# 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