# Highlights in Linear Algebraic Groups 6: Representations I

Posted by Jason Polak on 18. March 2013 · Write a comment · Categories: algebraic-geometry, group-theory · Tags: ,

Soon it will be time to explore some aspects of root systems and structure theory for reductive groups. Our goal is to understand everything in the classical setting over an algebraically closed field, and then explore reductive groups over arbitrary base schemes.

Before we do this, I will give a few examples for some of the technical machinery we shall rely on. In this post, we shall see how an algebraic group acting on a variety $X$ and a function $f\in k[X]$ gives rise to a representation of $G$, and in the next post we shall see an example. I learnt the material in this section mainly from Jim Humphrey's book "Linear Algebraic Groups".

### Finite Dimensional, Infinite Dimensional

Our setting is an arbitrary algebraic group $G$ over an algebraically closed field $k$.

Let $X$ be a variety over $k$ on which $G$ acts, so that $G\times X\to X$ is a group action and a morphism of varieties. If $g\in G$ then there is a translation algebra homomorphism $\tau_g:k[X]\to k[X]$ defined by $\tau_g(f)(y) = f(g^{-1}y)$. The inverse is there so that

$[\tau_g(\tau_h(f))](y) = \tau_h(f)(g^{-1}y) = f(h^{-1}g^{-1}y) = f((gh)^{-1}y) = \tau_{gh}(y)$

In other words, $G\to\mathrm{Aut}_{k}(k[X])$ is actually a group homomorphism. Now, $k[X]$ is a $k$-vector space and thus this gives a representation of $G$, but it is infinite dimensional. How can we get finite dimensional representations?

On the other hand, if we choose a finite-dimensional subspace $F\subseteq k[X]$ that happens to be invariant the homomorphisms $\tau_g$ for all $g\in G$ ("translation invariant"), then we will get a finite-dimensional representation, which will be quite useful later, both for examples and as a theoretical tool. Of course, not all subspaces of $k[X]$ are going to be translation invariant.

So what do we do? Pick subspaces at random and hope for the best? Definitely not. Actually, given a finite dimensional subspace $F\subseteq k[X]$, we can always find a larger finite dimensional subspace $E\supseteq F$ that contains $F$ that is translation invariant, and is minimal in the sense that every translation invariant subspace containing $F$ will also contain $E$.

The idea is quite simple: just choose a basis for $F$, and $E$ will be the span of all the translates of the basis elements! Of course, there is something to be checked: that for a fixed $f\in k[X]$ the span of the set $\{ \tau_g(f) \}_{g\in G}$ is finite-dimensional. That the space $E$ is translation invariant is clear.

### Finite Dimensional

It is not hard to accomplish the latter. First, let $f\in k[X]$. The morphism $\phi:G\times X\to X$ that gives the action of $G$ on $X$ is given by a comorphism $\phi^*:k[X]\to k[G]\otimes_k k[X]$ of $k$-algebras (of course, not any morphism of algebras here will do; it satisfies additional conditions that imply that the corresponding action is a group action: it's a good exercise to write down these extra conditions!).

Next we apply $\phi^*$ to $f$ to get

$\phi^*(f) = \sum g_i\otimes h_i$

Now, $f(g^{-1}y) = \sum h_i(g^{-1})k_i(y)$ so $\tau_g(f) = \sum h_i(g^{-1})k_i$; in other words, the span of all the translates of $f$ is the span of the functions $\{h_i\}$, and there are only finitely many of them! Hence the corresponding space $E$ is finite dimensional.