# Is the image of the map $A \to \bigwedge^{k}A$ a weakly embedded submanifold?


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

Question: Is $\text{Image}(\psi)=\psi(H_{>k})$ a weakly embedded submanifold of $\tilde H_{>k}$?

Note that the image is not closed. The reason is that the rank can fall in the limit to an "illegal value"- a value which is not obtainable by exterior powers of endomorphisms of $V$.

Weakly embedded here means that for every manifold $Q$ and for every smooth map $h:Q \to \tilde H_{>k}$, with $h(Q)\subset \psi(H_{>k})$,the associated map $h:Q\to \psi(H_{>k})$ is also smooth. In other words, it's always valid to restrict the range.

(Some authors call "weakly embedded" submanifolds, initial submanifolds, or diffeological submanifolds).

It is known that it suffices to prove that $h:Q\to \psi(H_{>k})$ is continuous.

Of course, $\psi(H_{>k})$ is an immersed submanifold, but it is not an embedded submanifold, since $\psi$ is not an embedding (I prove this at the end).

Here are some more thoughts:

1. Using group actions and equivariance:

There is a natural action on $H_{>k}$ by $\GL(V) \times \GL(V)$, given by $$(A,B) \cdot C=ACB^{-1}.$$

This action has a finite number of orbits; each orbit is the set of matrices of a certain rank. There is also a $\GL(V) \times \GL(V)$-action on $\End(\bigwedge^k V)$, given by $$(A,B) \cdot D=\bigwedge^k A \circ D \circ \bigwedge^k B^{-1}.$$ $\psi$ is equivariant w.r.t these actions. This might brings us close to the solution, but it's still not enough:

If the action was transitive on the source manifold $H_{>k}$, we were done: The image would then be the orbit of a Lie group action, which is always weakly embedded.

However, in our case the action is not transitive, so this argument cannot be applied directly. Indeed, there are equivariant maps, where the sources has finitely many orbits, whose images are not weakly embedded.

1. Using the "stratified/filtered" structure:

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 $\psi(H_r) \subseteq \tilde H_{\binom {r}{k}}$.

Each restriction $\psi|_{H_r}:H_r \to \tilde H_{\binom {r}{k}}$ is proper, hence $\psi(H_r)$ is an embedded submanifold, and we have $\psi(H_{>k})=\cup_{i=k+1}^d \psi(H_r)$.

The problem is that union of embedded submanifolds does not need to be weakly embedded in general.

Finally, note that in the case $k=d-1$, the answer is positive, since then $\psi:\text{GL}(V) \to \text{GL}(\bigwedge^k V)$ is proper.

Here is a proof showing $\psi$ is not an embedding:

Take $d=4,k=2$:

Let $A=\text{diag}(1,1,1,0) \in \text{End}( V)$. Then $\bigwedge^2 A =\text{diag}(1,1,1,0,0,0) \in \text{End}( \bigwedge^2 V)$. Let $B_r(A) \in H_{>2}$ be an open ball. Assume by contradiction $\psi(B_r(A))$ is open in $\tilde H_{>2}$; there exist an open set $U \subseteq \text{End}( \bigwedge^2 V)$ such that $\psi(B_r(A))=\tilde H_{>2} \cap U$. Define $A_n=\text{diag}(n,\frac{1}{n},\frac{1}{n},\frac{1}{n})$.

Then $\psi(A_n)=\bigwedge^2 A_n =\text{diag}(1,1,1,\frac{1}{n^2},\frac{1}{n^2},\frac{1}{n^2}) \in \tilde H_{>2} \cap U=\psi(B_r(A))$ for sufficiently large $n$.

Since $\psi$ is locally injective ($\psi(A)=\psi(B) \Rightarrow A=\pm B$), this implies $A_n$ or $-A_n$ lies in $B_r(A)$ for every sufficiently large $n$. This is a contradiction.

*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 $d=6,k=3$: Take $A=\text{diag}(1,1,1,1,1,0) \in \text{End}( V)$. Then $\bigwedge^3 A =\text{diag}(1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0) \in \text{End}( \bigwedge^3 V)$. Take $A_n=\text{diag}(n,\frac{1}{\sqrt n},\frac{1}{\sqrt n},\frac{1}{\sqrt n},\frac{1}{\sqrt n},\frac{1}{\sqrt n})$. Now the same argument as above applies.

• Here is an earlier published source of the fact that an orbit of a Lie group action is always initial: 5.14 (based on 3.20 which shows that leaves of integrable distributions of non-constant rank are always initial) in mat.univie.ac.at/~michor/kmsbookh.pdf – Peter Michor Jun 17 '18 at 9:13