Is the image of the map $A \to \bigwedge^{k}A $ an embedded submanifold of $\text{GL}(\bigwedge^{k}V)$?

Question

$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ Let $V$ be a real oriented $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed.

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

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

Is $\text{Image}(\psi)$ an embedded submanifold of $\text{GL}(\bigwedge^{k}V)$?

Edit:

By the closed subgroup theorem, it suffices to show the image is closed. (I think this might be proven using SVD, and the fact $k \le d-1$, but I am not sure. When $k=d$, $\det A_n$ can converge even when $A_n$ does not, of course).

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I know that $\psi$ is a smooth, locally injective homomorphism of Lie groups. In particular, it is an immersion.

I am particularly interested in the case where $k,d$ are not both even; in that case $\psi$ is injective.

If we can prove $\psi$ is an embedding in this case, then we are done. (However, I am not sure this is a necessary condition).

Comment: My motivation is connected to this question.

Answer 1

The answer is positive. The image is closed, hence embedded (by the closed subgroup theorem).

Let us prove it is closed: Let $A_n \in \text{GL}^+(\mathbb{R}^d)$ and suppose that $\bigwedge^k A_n$ converges to some element $D$ in $\text{GL}(\bigwedge^k\mathbb{R}^d)$. Since the map $A \to \bigwedge^kA$ is continuous, it suffices to show $A_n$ converges to some element in $\text{GL}^+(\mathbb{R}^d)$.

By using SVD, we can assume $A_n=\Sigma_n=\text{diag}(\sigma_1^n,\dots,\sigma_d^n)$ is diagonal. (Since the orthogonal group is compact, the isometric components surely converge after passing to a subsequence).

We know that $\bigwedge A_n$ is diagonal with eigenvalues $\Pi_{r=1}^k \sigma_{i_r}^n$, where all the $i_r$ are different. So, we know every such product converges when $n \to \infty$. Let $1\le i \neq j \le d$. Since $k \le d-1$, we can choose some $1 \le i_1,\dots,i_{k-1} \le d$ all different from $i,j$. Since both products $$(\Pi_{r=1}^{k-1} \sigma_{i_r}^n)\sigma_{i}^n,(\Pi_{r=1}^{k-1} \sigma_{i_r}^n)\sigma_{j}^n$$ converge to positive numbers, so does their ratio $C_{ij}^n=\frac{\sigma_i^n}{\sigma_j^n}$.

Now we know that $$\Pi_{r=1}^k \sigma_{r}^n=\Pi_{r=1}^k \sigma_{1}^n\frac{\sigma_r^n}{\sigma_1^n}=\Pi_{r=1}^k \sigma_{1}^nC_{r1}^n=(\sigma_{1}^n)^k \Pi_{r=1}^k C_{r1}^n$$ converges to a positive number. Since all the $C_{r1}^n$ converge, we deduce $\sigma_1^n$ converges. W.L.O.G the same holds for every $\sigma_i^n$, so $A_n$ indeed converges to an invertible matrix.

Answer 2

We'll sketch a proof that avoids the theory of algebraic grous.

Let $\phi\colon G=GL(V)\to H=GL(W)$ a morphism of Lie groups such that $\phi(a I_V)=a^k \cdot I_W$ for some $k$ real. Then $\phi(G)$ is closed.

Let's first show: the image of $SL(V)$ is closed. Indeed, the image of $SL(V)$ has as Lie subalgebra $h_1=d\phi(sl(V))$. Since we have $[sl(V), sl(V)]=sl(V)$, we conclude $[h_1, h_1]=h_1$. Now use

Lemma ( Malcev) If a connected Lie subgroup $H_1\subset H$ has Lie algebra satisfying $[h_1, h_1]=h_1$ then $H_1$ is closed ( see the book by Onishchik and Vinberg).

Now, enough to show that $\phi(GL_{+}(V)$ is closed. Let $g_n = a_n \cdot s_n$, $a_n$ scalars, $s_n\in SL(V)$ such that $\phi(g_n) \to h$. So $a_n^k \cdot \phi(s_n)\to h$. Now, $\phi(s_n)\in SL(V)$. So if $k=0$ we are done. Otherwise, taking $\det$, we get $\det h= \lim a_n^{k d}= (\lim a_n)^{kd}=a^{kd}$. So $\frac{h}{a^k} = \lim \phi(s_n) = \phi(s)$, and so $h=\phi( a s)$.

In fact, in all our cases the morphism $\phi$ is algebraic ( the components are polynomial), so the image will be given by some polynomial equations. This follows from general facts about algebraic groups over $\mathbb{R}$ and $\mathbb{C}$. However, it would be interesting to find out:

1.a system of explicit polynomial equations for the image

2. an inverse polynomial map (up to determinants).

In the case $k=n-1$, from $\wedge^{n-1}A$ we can get $\tilde{A}=Adj(A)$, by changing signs of components alternately. Now, $\det Adj(A)= \det(A)^{n-1}$. So we get $$A= \frac{Adj(Adj(A))}{\det A ^{n-2}}$$