$\DeclareMathOperator\GL{GL}$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.



Centralizers of semisimple subgroups
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.


