Posted by: Steven Sam | June 29, 2011

Do the rank k matrices form an affine variety?

This question on math.stackexchange caught my eye and I thought I would reproduce the nice argument given by David Speyer, possibly with more background explained.

The question is as follows: is the variety of n \times n matrices of rank exactly k an affine variety?

Those of rank at most k is an affine variety since they are defined by the vanishing of all minors of size k. This also shows that the rank k matrices form a quasi-affine variety. In fact, those of rank exactly n form an affine variety since this is the solution set of the equation \det(x_{ij}) t = 1 where t is a newly introduced indeterminant. However, this argument doesn’t work for smaller values of k, and we will see they cannot be written as the solution set of a system of equations, regardless of how many new indeterminants are introduced.

This follows from a more general theorem.

Theorem. Let X be an affine variety (or more generally, a Noetherian affine scheme) and let Y be a closed subvariety with an irreducible component of codimension >1. Then the complement of Y is not affine.

To apply this, we first note that the n \times n matrices of rank at most k have codimension (n-k)^2 in the space of all matrices and set X to the matrices of rank at most k, and Y to be the matrices of rank at most k-1. When k<n, we see that Y has codimension at least 2 in X.

To prove the theorem, we use the following cohomological characterization of affine varieties due to Serre:

Theorem (Serre). Let X be a variety (or more generally, a Noetherian scheme). Then the following are equivalent:

  1. X is affine
  2. For all quasi-coherent sheaves F and all i>0, one has H^i(X; F) = 0.
  3. For all coherent sheaves of ideals I, one has H^1(X; I) = 0.

We won’t need the third condition though. The next ingredient is cohomology with supports / local cohomology. Given any topological space X with an open subset U, let Y be the complement of U. For a sheaf F of Abelian groups on X, we define the group of sections of F with support in Y to be those sections s of F such that for each point P of Y, s is not identically 0 on any open neighborhood of P. Denote it by H^0_Y(X;F). This is left exact, so we get derived functors. In particular, there is a long exact sequence of the form

\cdots \to H^i_Y(X; F) \to H^i(X;F) \to H^i(U; F) \to \cdots

Now suppose that X is an affine scheme of the form Spec(R), Y is the subvariety cut out by an ideal I, and F is a quasi-coherent sheaf corresponding to an R-module M. There is an alternative description of H^i_Y(X;F) in terms of local cohomology. I wrote a little bit about local cohomology before, but let me define it again. We define

H^0_I(R;M) = \{ m \in M \mid I^n m = 0 \text{ for some } n > 0 \}.

This is left-exact and its derived functors agree with what we just defined:

H^i_I(R;M) = H^i_Y(X;F).

Furthermore, these groups only depend on the radical of I, and not I itself. In the case when R is local and I is the maximal ideal of R, Grothendieck’s nonvanishing theorem says that

H^i_I(R;M) \ne 0

where i = \dim(R/ann(M)). In particular, since local cohomology behaves well with respect to localization, one can say that

H^i_I(R;M) \ne 0

for an ideal I and arbitrary ring and i = \dim(R_P/ann(M_P)) where the subscript denotes localization at a prime P which is minimal over I.

Now let’s put it all together. Take F to be the structure sheaf \mathcal{O}_X of X. Suppose that Y has an irreducible component of codimension > 1. Since X is affine, we have H^i(X; \mathcal{O}_X) = 0 for all i>0 by Serre’s theorem. Then the long exact sequence above implies that

H^{r-1}(U; \mathcal{O}_X) = H^r_Y(X; \mathcal{O}_X) \ne 0

where the last part is Grothendieck’s nonvanishing theorem. But r-1>0, so Serre’s theorem implies that U is not affine. QED

One possible reference is: “Twenty-four hours of local cohomology” by Iyengar, Leuschke, Leykin, Miller, Miller, Singh, Walther.

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Responses

  1. Moral: the stratification of kxn matrices by rank is too coarse to be the right stratification.

    The right one (!) is actually this. Start with the hypersurface defined by the product of the ixi determinants in the NW corner, for i=1..k-1, then all the kxk determinants using columns j+1..j+k, for j=0..n-k, then the ixi determinants in the SE corner, for i=1..k-1.

    From that hypersurface (whose complement is affine) one can construct a stratification by decomposing it into components, intersecting them, decomposing the intersections, etc. The rank k open set will then be a union of some of these finitely many strata.

    Some hints that this is a good choice: all of these intersections are reduced, and all of these subvarieties have simultaneous Grobner degenerations to SR(vertex-decomposable balls).

  2. “Right” for what?

    Also, is this related to matrix Schubert varieties? I think you mentioned this in a talk at a MIT colloquium last year but I can’t remember.

  3. “Right” in the following sense. Let’s think about the complement, matrices of rank less than k. A priori to get at that one needs to specify many equations. My assertion is that once one specifies my hypersurface, it’ll pick out that variety (and finitely many others, including all the matrix Schubert varieties) automatically.

    Of course you could do this with any subscheme: just write down the product of its defining equations. But that would be of very high degree. I’m saying you can do it with a degree kn hypersurface (in kn variables); that’s very rare and special.

  4. [...] Steven Sam: Do the rank k matrices form an affine variety? [...]


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