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Let $M$ be a $p\times q$ matrix (say, $p\leq q$) with formal coefficients $m_{i,j}$. It has $pq$ entries, and ${q\choose p}$ minors of maximal size $p\times p$, which gives us a map $$\varphi_{p,q}: \Bbb Z^{pq}\to \Bbb Z^{q\choose p}$$

I'm interested in the surjectivity of this map. It is easy to see that $\varphi_{3,4}$, for instance, is surjective, meaning that we can adjust the coefficients of a $3\times 4$ matrix to give each of the $3\times 3$ minors a prescribed value.

In most cases, ${q\choose p}$ is much bigger than $pq$, so it feels unlikely that $\varphi$ will be surjective. But what if we keep only a relatively small amount of minors?

Question 1: For what values of $p,q,r\in\Bbb N$ is there a map $f:\Bbb Z^{q\choose p}\to \Bbb Z^r$ defined by keeping only some $r$ coordinates, such that $f\circ \varphi_{p,q}$ is surjective?

Question 2 (less general but the one I'm really interested in) : Take $p=3, q=6$ and forget about the three minors corresponding to columns $(1, 4, 5), (2, 4, 6), (3, 5, 6)$. Is the resulting map $\Bbb Z^{18}\to \Bbb Z^{17}$ surjective?

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  • $\begingroup$ In your very special case does it help that the 20 minors satisfy a quadratic equation (rather a la Klein quadric)? $\endgroup$ – ancientmathematician Jun 16 at 13:42
  • $\begingroup$ @ancientmathematician I think I see what you mean - stacking two copies of M one on top of the other. It does prove that $\varphi_{3,6}$ is not surjective, and it's certainly a relation to have in mind. I can't see yet how it would imply the non-surjectivity of the 17 minors map. Thanks for the idea. $\endgroup$ – Arnaud Mortier Jun 16 at 13:56
  • $\begingroup$ My geometric insight is poor, but I can't see how one could project even all of this $(10,-10)$ quadric onto the whole of a 17-space. $\endgroup$ – ancientmathematician Jun 16 at 14:47
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    $\begingroup$ @ancientmathematician Actually there are 35 independent such quadratic equations, and yes this does help a lot - see answer below. Thanks again. $\endgroup$ – Arnaud Mortier Jun 19 at 12:56
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Thanks to a comment by @ancientmathematician, I was able to focus my research and to answer Question 2: the $\Bbb Z^{18}\to \Bbb Z^{17}$ map from the question is not surjective. However, the way to the answer yields new interesting questions. First, here is the proof.

Theorem: an ordered collection of $q\choose p$ integers is the collection of (lexicographically ordered) maximal minors of some $p\times q$ matrix if and only if these numbers satisfy the so-called Plücker equations.

Context: see these lecture notes by Alexander Yong.

Proof: see Schubert Calculus by Kleiman and Lakso.

Plücker equations for $(p,q)=(3,6)$ can be displayed by typing Grassmannian(2,5) into Macaulay2 (the $2$ and $5$ come from projective reasons). Here is one of these equations: $$p_{2,3,4} p_{1,3,6} -p_{1,3,4}p_{2,3,6} +p_{1,2,3}p_{3,4,6}$$ This one, and five others, involve only determinants that I wanted to keep (none of $p_{1,4,5}, p_{2,4,6}, p_{3,5,6}$ is involved). Hence the result: the $\Bbb Z^{18}\to \Bbb Z^{17}$ map from the question is not surjective.


New problem as promised: what if we forget about enough minors that each of the Plücker equations involves at least one of them (basically making the argument above fail)? We can't conclude right away, but what tools would one use to conclude then? I have been thinking of using Groebner bases, but it doesn't seem straightforward.

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