# Why is the space of nilpotent spaces of matrices Zariski closed?

I'm reading the paper "Nilpotent subspaces of maximal dimension in semi-simple Lie algebras" by Jan Draisma, Hanspeter Kraft, and Jochen Kuttler. The authors are generalizing a theorem of Gerstenhaber on the maximal dimension of a vector space of nilpotent matrices to a Lie algebra setting.

I'm stuck on a detail at the start of Section 2. They're looking at the nilpotent cone $$\mathcal{N}_\mathfrak{g}$$, consisting of all nilpotent Lie elements. This is well-known to be Zariski closed. (In the motivating case from Gerstenhaber where the Lie algebra is the matrix algebra $$\mathfrak{gl}_n$$, I guess you can see this by setting the traces of powers to 0, since these are polynomial in the entries of the matrix.) I think I understand what they're saying, thus far.

They then go on to say that:

The space $$Z_m := \{V \in \operatorname{Gr}_m(\mathfrak{g}) : V \subseteq \mathcal{N}_\mathfrak{g}\}$$ is clearly Zariski closed in $$\operatorname{Gr}_m$$.

Here, $$\operatorname{Gr}_m$$ is the Grassmannian, consisting of all vector subspaces of $$\mathfrak{g}$$ having dimension $$m$$.

Why is the statement about $$Z_m$$ clear? Is there an elementary explanation, at least in the matrix case? (And what is the 'right' explanation for someone that really knows algebraic geometry?)

• I don't know all of the notation you are using, but $A\in\mathrm{Mat}_m(F)$ is nilpotent if and only if $A^m=0$. This equation can be expressed as $m^2$ polynomial equations in the entries of $A$ which means that it's a subvariety of $\mathrm{Mat}_m(F)$. – Alex Youcis Sep 11 at 13:00
• @Alex, yes, that's what I'm saying around the nipotent cone, possibly confusingly. The paper I'm looking at is written in terms of Lie algebras; the matrix case is a nice concrete and accessible example. – Russ Woodroofe Sep 11 at 14:11
• If you choose an embedding of $G$ into $\mathrm{GL}_n$ then isn't the 'nilpotent cone' the pullback of the nilpotent cone for $\mathrm{GL}_n$? Thus, doesn't it reduce to the $\mathrm{GL}_n$ case (since preimage of closed is closed)? Perhaps I'm not understanding. This should also work for Lie algebras I suspect (using Ado's theorem). – Alex Youcis Sep 11 at 14:55
• Yeah, the nilpotent cone part is fairly clear. I wasn't seeing how to translate Zariski closedness to the Grassmannian. Edited to hopefully clarify what I'm asking, and what Lazzaro answered below. – Russ Woodroofe Sep 11 at 19:57

This is a special case of the following lemma, which doesn't have anything to do with matrices. A reference is Section 6.1.1 of Eisenbud--Harris, 3264 And All That. (I am writing it in the projective setting, but you can projectivise everything in sight in your question to convert it to this form.)

Lemma: Let $$X \subset \mathbf P^n$$ be a Zariski-closed set. Let $$\mathbf G(k,n)$$ be the Grassmannian of $$k$$-dimensional linear subspaces of $$\mathbf P^n$$. Define

$$F_k(X) := \left\{ V \in \mathbf G(k,n) \mid V \subset X \right\}.$$

Then $$F_k(X)$$ is a closed subset of $$\mathbf G(k,n)$$.

You should look at Eisenbud--Harris for an idea of how to prove it. The basic idea is that $$X$$ is cut out by homogeneous polynomials $$F_1,\ldots,F_n$$ of degrees $$d_1,\ldots,d_n$$. Each of these defines a certain section $$\varphi_i$$ of a certain vector bundle $$\Sigma_i$$ on the Grassmannian, and $$F_k(X)$$ is exactly the common zero locus of the sections $$\varphi_i$$.

The set $$F_k(X)$$ is traditionally called the Fano variety or Fano scheme of $$k$$-planes on $$X$$.

• Thank you! I thought that there ought to be a result of that form, but am not familiar enough with the algebraic geometry literature to find it! Off to look in Eisenbud and Harris. – Russ Woodroofe Sep 11 at 14:12
• Similar material seems to be in Example 6.19 of Harris's Algebraic geometry: a first course. It's in the language of varieties there, rather than the scheme language of Eisenbud and Harris. – Russ Woodroofe Sep 11 at 19:57
• do you happen to know a reference for the analogous result for products of Grassmannians? That is, assuming I'm not mistaken, it looks like the same techniques can be used to show that if X is closed in P^m \times P^n, then the subset of all {V \times W in G(k,n) \times G(j,n) with every (v,w) in X} is closed. But I'm surprised not to find it in the literature! – Russ Woodroofe Sep 13 at 7:22
• Dear Russ, I don't know a reference for what you ask. As you say, it seems to follow in more or less the same way as the original claim. – Lazzaro Campeotti Sep 17 at 10:09