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

284 Views Asked by At

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?)

1

There are 1 best solutions below

4
On BEST ANSWER

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 $X$.