# Are the unipotent and nilpotent varieties isomorphic in bad characteristics?

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 here 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 here, 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. A paper here 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 $G$.

A relevant paper of Slodowy is LNM 815. On p29 he calculates the unipotent variety of $PGL_2$ 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.

• This adjoint case did bother me, but I was inclined at first to exclude it. What you suggest does make a lot of sense, since there is also a problem here with Springer's original formulation for groups which are not simply connected (already for good $p$). – Jim Humphreys May 2 '17 at 20:52
• P.S. Though all primes are technically good for type $A_\ell$, the treatment by Slodowy of type $A_1$ for $p=2$ does raise interesting questions. Presumably nilpotent varieties are isomorphic regardless of the isogeny type of $G$ in all Lie types, but for $G$ the unipotent variety can vary more (for example in type $A_\ell$ when $p|(\ell+1)$. This shows up in the revised hypothesis for Springer's theorem. But I'm still puzzled about what is happening for bad $p$. – Jim Humphreys May 3 '17 at 13:06
• I'm not sure what you mean by "nilpotent varieties are isomorphic regardless of the isogeny type of $G$", but it is not true in general that if $G_1$ and $G_2$ are isogenous then ${\mathcal N}({\rm Lie}(G_1))$ is isomorphic to ${\mathcal N}({\rm Lie}(G_2))$. David's example of $\mathfrak{pgl}_2$ vs. $\mathfrak{sl}_2$ shows this. – Paul Levy May 4 '17 at 20:40
• @Paul: I should have specified $p$ very good for this (still hypothetical) comment about isomorphism of nilpotent varieties for all isogeny types of group, since there is clearly a problem otherwise. Anyway, my main concern here is bad primes, where it's hard to compute even low rank examples when $G$ is simply connected. – Jim Humphreys May 4 '17 at 21:12
• I am not very familiar with the classification of orbits in bad characteristic, but do Lusztig's nilpotent and unipotent pieces help you? My memory is sketchy but I thought the nilpotent pieces were maximal connected smooth locally closed subsets, and similarly for unipotent pieces. So, while the number of pieces is the same (if I remember correctly) one could rule out the possibility of an isomorphism from ${\mathcal N}$ to ${\mathcal U}$ by showing that one of the pieces in ${\mathcal N}$ is not isomorphic to the corresponding piece in ${\mathcal U}$. – Paul Levy May 4 '17 at 21:30

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.

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" $\S$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".

• This does get complicated, but I'm asking mainly about simply connected groups $G$ for bad $p$ (as in Springer's theorem for good $p$). Your discussion illustrates the extra obstacles when $G$ isn't simply connected. [By the way, I'd expect similar phenomena for adjoint groups of type $A_\ell$ whenever $p | (\ell+1)$, though of course it's hard to be as concrete as in the case $\ell =1$.] – Jim Humphreys May 4 '17 at 14:34
• Maybe a first question is: if $G$ is simply connected case, are the "scheme of unipotent elements" and the "scheme of nilpotent elements" reduced? – George McNinch May 4 '17 at 17:05
• By the scheme of nilpotent elements, I assume you mean the zero locus of the set of homogeneous invariants of positive degree. But in characteristic 2 I don't think the ring of invariants is known in general, which would suggest that your question is quite non-trivial. Incidentally, I am fairly sure one can show that the nullcone of $\mathfrak{pgl}_n$ is non-reduced at the subregular locus for arbitrary $n$. – Paul Levy May 4 '17 at 21:19
• I checked in Magma and the nullcone of $\mathfrak{sp}_4$ in characteristic 2 is reduced... – Paul Levy May 4 '17 at 21:31
• @Paul: Indeed, "zero locus of the positive degree invariants" is surely what I should have intended. – George McNinch May 5 '17 at 0:50