $\mathfrak{g}=\mathfrak{sl}_2(\mathbb{C})$. As an abelian category, $\mathfrak{g}$-$\operatorname{mod}$ is rather dull. As a tensor category it has a huge amount of interesting structure, involving Catalan numbers, the Temperley-Lieb algebras of statistical physics, "spiders"... -- see section 2 of Kuperberg's http://arxiv.org/pdf/q-alg/9712003v1.pdf for example.

Answer 2

The $0$-dimensional Lie algebra $0$ is the terminal object in the category of Lie algebras; that is, every Lie algebra admits a unique morphism $\mathfrak{g} \to 0$. This morphism gives rise to a restriction functor $0\text{-Mod} \to \mathfrak{g}\text{-Mod}$ which is precisely the inclusion of trivial modules into modules (and once you have trivial modules you can isolate finitely generated trivial modules, and the $1$-dimensional trivial module is the unique generator of finitely generated trivial modules under direct sum). This functor in turn admits both a left and right adjoint, one of which sends a $\mathfrak{g}$-module to its submodule of invariants and one of which sends a $\mathfrak{g}$-module to its quotient module of coinvariants.

This is the parallel of a simpler story for groups, where the trivial group $1$ is the terminal object in the category of groups, etc.

Answer 3

To address the category O issue: no. Every finite dimensional simple in category O is "the same," in the sense there's a functor sending one to the other which isn't an equivalence on all of category O (you could cook such a thing up, but it wouldn't be very natural), but is an equivalence between the blocks of the two corresponding simples: the translation functors do this.