# Expressing invertible maps $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ as $\bigwedge^{d-1}A$ for some $A$

Let $V$ be a real $d$-dimensional vector space, let $\bigwedge^{d-1} V$ be its exterior power. Consider the following claim:

Proposition: If $d$ is even, then every invertible linear map $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ equals $\bigwedge^{d-1}A$ for some $A \in \text{GL}(V)$. If $d$ is odd, then every orientation-preserving* invertible map $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ equals $\bigwedge^kA$ for some $A \in \text{GL}(V)$.

I found a proof for the above proposition, but it is based on endowing $V$ with an inner product, which I don't like very much. Since there is no mention of products in the claim, it's natural to expect a metric-free proof.

Is there such a proof?

Edit:

Here is an argument for showing that when $d$ is odd, it is impossible to express orientation-reversing maps $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ as "$(d-1)$-wedge" of a map $V \to V$.

Let $A:V \to V$. Since $$\det (\bigwedge^k A)=(\det A)^{\binom{d-1}{k-1}},$$ we get for $k=d-1$ that $$\det (\bigwedge^{d-1} A)=(\det A)^{\binom{d-1}{d-2}}=(\det A)^{d-1},$$

so if $d$ is odd, we see that $\det (\bigwedge^{d-1} A)$ is always positive, whether or not $A$ was orientation-preserving to begin with.

*Note there is no need for a choice of orientation on $\bigwedge^{d-1} V$ to define which maps $\bigwedge^{d-1} V \to \bigwedge^{d-1} V$ are orientation preserving. (If you like you can put the same orientation on "both sides", it does not matter which).

## 1 Answer

Consider the perfect pairing $\left< \cdot, \cdot \right> \colon V \times \bigwedge^{d-1}(V) \rightarrow \bigwedge^d(V)$ given by the wedge product $\left<v, \omega \right> = v \wedge \omega$. The adjugate of a linear map $T \colon V \rightarrow V$ is characterized by the property that it is the adjoint map to $\bigwedge^{d-1}(T)$ with respect to $\left< \cdot, \cdot \right>$. That is, we have

$$\left< \operatorname{adj}(T)v, \omega \right> = \left<v, \bigwedge\nolimits^{d-1}(T)\omega \right>$$

for all $v \in V$ and $\omega \in \bigwedge^{d-1}(V)$. Using this definition, one can prove directly that $$\operatorname{adj}(T) \circ T = T \circ \operatorname{adj}(T) = \det(T) I$$ and $$\operatorname{adj}(\operatorname{adj}(T)) = \det(T)^{d-2} T.$$

I'll assume that $d$ is even and show that given any invertible map $S \colon \bigwedge^{d-1}(V) \rightarrow \bigwedge^{d-1}(V)$ we can find an invertible map $T \colon V \rightarrow V$ such that $\bigwedge^{d-1}(T) = S$. Since the pairing is perfect, there exists a (unique) map $R \colon V \rightarrow V$ which is adjoint to $S$ so that

$$\left< Rv, \omega \right> = \left< v, S\omega \right>$$

for all $v \in V$ and $\omega \in \bigwedge^{d-1}(V)$. Note that $R$ must also be invertible. Define $T = \det(R)^{\frac{2-d}{d-1}}\operatorname{adj}(R)$. Then we have

$$\operatorname{adj}(T) = \det(R)^{2 - d} \operatorname{adj}(\operatorname{adj}(R)) = \det(R)^{2-d} \det(R)^{d - 2} R = R$$

so

$$\left< v, S\omega \right> = \left< Rv, \omega \right> = \left< \operatorname{adj}(T)v, \omega \right> = \left< v, \bigwedge\nolimits^{d-1}(T) \omega \right>$$

for all $v \in V$ and $\omega \in \bigwedge^{d-1}(V)$ which shows that $S = \bigwedge^{d-1}(T)$.

In general, one can show that $\det(R) = \det(S)$ (where $R,S$ are adjoint with respect to $\left< \cdot, \cdot \right>$, just like one has with an inner product). When $d$ is odd, $d - 1$ is even so the previous argument works only if $\det(R) > 0$ (because we need to take an even square root) which will happen if and only if $\det(S) > 0$.

• Thanks! Your answers are amazing, as always. The $adj(adj(T))=\det(T)^{d-2}T$ looks like magic. After all $adj(T)$ encodes in some way the action of $T$ on $d-1$-dimensional parallelepipeds. It is truly a miracle that when you does this operation twice (i.e "encodes the encoding"), you recover your original map (up to the necessary scaling of course). I do not see any "prime reason" for why such a thing needs to hold ,though, since I really don't have a good intuition or interpretation of $adj(T)$ as a map in its own right (I really tends to view it as a smart-and very useful-encoding). – Asaf Shachar Aug 14 '17 at 15:15
• I understand the formal derivation of the miracle of course, but am still amazed by it. – Asaf Shachar Aug 14 '17 at 15:15
• @AsafShachar: Yeah, this is surprising. At least for invertible $T$, we can think of $\operatorname{adj}(T)$ just as $\det(T) T^{-1}$ and then all the relevant properties become obvious but this obscures the relation with the $d - 1$-dimensional parallelepipeds. Since the map $T \mapsto \Lambda^{d - 1}(T)$ is a non-trivial polynomial map, I thought that your property must come from some "nice identity" and once I remembered the relation between $\Lambda^{d-1}(T)$ and $\operatorname{adj}(T)$, it became clear that something involving $\operatorname{adj}(T)$ must work. – levap Aug 14 '17 at 17:04
• @levap: Very nice answer, thanks! – Hanno Aug 14 '17 at 17:09
• I've added some details about the odd dimensional case. – levap Aug 14 '17 at 17:37