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


Define $H_r=\{ A \in \text{End}(V) \mid \operatorname{rank}(A) = r \}$. $H_r$ is a submanifold of $\text{End}(V)$.

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

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

Denote $\tilde H_s=\{ B \in \text{End}(\bigwedge^kV) \mid \operatorname{rank}(B) = s \}$, and note that for $A \in \text{End}(V)$,
$$\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k} ,$$ that is $\psi(H_r) \subseteq \tilde H_{\binom {r}{k}}$

Is $\psi(H_r)$ a closed subset of $\tilde H_{\binom {r}{k}}$? (The topology on $\tilde H_{\binom {r}{k}}$ is the subspace topology induced by $\text{End}(\bigwedge^kV)$).

Comment: In the special case where $r=d$ the answer is positive; we have $H_d=\text{GL}(V)$ and $\tilde H_{\binom {d}{k}}=\text{GL}(\bigwedge^{k}V)$, and $\psi(\text{GL}(V))$ is closed in $\text{GL}(\bigwedge^{k}V)$.

The answer is positive, since $\psi:H_r \to \tilde H_{\binom {r}{k}}$ is proper, and every proper map is closed.

Here is a proof $\psi$ is proper:

Let $K \subseteq \tilde H_{\binom {r}{k}}$ be compact, and let $A_n \in \psi^{-1}(K)$. We shall prove $A_n$ has a convergent subsequence in $\psi^{-1}(K)$. It suffices to prove $A_n$ converges in $\text{End}(V)$; indeed, if $A_n \to A$, then $\bigwedge^k A_n \to \bigwedge^k A$, and the limit $\bigwedge^k A$ must be in $K$. In particular, $\binom {r}{k}=\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k}$, so $\operatorname{rank}(A)=r$, that is $A \in H_r$.

By using SVD, we can assume $A_n=\text{diag}(\sigma_1^n,\dots,\sigma_r^n,0,\dots,0)$ is diagonal, where the first $r$ diagonal elements are non-zero, and the last $d-r$ elements are zero. (Since the orthogonal group is compact, the isometric components surely converge after passing to a subsequence).

$\bigwedge^k A_n$ is diagonal, and its first $\binom {r}{k}$ elements are of the form $\Pi_{s=1}^k \sigma_{i_s}^n$, where all the $1 \le i_s \le r$ are distinct. So, every such product converges when $n \to \infty$ to a positive number. Indeed, $\psi(A_n)=\bigwedge^k A_n \in K \subseteq \tilde H_{\binom {r}{k}}$, so it converges (after passing to a subsequence) to an element $D \in K$. Since $\text{rank}(D)=\binom {r}{k}$, it follows that the products $\Pi_{s=1}^k \sigma_{i_s}^n$ must converge to positive numbers. (If even one of them converges to zero instead, the rank of the limit $D$ would be too low, which is a contradiction).

Now, let $1\le i \neq j \le r$. Since $r \ge k+1$, we can choose some $1 \le i_1,\dots,i_{k-1} \le r$ all different from $i,j$. Since both products $$(\Pi_{s=1}^{k-1} \sigma_{i_s}^n)\sigma_{i}^n,(\Pi_{s=1}^{k-1} \sigma_{i_s}^n)\sigma_{j}^n$$ converge to positive numbers, so does their ratio $C_{ij}^n=\frac{\sigma_i^n}{\sigma_j^n}$.

We know that $$\Pi_{s=1}^k \sigma_{s}^n=\Pi_{s=1}^k \sigma_{1}^n\frac{\sigma_s^n}{\sigma_1^n}=\Pi_{s=1}^k \sigma_{1}^nC_{s1}^n=(\sigma_{1}^n)^k \Pi_{s=1}^k C_{s1}^n$$ converges to a positive number. Since all the $C_{s1}^n$ converge to positive numbers, we deduce $\sigma_1^n$ converges. W.L.O.G the same holds for every $\sigma_i^n$, so $A_n$ indeed converges. (and we know the limit must have the right rank).