Linear transformations which keep wedge product

Question

Let $V$ be a $\mathbb K$ vector space.

A linear transformation $A \in \mathscr L(V)$ is called ''keeping the wedge product of two vectors $a$ and $b$ (i.e. $a\wedge b$)'' iff $(Aa) \wedge (Ab) = a\wedge b$. If it fixs the wedge product for every pair of vectors. It's called keeping the wedge product.

I want to figure out all $A$ keeping the wedge product. Intuitively, $A\in O(n)$ may keep. However, I cannot find all these linear transformations, since lacking enough training on wedge product. Anyone who can help with this?

Answer 1

$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-1mu#1}}} $Let $n$ be the dimension of our vector space $V$.

This has nothing to do with $\mathrm O(V)$. The condition $$ (Aa)\wedge(Ab) = a\wedge b \tag{$*$} $$ means precisely that $$ A(\mathrm{span}\{a,b\}) = \mathrm{span}\{a,b\}\quad\text{and}\quad \det A|_{\mathrm{span}\{a,b\}} = 1 \tag{$**$} $$ assuming $a, b$ are linearly independent and where the determinant of the restriction is well-defined precisely because $A$ preserves the span. If $n = 2$ this means $A \in \mathrm{SL}(V)$. If the above is true for all $a, b$ then it folllows that $n = 2$ or $A$ is $\pm$ the identity.

To understand this we need three theorems:

1. Define the support $[X]$ of a multivector $X \in \Ext V$ by $$ [X] = \{v \in V \;:\; v\wedge X = 0\}. $$ Then for decomposable multivectors $X, Y$, we have $[X] = [Y]$ if and only if there is a scalar $\alpha$ such that $X = \alpha Y$ or $\alpha X = Y$. (If we further assume $X \ne 0 \ne Y$ then we can just say $X = \alpha Y$.)
2. If $v_1,\dotsc,v_k \in V$ then $$ v_1\wedge\dotsb\wedge v_k = 0 \iff \text{ those vectors are linearly dependent} $$ and $$ [v_1\wedge\dotsb\wedge v_k] = \mathrm{span}\{v_1,\dotsc,v_k\}\text{ when those vectors are linearly independent.} $$
3. The determinant of $A : V \to V$ is the unique scalar such that $$ (Av_1)\wedge\dotsb\wedge(Av_n) = (\det A)v_1\wedge\dotsb\wedge v_n $$ for all $v_1,\dotsc,v_n \in V$. Combined with Theorem 2, we see it is also the unique such scalar for any fixed, linearly independent $v_1,\dotsc,v_n$.

All of these are possible to prove using only basic properties of the wedge product, and you should try to do so yourself.

Let us now use these theorems to prove the equivalence of ($*$) and ($**$) for fixed, linearly independent $a, b$. (Theorem 1 is strictly speaking unneeded here, but I felt like it should be mentioned.)

Assume ($*$). Then $Aa, Ab$ are linearly independent by Theorem 2 and also $$ A(\mathrm{span}\{a,b\}) = \mathrm{span}\{Aa,Ab\} = [(Aa)\wedge(Ab)] = [a\wedge b] = \mathrm{span}\{a,b\}. $$ Thus we can consider $A|_{\mathrm{span}\{a,b\}}$ as a map $\mathrm{span}\{a,b\} \to \mathrm{span}\{a,b\}$ and $\det A|_{\mathrm{span}\{a,b\}}$ is well defined. We can naturally identify $$ \Ext\mathrm{span}\{a,b\} \subseteq \Ext V $$ and it follows by Theorem 3 that $\det A|_{\mathrm{span}\{a,b\}} = 1$.

Now assume ($**$). It immediately follows from Theorem 3 and the identification $$ \Ext \mathrm{span}\{a,b\} \subseteq \Ext V $$ that $(Aa)\wedge(Ab) = a\wedge b$.

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Let me elaborate on why ($**$) holding for all $a, b$ (and assuming $n \geq 2$) implies that $n = 2$ or $A$ is $\pm$identity.

(If $n = 0,1$ then ($*$) says $0=0$ and $A$ is arbitrary.)

Since $n \geq 2$, choose linearly independent $a, b$. If $n = 2$ then ($**$) is equivalent to $A \in \mathrm{SL}(V)$ by definition, so assume $n > 2$. Choose $c$ linearly independent from $a, b$. Now ($**$) holds for the pair $a,b$ and the pair $a,c$ so $$ Aa \in \mathrm{span}\{a,b\}\cap\mathrm{span}\{a,c\} = \mathrm{span}\{a\}. $$ Thus $Aa = x_aa$ for all $a$ and some scalars $x_a$. This is only possible if $Aa = xa$ for all $a$ and some fixed $x$, and the determinant condition of ($**$) means $$ x^2 = \det A|_{\mathrm{span}\{a,b\}} = 1. $$ Thus $Aa = \pm a$ for all $a$.

Answer 2

Wedge products represent areas of integration locally without any reference to a euclidean metric.

Areas transform by the determinant of the Jacobian. In the case of a metric manifold, the determinant of the Jacobian can be expressed by the square root of the determinant of the metrics by $g=J^t \ J$