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

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

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

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    $\begingroup$ 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. $\endgroup$
    – Zach Teitler
    Commented Jun 6, 2018 at 19:26
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    $\begingroup$ 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? $\endgroup$
    – Asaf Shachar
    Commented Jun 6, 2018 at 19:48
  • $\begingroup$ 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. $\endgroup$
    – Zach Teitler
    Commented Jun 7, 2018 at 4:46
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    $\begingroup$ 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 $\psi(A)=0$. $\endgroup$
    – Daniel Litt
    Commented Jun 11, 2018 at 12:41
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    $\begingroup$ 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 $f(x,y)=(x^2,xy)$. $\endgroup$
    – François Brunault
    Commented Jun 12, 2018 at 9:15

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The answer is negative: In general $\psi({\rm End}(V))$ is not closed. Here is a proof when $d\ge4$ is even and $k=d-1$. Notice that $\Lambda^{d-1}V$ can be identified with $V$, so that $\psi(A)$ is just the cofactor matrix $\widehat{A}$.

${\rm End(V)}$ is the disjoint union of ${\rm GL}(V)$ and $\Delta$, the set defined by $\det A=0$. Because $d$ is even, $\psi$ is a homeomorphism from ${\rm GL}(V)$ onto itself, with inverse given by the formula $$A=(\det\psi(A))^\frac1{d-1}\psi(A)^{-T}.$$ If instead $A$ is singular, then either $\psi(A)=0_V$ if $A$ has rank $\le d-2$, or $\psi(A)$ has rank $1$ if the rank of $A$ is $d-1$. In conclusion, $\psi({\rm End}(V))$ contains ${\rm GL}(V)$, which is dense in ${\rm End}(V)$, but does not contain any element of rank between $2$ and $d-1$. In particular, it is not closed.

The same analysis works also when $d\ge3$ is odd (and still $k=d-1$). One obtains that $\psi({\rm End}(V))$ contains ${\rm GL}_+(V)$, but does not contain any element of rank between $2$ and $d-1$. Hence it is not closed.

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    $\begingroup$ I think your argument can be extended to show that the answer is negative for any $1 < k \leq d-1$. For instance, consider the case $d=4$ and $k=2$. Choose a decomposition $V=W \oplus L$ with $\dim W=3$ and $\dim L=1$. Let $\phi$ be any endomorphism of $\Lambda^2 W$. By your result $\phi$ can be approximated by linear maps $\Lambda^2 g_n$ with $g_n : W \to W$ invertible. Define $f_n \in \operatorname{End} V$ by $f_n=g_n$ on $W$, and $f_n=0$ on $L$. Then $\Lambda^2 f_n$ tends to the block matrix $(\begin{smallmatrix} \phi & 0 \\ 0 & 0 \end{smallmatrix})$. (...) $\endgroup$
    – François Brunault
    Commented Jun 11, 2018 at 18:37
  • $\begingroup$ (...) If $\phi$ has rank 2 then this block matrix cannot be of the form $\Lambda^2 f$ with $f \in \operatorname{End} V$ since the possible ranks of $\Lambda^2 f$ are $0,1,3,6$. $\endgroup$
    – François Brunault
    Commented Jun 11, 2018 at 18:39

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