Blog:Volume

Convex polytopes are point sets. As such, we can appeal to one of various equivalent notions of volume for sets of points and define their volume that way. However, this approach breaks down when we try to generalize volume to nonconvex polytopes. We'll present a few approaches we can use to try to properly define volume in this case.

There's three crucial properties that any definition of volume must have. We'll give vague descriptions of them so as to give leeway in the possible definitions.


 * 1) The volume of some simple family (say hypercubes or simplices) must coincide with what we expect.
 * 2) Volume must be invariant under translations (or more generally, under rigid transformations).
 * 3) Volume must in some sense be "additive". That is, the volume of a shape can be obtained by partitioning it into finitely or countably many other shapes and adding up their volume.

Volume as an integral
Perhaps the first approach most mathematicians learn when it comes to defining volume is the Riemann integral. This can be summarized as "the area under the curve" in the case of real-valued functions, and generalized to represent a sort of weighed volume in the case of functions from $$\mathbb{R}^n$$ into $$\mathbb{R}$$. For a detailed introduction to this integral, the Riemann integral and multiple integral articles on Wikipedia have you covered.

Armed with this definition, we can define the volume of a set $$A\subseteq\mathbb R^n$$ as
 * $$\int\int\ldots\int_{A}1\mathrm dV,$$

whenever this integral exists. We should be able to then prove that this is indeed well-defined in the case where $$A$$ is a convex polytope, though I'll admit I'm not sure about the details. Furthermore, we should be able to verify our three conditions for volume:


 * 1) The volume of a hyperrectangle coincides with what we expect.
 * 2) This notion of volume is easily seen to be translation-invariant.
 * 3) Furthermore, the volume of a countable union of sets with disjoint interiors will equal the sum of the individual volumes. (Not 100% sure on this one either, but I'm pretty sure it holds when each of the sets is a convex polytope).

So, how could we generalize this for nonconvex polytopes? A way to make sure that we properly account for the volume of a "union" of shapes where one is inscribed in the other is via the concept of density. Specifically, instead of integrating over the constant 1, we would integrate over the density of every point in the polytope. Unfortunately, although density is also a relatively intuitive concept, it turns out to not be trivial to define either.

Take an oriented polytope. For each flag $$f$$, we'll associate a sign $$\sigma(f)$$, either 1 or -1. We'll do this in such a way that adjacent flags have an opposite sign. There's at least two ways we can do this, and they'll in general lead to different results for the volume. We'll call these polytopes with extra info oriented concrete polytopes.

So now, take a planar oriented concrete polytope $$\mathcal P$$. For each element other than the minimal element, associate an arbitrary point in the subspace. Now, every flag can be associated with a sequence of points $$(P_0,\ldots,P_n),$$ where $$P_i$$ is the point in the subspace of the element of rank i. For each point $$P_i$$, create a vector $$\vec{v}_i$$ by taking the coordinates of $$P_i$$ and appending 1. We can thus define the signed volume of this flag as
 * $$V(f)=\frac{\sigma(f)}{n!}\det\begin{bmatrix}\vec{v}_0&\vec{v}_1&\ldots&\vec{v}_n\end{bmatrix},$$

where $$\det$$ is the determinant. This is really just the usual volume of the simplex, except that we've taken care to ensure that it has an appropriate sign.

Now, take a point $$Q$$ outside of all of the face planes of $$\mathcal P$$, and contained in the interior of all of the flag simplices. Take the number of flag simplices with a positive signed volume containing $$Q$$, and subtract the number of flag simplices with a negative signed volume containing it. We claim that this count will be independent of how the points in each of the subspaces of $$\mathcal P$$ are chosen. We define the density of $$\mathcal P$$ at $$Q$$ as this count.

Why is density well-defined? Why can we integrate over it? I honestly don't know. I've certainly verified it for a lot of examples, but I don't have anything that's mathematically certain. This result kinda reminds me a bit of Stokes' theorem, which I very much do not fully understand. Maybe once I get some more mathematical background, I'll be able to actually prove things here? I hope so.

Anyways, assuming this all works, we can once again check our three properties for oriented concrete polytopes.


 * 1) The volume of simplices corresponds to their usual volume.
 * 2) By basic properties of determinants, this volume is transltion-invariant.
 * 3) If we take a blend of two oriented concrete polytopes, its volume should be the sum of the individual volumes, since for every point we now need to account for both sets of simplices. Of course, this requires an appropriate definition of blending, which we do not yet have.

If we want to work with connected concrete polytopes, we may define the volume as the absolute value of the volume of either of the two oriented concrete polytopes that may be formed from it. This definition doesn't satisfy some of the nice properties of our signed volume though, namely additivity and continuity.

Volume as an algebraic expression
You might have noticed in the previous definition that the integral has now been made redundant, since we're pretty much just adding and subtracting the volumes of simplices. If we do this, we'll get a quite more general definition of volume, since we now don't depend on $$\mathbb R$$ to define our integrals, and can define the volumes of polytopes embedded in any vector space whatsoever.

I also don't know how to actually prove that this definition works. I know that it probably does, since it's what I implemented in Miratope, and it has consistently given the results that we would expect. This definition also has all of the cool requisites for a volume definition... again, under an appropriate definition of blends, probably. But until we can prove that it makes sense, we might have to stick to something else.

An indirect definition of volume?
Here's one last idea. Perhaps we could define volume in a more indirect way? That is, we precisely state the axioms that we want this function to satisfy, and then prove that there's a unique such function.

These axioms could be:


 * 1) The volume of any simplex corresponds to what we expect.
 * 2) Volume is translation invariant.
 * 3) The volume of a blend of polytopes equals the sum of the individual volumes.

Of course, there's various obstacles in such an approach. Once again, we're forced to define blends, which we haven't. Also, we would somehow need to prove that any two subdivisions of a polytope give the same result, which seems nontrivial at best. But hopefully, with such an approach, we could prove that the previous definition is equivalent to this one.