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

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?

• It's risky to talk aboout "the" category here. Probably you are thinking only about finite dimensional modules, but in any case there are many different module categories (such as "blocks") associated to a typical Lie algebra. So it's important to specify. Also, if there is additional structure, as occurs in a restricted Lie algebra in prime characteristic, it's especially important to say which modules are considered. – Jim Humphreys Sep 21 '13 at 18:21
• @JimHumphreys En, you're right! As m_t explained below, if $\mathfrak{g}$ is semisimple, the category structure of finite dimensional modules (as an abelian category) is not very interesting. However, what if we consider the BGG category $\mathcal{O}$? This category has some interesting features, for example, projective modules. On the other hand it is not a tensor category. Does the trivial $\mathfrak{g}$ plays some special role in the category $\mathcal{O}$? – Zhaoting Wei Sep 21 '13 at 18:52

## 3 Answers

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.

• Thank you for your excellent answer! I make a same comment as above. As Jim pointed out above, this question depends on what kind of $\mathfrak{g}$ modules do we consider. If $\mathfrak{g}$ is semisimple, the category structure of finite dimensional modules (as an abelian category) is not very interesting, as you said. However, what if we consider the BGG category $\mathcal{O}$? This category has some interesting features, for example, projective modules. On the other hand it is not a tensor category. Does the trivial $\mathfrak{g}$ plays some special role in the category $\mathcal{O}$? – Zhaoting Wei Sep 21 '13 at 18:55
• @ZhaotingWei while $\mathcal{O}$ is not a tensor category, it is closed under tensor products with finite-dimensional modules (ie, tensoring any module in $\mathcal{O}$ by a finite-dimensional module gives a module in $\mathcal{O}$), so the trivial module still plays the role of identity when we restrict like this. – Tobias Kildetoft Sep 23 '13 at 6:16

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.

• Thank you Qiaochu! This idea is very interesting, which means we consider the category of Lie algebras together with their modules. I wonder could we rephrase this in more formal language, for example fibered categories. – Zhaoting Wei Sep 21 '13 at 18:58
• @Zhaoting: another way to say what I'm saying from the perspective that $\mathfrak{g}\text{-Mod}$ is equivalent to $U(\mathfrak{g})\text{-Mod}$ is that the canonical map $\mathfrak{g} \to 0$ gives $U(\mathfrak{g})$ a canonical augmentation, which provides the trivial module. In fact, of course, $U(\mathfrak{g})$ is canonically a Hopf algebra and the augmentation is its counit, which provides the tensor unit of the monoidal product that the comultiplication provides. So this is a connection between my answer and m_t's. – Qiaochu Yuan Sep 28 '13 at 4:00

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.