# Is the image of the map $A \to \bigwedge^{k}A$ from matrices above rank $k$ closed?

$\newcommand{\Cof}{\operatorname{cof}}$ $\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\End}{\operatorname{End}}$ $\newcommand{\GL}{\operatorname{GL}}$

Let $V$ be a real $d$-dimensional vector space ($d \ge 4$). Let $2 \le k \le d-2$ be odd. Define $H_{>k}=\{ A \in \text{End}(V) \mid \operatorname{rank}(A) > k \}$. $H_{>k}$ is an open submanifold of $\text{End}(V)$.

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

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

where $\bigwedge^{k} V$ is the $k$-th exterior power of $V$.

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 but not an embedding. (The injectivity uses the fact $k$ is odd, since otherwise $\psi(A)=\psi(-A)$).

Question: Is $\text{Image}(\psi)=\psi(H_{>k})$ closed in $\tilde H_{>k}$?

Here are some thoughts:

Let's try prove the image is closed. Let $A_n \in H_{>k}$, and suppose $\psi(A_n)=\bigwedge^k A_n$ converges to some $D \in \tilde H_{>k}$. We can assume all the $A_n$'s have the same rank $r$. If $\text{rank}(D)=\binom {r}{k}$ then we are done, since the restriction of $\psi$ to the space of matrices of rank $r$ is proper, as I explain below.

The problem is that the rank of the limit $D$ can fall below the shared rank of the $\psi(A_n)$. We can ask whether or not $D \in \tilde H_{\binom {i}{k}}$ for some $i>k$? That is, if the rank of $D$ "falls", must it fall to another "legal" value, which is one of the values that are obtainable from endomorphisms of $V$?

Explanation regarding the stratified/filtered structure:

Note that $H_{>k}=\cup_{r=k+1}^dH_r$, where $H_{r}=\{ A \in \text{End}(V) \mid \operatorname{rank}(A) = r \}$.

Denote $\tilde H_s=\{ B \in \text{End}(\bigwedge^kV) \mid \operatorname{rank}(B) = s \}$. If $\operatorname{rank}(A) = r$, then $\operatorname{rank}(\bigwedge^kA) = \binom {r}{k}$, which means $\psi(H_r) \subseteq \tilde H_{\binom {r}{k}}$, so we have

$$\psi(H_{>k})= \cup_{r=k+1}^d \psi(H_r) \subseteq \cup_{r=k+1}^d \tilde H_{\binom {r}{k}}.$$

I know that each restriction $\psi|_{H_r}:H_r \to \tilde H_{\binom {r}{k}}$ is proper, hence $\psi(H_r)$ is an embedded closed submanifold of $\tilde H_{\binom {r}{k}}$.

• I suggest you try the case of $d=3, k=2$ to see that $\psi(H_k)$ is not closed. – Mohan Jun 4 '18 at 15:42
• Thanks, you are right. I meant to suggest $\psi(H_k)$ is closed in a certain subspace of $End(\bigwedge^k V)$. I have edited the question to reflect this. – Asaf Shachar Jun 5 '18 at 5:55

The answer is negative. The image of $\psi$ is not closed:
Let $d=5,k=3$:
Set $A_n=\text{diag}(n,\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}}) \in H_{>3}$. Then $\psi(A_n)=\bigwedge^3 A_n=\text{diag}(1,1,1,1,1,1,(\frac{1}{\sqrt{n}})^3,(\frac{1}{\sqrt{n}})^3,(\frac{1}{\sqrt{n}})^3,(\frac{1}{\sqrt{n}})^3)$ converges to $D=\text{diag}(1,1,1,1,1,1,0,0,0,0) \in \tilde H_{>3}$.
However, $D \notin \psi(H_{>3})$ since it does not have the right rank:
$\text{rank}(D)=6$, and $\psi(H_{>3})= \psi(H_4)\cup \psi(H_5) \subseteq \tilde H_{\binom {4}{3}} \cup H_{\binom {5}{3}}=\tilde H_4 \cup \tilde H_{10}$.
Similar examples exist when $k$ is even.