Posted by Jason Polak on 03. August 2013 · Write a comment · Categories: math

($ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.)

In the previous post, to prove that a two-dimensional connected group $ G$ is solvable, we used that if some Borel in it is either semisimple or unipotent, then it is actually all of $ G$.

Here is a short proof: We first prove that $ B$ is actually nilpotent. For $ B = B_u$, its unipotent part, this is true since connected unipotent groups are nilpotent (and thus solvable, but we already knew that). If $ B = B_s$, then we use the very handy result that a connected solvable group is the product of a torus and its unipotent part, but in this case the unipotent part is trivial, so $ B$ itself is a torus and so $ B$ is again nilpotent.

Now, we proceed by induction on the dimension of $ G$. If $ \dim G = 0$ then the result is clear, since $ G$ is then trivial since it is connected. Now suppose $ G$ has positive dimension, and so $ B$ has positive dimension as well (since it is connected too, and the conjugates of $ B$ cover $ G$). In this case, as we have observed, $ B$ is nilpotent and nontrivial, so it has a nontrivial center (this is true for any nontrivial nilpotent group, not just algebraic groups).

Now $ Z(G)^\circ$, being solvable and connected, must be contained in a Borel subgroup, and since it is normal, it is contained in every Borel subgroup so we have the inclusions $ Z(G)^\circ\subseteq Z(B)\subseteq Z(G)$. Since we have observed that $ Z(B)$ is nontrivial, and hence has positive dimension, it also must be that $ Z(G)^\circ$ (a normal subgroup) has positive dimension so that $ G/Z(G)^\circ$ has strictly smaller dimension than $ G$.

Since the surjective image of a Borel subgroup is a Borel (likewise for maximal tori, parabolics, and maximal connected unipotents), the group $ B/Z(G)^\circ$ is a nilpotent Borel of $ G/Z(G)^\circ$. But now we use the induction hypothesis which implies that $ G/Z(G)^\circ = B/Z(G)^\circ$ and hence $ B = G$. QED

Observe that the key to this proof really was the use of induction via the nontrivial $ Z(G)^\circ$ contained in both $ G$ and $ B$, and that the homomorphic image of a nilpotent group is still nilpotent.

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