# The Unapologetic Mathematician

## Parallelepipeds and Volumes II

Yesterday we established that the $k$-dimensional volume of a parallelepiped with $k$ sides should be an alternating multilinear functional of those $k$ sides. But now we want to investigate which one.

The universal property of spaces of antisymmetric tensors says that any such functional corresponds to a unique linear functional $V_k:A^k\left(\mathbb{R}^n\right)\rightarrow\mathbb{R}$. That is, we take the parallelepiped with sides $v_1$ through $v_k$ and represent it by the antisymmetric tensor $v_1\wedge\dots\wedge v_k$. Notice, in particular, that if the parallelepiped is degenerate then this tensor is ${0}$, as we hoped. Then volume is some linear functional that takes in such an antisymmetric tensor and spits out a real number. But which linear functional?

I’ll start by answering this question for $n$-dimensional parallelepipeds in $n$-dimensional space. Such a parallelepiped is represented by an antisymmetric tensor with the $n$ sides as its tensorands. But we’ve calculated the dimension of the space of such tensors: $\dim\left(A^n\left(\mathbb{R}^n\right)\right)=1$. That is, once we represent these parallelepipeds by antisymmetric tensors there’s only one parameter left to distinguish them: their volume. So if we specify the volume of one parallelepiped linearity will take care of all the others.

There’s one parallelepiped whose volume we know already. The unit $n$-cube must have unit volume. So, to this end, pick an orthonormal basis $\left\{e_i\right\}_{i=1}^n$. A parallelepiped with these sides corresponds to the antisymmetric tensor $e_1\wedge\dots\wedge e_n$, and the volume functional must send this to ${1}$. But be careful! The volume doesn’t depend just on the choice of basis, but on the order of the basis elements. Swap two of the basis elements and we should swap the sign of the volume. So we’ve got two different choices of volume functional here, which differ exactly by a sign. We call these two choices “orientations” on our vector space.

This is actually not as esoteric as it may seem. Almost all introductions to vectors — from multivariable calculus to vector-based physics — talk about “left-handed” and “right-handed” coordinate systems. These differ by a reflection, which would change the signs of all parallelepipeds. So we must choose one or the other, and choose which unit cube will have volume ${1}$ and which will have volume $-1$. The isomorphism from $\Lambda(V)$ to $\Lambda(V)^*$ then gives us a “volume form” $\mathrm{vol}\left(\underline{\hphantom{X}}\right)=n!\langle e_1\wedge\dots\wedge e_n,\underline{\hphantom{X}}\rangle$, which will give us the volume of a parallelepiped represented by a given top-degree wedge.

Once we’ve made that choice, what about general parallelepipeds? If we have sides $\left\{v_1\right\}_{i=1}^n$ — written in components as $v_i^je_j$ — we represent the parallelepiped by the wedge $v_1\wedge\dots\wedge v_n$. This is the image of our unit cube under the transformation sending $e_i$ to $v_i$, and so we find

\displaystyle\begin{aligned}\mathrm{vol}\left(v_1\wedge\dots\wedge v_n\right)&=n!\langle e_1\wedge\dots\wedge e_n,v_1\wedge\dots\wedge v_n\rangle\\&=\det\left(\langle e_i,v_j\rangle\right)\\&=\det\left(v_j^i\right)\end{aligned}

The volume of the parallelepiped is the determinant of this transformation.

Incidentally, this gives a geometric meaning to the special orthogonal group $\mathrm{SO}(n,\mathbb{R})$. Orthogonal transformations send orthonormal bases to other orthonormal bases, which will send unit cubes to other unit cubes. But the determinant of an orthogonal transformation may be either $+1$ or $-1$. Transformations of the first kind make up the special orthogonal group, while transformations of the second kind send “positive” unit cubes to “negative” ones, and vice-versa. That is, they involve some sort of reflection, swapping the choice of orientation we made above. Special orthogonal transformations are those which preserve not only lengths and angles, but the orientation of the space. More generally, there is a homomorphism $\mathrm{GL}(n,\mathbb{R})\rightarrow\mathbb{Z}_2$ sending a transformation to the sign of its determinant. Transformations with positive determinant are said to be “orientation-preserving”, while those with negative determinant are said to be “orientation-reversing”.

November 3, 2009 - Posted by | Analytic Geometry, Geometry

1. [...] and Volumes III So, why bother with this orientation stuff, anyway? We’ve got an inner product on spaces of antisymmetric tensors, and that should [...]

Pingback by Parallelepipeds and Volumes III « The Unapologetic Mathematician | November 4, 2009 | Reply

2. [...] use our new tools. We represent the parallelogram as the wedge — incidentally choosing an orientation of the parallelogram and the entire plane containing it — and calculate its length using the [...]

Pingback by An Example of a Parallelogram « The Unapologetic Mathematician | November 5, 2009 | Reply

3. [...] of all, we should see why there may be such a correspondence. We’ve identified -dimensional parallelepipeds in an -dimensional vector space with antisymmetric tensors of degree [...]

Pingback by The Hodge Star « The Unapologetic Mathematician | November 9, 2009 | Reply

4. [...] recall that if is an -dimensional manifold than the space of “top forms” on is [...]

Pingback by Oriented Manifolds « The Unapologetic Mathematician | August 25, 2011 | Reply

5. [...] seen this before when talking about the volume of a parallelepiped, but it still feels like this should have volume . For this reason, many authors will rescale the [...]

Pingback by Inner Products on Differential Forms « The Unapologetic Mathematician | October 4, 2011 | Reply