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I read a book related to determinantal variety. In the book, I studied fact that $M_{k-1}$is singular locus of $M_{k}$, (here $M_k$ means $m*n$ matrices with rank at most $k$) my question is : I know that $M_1$ is segre variety. then what singularities of $M_1$? if $m=3$, $n=4$ $M_1$ may be expressed by {$3 *4$ matrices}/($2*2$ minors) and singular locus be {$3*4$ matrices}/($3*3$ minors), but I cannot imagine geometric shape of this one and how can blowup at singularities.

Any advice must be helpful for me, thanks!

victoria
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  • $M_1$ is singular along $M_0$, which is a single point corresponding to the zero matrix (matrix of rank $0$). In projective space, the projective variety corresponding to $M_1$ is nonsingular.

    You are making a mistake: the singular locus of $M_1$ is not defined by $3 \times 3$ minors. It is defined by the $1 \times 1$ minors.

    In general, every $M_k$ is defined by the vanishing of the $(k+1) \times (k+1)$ minors. The first derivatives of $(k+1) \times (k+1)$ minors are $k \times k$ minors. The singular locus of $M_k$ is $M_{k-1}$, defined by the vanishing of the $k \times k$ minors.

     – Zach Teitler Nov 09 '17 at 04:58

1 Answers1

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Disclaimer: I'm not an expert on the topic.

  1. $M_{k-1} \subset M_{k}$, not the other way around!
  2. Your notation $\{3\times 4 \text{ matrices}\}/(2\times 2 \text{ minors})$ is misleading, because we don't consider quotient spaces. Better: $$ M_k = M_k(m,n) = \{ a \in Mat(m,n) \mid \textrm{rank}(a) \leq k \} = \{ a \in Mat(m,n) \mid \text{all } (k+1)\times(k+1) \text{ minors of } a \text{ vanish}\} $$
  3. $M_3(3,4) = Mat(3,4) = \mathbb{A}_{3\times 4}$ is the 12-dimensional affine space. The blowup $\phi : Bl(\mathbb{A}_{12}) \to \mathbb{A}_{12}$ along the origin $M_0 = \{ 0 \}$ is well-known (see Hartshorne p.28f, or whatever you're working with), so you can look that up. Now restricting to $M_1$ gives you the blowup of $M_1$ along $M_0$: $$\phi : \phi^{-1}(M_1) \to M_1$$
Long
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