# Number of points of the nilpotent cone over a finite field and its cohomology

Let $k=\Bbb F_q$ be a finite field, $G$ be a connected reductive group over $k$, $\mathcal{N}$ be the nilpotent cone of $G$ which consists of nilpotent elements in the lie algebra of $G$.

Motivation & example: If $G=GL_n$, then $\#\mathcal{N}(k)=q^{n^2-n}$ (here is a direct proof) i.e there are exactly $q^{\text{dim}\mathcal{N}}$ nilpotent matrices over $k$.

The direct proof uses no geometry while the result is simple, so I am interested in an explanation by Grothendieck trace formula. However, $\mathcal{N}$ is not smooth ( $\mathcal{N} \cong \{xy=z^2\}$ if $G={GL}_2$ ) in general, so interesection cohomology might be better than etale cohomology (The nilpotent cone is rationally smooth when using intersection cohomology at least in characterestic zero).

Q1. What is $\#\mathcal{N}(k)$ for general $G$?

Q2 What is the cohomology (etale/intersection cohomology) of $\mathcal{N}_{\bar k}$ and the trace of geometric Frobenius action on it?

Q3 Do we have Grothendieck trace formula for intersection cohomology?

My vague idea for Q1 is to consider Springer resolution and reduce to counting size of the flag variety and Springer fiber.

• You only have trace formula for the compactly supported etale cohomology in general, not the intersection cohomology. However in the rationally smooth case the etale cohomology with compact supports agrees with the intersection cohomology with compact supports. (I don't think there is a difference between rationally smooth in the usual and intersection cohomology senses.) May 26, 2018 at 13:26
• @WillSawin Thank you, I am not familiar with the positive characterestic case. I am mostly interested in Q2 and compute the case $n=2$. May 26, 2018 at 16:06
• If it's a rational homology manifold, then its cohomology with compact supports is dual to its cohomology, which is one-dimensional in degree zero and zero-dimensional in all higher degrees. Hence the compactly supported cohomology lives in degree $2 ( \dim G - \operatorname{rank} G)$. If this is also true in characteristic $p$, this implies that the number of points is $q^{ \dim G - \operatorname{rank} G}$. May 26, 2018 at 18:19
• One can compute the ordinary cohomology in every characteristic directly from the fact that it is an affine cone. In characteristic zero, it follows from the fact that affine cones are contractible. In characteristic $p$, one can reason like this (don't know if it's the simplest argument): Sufficient to show the cohomology relative to the origin is zero. This is equivalent to the cohomology of the blow-up at the origin relative to the exceptional divisor. But the blow-up at the origin is an $\mathbb A^1$-bundle on the exceptional divisor, so they have the same cohomology. May 26, 2018 at 19:05
• If I remember correctly precisely these questions are addressed in "Partial resolutions of nilpotent varieties" by Borho-MacPherson. In any case it would be worth having a look at this paper. May 27, 2018 at 14:03

When formulating a question about the cohomology of a variety it's important to determine which cohomology group you want to ask about. One rule of thumb is:

If you want to understand the number of $\mathbb F_q$-rational points, or you already understand the number of $\mathbb F_q$-rational points and want to extend that understanding to cohomology, you almost always want to study the etale cohomology with compact supports of the constant sheaf.

This is for the simple reason that the Grothendieck-Lefschetz fixed point formula relates the etale cohomology with compact supports to a sum over points, where the summand is $1$ in the case of the constant sheaf.

With regards to your question 3, in the etale setting we very rarely think of intersection cohomology as a different cohomology theory, but rather as a choice of coefficient sheaf (the intersection cohomology complex). In particular, there exists a Grothendieck-Lefschetz fixed point formula for the intersection cohomology with compact supports, but it relates to the sum over points of the trace function of the intersection cohomology complex. (This is often a very interesting function to sum, more interesting than the constant function 1, e.g. in the case of a Schubert cell in a Grassmanian or affine Grassmanian.)

Anyways, with regards to your question 2 we have the general result that if $X$ is any closed, scaling-invariant subset of $\mathbb A^n$ containing the origin, then $H^*(X, \mathbb Q_\ell) = \mathbb Q_\ell$ in degree zero.

Indeed, let $j: X - \{0 \} \to X$ be the open immersion, then we have an exact sequence $0 \to j_! \mathbb Q_\ell \to \mathbb Q_\ell \to \delta_0\to0$, where $\delta_0$ is the skyscraper sheaf, hence a long exact sequence $H^*(X, j_! \mathbb Q_\ell) \to H^*(X,\mathbb Q_\ell) \to H^*( 0, \mathbb Q_\ell)$, so it is sufficient to prove that the second map is an isomorphism by proving that $H^*(X, j_! \mathbb Q_\ell)=0$.

Then if we let $Y$ be the blowup of $X$ at $0$, $u: X-\{0\} \to Y$ the open immersion, $\pi: Y \to X$ the projection map, then one can check that $\pi_* u_! \mathbb Q_\ell= j_! \mathbb Q_\ell$, so by the Leray spectral sequence it suffices to show that $H^* (Y, j_! \mathbb Q_\ell)=0$. Applying Leray again, it suffices to show that the derived pushforward from $Y$ to the projective variety on which $X$ is an affine cone of $u_! \mathbb Q_\ell$ vanishes. This is a locally trivial fibration by $\mathbb A^1$, on which $j_! \mathbb Q_\ell$ is locally the extension by zero of the constant sheaf on $\mathbb A^1-\{0\}$, so by smooth base chance we can reduce to $H^* (\mathbb A^1, j_! \mathbb Q_\ell)=0$ for $j : \mathbb A^1 - \{0\} \to \mathbb A^1$ the open immersion, where it is straightforward.

It follows that if $X$ is a rational homology manifold of dimension $d$, in the sense that its dualizing complex is a constant sheaf in degree $-2d$, then $H^*_c(X,\mathbb Q_\ell) = \mathbb Q_\ell(-d)$ in degree $2d$, and so the trace of Frobenius on $H^*_c(X,\mathbb Q_\ell)$ is $q^d$, and thus $X$ has $q^d$ rational points.

As you note, this is only possible if $X$ is the affine cone on a projective variety with $\frac{q^d-1}{q-1}$ points. So the condition of being a rational homology manifold is quite restrictive, already at the level of counting rational points!

However, the vanishing for ordinary cohomology is no restriction on $X$ because there is no Grothendieck-Lefschetz formula for ordinary cohomology.

My understanding (which I didn't check) is that the foundational papers on the Springer resolution and its cohomology do prove that the nilpotent cone is a rational homology manifold in characteristic $p$ in general. If this is the case, then indeed the number of points must be $q^{ \dim G - \dim T}$ with $T$ the maximal torus of $G$.

• Now I understand, thank you. Besides, I find a proof using Fitting lemma math.stackexchange.com/questions/1298453/… May 27, 2018 at 9:15
• Note that the computation at the end was done earlier by Steinberg using less sophisticated methods. May 27, 2018 at 16:55

Concerning your Q1, it may be of interest to fill in some of the background due to Steinberg (which in turn had a lot of influence on Springer's work).

In his 1966 ICM talk, Steinberg formulated quite a few "problems" (not all of which have the solutions he expected). When considering a finite field of definition for a connected and simply-connected semisimple algebraic group $G$ (the relevant case here), he pointed out that certain numbers were pure powers of $q$ (the number of elements in the field $k$). For example, there are $q^r$ semisimple conjugacy classes, $r$ being the rank of $G$.

He also proved in Theorem 15.1 of his soon-published AMS Memoir Endomorphisms of Linear Algebraic Groups (No. 80, 1968) that the total number of unipotent elements in $G$ is $q^{\dim \mathcal{U}}$, the square of the number of unipotents in a maximal unipotent subgroup of the finite group attached to $G$ and an endomorphism $\sigma$; here $\mathcal{U}$ is the unipotent variety. (Springer had shown for good primes that $\mathcal{U} \cong \mathcal{N}$, the nilpotent variety; in any case, it's easy to see that these varieties have the same dimension $\dim G -r)$.)

Steinberg's ICM Problem (22) stated: Find a natural action of $W$ on affine $(\dim G -r)$-space so that the quotient variety is isomorphic to $\mathcal{N}$.

As usual $W$ is the Weyl group of $G$. But there turn out to be counterexamples, found first in a special case by Popov in 1977 and then more generally by Panyushev in 1985 here. Even so, both Steinberg and Springer were actively looking at the unipotent and nilpotent varieties during that period, though as Will points out the underlying questions required more sophisticated geometry.

This paper of Springer gives a quick proof that indeed $$\#\mathcal{N}(\mathbb{F}_q)=q^{\dim(G)-\dim(T)}$$.