Is the map $A \to \bigwedge^{k}A $ from matrices above rank $k$ proper?

Question

$\newcommand{\End}{\operatorname{End}}$

Let $V$ be a $d$-dimensional real vector space. ($d \ge 3$). Fix an odd $2 \le k \le d-1$. Define $H_{>k}=\{ A \in \End(V) \mid \operatorname{rank}(A) > k \}$. $H_{>k}$ is an open submanifold of $\End(V)$.

We also define, for a given number $s$, the open submanifold

$$\tilde H_{>s}=\{ B \in \End(\bigwedge^k V) \mid \operatorname{rank}(B) > s \} \subseteq \End(\bigwedge^k V).$$

Consider the map $$ \psi:H_{>k} \to \tilde H_{>k} \, \,, \, \, \psi(A)=\bigwedge^{k}A, %\psi:H_r \to \text{End}(\bigwedge^{k}V) \, \,, \, \, \psi(A)=\bigwedge^{k}A, $$

$\psi$ is a smooth injective immersion. (The injectivity uses the fact $k$ is odd, since otherwise $\psi(A)=\psi(-A)$).

Question: Is $\psi$ is a proper map?

(In the case $k=d-1$, the answer is positive, since then we ask if $\psi:\text{GL}(V) \to \text{GL}(\bigwedge^k V) $ is proper. See more details below).

Explanation on the codomain of $\psi$:

Let $H_{i}=\{ A \in \End(V) \mid \operatorname{rank}(A) = i \}$, and $\tilde H_i$ its analog on the exterior algebra $\bigwedge^kV$.

For $A \in \text{End}(V)$,
$$\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k} ,$$ that is $\psi(H_r) \subseteq \tilde H_{\binom {r}{k}}$.

In particular, this implies $\psi(H_{>k}) \subseteq \tilde H_{>k}$.

I know that each restriction $\psi|_{H_r}:H_r \to H_{\binom {r}{k}}$ is proper, but I am having trouble with handling the case where the domain is the union $H_{>k}=\cup_{i=k+1}^d H_i$.

In particular, I am not sure about the following related sub-question:

Let $A_n \in H_{>k}$, and suppose that $\psi(A_n)=\bigwedge^k A_n$ converges to some $D \in \tilde H_{>k}$. Is it true that $D \in \tilde H_{\binom {i}{k}}$ for some $i>k$? (We can assume all the $A_n$'s have the same rank, hence all the $\psi(A_n)$ also have the same rank. In general, the rank of the limit $D$ can fall below the shared rank of the $\psi(A_n)$. The question is if it must fall to another "legal" value, which is one of the values that are obtainable from endomorphisms of $V$).

Answer 1

The answer is negative:

Let $d=4,k=2$: Let $A_n=\text{diag}(n,\frac{1}{n},\frac{1}{n},\frac{1}{n}) \in \text{End}( V)$. (We choose a basis and let $A_n$ be diagonal w.r.t this basis).

Then $\bigwedge^2 A_n =\text{diag}(1,1,1,\frac{1}{n^2},\frac{1}{n^2},\frac{1}{n^2}) \in \text{End}( \bigwedge^2 V)$ converges to $D=\text{diag}(1,1,1,0,0,0) \in \tilde H_3$.

Note that in this case we consider $\psi$ as a map $$ \psi:H_{>2}= H_3 \cup H_4 \to \tilde H_{>2}=\tilde H_3 \cup \tilde H_4 \cup \tilde H_5 \cup \tilde H_6.$$

Define $K=\{\bigwedge^2 A_n\}\cup D \subseteq \tilde H_{>2}$. $K$ is compact, but $A_n \in \psi^{-1}(K)$ does not contain a convergent subsequence in $H_{>2}$, so $\psi^{-1}(K)$ is not compact.

This shows the map $\psi$, as defined with the given domain and co-domain, is not proper. I required $k$ to be odd, but this example can be easily adapted to the odd case.

---

Indeed, here is an example for the case $k$ is odd:

Set $d=5,k=3$:

$A_n=\text{diag}(n,\frac{1}{\sqrt n},\frac{1}{ \sqrt n},\frac{1}{ \sqrt n},\frac{1}{ \sqrt n}) \in \text{End}( V)$.

Then $\bigwedge^3 A_n =\text{diag}(1,1,1,1,1,1,\frac{1}{(\sqrt n)^3},\frac{1}{(\sqrt n)^3},\frac{1}{(\sqrt n)^3},\frac{1}{(\sqrt n)^3}) \in \text{End}( \bigwedge^3 V)$ converges to $D=\text{diag}(1,1,1,1,1,1,0,0,0,0) \in \tilde H_6 \subseteq \tilde H_{>3}$.

Note that in this case we consider $\psi$ as a map $$ \psi:H_{>3}\to \tilde H_{>3}.$$

Define $K=\{\bigwedge^3 A_n\}\cup D \subseteq \tilde H_{>3}$. $K$ is compact, but $A_n \in \psi^{-1}(K)$ does not contain a convergent subsequence in $H_{>3}$, so $\psi^{-1}(K)$ is not compact.