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


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 $V$).



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


