4
$\begingroup$

$\DeclareMathOperator\GL{GL}$If $G$ is a simple Lie group, and $\rho: G \to \GL(V)$ is a representation, then by Schur's lemma, the group of automorphisms of $\rho$ is a reductive subgroup of $\GL(V)$. I'm wondering whether this generalizes to the case where $\GL(V)$ is replaced by an arbitrary reductive group?

More generally: if $G$ is a semisimple algebraic group (or even reductive), $H$ is a reductive algebraic group, and $\rho : G \to H$ is a homomorphism, is the centralizer of the image of $\rho$ in $H$, $C_{H}(\rho(G))$, a reductive subgroup of $H$?

Edits:

  1. The case I'm most interested is when the field is $\mathbb{C}$, but I would be interested in hearing about other cases as well.

  2. As pointed out in the comments, in the case $G = H$, the centralizer is the centre, which is reductive but not necessarily connected. So the centralizer may not be connected in the general case as well.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Note that there is no difference between handling the reductive, the semisimple, and the simple cases: $G$ is covered by the product of its simple subgroups and a central torus, and so we may as well regard the representation as coming from that product; the centraliser of a torus is always reductive; and then we reduce the case of $G$ a product of simple factors to the simple case by induction. (One still has to worry about simple vs absolutely simple, though.) $\endgroup$
    – LSpice
    Commented Dec 28, 2020 at 4:37
  • 3
    $\begingroup$ You probably also need to specify whether you want to require characteristic $0$, or algebraic closure (which I guess you don't since you begin by talking about (presumably real) Lie groups). In positive characteristic I think you'll run into trouble with non-completely reducible representations. $\endgroup$
    – LSpice
    Commented Dec 28, 2020 at 4:38
  • 1
    $\begingroup$ algebraic closure of a ground field is irrelevant because being reductive (in char. 0) is absolute. $\endgroup$
    – YCor
    Commented Dec 28, 2020 at 7:20
  • 1
    $\begingroup$ Your definition of 'reductive' doesn't require 'connected', right? Otherwise it is easy to come up with counterexamples (e.g., $G = H = \operatorname{SL}_2$). $\endgroup$
    – LSpice
    Commented Dec 28, 2020 at 18:01
  • 1
    $\begingroup$ Centralizers of semisimple elements are reductive. Let's suppose that H is connected. Semisimple elements are Zariski-dense in H. By dimension reasons, finitely many suffice, of which you even can assume that they are pairwise conjugate, as you can take one regular element and use the fact that all maximal tori are conjugate. Hence all you need to show is that the intersection of finitely many conjugates of a reductive group is reductive. $\endgroup$
    – user130903
    Commented Dec 30, 2020 at 6:34

1 Answer 1

Reset to default
3
+50
$\begingroup$

In characteristic zero, the connected closed subgroups of $G$ are in 1-1 correspondence with the Lie subalgebras of ${\rm Lie}(G)={\mathfrak g}$, and the Killing form a suitably chosen symmetric bilinear $G$-equivariant form on ${\mathfrak g}$ is non-degenerate. In fact we can choose this form to be the trace form for a rational representation for $G$. Since the Killing form this form is $G$-equivariant, it is $H$-equivariant, and irreducible summands for non-isomorphic $H$-submodules are orthogonal. Hence the Killing form on ${\mathfrak g}$ restricts to a non-degenerate form on ${\mathfrak g}^H={\rm Lie}(Z_G(H)^\circ)$.

I now claim that if $K$ is a connected algebraic group with a non-degenerate trace form $\kappa$ arising from a rational representation $\rho:K\rightarrow {\rm GL}(V)$, then $K$ is reductive. Indeed, if the unipotent radical $R_u(K)$ is non-trivial then $\rho(R_u(K))$ is a unipotent subgroup of ${\rm GL}(V)$, so after conjugation is contained in the subgroup of upper-triangular unipotent matrices, so the restriction of $\kappa$ to ${\mathfrak u}={\rm Lie}(R_u(K))$ is zero. Since ${\mathfrak u}$ is an ideal of ${\rm Lie}(K)$, this contradicts the non-degeneracy of the form. In particular, $Z_G(H)^\circ$ is reductive.

Improve this answer
$\endgroup$
14
  • 1
    $\begingroup$ What is the Killing form on a reductive (as opposed to semisimple) group? Do you restrict from some embedding into $\operatorname{SL}(V)$? $\endgroup$
    – LSpice
    Commented Sep 5, 2021 at 13:25
  • 1
    $\begingroup$ Good point - I guess I just mean that there exists a $G$-equivariant symmetric bilinear form on ${\mathfrak g}$. Any such form (it doesn't have to be the Killing form) will work for this argument. $\endgroup$
    – Paul Levy
    Commented Sep 5, 2021 at 14:27
  • $\begingroup$ I think another step is needed, because the Lie algebra of $\mathbb G_a$ also has a nondegenerate form. $\endgroup$
    – Will Sawin
    Commented Sep 5, 2021 at 18:35
  • $\begingroup$ Ok, I want the form to be a trace form - I'm fairly sure ${\rm Lie}({\mathbb G}_a)$ can't have a non-degenerate form coming from a rational representation for ${\mathbb G}_a$. $\endgroup$
    – Paul Levy
    Commented Sep 5, 2021 at 21:39
  • 1
    $\begingroup$ Sorry, scratch that. Easy explanation: let $x\in{\rm Lie}(K)$ and $y\in{\mathfrak u}$. Then there's a Borel subalgebra ${\mathfrak b}$ of ${\rm Lie}(K)$ containing $x$, and then ${\mathfrak u}$ is contained in the nilradical of ${\mathfrak b}$, and we may assume ${\mathfrak b}$ consists of upper triangular matrices, then it's easy to see ${\rm tr}(xy)=0$. $\endgroup$
    – Paul Levy
    Commented Feb 5, 2022 at 23:08

You must log in to answer this question.

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