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Steinberg's lecture at the 1966 ICM in Moscow here surveyed his work on regular elements of semisimple algebraic groups, while also formulating a number of then-open questions as "problems" (with positive solutions expected). For example, he wrote: "A more modest problem is as follows. (10) Problem. Prove that the number of unipotent classes of $G$ is finite." [This was later proved by Lusztig, using an ingenious argument based on Deligne-Lusztig character theory for Chevalley groups; the problem is not so "modest".] Another question (22) was later disproved. But by now most of the problems have well-documented outcomes. In one case I don't recall any relevant sources, which prompts my question:

Is Steinberg's Problem (12), described below, ever referred to in the subsequent literature?

To fill in essential background here (using Steinberg's notation), take $G$ to be a connected semisimple and simply connected algebraic group over an algebraically closed field $K$, with Lie algebra $L$. It's enough to assume $G$ is simple. [Here $L$ is in fact the Lie algebra over $K$ obtained by Chevalley's process from a Chevalley basis in a simple Lie algebra of the same type over $\mathbb{C}$.]

Now Steinberg considers the adjoint representation of $G$ on $L$, with $G_x$ denoting the centralizer of $x$ in $G$; its Lie algebra Lie($G_x$) is contained in the subalgebra $L_x$ of $L$ fixed pointwise by Ad $x$, as is the center of $L$. When $p$ is good (leaving aside type $A$), Steinberg shows in (11) that $L_x$ is just the Lie algebra of $G_x$. In (12) he asks for a proof that $L_x$ is the sum of Lie($G_x$) and the center of $L$ for all $p$ and all $G$, noting that the case of regular $x$ follows from his own study of regular elements. [Recall that $p$ is bad if $p=2$ for types other than $A$, or $p=3$ for exceptional types, or $p=5$ for type $E_8$.].

It turns out that (12) sometimes fails. One example, pointed out by Sasha Premet in a 2015 email, involves the unique group $G$ of type $E_8$ and the bad prime $p=5$. It's been known, since early work by Dieudonne and also by Steinberg, that $L$ here is simple for all $p$ (and thus is centerless). For $p=5$, the 1980 paper of Mizuno shows that the list of unipotent classes and their dimensions is the same as for p=0. For the class labelled $A_4+A_3$ by Bala-Carter, the dimension of $G_x$ is therefore 48. On the other hand, 1995 computations by Lawther in his Comm. Algebra paper show that in the smallest faithful representation of the group of type $E_8$ (the adjoint representation) one gets $\dim L_x=50$ (Table 9, p. 4148). This follows from Lawther's computation of Jordan blocks for a conjugate of $x$: the number of blocks is the number of independent fixed points of Ad $x$. Because $L$ is centerless, this disagrees with Steinberg's expectation in (12).

[EDIT] I'm still curious as to whether there is any follow-up to that problem in the literature.

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    $\begingroup$ I don't know if it's subsequently referred to again, but it's probably worth a comment that at least in the exceptional types, Liebeck and Seitz's book gives a list of the centraliser dimensions dim $G_x$ for x unipotent. Together with Lawther you can presumably get a list of all conjugacy classes where the answer is no. I guess there are a few. On the other hand work of Bate, Martin, Roehrle tell you that all subgroups are separable in very good characteristic---uses an argument of Richardson. So the answer will be yes there (well, also there's no centre $Z(L)$ to worry about!). $\endgroup$ – David Stewart Jul 18 '16 at 21:33
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    $\begingroup$ Yes, these later theoretical refinements are more precise. But Steinberg himself worked out many details for good prime characteristics, leaving some analogues for bad $p$ as problems. (Recent computations have filled in a lot more of the specifics for bad primes. But for example it's hard to find a nice proof that there are only finitely many nilpotent orbits without just pointing to case-by-case work.) $\endgroup$ – Jim Humphreys Jul 19 '16 at 15:52
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    $\begingroup$ I don't think there's any proof of the finiteness of the list of nilpotent orbits in bad characteristic which doesn't involve (i) working out conjugacy class dimensions for a putative list, (ii) using those to write down the number of elements in each orbit in $\mathfrak{g}(\mathbb F_q)$ in terms of a polynomial in $q=p^r$, (iii) then checking you have all of them by adding it up and praying that you get $q^{|\Phi|}$ elements. $\endgroup$ – David Stewart Jul 20 '16 at 11:09
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    $\begingroup$ @David: Yes, that's apparently the status at present. However, Lusztig's less direct proof of finiteness for the number of unipotent classes makes it tempting to look for an analogous proof. Of course, the "number" of classes or orbits can vary for bad $p$, which complicates matters. $\endgroup$ – Jim Humphreys Jul 20 '16 at 13:31

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