# The Unapologetic Mathematician

## Parallelepipeds 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 give us a concept of length. Why can’t we just calculate the size of a parallelepiped by sticking it into this bilinear form twice?

Well, let’s see what happens. Given a $k$-dimensional parallelepiped with sides $v_1$ through $v_k$, we represent the parallelepiped by the wedge $\omega=v_1\wedge\dots\wedge v_k$. Then we might try defining the volume by using the renormalized inner product

$\displaystyle\mathrm{vol}(\omega)^2=k!\langle\omega,\omega\rangle$

Let’s expand one copy of the wedge $\omega$ out in terms of our basis of wedges of basis vectors

$\displaystyle k!\langle\omega,\omega\rangle=k!\langle\omega,\omega^Ie_I\rangle=k!\langle\omega,e_I\rangle\omega^I$

where the multi-index $I$ runs over all increasing $k$-tuples of indices $1\leq i_1<\dots. But we already know that $\omega^I=k!\langle\omega,e_I\rangle$, and so this is squared-volume is the sum of the squares of these components, just like we’re familiar with. Then we can define the $k$-volume of the parallelepiped as the square root of this sum.

Let’s look specifically at what happens for top-dimensional parallelepipeds, where $k=n$. Then we only have one possible multi-index $I=(1,\dots,n)$, with coefficient

$\displaystyle\omega^{1\dots n}=n!\langle e_1\wedge\dots\wedge e_n,v_1\wedge\dots\wedge v_n\rangle=\det\left(v_j^i\right)$

$\displaystyle\mathrm{vol}(\omega)=\sqrt{\left(\det\left(v_j^i\right)\right)^2}=\left\lvert\det\left(v_j^i\right)\right\rvert$

So we get the magnitude of the volume without having to worry about choosing an orientation. Why even bother?

Because we already do care about orientation. Let’s go all the way back to one-dimensional parallelepipeds, which are just described by vectors. A vector doesn’t just describe a certain length, it describes a length along a certain line in space. And it doesn’t just describe a length along that line, it describes a length in a certain direction along that line. A vector picks out three things:

• A one-dimensional subspace $L$ of the ambient space $V$.
• An orientation of the subspace $L$.
• A volume (length) of this oriented subspace.

And just like vectors, nondegenerate $k$-dimensional parallelepipeds pick out three things

• A $k$-dimensional subspace $L$ of the ambient space $V$.
• An orientation of the subspace $L$.
• A $k$-dimensional volume of this oriented subspace.

The difference is that when we get up to the top dimension the space itself can have its own orientation, which may or may not agree with the orientation induced by the parallelepiped. We don’t always care about this disagreement, and we can just take the absolute value to get rid of a sign if we don’t care, but it might come in handy.

November 4, 2009 - Posted by | Analytic Geometry, Geometry

1. [...] we could calculate it by expanding in terms of basic wedges. That is, we can [...]

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

2. [...] sampling the function at the specified point in the subinterval , multiplying by the -dimensional volume of the subinterval (which is just the product of its side-lengths), and summing this over all the [...]

Pingback by Higher-Dimensional Riemann Integrals « The Unapologetic Mathematician | December 1, 2009 | Reply

3. [...] all the subintervals in the partition that are completely contained in the interior and add their volumes together. This is the volume of some collection of boxes which is completely contained within , and [...]

Pingback by Jordan Content « The Unapologetic Mathematician | December 3, 2009 | Reply

4. When might we need to know the volume of an n-dimensional parallelpiped? I’m trying to figure out what the applications are. Something in Quantum Mechanics, maybe to do with Hilbert space? Crystallography?

Comment by Ellie | September 23, 2010 | Reply

5. Take a step back, Ellie: the more fundamental question is “when does $n$-dimensional space become meaningful in an application?”

And to that I’d point out that the configuration space of two particles floating around in three-dimensional space is six-dimensional — three to describe the position of each particle. If spatial orientation matters, then even one object has a six-dimensional configuration space, as video game designers are fond of hyping. It should be clear how to get higher and higher dimensional spaces in meaningful applications.

So, if we have a function that we need to integrate over such a space, then we do it by sampling the function at a bunch of points in the space and weighting each sample by the $n$-dimensional volume of a small piece of the space near the sample point. And we calculate those volumes by building them up from $n$-dimensional parallelepipeds!

Comment by John Armstrong | September 23, 2010 | Reply