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?

  • $\begingroup$ Is "Richardson isomorphisms" in the header an oversight? Aside from that, the non-uniqueness of Springer isomorphisms (especially for exceptional root systems) needs to be built more carefully into the question, since there might be multiple maps $\phi$ of the type you consider. For the classical root systems there are explicit maps available (Cayley, etc.); have you looked at these cases? It's also good to add a reference, such as McNinch-Testerman, J. Pure Appl. Algebra 213 (2009), 1346–1363. $\endgroup$ Dec 22, 2011 at 22:28
  • $\begingroup$ Oh yes, that should have been "Springer isomorphisms." Somehow I was thinking about Richardson elements and got mixed up. $\endgroup$ Dec 23, 2011 at 5:10

1 Answer 1


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.


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