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Consider an $m$-by-$n$ matrix $A$ with entries in a field $k$; we can see $A$ as a point in the affine space $\mathbb{A}^{m n}$. The rank of $A$ will be $\leq r$ (where $r<\min(m,n)$) if and only if every $(r+1)$-by-$(r+1)$ minor of A is $0$. That tells us that the set of all such matrices $A$ forms a variety $V$. Moreover, by higher-dimensional Bézout, we obtain a bound on the sum of the degrees of the components $V_i$ of $V$: it is at most $$(r+1)^{k_{r,m,n}},$$ where $k_{r,m,n} = \binom{n}{r+1} \binom{m}{r+1}$ is the number of $(r+1)$-by-$(r+1)$ minors.

Unfortunately, that's quite a large upper bound. Is there another way to characterize $V$, resulting in a better upper bound for $\sum_i \deg(V_i)$?

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The affine variety you described are called determinantal rings, and just about everything is known about them: dimension, singular loci, etc. The degree is also called the Hilbert-Samuel multiplicity, and it was computed as determinant of the matrix $a_{ij}= \binom{m+n-i-j}{m-i}$ for $i,j=1,...,r$. See this paper.

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One can compute the degree explicitly, by using a natural resolution of singularities of the associated projective variety. Indeed, to give a matrix of a rank $r$ one needs to fix an $r$-dimensional subspace of $k^n$, i.e., a point of $Gr(r,n)$, and a surjective map from $k^m$ to this subspace, i.e., a full-rank vector in the fiber of the vector bundle $k^m \otimes U \cong U^{\oplus m}$, where $U$ is the tautological bundle of the Grassmannian. If we relax the surjectivity (full-rank) condition and projectivize, we obtain a map $$ P_{Gr(r,n)}(U^{\oplus m}) \to P^{mn-1}, $$ which is birational onto the projectivization of the variety of matrices of rank at most $r$. Now, if $H$ is the relative hyperplane class for the projective bundle, the required degree is equal to $H^N$, where $$ N = r(n-r) + mr - 1 = r(n + m - r) - 1 $$ is the dimension. The integer $H^N$ can be explicitly computed using the Grothendieck description of the cohomology of a projective bundle and Schubert calculus on $Gr(r,n)$.

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