5
$\begingroup$

Let

  • $k$ be an algebraically closed field,
  • $G$ a (smooth, connected) reductive algebraic group over $k$,
  • $H$ a (smooth, connected) reductive group of semisimple rank 1, and
  • $T$ a maximal torus in $H$.

I am specifically interested in the case where the characteristic of $k$ is a bad prime for $G$.

Suppose that we are given a group embedding $F_T : T \to G$ and a $T$-equivariant Lie-algebra embedding $f_H : \mathfrak h \to \mathfrak g$, such that the restrictions to $\mathfrak t$ of $f_H$ and the derivative of $F_T$ agree. Can we extend $F_T$ to an embedding $H \to G$ whose derivative is $f_H$?

(I know that most results in this area are stated under the assumption of good characteristic, or even of characteristic 0, but I don't actually know any of the counterexamples or where to look for them, so I can't tell if the hypothesis of the existence of $T$ is strong enough to allow me to overcome bad characteristic.)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Typically not, even for good characteristic. This comes up indirectly in my paper with Adam Thomas The Jacobson–Morozov theorem and complete reducibility of Lie subalgebras, JLMS arxiv version.

The nilpotent orbits of $G$ on $\mathfrak g$ in characteristic $0$ have representatives $e_{\mathbb Z}$ which are sums of root elements and which remain in distinct orbits when base-changed to a field of good characteristic. Furthermore, each can be completed to an $\mathfrak{sl}_2$-triple $(e_{\mathbb Z},h_{\mathbb Z},f_{\mathbb Z})\in\mathfrak{g}_{\mathbb Z}^3$—this is a result of Pommerening, though Thomas and I extend it slightly. Denote by $(e,h,f)$ the elements $(e\otimes 1,h\otimes 1,f\otimes 1)$ of $(\mathfrak{g}_{\mathbb Z}\otimes k)^3$. We have $h$ is toral, so $\mathfrak t=\langle h\rangle$ is a torus of $\mathfrak{g}$. Since $h$ can be expressed as an integral combination of coroots, $\mathfrak t$ can be lifted to a torus $T$ such that $T$ normalises $\langle e\rangle$ and $\langle f\rangle$. Let $\mathfrak{h}=\langle e,h,f\rangle\cong \mathfrak{sl}_2$. We have that $T$ normalises $\mathfrak{h}$, so that the embedding $\mathfrak{h}\to\mathfrak{g}$ is $T$ equivariant. Suppose we could lift this to $F_H:H\to G$ for some reductive $H$. (So $H\cong \operatorname{SL}_2$ or $\operatorname{PGL}_2$.)

Take the case that $e$ is a regular nilpotent element of $\mathfrak g$. Then a non-trivial unipotent element of $H$ must be regular too. But for $p$ less than the Coxeter number of $G$, a regular unipotent element has order at least $p^2$, so cannot be embedded into an $SL_2$ or $PGL_2$ (whose unipotents have order $p$).

EDITED: I initially expected this problem goes away if you insist $p$ is bigger than the Coxeter number of $G$, because I said there is a 1-1 correspondence between conjugacy classes of $\mathfrak{sl}_2$-subalgebras of $\mathfrak g$ and subgroups of $G$ of type $A_1$. This is obviously wrong however, since in positive characteristic one gets many more irreducible representations for $SL_2$ than for $\mathfrak{sl}_2$, giving many more $SL_2$ or $PGL_2$ subgroups of $GL_n$ than $\mathfrak{sl}_2$ subalgebras, c.f. Steinberg's tensor product theorem.

By way of correction, if $p$ is bigger than the Coxeter number, all unipotent elements have order $p$. Then one can cite Theorem 4.2 of Lawther, R.; Testerman, D. M., (A_1) subgroups of exceptional algebraic groups, Mem. Am. Math. Soc. 674, 131 p. (1999). ZBL0936.20039. in which an $A_1$ subgroup is guaranteed such that the Lie algebra of its root group contains nilpotent elements of the same Bala–Carter type; a fortiori one that corresponds under a Springer isomorphism.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ It seems that at some point you switch from characteristic $0$ to characteristic $p$, but I guess characteristic $0$ was just motivation. Do you have a reference for the fact that about regular unipotent elements when $p$ is less than the Coxeter number? $\endgroup$
    – LSpice
    Commented Jun 5, 2021 at 19:58
  • 2
    $\begingroup$ One reference is [Liebeck-Seitz, Unipotent and Nilpotent Classes in Simple Algebraic Groups and Lie Algebras] and look in the index for 'Orders …'. This probably goes back to the Order Formula 0.4 in [Testerman, A_1-type overgroups of elements of order p]. (Will clarify where I base-change to char p.) $\endgroup$
    – David Stewart
    Commented Jun 5, 2021 at 20:28
  • 1
    $\begingroup$ @spin. There must be a better way to do this, maybe using a Springer isomorphism and/or associated cocharacters, but here is an answer that I think works. Let $U$ be the unipotent radical of a Borel of $H$. By the Borel–Tits theorem, we can conjugate $U$ into the unipotent radical of a Borel of $G$. Since $e$ is regular, it must be supported on all the simple roots. Since $Lie(U)=\langle e\rangle$, the same must be true of $1\neq u\in U$. Hence $u$ is a regular unipotent of $U$. $\endgroup$
    – David Stewart
    Commented Jun 6, 2021 at 18:35
  • 3
    $\begingroup$ I'm not sure what the Borel–Tits theorem is (maybe just the statement that any connected unipotent subgroup can be so conjugated?), but conjugacy of $U$ into the unipotent radical of a Borel is elementary: the chosen Borel of $H$ is a connected soluble subgroup of $H$, hence of $G$, hence is contained in a maximal connected soluble subgroup $B$ of $G$, which is a Borel; and, since it is unipotent, it has trivial projection on the torus $B/\operatorname{Rad}_u(B)$, so is contained in $\operatorname{Rad}_u(B)$. $\endgroup$
    – LSpice
    Commented Jun 7, 2021 at 0:02
  • 1
    $\begingroup$ @LSpice. Agreed, that's reasonable. The B-T thm says that any closed subgroup $H$ of reductive $G$ lives in a parabolic $P$ of $G$ such that $R_u(H)\subseteq R_u(P)$. $\endgroup$
    – David Stewart
    Commented Jun 7, 2021 at 11:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .