Posted by Jason Polak on 14. June 2012 · 1 comment · Categories: algebraic-geometry · Tags: , , ,

I’ve decided to start this series with a few posts on the Lie algebra of an algebraic group. This seems to me the first real technical aspect of the classical theory that arises in Humphreys’ book. We shall loosely follow this book as a guide, but we shall also deviate and look at more scheme-theoretic treatments.

We will define a functor from the category of algebraic groups over a commutative ring $ k$ to Lie algebras over $ k$. The idea is that Lie algebras are often easier to work with than algebraic groups directly, so the Lie algebra will help us with things (such as classification problems). For now, what we do will be general enough so that we do not need to assume that $ k$ is a field.

We shall look at several definitions of a Lie algebra of an algebraic group, and prove that they are all equivalent. After this, we shall examine what this functor does to morphisms (the “differential of a morphism”) and then give a few examples of why this process is useful.

Preliminaries

Let $ k$ be a commutative ring, let $ G$ be our algebraic group and $ k[G]$ be the representing Hopf algebra of $ G$. In the classical language, $ k[G]$ is just the coordinate algebra of the $ k$-points $ G(k)$ of the algebraic group scheme $ G$.

A good example for the reader to keep in mind is $ \mathbb{G}_m$, which associates to every $ k$-algebra $ A$ the group of units of $ A$. In particular $ \mathbb{G}_m(k) = k^*$. This is an algebraic group scheme represented by the algebra $ k[x,x^{-1}]$ that has a Hopf algebra structure induced by the multiplication in the group of units. Of course, eventually we will want to consider more interesting algebraic groups, but this will be a good test case for our initial definition of the Lie algebra, which will definitely not be the easiest to work with!

The First Definition

A derivation $ D:k[G]\to k[G]$ is a $ k$-module homomorphism that satisfies the Leibniz rule

$ D(fg) = fD(g) + gD(f)$

for all $ f,g\in k[G]$. We denote the set of derivations by $ \mathrm{Der}(k[G])$.

If $ x\in G(k)$, then the morphism of varieties $ G(k)\to G(k)$ given by $ y\mapsto x^{-1}y$ induces a morphism $ \lambda_x: k[G]\to k[G]$ of $ k$-algebras, which we call left translation. The term “translation” refers to translation of functions because $ k[G]$ is the $ k$-algebra of $ k$-valued functions on $ G(k)$. We call a derivation $ D:k[G]\to k[G]$ left-invariant if it satisfies $ D\lambda_x = \lambda_xD$ for all $ x\in G(k)$.

Now, given any $ k$-algebra $ A$ we can associate to it a $ k$-Lie algebra $ \mathrm{Lie}(A)$ by defining the Lie bracket as $ [x,y] = xy – yx$ for all $ x,y\in A$. It is left as an exercise for the reader to show:

  1. The bracket of two derivations is a derivation, and hence the $ k$-module of derivations $ \mathrm{Der}(k[G])$ is a $ k$-Lie algebra.
  2. If $ D,E$ are left-invariant then $ [D,E]$ is also left-invariant.

Thus the Lie subalgebra $ \mathcal{L}(G)$ of all left-invariant derivations in $ \mathrm{Der}(k[G])$ is a Lie algebra via the formula $ [x,y] = xy – yx$, and we call it the Lie algebra of $ G$.

A Computation

Despite the relative simplicity of the above definition, computing the Lie algebra of $ G$ is quite tricky. The reader should work out the Lie algebra for the example of $ \mathbb{G}_m$, using the following sketch if desired. In this case, the representing Hopf algebra is $ k[x,x^{-1}]$. It follows from the Leibniz rule that a derivation $ D$ in this case is determined by $ D(x)$ and that $ D(x) = x$ is in fact the only possibility, which follows from left-invariance.

Thus the Lie algebra $ \mathcal{L}(G)$ is $ 1$-dimensional, and so the Lie bracket is identically equal to $ 0$. Notice how the dimension of the Lie algebra associated to $ \mathbb{G}_m(k)$ is the same as the dimension of $ \mathbb{G}_m(k)$ as an algebraic variety. This in fact will always turn out to be the case, as we shall see later.

This computation was fairly straightforward, but for more complicated algebraic groups, working out the Lie algebra could be rather tricky. It’s not even clear from this definition that the Lie algebra is in fact finite dimensional in all cases.

What About Morphisms?!

Now, let us see how the above definition is actually functorial. For this, we should explain how an algebraic group morphism gives rise to a corresponding morphism of Lie algebras.

Keeping with the functorial language, a morphism of algebraic groups $ \varphi:G\to H$ is a natural transformation that is also a group homomorphism for each $ k$-algebra $ A$. Via Yoneda, these natural transformations correspond naturally bijectively to Hopf algebra homomorphisms $ \varphi:k[H]\to k[G]$. This is just a generalisation of the classical case where a morphism of algebraic groups $ G(k)\to H(k)$ is determined by the corresponding Hopf algebra morphism.

Thus we may as well just consider a morphism $ \varphi:k[H]\to k[G]$ of Hopf algebras. (Of course, if we’re coming from the group side, we don’t need to verify that it is a Hopf-algebra morphism in the first place!). Now the question I pose to the dear reader is: given a left-invariant derivation $ D\in \mathcal{L}(G)$, how can we use $ \varphi$ to get a left-invariant derivation on $ \mathcal{L}(H)$? I encourage the reader to try a few things before moving onto the answer, which will be contained in the next post.

1 Comment

  1. Hi, thanks for this lovely blog! I found exactly what I was looking for (and more!).

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