$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ 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: $$\psi:\text{GL}^+(V) \to \text{GL}(\bigwedge^{k}V,\bigwedge^{k}V) \, \,, \, \, \psi(A)=\bigwedge^{k}A,$$

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

Is $\psi$ an immersion?

Note that $\psi$ is injective and smooth. My motivation is connected to this question.


Here are some observations: First, using the multiplicative nature of $\psi$, we can reduce everything to the identity. (Since $d\psi_A(B)=\psi(A) \circ d\psi_{\id}(A^{-1}B)$).

Let's consider for a moment the case $k=2$: In that case $$ \big(d\psi_{\id}(B)\big)(e_i \wedge e_j)=e_i \wedge Be_j+Be_i \wedge e_j. \tag{1}$$

So, we have reduced the question into showing that if equation $(1)$ holds for every two vectors $e_i,e_j$, then $B=0$. (Of course, it suffices to take $e_i,e_j$ to be part of a given basis for $V$).

I have also proved that $\text{trace} B=0$ (In general $d\psi_A(B)=0 \Rightarrow \langle \Cof A,B \rangle=0$ and $\Cof A=\id$).

Since the answer is positive for $k=d-1$ (see explanation below), we see that the first non-obvious case is $k=2,d=4$.

A proof the answer is positive for $k=d-1$:

In that case $\bigwedge^{d-1}A$ is essentially the cofactor matrix of $A$, $\Cof A$. Then we have the following formula for its derivative:

$$d(\Cof)_A(B) = (A^{T})^{-1}\big(\langle \Cof A , B\rangle \cdot \id - B^T \circ \Cof A \big),$$

so $d(\Cof)_A(B)=0$ implies $\langle \Cof A , B\rangle \cdot \id = B^T \circ \Cof A$. Taking traces we get $$d \langle \Cof A , B\rangle = \langle \Cof A , B\rangle \Rightarrow \langle \Cof A , B\rangle =0,$$ which in turn implies $$ B^T \circ \Cof A =0 \Rightarrow B=0.$$

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    $\begingroup$ If a morphism $\psi$ of Lie groups is injective then the tangent map at the origin is also injective ( $d\psi(X)=0$ implies $\phi(\exp X)=e$.). $\endgroup$ – Orest Bucicovschi May 10 '18 at 13:37
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    $\begingroup$ your map is only locally injective if both $n$ and $k$ are even ($A$, $-A$ map to the same operator), but the conclusion still holds for the tangent map at the origin. $\endgroup$ – Orest Bucicovschi May 10 '18 at 14:33
  • $\begingroup$ Hmmm... you are absolutely right. I need to be more careful here. $\endgroup$ – Asaf Shachar May 10 '18 at 18:27
  • $\begingroup$ It is not a problem though, since locally it is injective, and that guarantees that you have an inverse locally, enough for smoothness. For instance, $e^{i\phi(t)}$ smooth implies $\phi$ smooth, since we have a local inverse for $e^{ix}$. $\endgroup$ – Orest Bucicovschi May 10 '18 at 18:35

Note that $\psi(A \circ B) = \psi(A) \circ \psi(B);$ so $\psi$ is a homomorphism of Lie groups.

Proposition. Any injective Lie group morphism $\psi : G \to H$ is an immersion.

Proof. Since $\exp_H \circ D\psi_\mathrm{id} = \psi \circ \exp_G$ and exponential maps are diffeomorphisms near the identity, injectivity of $\psi$ implies injectivity of $D\psi_\mathrm{id}.$ By homogeneity we conclude that $D\psi$ is injective everywhere.

  • $\begingroup$ the map in question may only be locally injective but your argument takes care of that too. I guess the OP would very much like an explicit inverse map ( find the elements of a matrix given the $k\times k$ minors, up to a constant), interesting what those formulas would be $\endgroup$ – Orest Bucicovschi May 10 '18 at 14:38

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