**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.