# Borel subgroups of centralisers of Lie algebra elements in bad characteristic

Let $G$ be a simple linear algebraic group over an algebraically closed field $k$ of characteristic $p>0$, and let $\mathfrak{g}=\mathrm{Lie}(G)(k)$ denote (the $k$-points of) the Lie algebra.

Question. Assume that $p$ is not very good for $G$. For $X\in\mathfrak{g}$, is there a Borel subgroup $B_{0}\subseteq C_{G}(X)_{\mathrm{red}}^{\circ}$ and a Borel subgroup $B\subseteq G$ such that $X\in\mathrm{Lie}(B)(k)$ and $B_{0}\subseteq B$? (Here $C_{G}(X)_{\mathrm{red}}^{\circ}$ denotes the connected component of the reduced subgroup of the centraliser $C_{G}(X)$.)

If $p$ is very good for $G$ it is known that $C_{G}(X)$ is reduced, so $C_{\mathfrak{g}}(X)=\mathrm{Lie}(C_{G}(X))(k)$, and hence $X\in\mathrm{Lie}(C_{G}(X))(k)$. By a theorem of Grothendieck (see 14.25 in Borel's book on linear algebraic groups), the Lie algebra of any smooth algebraic group is covered by Borel subalgebras (i.e., Lie algebras of Borel subgroups), so there exists a Borel subgroup $B_{0}\subseteq C_{G}(X)^{\circ}$ such that $X\in\mathrm{Lie}(B_{0})(k)$. Since $B_{0}$ lies in some Borel subgroup of $G$, the desired conclusion follows.

When $p$ is not very good for $G$, it is not true in general that there is a Borel $B_{0}\subseteq C_{G}(X)_{\mathrm{red}}^{\circ}$ such that $X\in\mathrm{Lie}(B_{0})(k)$. For example, take $G=\mathrm{SL}_{p}$ and $X=\lambda+Y$, where $\lambda$ is any non-zero scalar matrix and $Y$ is regular nilpotent. However, the above question has a positive answer in this case since $C_{G}(X)_{\mathrm{red}}^{\circ}$ is its own Borel and $C_{\mathfrak{g}}(X)=\mathrm{Lie}(C_{G}(X)_{\mathrm{red}})(k)\oplus\mathfrak{z}$ (where $\mathfrak{z}$ is the centre of $\mathfrak{g}$), which embeds in the Lie algebra of a Borel of $G$.

Added. Since the question can (in principle) be decided by going through the nilpotent $X$, I would also be interested in references to sources which may be helpful in carrying out the case by case analysis (i.e., containing explicit information about the Borels in $C_G(X)_{\mathrm{red}}^{\circ}$ and the structure of $C_{\mathfrak{g}}(X)$).

• The case of type $A_n$ when $p|n$ might be treatable separately. Beyond this, it's probably worth emphasizing that any question which focuses on bad primes necessarily requires case-by-case checking. One possibly helpful source may be the detailed treatment in the 2012 AMS monograph by M.W. Liebeck and G.M. Seitz, Unipotent and Nilpotent Classes in Simple Algebraic Groups and Lie Algebras. Most of what they do concerns bad primes, and their tables may be useful for your purpose. [By the way, adding a tag 'reference-request' might help.] – Jim Humphreys Jul 12 '17 at 14:36
• P.S. For exceptional types when $p$ is bad, the tables computed by Ross Lawther are useful to consult: ams.org/mathscinet-getitem?mr=1351124 and ams.org/mathscinet-getitem?mr=1627924 (keeping in mind that his later adjustments were mainly a response to a few errors in the older tables of Mizuno). Jordan blocks of centralizers of nilpotent elements allow one to compare centralizers in $G$ and $\mathfrak{g}$, whether or not $\mathfrak{g}$ has a nontrivial center. – Jim Humphreys Jul 12 '17 at 18:41
• Small correction to my first comment: I should write $p|(n+1)$, or specify $\mathfrak{sl}_n$ when $p|n$. – Jim Humphreys Jul 12 '17 at 22:12

I think the positive answer to this question follows from some results obtained in the paper The Hesselink stratification of nullcones and base change, Invent. Math., 191 (2013), 631-669, by M. Clarke and A. Premet. It is proved in the paper among other things that for any nonzero nilpotent element $$e\in \mathfrak{g}$$ there exists an optimal cocharacter $$\lambda\in X_*(G)$$ for $$e$$ such that $$e\in \bigoplus_{i\ge 2}\,\mathfrak{g}(\lambda,i)$$ and $$C_G(e)_{\rm red}\subseteq P(\lambda)$$, where $$P(\lambda)$$ is the parabolic subgroup of $$G$$ with Lie algebra $$\bigoplus_{i\ge 0}\,\mathfrak{g}(\lambda,i)$$. If we choose $$e\in\mathfrak{g}$$ nicely then this cocharacter is obtained by base-changing one of the Dynkin cocharacters (defined over the integers). Any Borel subgroup $$B$$ of $$P(\lambda)$$ has that property that $$\bigoplus_{i\ge 1}\,\mathfrak{g}(\lambda,i)\subset {\rm Lie}(B)$$ and hence $${\rm Lie}(B)$$ contains $$e$$. So any $$B$$ as above containing a Borel subgroup of $$C_G(e)_{\rm red}\subset P(\lambda)$$ would do.