Lately I’ve been reading a bit of Robert Steinberg’s book, “Endomorphisms of Algebraic Groups”, so I’ve decided to explain what I am reading for this series on linear algebraic groups, which will take us away from Lie algebras for a bit of time.

Let $ G$ be a linear algebraic group over an algebraically closed field $ k$. In this case we can choose a Borel subgroup $ B$ and a maximal torus $ T\subseteq B\subseteq G$; in fact, all the maximal tori reside in Borel subgroups because tori are solvable. What about if we are given an automorphism $ \sigma:G\to G$? This prompts a fairly natural question:

Is it possible to choose $ B$ and $ T$ such that they are stable under $ \sigma$?

If so, this would be very convenient in working with automorphisms, and pretty useful for working with $ \theta$-groups for instance (more on this later!). It turns out that if $ \sigma$ can be realised as conjugation by a semisimple element after an embedding of $ G\hookrightarrow G’$, then this is possible. This was proven by Robert Steinberg and also by David J. Winter.

Let us call an automorphism $ \sigma:G\to G$ semisimple if it satisfies this nice property: $ \sigma$ can be realised as conjugation by a semisimple element after embedding $ G$ into another group. Thus the result is:

Steinberg-Winter Theorem. If $ \sigma:G\to G$ is a semisimple automorphism, then there is a $ \sigma$-stable pair $ (B,T)$ where $ B$ is a Borel subgroup and $ T\subseteq B$ is a maximal torus.

We shall sketch the proof of this result, leaving out some technical details and explaining the more elementary ones, and the interested reader is urged to peruse Steinberg’s book, “Endomorphisms of Algebraic Groups”, Section 7 for more details. Winter’s proof, which is a bit more terse and uses different techniques, appears in his paper “On Automorphisms of Algebraic Groups” (Project Euclid link).

A Lemma

Let us take a step back and look at endomorphisms. Given an endomorphism $ \sigma:G\to G$, when does it fix a Borel subgroup? It turns out that we just need to assume that $ \sigma$ is surjective. If we can prove this, then we will be closer to proving the main theorem. Most of the hard work in this game is contained in the following technical lemma, which is the part that is the “sketch”:

Theorem 1. Let $ G$ be a connected and $ \sigma:G\to G$ be a surjective endomorphism. If $ B$ is a Borel subgroup fixed by $ \sigma$ then the map $ \alpha:G\times B\to G$ defined by $ (x,b)\mapsto xb\sigma(x^{-1})$ is surjective.
Sketch of Proof. We can reduce to the case where $ G$ is semisimple since $ B$ contains the radical. We take for granted that at some point of $ G$, the differential on tangent spaces is surjective. Then, $ \mathrm{Im}(\alpha)$ contains a nonempty open subset. Since $ G$ is connected, $ \mathrm{Im}{\alpha}$ is dense in $ G$. Thus in order to show that $ \alpha$ is surjective, it is sufficient to prove that $ \mathrm{Im}(\alpha)$ is closed in $ G$.

To do this, let $ K = \{ (x,y) \in G\times G : x^{-1}y\sigma(x)\in B\}$. Then $ K$ is closed since $ K$ is the inverse image of the closed subgroup $ B$ under the morphism $ G\times G\to B$ given by $ (x,y)\mapsto x^{-1}y\sigma(x)$. Moreover, $ K$ consists of complete cosets with respect to the subgroup $ B\times 1$. Why is this? If $ (x,y)\in (g,h)B\times 1$ and $ (x,y)\in K$, suppose that $ b\in B$ is arbitrary, so that $ (gb,h)$ is another element of the coset $ (g,h)B$. If $ (x,y) = (gb_1,h)$ then setting $ xb_1^{-1} = g$ in $ (gb,h)$ shows that $ (gb,h)$ is also an element of $ K$.

Since $ K$ is closed and consists of complete cosets, its projection $ K’$ in $ G/B\times G$ is also closed. Finally, the map $ G/B\times G\to G$ is closed because $ G/B$ is complete, and so the image of $ K’$ which is $ \mathrm{Im}{\alpha}$, is closed. QED

Endomorphisms!

Remember, we were trying to prove a piece of the puzzle:

Theorem 2. If $ \sigma:G\to G$ is a surjective endomorphism, then $ \sigma$ fixes a Borel subgroup.
Proof. How does this Theorem 1 imply that an arbitrary surjective $ \sigma$ actually fixes a Borel $ B$?. We first need to construct a surjective endomorphism that actually does fix a Borel. So let us start with an arbitrary Borel subgroup $ B\subseteq G$. Then $ \sigma(B)$ is also a Borel because surjective endomorphisms of algebraic groups preserve dimension (they have finite kernels). Since all Borel subgroups are conjugate, there exists a $ y\in G$ such that $ \sigma(B) = y^{-1}By$. If $ \mathrm{Int}_y$ denotes conjugation by $ y$ then clearly $ \mathrm{Int}_y\circ\sigma$ fixes the Borel subgroup $ B$. Thus by Theorem 1, the map $ G\times B\to G$ given by

$ (x,b)\mapsto xb(\mathrm{Int}_y\circ\sigma)(x)$

is surjective, so there is some $ (x,b)$ such that $ xby\sigma(x^{-1})y^{-1} = y^{-1}$. Rearranging gives $ y\sigma(x^{-1}) = b^{-1}x^{-1}$. Applying $ \sigma$ to the Borel $ xBx^{-1}$ gives $ \sigma(xBx^{-1}) = xbBb^{-1}x^{-1} = xBx^{-1}$ and hence $ \sigma$ fixes $ xBx^{-1}$. QED

Automorphisms

We now turn to the case that $ \sigma:G\to G$ is a semisimple automorphism. Recall that this means we can embed $ G$ in a larger group so that $ \sigma$ is realised by conjugation by some semisimple element $ s$.

By our work above, we already know that $ \sigma$ fixes a Borel subgroup $ B$. Thus, in order to finish our investigations into the wonderful world of automorphisms, we just need to show that there is a maximal torus in $ B$ also fixed by $ \sigma$. This follows from the following theorem because $ B$ is solvable and because $ \sigma$ restricts to $ B$:

Theorem 3. If $ G$ is solvable and $ \sigma:G\to G$ is a semisimple automorphism then $ \sigma$ fixes a maximal torus.
Proof. Since a maximal torus will lie in the identity connected component, we may assume that $ G$ is connected. Choose some maximal torus $ T\subseteq G$. Recall that a basic structure result for connected solvable groups: we can write $ G$ as a semidirect product $ G = T\ltimes G_u$ where $ G_u$ is the unipotent closed subgroup of $ G$. Now embed $ G$ in some larger group so that $ \sigma$ is realised as conjugation by an element $ s$. Now the tori $ \sigma(T) = sTs^{-1}$ and $ T$ are maximal tori in $ G$ and hence $ usTs^{-1}u^{-1} = T$ for some $ u\in G$, and in fact we may assume that $ u\in U$ by our semidirect product decomposition.

Now, suppose that we could show that $ us$ were conjugate to $ s$ (in $ G$); i.e. $ us = xsx^{-1}$; then this would mean that $ \sigma$ fixes $ x^{-1}Tx$, a maximal torus. Hence we need to show that $ us$ and $ s$ are conjugate. We can now assume that $ us$ is semisimple. (I was confused by this point in the proof; I asked MathOverflow Question 111916, in which Jim Humphreys kindly explained the missing details).

Now we can assume that $ U$ is commutative, and the general case will follow by induction on the derived series (exercise!). According to a theorem of Chevalley, we can then write $ u = u_1^{-1}u_2su_1s^{-1}$ with $ u_1\in U$ and $ u_2\in U_s$ where $ U_s$ is the set of fixed points of $ U$ under $ s$. Multiplying by $ s$ on the right gives that $ us$ is conjugate to $ u_2s$. Since $ u_2$ commutes with $ s$, $ u_2s$ is the Jordan decomposition of $ u_2s = us$, which is semisimple, and hence $ u_2 = 1$. Thus $ us$ is conjugate to $ s$. QED

Thus if $ \sigma:G\to G$ is a semisimple automorphism, then it fixes a Borel subgroup $ B$ (Theorem 2) and within this $ B$ there is a maximal torus $ T\subseteq B$ that is also fixed by $ \sigma$ (Theorem 3), so we have shown the Steinberg-Winter theorem.

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