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

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?

• What you are trying to describe is called a top form in manifold theory. This is very important in integration on manifolds. In your definition, you should be applying the alternating function to a $k$-linear function, not a $1$-linear function. – Michael Burr Mar 23 '16 at 10:43
• To me it looks like $\alpha=0$ whenever $n\ge 2$ -- all of the $\varphi(v_{\sigma(i)})$ factors are the same in each term (just in different orders) and can therefore be pulled out in front of the sum. It might be more interesting if the codomain of $\varphi$ was not commutative... – Henning Makholm Mar 23 '16 at 10:43
• By the way, you don't mean "unique" in the usual sense that $\varphi$ can be reconstructed from $\alpha$, do you? – Henning Makholm Mar 23 '16 at 10:45
• Oh yes, I see this construction is senseless. Totally overlooked it, thank you. I will going to delete this question in a few minutes. – StefanH Mar 23 '16 at 10:46
• You can make the construction make more sense if you use a $n$-linear function in the alternating function. – Michael Burr Mar 23 '16 at 10:48

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.

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 $f$.

• How exactly is the map $\Lambda^k f$ induced by $f$? I just can think of, that for $\varphi \in \Lambda^k W$ we set $\psi(v_1, \ldots, v_k) := \varphi(f(v_1), \ldots, f(v_k)) \in \Lambda^k V$, and this induced map is $\Lambda^k W \to \Lambda^k V$, but you wrote it the other way!? – StefanH Mar 23 '16 at 15:11
• You are confusing $\Lambda^k V$ and its dual. The space of $k$-linear alternating forms on $V$ is not $\Lambda^k V$, but $(\Lambda^k V)^*$. So take the transpose of your idea. – Captain Lama Mar 23 '16 at 15:21
• I guess you are talking about the wedge-product as applied to vectors... but in my references (Spivak, Calculus on Manifolds and others) it is just defined for alternating $k$-linear maps as $\varphi \wedge \psi := \frac{(k+l)!}{k!l!} A(\varphi \otimes \psi)$. So could you please add the necessary definitions? I can image that then the induced map is simply $v \wedge w \mapsto f(v) \wedge f(w)$ and so on; but still without the definitions I cannot connect that to what I have written in my question. Btw note that I defined $\Lambda^k(V)$ as the set of $k$-linear maps, so I confused that here. – StefanH Mar 23 '16 at 20:02
• I will edit my answer to be clearer on that. – Captain Lama Mar 23 '16 at 20:12
• You are perfectly right, I made a silly mistake translating your construction in the language of exterior algebra, I will edit my answer to make it correct (hopefully !). The bottom line is still that the construction gives $0$, though. – Captain Lama Mar 29 '16 at 19:32

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.

• On the other hand, you still won't get a canonically defined number ; you have to choose beforehand a determinant in $V$ (for instance, by choosing a basis of $V$). – Captain Lama Mar 23 '16 at 11:19
• @CaptainLama Thas is why I supposed $V = \mathbb R^n$ with the usually chosen determinent function. So now the same question, if $A(\phi) = \alpha\cdot \det$ for $\phi \in \Lambda^n(\mathbb R^n)$, has $\alpha$ a geometrical meaning? PS: You wrote $A(f)$ instead of $A(\phi)$ in your definition of $A$, I edited it if its okay for you. – StefanH Mar 23 '16 at 11:21