Talk (?) -- Universal Enveloping Algebras

14 Feb 2022 - Tags: my-talks

This barely counts as a talk, but I want it catalogued with the rest of the talks I’ve given, because this is going to have a retrospective aspect to it like all of my post-talk posts do. A while ago now, I gave a 30 minute presentation in my Lie Algebras class where I answered two questions that I’d brought up over the course of the class, with the unifying thread being this: both questions are naturally answered by the Universal Enveloping Algebra.

As a brief aside, I gave a talk last week about a concrete application of topos theory. I want to write up a post about it, but unfortunately I screwed up the last ten minutes… I’ll say more about it when I get around to posting it, but the moral of the story is that I don’t want to write that post until I really understand what the last ten minutes should have been if I’d done it correctly.

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So: what did I talk about in the presentation? Well, a lie algebra is a vector space with bonus structure, and we talk about short exact sequences of lie algebras. So it’s reasonable to wonder if the category of lie algebras is abelian.

In hindsight the answer is obviously “no”, and I briefly mentioned why. It comes down to the existence of monos which aren’t normal. In the category of abelian groups (the prototypical abelian category) every subgroup is normal. Rephrased categorically, this says that every monomorphism is the kernel of some morphism. In an abelian category this condition (and its dual) must be satisfied, but we know there is a distinction between ideals (which we can quotient by) and subalgebras (which, in general, we can’t) of a given lie algebra. So not every mono is normal, and the category of lie algebras cannot be abelian.

This leads us to a related question: Is the category of $\mathfrak{g}$-reps abelian for a lie algebra $\mathfrak{g}$? I was fairly sure the answer would be “yes” asked this, but I wanted to bring it up in class because the class has been entirely devoid of category theory, despite representation theory having a fairly categorical reputation in my mind.

The answer, of course is yes, and the easiest way to see this (imo) is via the universal enveloping algebra $U(\mathfrak{g})$. Before we go into that, though, let’s briefly say what the other question I wanted to answer was.

We know that $\mathfrak{g}$ naturally acts on itself by “left multiplication”, by which I mean $[g,-]:\mathfrak{g}\to \mathfrak{g}$. However this action need not be faithful!

With this in mind, it’s natural to ask if there is an space which admits a natural, faithful $\mathfrak{g}$ action. The answer, again, is yes, and the vector space in question is the universal enveloping algebra¹!

So then, at this point you should be sold on $U\mathfrak{g}$ being an interesting object… but what exactly is it? I’ll start with a motivating categorical approach, then (for people whole aren’t as well versed in the way of adjoint functors), we’ll go over how one might build it by hand. This follows the talk fairly closely², but it should go better here since I expect a bit more categorical maturity from readers of my blog³.

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So then, what’s the goal of the universal enveloping algebra? We know that matrix algebras are automatically lie algebras, when we interpret $[A,B]$ as $AB-BA$. More generally we can do this with the ring of endomorphisms of your favorite vector space $\text{End}(V)$. More generally still, we can do this for your favorite algebra over a field.

This is obviously functorial, and sends the ring $\text{End}(V)$ to the lie algebra $\mathfrak{gl}(V)$, but I’m going to call this $\mathsf{Lie}(\text{End}(V))$ instead, to emphasize the fact that $\mathsf{Lie}$ is a functor from $\mathbb{F}\text{-alg}$ to $\mathbb{F}\text{-lie-alg}$.

Here, as usual, $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$. I’m not sure what happens in positive characteristic (it seems to be subtle. See here) so I’m restricting attention to particularly nice fields, haha.

Now, recall a Representation of a lie algebra $\mathfrak{g}$ is a homomorphism $\alpha :\mathfrak{g}\to \mathsf{Lie}(\text{End}(V))$. These obviously assemble into a category whose objects are pairs $(V,\alpha )$ and whose morphisms are linear maps $V\to W$ which commute with the $\mathfrak{g}$-action.

If this sounds like a category of modules, you’re right! That’s some good justification for this being an abelian category, and it’s not super hard to verify the axioms by hand… But wouldn’t it be nice if we didn’t have to?

Let’s say we had a left adjoint $U⊣\mathsf{Lie}$. Then we would have

$$\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\mathfrak{g}\text{-reps}\end{array}\\ \alpha :\mathfrak{g}\to \mathsf{Lie}(\text{End}(V))\end{array}\end{array}\\ \tilde{\alpha }:U\mathfrak{g}\to \text{End}(V)\end{array}\end{array}\\ U\mathfrak{g}\text{-modules}\end{array}$$

so, if we can find a left adjoint $U$ for $\mathsf{Lie}$, the category of $\mathfrak{g}$-representations will be equivalent to the category of $U\mathfrak{g}$-modules.

Now, let’s get all the abstract nonsense out of our system right now. Since it’s clear that $\mathsf{Lie}$ preserves limits (indeed, it’s basically not doing anything at all), we can apply the Special Adjoint Functor Theorem to quickly see that $\mathsf{Lie}$ has a left adjoint. Unfortunately this tells us nothing about what $U$ looks like, so while true, this proof is not as useful as it could be.

There’s an art to finding left adjoints, where we try to figure out what the “freest” way to build an object should be. Usually to introduce free structure, you just add the syntax, and declare it has to satisfy all the rules you want.

So for us, we want to add multiplication, to turn our lie algebra into a traditional algebra. Following the hint I gave in the last paragraph, we might look at the space of all sums of formal strings $x_1{x}_2\cdots x_n$, where we multiply by concatenation. This construction already has a name, if we recognize $x_1{x}_2\cdots x_n$ as an element of the $n$th tensor power of $\mathfrak{g}$.

That is, we look at

$${\mathfrak{g}}^{\otimes n}\triangleq \underset{n\text{ times}}{\underbrace{\mathfrak{g}\otimes \mathfrak{g}\otimes \cdots \otimes \mathfrak{g}}}$$

and we think of the element $x_1\otimes x_2\otimes x_3\cdots \otimes x_n$ as the product of the $x_i$.

So ${\mathfrak{g}}^{\otimes n}$ consists of the free $n$-fold products of elements of $\mathfrak{g}$, and we want to allow arbitrary products. What’s the obvious thing to do? We look at the tensor algebra

$$\underset{n\ge 0}{⨁}{\mathfrak{g}}^{\otimes n}$$

Now elements of this are exactly sums of products of elements of $\mathfrak{g}$!

There’s one piece of data we’re missing, though. We want $\mathsf{Lie}$ of this algebra to be related to $\mathfrak{g}$ somehow. We know $\mathfrak{g}\hookrightarrow \underset{n\ge 0}{⨁}{\mathfrak{g}}^{\otimes n}$ since $\mathfrak{g}={\mathfrak{g}}^{\otimes 1}$, so it’s reasonable to want $[x_1,x_2]$ as computed in the tensor algebra to be the same as $[x_1,x_2]$ as computed in $\mathfrak{g}$.

How do we do that? Well, we just force it to be true! For every $x_1,x_2\in \mathfrak{g}$ we add a rule telling us how to evaluate the syntax $[x_1,x_2]=x_1\otimes x_2-x_2\otimes x_1$. It should be $[x_1,x_2{]}_{\mathfrak{g}}$, and so we define

$$U\mathfrak{g}\triangleq \frac{\underset{n\ge 0}{⨁}{\mathfrak{g}}^{\otimes n}}{x_1\otimes x_2-x_2\otimes x_1=[x_1,x_2{]}_{\mathfrak{g}}}$$

Remember at the start of all this, we had two questions that $U\mathfrak{g}$ is supposed to help us answer.

We’ve already spent some time showing that the category of $\mathfrak{g}$-reps is abelian. This is because the category of $\mathfrak{g}$-reps is equivalent to the category of $U\mathfrak{g}$-modules, and we know that module categories are always abelian.

But how does this provide us with a canonical faithful $\mathfrak{g}$-rep? Well, I’ll leave it to you to check that the obvious map $\mathfrak{g}\hookrightarrow \mathsf{Lie}U\mathfrak{g}$ (if you like, the unit of the adjunction) really is an embedding. So $\mathfrak{g}$ acts faithfully on $U\mathfrak{g}$.

For extra concreteness, what is this action? It’s exactly the left-multiplication action! So

$$g\cdot (x_1\otimes x_2\otimes \cdots \otimes x_n)\triangleq g\otimes x_1\otimes x_2\otimes \cdots \otimes x_n$$

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Let’s work this out the only example in lie theory I know, ${\mathfrak{sl}}_2(\mathbb{C})$⁴.

This lie algebra is three dimensional, with a basis

$$x=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)y=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

Moreover, the lie structure is generated by the equations

• $[x,y]=h$
• $[h,x]=2x$
• $[h,y]=-2y$

Now the tensor algebra is going to be all the formal products of $x$, $y$, and $h$. That is, we get $\mathbb{C}⟨x,y,h⟩$, the space of (noncommutative!) polynomials in $x$, $y$, and $h$. The subspace of degree $1$ polynomials can be identified with ${\mathfrak{sl}}_2$, since the degree $1$ polynomials are linear combinations $x$, $y$, and $h$.

Now we have to quotient the tensor algebra by the generating equations for the lie bracket. So we end up with

$$U{\mathfrak{sl}}_2(\mathbb{C})\triangleq \frac{\mathbb{C}⟨x,y,h⟩}{xy-yx=hhx-xh=2xhy-yh=-2y}$$

A fairly concrete object!

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