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Jun 12, 2018 at 9:15 comment added François Brunault A simple example where the image of a homogenous polynomial map is not closed: consider $f : \mathbf{C}^2 \to \mathbf{C}^2$ given by $f(x,y)=(x^2,xy)$.
Jun 12, 2018 at 1:54 comment added Zach Teitler My first comment above has the same error—we can't simply operate in projective space, for the reason @DanielLitt said. Sorry!... but perhaps elimination theory could still provide some helpful information.
Jun 11, 2018 at 12:43 comment added Asaf Shachar @DanielLitt Thanks, you are right. That is why I removed it...
Jun 11, 2018 at 12:41 comment added Daniel Litt Just to be clear, the error with the proposed proof above is that the projectivization of $\psi$ is not defined everywhere, since there are non-zero matrices $A$ with $\psi(A)=0$.
Jun 11, 2018 at 12:21 history edited Asaf Shachar CC BY-SA 4.0
I have deleted many irrelevant parts and focused the question considerable, eliminating all the chatter.
Jun 11, 2018 at 12:17 vote accept Asaf Shachar
Jun 11, 2018 at 11:52 answer added Denis Serre timeline score: 13
Jun 11, 2018 at 11:00 comment added Suvrit For the case of principal minors (not all minors, even though some notation in the comments suggests a focus on principal minors?), you may find the following paper interesting: arxiv.org/pdf/math/0604374.pdf
Jun 11, 2018 at 7:36 history edited Asaf Shachar CC BY-SA 4.0
I have added a suggested solution.
Jun 8, 2018 at 5:03 comment added Asaf Shachar Thanks. So, if $A_n$ are real and $\bigwedge^k A_n \to D \in \mathbb{R}^{\binom{k}{d}^2}$, we know that $D=\bigwedge^k B$ for some complex matrix, right? So, the only question is the following: suppose $D=\bigwedge^k B \in \mathbb{R}^{\binom{k}{d}^2}$ where $B$ is complex. Is $D=\bigwedge^k A$ for some real matrix $A$? Of course, $B$ itself does not have to be real, since if $k$ is even, $\bigwedge^k A=\bigwedge^k (iA)$.
Jun 7, 2018 at 4:46 comment added Zach Teitler Yes, Zariski-closed does imply closed in the usual topology (from the Euclidean metric). BUT on the other hand I did not notice before that you are interested in $\mathbb{R}$. I think that everything I said is okay for an algebraically closed field (like $\mathbb{C}$) but there are problems over $\mathbb{R}$. I should be a little more careful. The most we get from general theory over $\mathbb{R}$ is that the image of a polynomial map is semi-algebraic, i.e., defined by some equations and inequalities (this is the Tarski-Seidenberg theorem); I'm not sure if it has to be closed.
Jun 6, 2018 at 19:48 comment added Asaf Shachar Thanks. So, in particular does this mean that the image is closed in the standard topology (considering $\text{Hom}(\bigwedge^kV,\bigwedge^kV) \simeq \mathbb{R}^{\binom{k}{d}^2}$) ? In other words does Zariski-closed imply closed in the usual sense?
Jun 6, 2018 at 19:26 comment added Zach Teitler The $k$-minors of $A$ are given by polynomials in the entries of $A$. By the method of elimination (e.g., using Gröbner bases) it is possible to express the defining ideal of the Zariski closure of the $k$-minors map. In this case, the map is homogeneous (all the polynomials are homogeneous of degree $k$) so we can operate in projective space. This means the map is proper, and we can drop the Zariski closure step. By elimination theory, the image of the $k$-minors map is a Zariski-closed algebraic variety. The generators of the defining ideal give the conditions you ask for.
Jun 6, 2018 at 18:49 history edited Asaf Shachar CC BY-SA 4.0
I defined more clearly the question, and made the focus on the closedness sharper.
Jun 5, 2018 at 15:09 history edited Asaf Shachar CC BY-SA 4.0
I have stated the question more clearly
Jun 5, 2018 at 11:10 history asked Asaf Shachar CC BY-SA 4.0