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

$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ 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).

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.

• Nice question. Btw, maybe you should write $GL(\wedge^k V)$ instead. The image is always a closed subgroup. It's true for instance anytime you have a morphism of algebraic groups ( whether real or complex) – Orest Bucicovschi May 11 '18 at 15:44
• Thanks for your comment, I will change the notation to $GL(\wedge^k V)$. Do you have a reference for your claim? (or elementary proof? Unfortunately I am not very familiar with algebraic geometry). If what you are saying is true, then image is automatically an embedded submanifold, by the closed subgroup theorem. – Asaf Shachar May 11 '18 at 17:57
• There is a book by Onishchik & Vinberg Lie Groups and Algebraic Groups that would take care of this result. Let me try to sketch a proof in the answers. – Orest Bucicovschi May 11 '18 at 18:26
• Thanks, I really should learn some more algebraic geometry.... (I guess you are talking about fairly elementary stuff, right?). I think I also found a "direct" (linear-algebraic) proof, you can see my answer below. – Asaf Shachar May 11 '18 at 18:39
• @orangeskid BTW, I am planning to publish a small note regarding the "regularity via minors" question. Your comments here about the image being closed really helped me. If you want any acknowledgements, please tell me. My email address shows in my profile page, if you want to contact me personally (since I guess you might want to keep your real name private). Thanks anyway for interesting discussion. – Asaf Shachar May 18 '18 at 10:08

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.

• That's a very clever proof! – Orest Bucicovschi May 11 '18 at 19:25

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

1. 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 $$A= \frac{Adj(Adj(A))}{\det A ^{n-2}}$$

• Thanks. It will take me some time to digest your answer. Regarding the "search for the inverse formula"-I agree it's very interesting. (Actually, I already knew the inverse formula in the case $k=n-1$). Are you sure that the inverse will be polynomial in general (for $k \neq d-1$)? (e.g there won't be any roots involved? Does this follow from some facts on algebraic/regular morphisms and their inverses?). Thanks again for all your thoughts on this matter, this discussion is really interesting for me. – Asaf Shachar May 11 '18 at 20:34
• @Asaf Shachar: Yes, there could be some roots.So the formula would be modulo some power of the determinant. In fact, if we consider $PGL(V)$ to $PGL(\wedge^k V)$ now this map is a true imbedding. Here we should have some polynomial inverse. We could consider over $\mathbb{C}$, then we have a regular morphism, bijective, so it must have a regular inverse, as you said. Will get some more thinking on this. – Orest Bucicovschi May 11 '18 at 21:51