Let $G$ be a semisimple, simply-connected algebraic group over an algebraically closed field $k$ of positive characteristic. Fix a Borel subgroup $B \subseteq G$ with unipotent radical $U$. Also let $P$ be a parabolic subgroup of $G$ containing $B$ and let $L$ be its Levi factor. Denote by $U\_P \subseteq U$ the unipotent radical of $P$ and set $U\_L := U \cap L$.


Let $\mathfrak U \subseteq G$ denote the unipotent variety and let $ \mathfrak R \subseteq \textrm{Lie}(G) $ denote the nilpotent cone. If $p$ is a good prime for $G$ then there is a $G$-equivariant Springer isomorphism $\phi : \mathfrak U \to \mathfrak R$ that restricts to an isomorphism $ U \to \textrm{Lie}(U) $. My question is: Does $\phi$ restrict to isomorphisms $ U\_P \to \textrm{Lie}(U\_P) $ and $U\_L \to \textrm{Lie}(U\_L)$? If this does not always happen, are there conditions on the parabolic $P$ under which it will be true?



Springer isomorphisms and parabolics
Any Springer isomorphism has the desired property. 
(Here I'm working over an algebraically closed field, else one should be more careful with the language)


Indeed, let $P$ be any parabolic subgroup
of $G$. Then there is a cocharacter $\lambda:\mathbf{G}\_m \to G$ for which $P = P(\lambda)$
is the parabolic subgroup determined by $\lambda$ -- see [Springer, Linear Alg. Groups Prop. 8.4.5].


Explicitly, $P =$ {$x \in G \mid \operatorname{lim}\_{t \to 0} \lambda(t) x \lambda(t)^{-1}$ exists} 
and the unipotent radical $U = R\_u(P)$ is given by $U = $ {$x \in P \mid \operatorname{lim}\_{t \to 0} \lambda(t) x \lambda(t) ^{-1} = 1$}; see [Springer, Linear Algebraic Groups, 3.2.13] for more on these limits.


In particular, it follows that $U$ is the set of all $x$ in the unipotent variety for which $\operatorname{lim}\_{t \to 0} \lambda(t) x \lambda(t)^{-1}$ exists and is equal to $1$.


Moreover, $\operatorname{Lie}(P)$ and $\operatorname{Lie}(U)$ have similar descriptions
-- e.g. $\operatorname{Lie}(P)$ consists of all $X \in \operatorname{Lie}(G)$ such
that $\operatorname{lim}\_{t \to 0} \operatorname{Ad}(\lambda(t))X$ exists. 


As before, one finds that $\operatorname{Lie}(U)$ consists in all the $X$ in the nilpotent
variety for which $\operatorname{lim}\_{t \to 0} \operatorname{Ad}(\lambda(t))X$ exists and is equal to $0$.


Since a Springer isomorphism $\phi$ is $G$-equivariant (and maps $0 \mapsto 1$), it follows
from these descriptions that $\phi$ maps $\operatorname{Lie}(U)$ isomorphically onto $U$.


I suspect (hope?!) that something like this argument is given in some paper I've written; maybe the one with Donna that Jim mentioned in his comment?


*EDIT:* Actually I wrote down the required statement in section 4 (remark 10) of "Optimal SL(2)-homomorphisms,"
Comment. Math. Helv. 80 (2005), no. 2, 391–426. 


