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I defined more clearly the question, and made the focus on the closedness sharper.
Asaf Shachar
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Existence of relations characterizing the $k$-minors of a matrix

Let $V$ a real vector space of dimension $d$. Let $1<k < d-1$. Consider the map induced by the exterior algebra functor:

$$ \psi:\text{Hom}(V,V) \to \text{Hom}(\bigwedge^kV,\bigwedge^kV) \, \, \, \, , \, \, \,\psi(A)=\bigwedge^k A$$

Is the image of $\psi$ closed in the standard topology on the $\text{Hom}$-space?

I know that $\psi(\text{GL}(V)) $ is closed in $ \text{GL}(\bigwedge^{k}V)$ but this does not seem to help. (This follows from the fact $\psi|_{\text{GL}(V)}$ is a morphism of algebraic groups. There is also a simple direct argument here).


I think this question is equivalent to the following question:

Do there exist polynomial relations which characterize the $k$-minors of a square $d \times d$ matrix over $\mathbb{R}$? i.e. is $\psi(\text{Hom}(V,V))$ the zero set of some polynomials defined on $\text{Hom}(\bigwedge^kV,\bigwedge^kV)$?

(By "characterizing" I mean necessary and sufficient conditions for a given sequence of numbers to be the minors of some matrix. See an explicit statement below).

It seems that finding explicit relations is an open problem, e.g. see this paper. However, I am not asking about an explicit construction, only on mere existence.


Here is a more concrete phrasing:

Let $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)} \in \mathbb{R}$, where $(i_1,\dots,i_k),(j_1,\dots,j_k)$ are increasing multi-indices of order $k$ ( $1 \le i_1 <i_2 < \dots<i_k\le d$).

Are there necessary and sufficient conditions for the existence of a $d \times d$ matrix $A$, such that $ b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ is the $k$-minor of $A$, corresponding to rows $(i_1,\dots,i_k)$, and to columns $(j_1,\dots,j_k)$?

In "conditions" I mean here a set of continuous functions $\{ f_{\lambda} \}$ in the $\binom{d}{k}^2$ variables $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ such that the $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ are minors if and only if $ f_{\lambda} (b_{(i_1,\dots,i_k),(j_1,\dots,j_k)})=0$ for every $ f_{\lambda} $.

Asaf Shachar
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