# A canonical G_m (or G) action on the Slodowy slice

Question

By Slodowy slice I mean a transverse slice at a subregular nilpotent orbit in a simple Lie algebra $\mathfrak{g}$ (in particular I am not intersecting with the nilpotent cone). Consider the usual eigenvalue map $$\mathfrak{g} \rightarrow \mathfrak{h}//W$$ which is equivariant with respect to the scaling $\mathbb{G}_m$ action upstairs and a weighted action (by degrees) downstairs. One can restrict this map to the Slodowy slice but since the Slodowy slice is not closed under scaling it no longer makes sense to ask whether it is equivariant.

My question is whether there is some $\mathbb{G}_m$ action one can put on the Slodowy slice such that the restriction of the eigenvalue map is equivariant with respect to this action and how canonical this is (how many choices are involved). I'm also curious about what can be said for a $G$ action (where $G$ is the Lie group of $\mathfrak{g}$, or a subgroup).

Explicitly worked example

For $SL_2$ the question is vacuous. For $SL_3$, if one fixes coordinates for the Slodowy slice (here one makes a choice of subregular $\mathfrak{sl}_2$ triple) $$\left(\begin{array}{ccc} a &1&0\\d&a&f\\g&0&-2a\end{array}\right)$$ then one finds the eigenvalue maps are given by (using $t_2, t_3$ as generators of degree $2, 3$ for $\mathbb{C}[\mathfrak{h}//W]$) $$3a^2 + d = t_2$$ $$-2a^3 + 2ad + fg = t_3$$ So it seems like one should assign $\deg(a) = 1$, $\deg(d) = 2$, and $\deg(fg) = 3$, so there is some choice involved in the last assignment. But this is not a very satisfactory answer even for $SL_3$ for me, since it doesn't appeal to much of the structure in these objects.

Well, it's not completely canonical, but it's close: the $$\mathbb{G}_m$$ action you're looking for is given by the Kazhdan grading (that's the induced grading on the polynomial ring $$\mathbb{C}[\mathfrak{g}^*]$$), defined (I kind of assume not for the first time) in 4.1 of Gan and Ginzburg's paper Quantization of Slodowy slices (Int. Math. Res. Not. 2002, no. 5, 243-255, doi:10.1155/S107379280210609X). The trick is this: let $$e$$ be your fixed nilpotent, by Jacobson-Morozov, we can choose $$f,h$$ that satisfy the relations of $$\mathfrak{sl}_2$$ (you need to do this to define the Slodowy slice anyways). In particular, $$e$$ has $$h$$-weight 2 under the adjoint action. Furthermore, the subspace $$\mathrm{ker}(f)$$ is invariant under $$h$$ (since $$[h,f]=-2f$$). Thus, if we let $$H(t)=\mathrm{exp}(h\log t)$$, then $$H(t)\cdot e=t^2 e$$, so the square of scalar multiplication times $$H(t^{-1})$$ preserves $$e$$, that is $$t^2H(t^{-1})\cdot e=e$$. Since $$\mathrm{ker}(f)$$ is invariant under both these actions, we have that $$e+\mathrm{ker}(f)$$, the Slodowy slice is invariant as well.
Since $$H(t)$$ doesn't change eigenvalues, under the eigenvalue map, this goes to the square of scalar multiplication (I had to because $$h$$ might have odd eigenvalues; you can actually fix this by choosing a slightly different $$h$$, corresponding to a good grading. I believe this is always possible for the subregular, but not for all nilpotents). This action is especially nice because it has positive weights on the Slodowy slice; after all, $$\mathrm{ker}(f)$$ is the lowest weight vectors, which thus have non-positive weight, and so positive weight for $$t^2H(t^{-1})$$ (because of the inverse).
Incidentally, I don't think there's any nice $$G$$ action on the slice. In general, the maximal semi-simple Lie group acting on the slice (commuting with the $$\mathbb{G}_m$$ action above, and preserving the Poisson structure) is the elements of $$G$$ which commute with $$e,h,f$$. Unfortunately, for the subregular, there aren't many of these.