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).



On components of centralisers in unipotent groups
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 $u\notin Z\_U(u)^\circ$. 


