Assign a unique number to every linear form $\varphi : \mathbb R^n \to \mathbb R$, has this number a geometric interpretation?

Question

If $V$ is a finite-dimensional vector space of dimension $n$, denote the space of all alternating $k$-fold multilinear maps (also called alternating $k$-tensors) by $\Lambda^k(V)$. Then $\dim \Lambda^n(V) = 1$.

Now let $V = \mathbb R^n$, and consider $\det \in \Lambda^n(\mathbb R^n)$. Every $n$-fold alternating map is a scalar multiple of the determinant function. Now let $\varphi : V \to \mathbb R$ be a linear form, then $T : V^n \to \mathbb R$ given by $T(v_1, \ldots, v_n) := \sum_{\sigma \in S_n} \mbox{sgn}(\sigma) \varphi(v_{\sigma(i)}) \cdots \varphi(v_{\sigma(n)})$ is an $n$-fold alternating multilinear map, i.e. $T \in \Lambda^n(V)$. Hence $$ T(v_1, \ldots, v_n) = \alpha \cdot \det(v_1, \ldots, v_n) $$ for some $\alpha \in \mathbb R$.

So for each linear form $\varphi : V \to \mathbb R$ we have a unique $\alpha \in \mathbb R$. Now my question has this number any geometric meaning, does it occur somewhere?

Answer 1

As @HenningMakholm points out, your definition makes sense but is trivial.

In all generality, if $V$ is a $K$-vector space, then for any $k\in \mathbb{N}$ you can consider the vector space $\Lambda^k V$ : it is defined as being a vector space equipped with a $k$-linear alternating map $\varphi: V^k\to \Lambda^k V$ such that this map is universal, in the sense that any $k$-linear alternating map $\psi: V^k\to W$ factors through $\varphi$, ie there is a unique linear map $\psi': \Lambda^k V\to W$ such that $\psi = \psi'\circ \varphi$. It can be constructed as a quotient of the tensor product $V^{\otimes k}$.

In particular, $(\Lambda^k V)^*$ is canonically identified to the space of $k$-linear alternating forms on $V$. So when you say that $\Lambda^n V$ is the set of $n$-linear alternating forms on $V$, it's actually a confusion : what you really mean is the dual of this space. Of course it doesn't change the fact that it has dimension $1$.

The space $\Lambda^k V$ is spanned by elements of the form $v_1\wedge\cdots\wedge v_k$ with $v_i\in V$, and such a product is zero if and only if the family $(v_i)$ is linearly dependent. In particular, $v_1\wedge\cdots\wedge v_n\in \Lambda^n V$ is non-zero if and only if $(v_i)$ is a basis of $V$. You can see that the lines in $\Lambda^k V$ are in canonical bijection with the $k$-dimensional subspaces in $V$.

In this context, what happens when you choose a basis ? If you have a basis $(v_i)$ then you have a basis (with one element) $v_1\wedge\cdots\wedge v_n$ of $\Lambda^n V$, so the usual dual basis trick gives you a basis (ie a non-zero element) of $(\Lambda^n V)^*$, ie it gives you a non-zero $n$-linear alternating form on $V$ : we call that the determinant associated to the basis $(e_i)$.

If $\alpha\in \Lambda^p V$ and $\beta\in \Lambda^q V$ you can define $\alpha\wedge \beta\in \Lambda^{p+q}V$ by concatenating the wedge products. You seem to be familiar with the wedge product of alternating forms : it is actually the same as this. In fact, there is a canonical isomorphism $\Lambda^k (V^*) \simeq (\Lambda^k V)^*$, given by the perfect coupling $\Lambda^k (V^*)\times \Lambda^k V\to K$ that sends $(\varphi_1\wedge\cdots\wedge \varphi_k, v_1\wedge\cdots\wedge v_k)$ to $Det(\varphi_i(v_j))_{i,j}$ (this upper-case $Det$ is well-defined because it's the determinant of a matrix, there is no choice of a basis to make here). So you can see elements of $\Lambda^k (V^*)$ as $k$-linear alternating forms on $V$, and through this identification the wedge product that you know is the same as the one I define.

Now what you do is the following :

You take $\varphi\in V^*$, and consider $\varphi\wedge \cdots \wedge \varphi\in \Lambda^n(V^*)$, and then taking the corresponding element of $(\Lambda^n V)^*$ through the isomorphism above, you say that you have a $n$-linear alternating form $T$ on $V$, which is true.

Except that of course $\varphi\wedge \varphi = 0$ by the most elementary property of the exterior product. So your map is zero.

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EDIT : this part was due to an error of interpretation on my part, but now that it's written, it may be interesting to some people, so I'll leave it.

If $f:V\to W$ is any linear map, then for any $k\in \mathbb{N}$ you get an induced map $$\Lambda^k f: \Lambda^k V\to \Lambda^k W$$ (we say that taking exterior powers is functorial).

Let me explain this $\Lambda^k f$ : you just send $v_1\wedge\cdots\wedge v_k$ to $f(v_1)\wedge\cdots\wedge f(v_k)$. And for $k=n$ you get something nice : if $f:V\to V$ is a linear map, then you have $\Lambda^n f: \Lambda^n V\to \Lambda^n V$ ; but since $\Lambda^n V$ is a line, $\Lambda^n f$ must be the multiplication by some constant $\delta$. We call this constant (canonically defined, without any choice of basis) the determinant of $f$.

Answer 2

You can correct your construction as follows: Let $\phi:V^k\rightarrow\mathbb{R}$ be a $k$-linear map. Then, define $$ A(\phi)=\sum_{\sigma\in S_k} \text{sgn}(\sigma)\sigma(\phi). $$ In this case, $\wedge^n(\mathbb{R}^n)$ is one dimensional and you can ask the second part of your question.