# On components of centralisers in unipotent groups

Let $$k$$ be an algebraically closed field with $$\mathrm{char}(k)=p>0$$. Let $$U$$ be a connected unipotent algebraic group over $$k$$.

Question: When $$p$$ is big enough, is it true that $$Z_U(u)$$ is connected for any $$u\in U$$, or at least $$u\in Z_U(u)^o$$ for any $$u\in U$$?

Remark: This is true when $$U$$ is the unipotent radical of a Borel subgroup of a reductive group with $$p$$ being good (so that a Springer homeomorphism exists).

• Did you mean to ask whether, for a unipotent group scheme $U$ over $\mathbb{Z}$, the fibre $U_{\mathbb{F}_p}$ has connected centralisers for all sufficiently large $p$? This seems to me to be the more interesting question because one usually expects the fibres $U_{\mathbb{F}_p}$, for large $p$, to behave like groups over characteristic $0$ fields. There is no reason to expect this if $U$ is allowed to vary with $p$. Jan 31 '20 at 9:21
• @AStasinski Yes, $p$ is expected to be irrelevant to the defining equations of $U$. The question stated in the current form is a bit misleading. Jan 31 '20 at 10:45
• In that case kneidell's answer is not an answer to the intended question. Jan 31 '20 at 10:54
• If your element and your group are globally defined (say, over $\mathbb Z$) then both $Z_U(u)$ and $Z_U(u)^\circ$ would be $\mathbb Z$ defined, hence so would the quotient group. Taking $p$'s larger than the exponent of this finite group, shouldn't this finite algebraic group have no points over $k$? Feb 1 '20 at 15:06

The following is a counterexample which can be defined for arbitrarily large $$p$$'s.
Consider $$U=\left\{ \begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}:a,b\in k\right\}\subseteq\mathrm{GL}_3(k)$$ and take $$u=u_\lambda=\begin{pmatrix}1&\lambda\\&1&\lambda\\&&1\end{pmatrix}$$ with $$0\ne \lambda\in\mathbb{F}_p$$ (i.e. $$\lambda=\lambda^p$$). Then one easily computes that $$\begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}\in Z_U(u)\quad\iff\quad a=a^p,$$ and thus $$Z_U(u)\simeq \mu_p\ltimes \mathbb G_a$$ and $$Z_U(u)^\circ=\begin{pmatrix}1&&*\\&1&\\&&1\end{pmatrix}\simeq \mathbb G_a$$ (here $$\mu_p$$ is the group of $$p$$-th roots of $$1$$). In particular $$u\notin Z_U(u)^\circ$$.
• Note that this gives a family of groups $U$ depending on $p$. See also my comment on the question. Jan 31 '20 at 9:14