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As the question says, is a connected unipotent subgroup $U$ of a linear algebraic group scheme $G$ always contained in a Borel subgroup of $G$? I have an argument for why the answer is yes, and I have a somewhat confusing example that the answer is no. Obviously one (or both!) of these is wrong but I can't spot the error.

Argument for yes: Unipotent subgroups are solvable and a solvable connected group must be contained in a maximal solvable connected group, i.e., a Borel subgroup.

That seems too simple to be false, so I suspect the problem is with the counterexample which is not so simple.

Counterexample: Let $G = \mathrm{PGL}_2$ over an algebraically closed field $k$ of characteristic $2$. The first Frobenius kernel $G_1 \leq G$ is a closed subgroup whose group ring $kG_1$, which is the dual of the coordinate ring $k[G_1]$, is isomorphic as a Hopf algebra to the universal enveloping algebra of $\mathrm{Lie}(G) = \mathfrak{pgl}_2$. This means any Lie subalgebra of $\mathfrak{pgl}_2$ which consists of nilpotent elements is, by Engel's theorem, a unipotent subgroup of $\mathrm{PGL}_2$. Well, take the subalgebra $$\mathfrak e = \begin{bmatrix}0 & k \\ k & 0\end{bmatrix}.$$ The enveloping algebra of $\mathfrak e$ is $k[x, y]/(x^2, y^2)$. This is self dual so as a group scheme $\mathfrak e$ is connected. This means it should lie in a Borel subgroup of $\mathrm{PGL}_2$, and hence lie in the Lie algebra of that Borel subgroup. But $\mathrm{PGL}_2$ is semisimple of type $A_1$, the unipotent radical of any Borel has a $1$-dimensional Lie algebra so $\mathfrak e$ can't be contained in it.

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  • $\begingroup$ The answer in the affirmative is Corollary 11.3 (2) of Borel's book. In fact, the maximal such groups are each the unipotent part of a Borel subgroup and they are all conjugate. I don't know where the counterexample fails. $\endgroup$ Aug 23 '17 at 16:41
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I assume that you're working over $k$ algebraically closed to avoid existence of Borel issues. If not, let me know, and I can rewrite this.

Let $G$ be a connected linear algebraic group over $k$. Then, recall that a Borel subgroup of $G$ is a maximal connected smooth unipotent subgroup of $G$. In particular, clearly if $U$ is a smooth connected unipotent subgroup of $G$, then $U$ is contained in a Borel. But, if $U$ is not smooth then there is clearly no need. I don't understand your counterexample, but I have a hunch that this is the issue.

Here's a silly example to show that without smoothness hypotheses unipotent subgroups needn't be contained in Borels:

Example: Let $k$ be an algebraically closed field of characteristic $p$. Let $G=\alpha_p$, which is the kernel of the Frobenius map $\mathbb{G}_a\to\mathbb{G}_a$. Then, $\alpha_p$ is certainly unipotent, but the only Borel subgroup of $G$ is the trivial subgroup.

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  • $\begingroup$ The definition of a Borel subgroup in Borel's book does not include smooth as a condition, and as the other commenter noted 11.3(2) in that book proves that the answer to my question is yes. $\endgroup$
    – Jim
    Aug 17 '20 at 21:28
  • $\begingroup$ As for your example, it's been some time since I've thought of these things, but isn't $\mathbb G_a$ itself connected, smooth, and unipotent? $\endgroup$
    – Jim
    Aug 17 '20 at 21:28
  • $\begingroup$ @jim Yes, but the point is that we're talking about $\alpha_p$, we're not talking about $\mathbb{G}_a$. If you look in Borel he DEFINES an algebraic group to be smooth (e.g. see the Proposition in Section 1.2). I'm not sure if you ever resolved your counterexample, but this is an actual techincal point. $\endgroup$ Aug 17 '20 at 21:31
  • $\begingroup$ Ah, I see you are right. When he defines schemes he defines them using max spec instead of the usual spec from hartshorne. So he's really still doing group varieties and not group schemes. $\endgroup$
    – Jim
    Aug 17 '20 at 21:39

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