Can the exterior algebra be seen as a "model for area/volume"?

Question

I am tying to understand the exterior product. Unfortunately I am not sure what it is supposed to describe. I found a wikipedia article, which I think says that the exterior algebra is sort of a formalisation/model of the notion of (signed) area. https://en.wikipedia.org/wiki/Exterior_algebra

I find this point of view really convincing. However what makes me wonder is that I haven't seen any such explanation before. Neither in the lecture nor in any book I looked at. As a matter of fact all I found regarding the exterior algebra are abstract definitions and not a single note to how one might come up with it or some other "intuitive" explanation.

Is this a valid way of looking at the exterior algebra?

The article then defines the exterior algebra in terms of a quotient.

From what I understand the above construction is/can be viewed as a realisation of a "model/formalisation of area". Is that correct?

Furthermore I would like to make sense of why we "mod out" by $I$.

If we only look at $v\otimes w$ it does not satisfy all of properties 1 to 5. In fact only the first and the 4th property is met only partially. In order to have property 4 we would need $w\otimes w$ to be zero. $w\otimes w=0$ is property 2. I take this as a hint to consider $(V\otimes V)/I$, since for the equivalence class $[v\otimes v]=\pi(v\otimes v)=[0]$, since $v\otimes v\in\ker\pi$. The other properties follow if we look at $\pi(v\otimes w)$. Hence, it is reasonable to consider it a model of area/volume.

It is all a bit vague, but I hope it still makes some sense. Many thanks in advance.

Answer 1

The first wikipedia excerpt is a good explanation of why the signed area of a parallelogram is an antisymmetric bilinear function of the two vectors giving two adjacent sides. But I consider the second excerpt irrelevant.

First, given a vector space $V$, which we visualize as a set of arrows, we always, in practice, use only the exterior algebra of the dual vector space, $V^*$, and almost never the exterior algebra of $V$ itself.

So here's how it goes: Given an abstract vector space $V$, there is no way to define the area of a parallelogram, if $\dim V=2$, or the volume of a parallelotope in higher dimensions, because there are no units of measurement yet. For example, the length of a vector or the angle between two vectors is undefined.

The first cool thing is that you can define the concept of volume without using lengths or angles. What you do is choose a parallelotope in $V$, spanned by a basis $(e_1,\dots, e_n)$ of $V$, and declare it to have volume $1$. We can now try to define a function $f(v_1, \dots, v_n)$ to be the volume of the parallelotope spanned by the vectors $v_1, \dots, v_n$.

You do this by studying what properties you want a volume function $f$ to have. After some effort (as described in the first wikipedia excerpt), you find that it should be the absolute value of an antisymmetric multilinear function of the vectors $v_1, \dots, v_n$. The absolute value is hard to work with both algebraically and analytically, so we dump it. This requires allowing volume to be negative, so we call it signed volume. So the set of signed volume functions is the set of antisymmetric multilinear functions of an ordered set of $n$ vectors. This space is written as $\bigwedge^nV^*$.

The space $\bigwedge^nV$ is not as useful in geometry or differential geometry and, in general, should be avoided. That also means you can ignore its definition as the quotient of the tensor product by the ideal $I$.

Instead, you should think of $\Lambda^nV^*$ as the subspace of $\bigotimes^nV^*$ (which is the space of all multilinear functions of $n$ vectors) containing only multilinear functions that satisfy the antisymmetry property.

There is also a subtlety in the definition of the wedge product. Let me restrict to $n=2$; it is easily generalized to higher dimensions. Suppose you have $\theta^1, \theta^2 \in V^*$ and want to define their wedge product $\theta^1\wedge\theta^2 \in \bigwedge^2V^*$. What should it be? Well, if the forms are linearly independent and $(e_1,e_2)$ is the basis of $V$ dual to $(\theta^1,\theta^2)$, then you want $$ (\theta^1\wedge\theta^2)(e_1,e_2) = 1. $$ So, given an arbitrary ordered pair of vectors $(v_1,v_2)$, we define $$ (\theta^1\wedge\theta^2)(v_1,v_2) = \theta^1(v_1)\theta^2(v_2)-\theta^1(v_2)\theta^2(v_1). $$ Note the lack of a normalizing constant like $\frac{1}{2!}$. When $(v_1,v_2)=(e_1,e_2)$, the second term is zero, and we get what we want.

Finally, I advise against ever writing something like $(\theta^1\wedge\theta^2)(e_1\wedge e_2)$. You run into a lot of confusion about when to include a normalizing constant and when not to. Use only the exterior algebra of $V^*$ and avoid using the exterior algebra of $V$ unless there is somehow no way to avoid it. I've never seen any need for $\bigwedge^nV$ in differential geometry.

Answer 2

This point of view is adopted implicitly in multiple situations in linear algebra, analysis and differential geometry.

Linear algebra: The determinant of a linear transformation $T$ measures the signed volume of a unit parallel-epiped after applying $T$ to it. Also, the determinant is an alternating multilinear form on the columns of $T$ (identified with its matrix representation). Or more abstractly, a determinant map is a linear map $\det:\bigwedge^n V\to K$, and the determinant $\det T$ of a linear transformation $T:V\to V$ is defined as the unique number satisfying $\det(Tv_1\wedge\dots\wedge Tv_n)=(\det T)\det(v_1\wedge\dots\wedge v_n)$ for all exterior products $v_1\wedge\dots\wedge v_n\in\bigwedge^n V$. That's the linear algebraic connection between signed volume and the exterior product.

Analysis: The transformation theorem for volume integrals says that if $\Omega\subseteq\mathbb R^n$ is open, $\Phi:\Omega\to\Phi(\Omega)$ a diffeomorphism, and $f:\Phi(\Omega)\to\mathbb R$. Then $f$ is integrable if $\vert\det\mathrm D\Phi\vert f\circ\Phi$ is integrable, and it holds that $$\int_{\Phi(\Omega)} f\mathrm dV=\int_\Omega \vert\det\mathrm D\Phi\vert f\circ\Phi\mathrm dV.$$

Here, $\vert\det\mathrm D\Phi\vert$ is a measure of how much $\Phi$ distorts volumes: $\mathrm D\Phi$ is a linear approximation of $\Phi$, and $\det\mathrm D\Phi$ measures how this linear approximation distorts signed volumes, and accordingly, $\vert\det\mathrm D\Phi\vert$ measures how much it distorts unsigned volumes. Since we already saw that determinants can be defined through exterior powers, this also gives way to understanding exterior powers in terms of signed volume.

Differential Geometry: On smooth manifolds, there is no intrinsic concept of volume. But we can give a manifold such a concept by defining a volume form: If $M$ is an $n$-dimensional smooth manifold, then a volume form is a nowhere vanishing differential $n$-form $\omega$. On tangent spaces, it acts exactly as the determinant map from the linear algebraic part at the beginning of my answer. And we can now define the volume of an open subset $\Omega\subseteq M$ as $$V(\Omega):=\int_\Omega\omega.$$ Essentially, $\omega$ behaves locally like a signed volume does on the tangent space, but globally, it gives different portions of the manifold different scales. For instance, if we have the unit sphere $S^2$, then the standard volume form would yield $\int_{S^2}\omega=4\pi$, the surface area of the unit sphere. But if we want to model a sphere with radius $r$, we could instead take the volume form $\omega':=r^2\omega$, which would give our manifold the "volume" $\int\omega'=4\pi r^2$. Or we could multiply $\omega$ by some other nowhere vanishing smooth function $f:S^2\to\mathbb R$ to get a volume form $\omega'':=f\omega$, which assigns different parts of the sphere different "area densities", so it essentially distorts the sphere.

Why the exterior product? A bit more abstractly, you shouldn't view the exterior product/power/algebra as a quotient of the tensor product/power/algebra. You should instead consider what properties it has. The tensor power $V^{\otimes n}$, for instance, has the property that for all multilinear maps $m:V^n\to W$, there is a unique linear map $\tilde m:V^{\otimes n}\to W$ such that $m(v_1,\dots, v_n)=\tilde m(v_1\otimes\dots\otimes v_n)$. This is nice, since it makes multilinear things linear, and linear is easy. But for signed volumes, we are only interested in alternating multilinear maps, not in general multilinear ones. So it would be nice to have a similar constuct to the tensor power, but one which works with alternating maps instead. This is the exterior power: $\bigwedge^n V$ is a vector space with the property that for all alternating multilinear maps $a:V^n\to W$, there is a unique linear map $\tilde a:\bigwedge^n V\to W$ such that $a(v_1,\dots, v_n)=\tilde a(v_1\wedge\dots\wedge v_n)$. So the exterior power allows us to make alternating multilinear things linear. Signed volumes included! This is the reason why I defined the determinant map above as a linear map on the exterior power of $V$. It's also the reason why we define differential forms as smooth sections of an exterior power of the cotangent bundle: Differential forms are meant to be integrated, but to integrate, we need to give a certain scale to our volume elements.