I'm trying to understand explicitly a construction of Springer isomorphisms for adjoint exceptional groups given by Bardsley and Richardson. Their construction is as follows. Let $G$ be an adjoint simple exceptional algebraic group over an algebraically closed field of characteristic $p$, which is a good prime for $G$. Consider the adjoint representation $\mathrm{Ad} : G \to \mathrm{GL}(V)$, where $V$ is the Lie algebra $\mathfrak{g}$ of $G$, then we have $\mathfrak{gl}(V) = \mathrm{ad}(\mathfrak{g}) \oplus \mathfrak{m}$, where $\mathfrak{m}$ is an $\mathrm{Ad}(G)$-invariant complement of $\mathrm{ad}(\mathfrak{g})$ containing the identity. We then have a $G$-equivariant map $\phi : G \to \mathfrak{g}$ given by the composition

$$G \overset{\mathrm{Ad}}{\longrightarrow} \mathrm{GL}(V) \overset{i}{\longrightarrow} \mathfrak{gl}(V) \overset{p}{\longrightarrow} \mathrm{ad}(\mathfrak{g}) \overset{\mathrm{ad}^{-1}}{\longrightarrow} \mathfrak{g}$$

Here $i$ is the natural inclusion map and $p$ is the natural projection map. According to Bardsley and Richardson this restricts to a Springer isomorphism between the unipotent variety and the nilpotent cone.

Now let $T \leqslant B\leqslant G$ be a maximal torus and Borel subgroup of $G$ and denote by $U$ the unipotent radical of $B$. Fix an ordering $\alpha_1,\dots,\alpha_r$ of the set of positive roots (with respect to $T\leqslant B$) and for any root $\alpha$ choose an isomorphism $x_{\alpha} : \mathbb{G}_a \to X_{\alpha}$, where $X_{\alpha} \leqslant G$ is the corresponding root subgroup. Then any element of $U$ can be written uniquely as $x_{\alpha_1}(k_1)\cdots x_{\alpha_r}(k_r)$ for some $k_i \in \mathbb{G}_a$. Assume $e_{\alpha} = dx_{\alpha}(1)$ then inspecting the formula for the adjoint action of $G$ it seems that in the simply laced case the following inductive formula should be true. Let $y = x_{\alpha_2}(k_2)\cdots x_{\alpha_r}(k_r)$ then

$$\phi(x_{\alpha_1}(k_1)y) = \begin{cases} (k_1e_{\alpha_1}+1)\phi(y) &\text{if }y\neq 1,\\ k_1e_{\alpha_1} &\text{if }y = 1. \end{cases}$$

This also seems to follow from section 10.2.7 of Springer's "Linear Algebraic Groups".

My question is, does something like this hold for the non-simply laced case? Also, has anyone seen this come up in the literature before?

  • $\begingroup$ Maybe it works here and even in types $G_2, F_4$, though I'm a bit skeptical about getting this simple algorithm from such arbitrary choices of root ordering. Anyway, aside from type $A_n$ the isogeny class doesn't seem to affect anything; does "exceptional" matter? Note too that there are numerous "Springer isomorphisms", considered by McNinch-Testerman: front.math.ucdavis.edu/0805.2574 $\endgroup$ – Jim Humphreys Jul 15 '14 at 18:44
  • $\begingroup$ Something's not quite right - what do you mean by $(k_1e_{\alpha_1}+1)\phi(y)$? $\endgroup$ – Paul Levy Jul 15 '14 at 19:13
  • $\begingroup$ @PaulLevy This is somewhat sloppy. The product is being carried out in $\mathfrak{gl}(V)$. So really this should be $(k_1\mathrm{ad}(e_{\alpha_1})+1)\phi(y)$ if we remove the final $\mathrm{ad}^{-1}$ in the definition of $\phi$. $\endgroup$ – Jay Taylor Jul 15 '14 at 19:35
  • $\begingroup$ @JimHumphreys Exceptional doesn't necessarily matter but then one can't use the adjoint representation in the other types. For the classical cases, not of type $\mathrm{A}$, one could play the same game with the natural representation. It is true that there are numerous Springer isomorphism but I am looking for just one with good properties. Thus I need something I can get my hands on to show that these good properties hold. $\endgroup$ – Jay Taylor Jul 15 '14 at 19:38

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