Connectedness of centralizers and regular elements in unipotent groups

Question

Let $G$ be a connected linear algebraic group over an algebraically closed field $k$ of characteristic $p$. An element $x\in G$ is called regular if its centralizer has minimal dimension among all the elements of $G$. Suppose now that $G$ is connected simple, let $u\in G$ be a regular unipotent element, and let $U$ be the unipotent radical of the Borel containing $u$. Then Springer [Some arithmetical results on semi-simple Lie algebras] has shown that the centralizer $C_{U}(u)$ is connected, provided $p$ is a good prime for $G$.

1. Is it true that if $p$ is a good prime for $G$, then $C_{U}(x)$ is connected for all $x\in U$?

2. Let $V$ by an arbitrary connected unipotent group over $k$. Is it possible to describe some 'small' subset $S$ of $V$, such that the connectedness of $C_{V}(x)$ for all $x\in S$ implies the connectedness of $C_{V}(x)$ for all $x\in V$? In particular, can $S$ be taken to be the set of regular elements in $V$?

Answer 1

Suppose that $G$ is semisimple and that the characteristic is very good for $G$. This means that the characteristic is good, and doesn't divide $n$ if $A_{n-1}$ is an irreducible component of the root system of $G$. [I'll ignore the issue of whether or not "good but not very good" primes present any real issue -- anyhow if you take GL(n) instead of SL(n) any problem should go away.]

Under this assumption, there is a $G$-equivariant isomorphism between the nilpotent variety and the unipotent variety of $G$ (a "Springer isomorphism"). This isomorphism gives by restriction an isomorphism $\operatorname{Lie}(U) \to U$ which is $B$-equivariant.

Now observe that $C_{\operatorname{Lie}(U)}(x)$ is equal to the fixed points of the action of $Ad(x)$ on $\operatorname{Lie}(U)$, and is connected (since a linear space)

The [choice of a] Springer isomorphism identifies $C_{\operatorname{Lie}(U)}(x)$ with the fixed points of $\operatorname{Int}(x)$ on $U$, namely with $C_U(x)$. So indeed $C_U(x)$ is connected.

Answer 2

To give a partial answer, note first that the relationship between $C_G(u)$ and $C_U(u)$ gets complicated to work out when $u$ is not regular, while the structure of $C_G(u)$ itself is known only through case-by-case analysis. The regular unipotent elements of $G$ form a single conjugacy class and are dense in the unipotent variety, with a typical centralizer $C_G(u) = Z(G) C_U(u)$ (which equals $C_U(u)$ when $G$ is simple in the group-theoretic sense). The group $C_G(u)$ turns out to be always commutative and (as noted) $C_U(u)$ is disconnected precisely when $p$ is bad for $G$, in which case $u$ itself fails to lie in the identity component.

The subregular elements form the next largest class and have also been well studied. Except in rank 1, the centralizer is solvable and often unipotent. But in general the study of $C_G(u)$ gets more complicated (and even the existence of a Levi factor is not obvious). Centralizers sometimes fail to be connected even for good $p$, but have been pinned down case-by-case. Component groups, important in Springer theory, can be elementary abelian 2-groups or symmetric groups $S_3, S_4, S_5$ (the latter just once, in type $E_8$); here only bad primes can divide the orders of component groups.

So without having all the data at hand I'm not sure about the answer to your first question but expect it will be yes.

The second question is more problematic (and open-ended). In prime characteristic the structure of connected unipotent groups can get very complicated. Moreover, the notion of "regular" element probably isn't useful here. This question probably needs more refinement and motivation.