## Lie algebras: fundamentals July 16, 2009

Posted by Akhil Mathew in algebra, representation theory, Uncategorized.
Tags: , ,

The following topic came up in the context of my project, which has been expanding to include new areas of mathematics that I did not initially anticipate. Consequently, I have had to learn about several new areas of mathematics; this is, of course, a common experience at RSI. For me, the representation theory of Lie algebras has been one of those areas, and I will post here about it to help myself understand it. Right here, I’ll aim to cover the groundwork necessary to get to the actual representation theory in a future post.

Lie Algebras

Throughout, we work over ${{\mathbb C}}$, or even an algebraically closed field of characteristic zero.

Definition 1 A Lie algebra is a finite-dimensional vector space ${L}$ with a Lie bracket ${[\cdot, \cdot]: L \times L \rightarrow L}$ satisfying:

• The bracket ${[\cdot, \cdot]: L \times L \rightarrow L}$ is ${{\mathbb C}}$-bilinear in both variables.
• ${[A,B] = -[B,A]}$ for any ${A,B \in L}$.
• ${[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0}$. This is the Jacobi identity.

To elucidate the meaning of the conditions, let’s look at a few examples.

Here are two easy ones:

Example 1 Any vector space is a Lie algebra with the bracket the zero map. Such Lie algebras are called commutative.

Example 2 Suppose ${L_1, L_2}$ are Lie algebras. Then ${L_1 \oplus L_2}$ is a Lie algebra as well, if the Lie bracket is defined as:

$\displaystyle [(x_1, x_2), (y_1, y_2)] = ( [x_1, x_2], [y_1, y_2] ).$

This is the direct sum

Lie algebras arise frequently in a more interesting way:

Definition 2 Let ${R}$ be a finite-dimensional associative ${{\mathbb C}}$-algebra. Then the Lie bracket of ${a,b \in R}$ is defined as ${[a,b] = ab - ba}$; this makes ${R}$ into a Lie algebra, as is easily checked (but associativity is important).

Note in particular the resemblance of the Lie bracket to commutators in group theory. For instance, ${[a,b]=0}$ iff ${a,b}$ commute with each other.

Many of the important Lie algebras in fact arise in this way, from rings, especially matrix rings:

Definition 3 Fix ${n \in \mathbb{N}}$. We define ${\mathfrak{gl}_n}$ as the Lie algebra coming as above from the ring of ${n}$-by-${n}$ complex-valued matrices, with the usual Lie bracket.

It is often of interest to consider Lie subalgebras of ${\mathfrak{gl}_n}$. One of the most important is ${\mathfrak{sl}_n}$:

Definition 4 ${\mathfrak{sl}_n}$ is the space of complex ${n}$-by-${n}$ matrices whose trace is zero.

One must actually check that ${\mathfrak{sl}_n}$ is indeed a Lie subalgebra. But this follows from the identity

$\displaystyle Tr(AB) = Tr(BA)$

if ${A,B}$ are ${n}$-by-${n}$ matrices. So ${Tr([A,B])=0}$, for any ${A,B}$.

Representations

In general, a representation of some algebraic object is an action of that object on a vector space in a manner compatible with its algebraic structure. In our case, this will mean respecting the bracket.

Definition 5 A representation of ${L}$ is a ${{\mathbb C}}$-linear map ${\rho: L \rightarrow M_n({\mathbb C})}$ (${M_n({\mathbb C})}$ is the space of ${n}$-by-${n}$ matrices, which is an associative algebra) such that

$\displaystyle \rho([X,Y]) = [\rho(X), \rho(Y)]. \ \ \ \ \ (1)$

In other words it is a Lie algebra-homomorphism ${L \rightarrow \mathfrak{gl}_n}$. This resembles the notion of group representations, where we had group-homomorphisms ${G \rightarrow GL_n}$.

One can also phrase the above definition in the following way.

Example 3 A (finite-dimensional) representation of ${L}$ is a finite-dimensional vector space ${V}$ with a ${{\mathbb C}}$-bilinear map ${L \times V \rightarrow V}$, say denoted by ${M, v \rightarrow Mv}$, such that

$\displaystyle [M_1, M_2] v = M_1 (M_2 v) - M_2( M_1 v).\ \ \ \ \ (2)$

The equivalence of this phrasing with the previous one follows from the correspondence between linear transformations and matrices. Looking at the (initial) definition, a representation of ${L}$ on ${V}$ means a map ${\rho: L \rightarrow \mathfrak{gl}_n}$ if ${V}$ has dimension ${n}$. Then, we define for ${X \in L, v \in V}$,

$\displaystyle Xv = \rho(X) v;$

this gives a map ${L \times V \rightarrow V}$, and since ${\rho}$ is a Lie homomorphism, (2) is satisfied. Similarly, one can work the other way.

There is a parallel here with the notion of modules and even group representations, which arise if one takes an “enveloping algebra” of ${L}$; this, however, is better saved for a later post with more generality.

So, I’m planning to continue this series with a post on ${\mathfrak{sl}_2}$ and its irreducible representations, and then possibly talk about Engel and Lie, semisimple Lie algebras, and further into representation theory.

1. hilbertthm90 - July 16, 2009

In definition 1, you have $\frak{sl}_2$, and I believe you just mean L.

Here is a question I should probably know the answer to, and I feel sort of dumb for asking, but…every Lie group generates a Lie algebra. Is there some sort of relationship between the representation theory of G and say, $\frak{g}$ the Lie group generated by G? My guess is that you lose information, since non-isomorphic Lie groups can generate the same Lie algebra (if my memory serves me correctly…). I haven’t gotten to the point in the summer where I review this for my qualifying exams, yet. So this is a nice jump start.

2. Akhil Mathew - July 16, 2009

Thanks for the correction! I was initially intending to write something up on just $\mathfrak{sl}_2$, because it’s more relevant to my project to understand the representation theory, but then realized it made more sense to do the general case, and tried to fix an older version of the post. $\mathfrak{sl}_2$ will likely be the next post I write.

As for your question: my understanding here is weaker than I would like, which is partially why I am writing these posts, but given a representation of a Lie group G, then one gets a representation of its Lie algebra, by taking differentials at the identity. One example of how to lose information could be a Lie group with infinitely many components, which has the same Lie algebra as one of the components. One might also lose information between covering spaces of Lie groups.

Still, given a representation of the Lie algebra of the simply connected Lie group G, i.e. a Lie-homomorphism $\mathfrak{g} \to \mathfrak{gl}_n$, one gets a Lie group-homomorphism $G \to GL_n$ by the “second principle” that Fulton and Harris refer to; as far as I know, this is just an application of the exponential map, and the universality of the Campbell-Hausdorff formula. Thus we should indeed get a bijection between representations of a simply connected Lie group and its Lie algebra. This is actually an idea for a future post, probably after July though.

3. hilbertthm90 - July 17, 2009

Ah, right. If I wasn’t so lazy, that probably would have been one of the first things I encountered when looking for theorems of the sort I was referring to. I did learn that at one point, but forgot it.

4. Qiaochu Yuan - July 17, 2009

This is an area I know nothing about, so I’m wondering what the functorial picture is for Lie algebra representations. For group representations and actions it’s clear: treating a group $G$ as a one-object category, a representation of $G$ is just a functor into ${K-Vect}$ and a group action of $G$ is just a functor into $\text{Set}$. Presumably one can also talk about representations of algebras in this way by enriching everything appropriately.

So what’s the functorial picture for Lie algebras? In other words, why is the commutator “universal” for Lie algebras? (This probably has a well-known explanation, but I’ve never seen it. Etingof’s notes on the subject introduce the definition of the universal enveloping algebra with no motivation at all.)

My best guess is that the process of finding the Lie algebra of a Lie group is also functorial and that all this should come out of the corresponding abstract nonsense, but again, I’d really appreciate someone clearing this up for me.

5. Akhil Mathew - July 17, 2009

Thanks for the comment!

Are you talking about a “categorification” of Lie algebra representations?
First, I intend sometime to talk about enveloping algebras, which will play the analog of the group ring, and which I will try to motivate with some abstract nonsense at least. I would suspect that there may indeed be a categorification of this notion, at least for specific Lie algebras, in suitable tensor categories (e.g. if L is the algebra and M the object, there should be a map $L \otimes M \to M$ satisfying some compatibility conditions), although I will need to think about it. I can ask my mentor about it today (it’s fairly relevant to my project anyway).

“My best guess is that the process of finding the Lie algebra of a Lie group is also functorial”

This is true because any morphism of Lie groups induces a map on the tangent spaces at the identity. Since this is a homomorphism, it commutes with the adjoint action and thus with the Lie bracket.

6. Akhil Mathew - July 17, 2009

I brought this up today morning, and apparently yes, the notion of Lie algebra can be categorified to some extent at least.

Let’s say we are in an additive tensor category. We can say $L$ is a Lie algebra if there is a morphism $L \otimes L \to L$ such that the morphism $L \otimes L \otimes L \to L$ coming from commutators satisfies some compatibility condition (i.e. should factor through the alternating product, and satisfy some analog of the Jacobi identity). Then we can define a representation of $L$ to be an object $M$ of the category with a morphism $L \otimes M \to M$ satisfying some compatibility conditions, etc. I think this can probably become a full post at some point (when I talk about my project, which is related to this kind of “categorification”* anyway).

*Strictly speaking, this may be the wrong word.

7. Qiaochu Yuan - July 17, 2009

Partially, but maybe what I’m looking for is a lifting of a Lie group representation to a Lie algebra representation that makes it obvious why the commutator is the “right” choice of bracket on $\mathfrak{gl}_n$.

8. Akhil Mathew - July 17, 2009

Well, I think the action $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g} = T_e(G)$ is derived as follows: take the map $G \times G \to G$ given by $(x,y) \to xyx^{-1}$, differentiate with respect to $y$ to get a map $G \to Aut(T_e(G))$, then with respect to $x$ to get a map $T_e(G) \to End(T_e(G))$. I believe the fact that this gives the usual bracket on $\mathfrak{gl}_n$ follows from a computation; I’m somewhat fuzzy on the details right now, as I haven’t really thought about Lie groups for a while now. I’ll probably post more about Lie groups later.

9. Representations of sl2, Part I « Delta Epsilons - July 17, 2009

[…] representation theory, semisimplicity, sl2 trackback is a special Lie algebra, mentioned in my previous post briefly. It is the set of 2-by-2 matrices over of trace zero, with the Lie bracket defined […]

10. hilbertthm90 - July 18, 2009

I think that no matter what, there has to be a “weird” step. I have a proof that $(Ad_* X)Y=[X, Y]$ by just computing at $e$ (which seems to be your method). It turns out that $((Ad_*X)Y)_e=\frac{d}{dt}\Big|_{t=0} d(\theta_{-t})(Y_{\theta_t(e)})$.

So now the intuition gets weird unless you are really good with the geometric intuition of flows, since that is precisely the definition of the Lie derivative of $Y$ with respect to $X$ at e, which is the Lie bracket at e. (It is still rather strange intuitively that the Lie derivative of vector fields is the Lie bracket).

11. Akhil Mathew - July 18, 2009

I believe this can be done directly: Let’s first fix the initial argument $x$, and take the derivative of $G \to G, y \to xyx^{-1}$, in the setting where $G = GL_n$. We can let $y$ trace out the path $1 + tM$ for some matrix $M$, which is the tangent vector as $t$ traces out some interval (and then take difference quotients, let $t \to 0$). The derivative must thus send $M \to x M x^{-1}$. Now we can differentiate w.r.t $x$ in the direction $N$, say $x = 1 + uN$, i.e. $(1 + uN) M (1 + uN)^{-1}$, and we apply the product rule: this is $NM(1) - MN(1)$. We have to take as a lemma the derivative of $(1 + uN)^{-1}$ at $u=0$, but this can be checked using e.g. the geometric series.

I don’t think we need the geometric intuition of flows here. Though alternatively, we could use the definition of the Lie bracket via left-invariant vector fields, but that would require a separate proof, which I should probably write up at some point to force myself to learn it properly.

12. Representations of sl2, Part II « Delta Epsilons - July 18, 2009

[…] The proof in our case runs as follows. Suppose . We must show , which is what the first assertion states. Indeed, by the definition of representations: […]

13. Lie algebras II « Delta Epsilons - July 20, 2009

[…] I’m going to get back eventually to the story about finite-dimensional modules, but for now, Lie algebras are more immediate to my project, so I’ll talk about them […]