Are the unipotent and nilpotent varieties isomorphic in bad characteristics?

Question

In characteristic 0 or good prime characteristic, there are standard ways to relate the unipotent variety $\mathcal{U}$ of a simple algebraic group $G$ and the nilpotent variety $\mathcal{N}$ of its Lie algebra $\mathfrak{g}$. Recall that $p$ is good just when it fails to divide any coefficient of the highest root (the root system being irreducible), bad otherwise. The only possible bad primes are $2,3,5$. In characteristic 0, algebraic versions of the exponential and logarithm maps provide Ad $G$-equivariant isomorphisms in both directions, whereas in good characteristic $p>0$, the less direct arguments of Springer in The unipotent variety of a semi-simple group yield similar results. [The isogeny type of $G$ adds some complications here.]

There is scattered literature on the varieties $\mathcal{U}$ and $\mathcal{N}$ when $p$ is bad, often treated case-by-case, e.g., four papers by Lusztig posted on arXiv starting with Unipotent elements in small characteristic, along with papers by his student T. Xue. A serious challenge when $p$ is bad is to find a uniform explanation for the failure of the numbers of unipotent classes and nilpotent orbits to agree in some cases: the details were worked out by Holt–Spaltenstein and others. In spite of this breakdown in $G$-equivariance, a natural question can be raised:

Are the two varieties $\mathcal{U}$ and $\mathcal{N}$ isomorphic in all characteristics, for example when $G$ is simply connected?

The answer does not seem to be written down explicitly (?), but for example one can see indirectly that both varieties have the same dimension in all characteristics: the number of roots. Existence of regular nilpotent elements in the Lie algebra of a simple algebraic group in bad characteristics by S. Keny, a former student of Steinberg, showed case-by-case that regular nilpotent elements always exist and form a dense orbit in $\mathcal{N}$. By definition, the isotropy group in $G$ of such an element has dimension equal to the rank of $G$.

Answer 1

A relevant paper of Slodowy is LNM 815, Simple singularities and simple algebraic groups. On p29 he calculates the unipotent variety of $\operatorname{PGL}_2$ in characteristic $2$ as being given by the three equations \begin{align*}X^2+YZ&=0\\ Y(X+1)&=S^2\\ Z(X+1)&=T^2\end{align*} in the five variables.

But (again in characteristic $2$) the nilpotent variety in $\mathfrak{pgl_2}$ is, I believe, just the $2$-dimensional vector space spanned by the root spaces, on which the toral subalgebra acts by derivations. So they are not isomorphic.

Yet another question is when they are isomorphic as schemes. Slodowy points out that the equations above define a non-reduced scheme. But of course the nilpotent variety of $\mathfrak{pgl}_2$ is reduced.

By contrast, I think the unipotent and nilpotent varieties of $\mathrm{SL}_2$ are isomorphic (and are both reduced). It would be reasonable to conjecture that the answer is 'yes' when $G$ is simply-connected, and 'no' when the covering map from the simply-connected cover is not smooth.

Answer 2

Update: Paul Levy points out in the comments that a reasonable way of defining the nullcone in $\operatorname{Lie}(G) = \mathfrak{g}$ is as the zero set of the homogeneous invariants of positive degree — i.e. of $(k[\mathfrak{g}]_+)^G = (S^+\mathfrak{g}^*)^G$.

But with this definition, my original comment isn't valid. Indeed, if $G = \operatorname{PGL}_2$ then the co-adjoint representation of $G$ on $\mathfrak{g}^*$ has a fixed vector, which is a linear invariant in $(k[\mathfrak{g}]_+)^G$ whose zero locus is — as in Dave Stewart's original answer — the span of the root vectors.

I suppose all my original objection (below) really amounted to was that if $X$ is the affine scheme defined by the $\mathbf{Z}$-algebra $R=\mathbf{Z}[A,B,C]/\langle A^2 + 4BC\rangle$, then for any field $k$ of char. not 2, $X_k$ identifies with the nilpotent variety of $\mathfrak{pgl}_{2,k}$, but if $k$ has char. 2, $X_k$ is not reduced.

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Original post: I'm going to write this as an "answer", though I think it mostly amounts to a comment on Dave Stewart's answer.

It is not completely clear to me that the statement "the nilpotent variety of $\mathfrak{pgl}_2$ is reduced" is correct when $p=2$. Well, I suppose that more precisely I mean: it isn't clear that the scheme of nilpotent elements should be viewed as reduced.

Taking a basis $x,y,h$ of $\mathfrak{pgl}_2$ (say, in its 3-dimensional representation), one finds that $ah + bx + cy$ is nilpotent just in case $a^2+4bc=0$. Of course, in char. 2 this amounts to $a^2=0$, which suggests that the scheme of nilpotent elements shouldn't be viewed as a reduced subscheme. (If you don't want to write down the matrices, see e.g. Jantzen "Nilpotent Orbits in Representation Theory" §2.7.)

I do doubt (?) that this nilpotent scheme is isomorphic to the scheme of unipotent elements of $\operatorname{PGL}_2$, but (assuming that doubt is correct — I didn't think too carefully about it) I think the reason is more complicated than the statement "one is reduced and the other isn't".