Scissors integration

On Twitter Sarah Griffith made the following wonderful observation:

Maybe this doesn't seem strange to anyone else, but I just realized I find it very strange that if you integrate a polynomial over Q between rational bounds you always get a rational number, even though the area under the curve can be quite complicated

— Canonically Singular Sarah (@SC_Griffith) April 29, 2020

Indeed, I think this fact — that the integral of a rational polynomial in one variable between rational bounds is rational — is very non-obvious geometrically, even though it’s clear from the fundamental theorem of calculus!

Why does this region have rational area?

I came up with a synthetic proof, which I want to share here. It give a geometric proof of the fundamental theorem of calculus for polynomials, which I think is pretty cool, but it also proves more, and points in the direction of some interesting generalizations.

Here’s the argument. By linearity of the integral, it’s enough to show that

$$\int_0^a x^n dx=\frac{a^{n+1}}{n+1}$$ for all rational $a$ and all non-negative integers $n$. Now $x^n$ is the volume of an $n$-cube with side length $x$; hence $$\int_0^a x^n dx$$ is the volume of a figure swept out by a growing $n$-cube, with side length ranging from $0$ to $a$.

So for example the figure below, swept out by an expanding square, computes the integral

$$\int_0^a x^2 dx.$$

Integating $x^2$

I claim that this figure can be dissected into $n!$ $n+1$-simplices, each of which has volume

$$\frac{a^{n+1}}{(n+1)!}.$$ Hence its total volume is $$n!\cdot \frac{a^{n+1}}{(n+1)!}=\frac{a^{n+1}}{n+1}$$ as claimed.

We first show that the volume of an $n+1$-simplex with side length $a$ is $\frac{a^{n+1}}{(n+1)!}$, as claimed. It suffices to show that an $n+1$-cube of side length $a$ (which has volume $a^n$) can be dissected into $(n+1)!$ congruent $n+1$-simplices. But we can view the cube as the region

$$0\leq x_1, \cdots, x_{n+1} \leq a.$$ Then for any element $\sigma$ of the symmetric group on $n+1$ letters, the region $$0\leq x_{\sigma(1)}\leq \cdots \leq x_{\sigma(n+1)}\leq a$$ is a simplex with side length $a$, and these $(n+1)!$ simplices evidently fill up the $n+1$-cube and intersect only along their faces. The action of the symmetric group shows the simplices are all congruent.

Now why does our figure (swept out by a growing $n$-cube) admit a dissection into $n!$ $n+1$-simplices? It’s because the $n$-cube can be dissected into $n!$ simplices by what we just did, and each of these sweeps out an $n+1$-simplex! So we’re done.

A dissection of our region into $2!$ $3$-simplices

By the way, @yougetaquote also found a beautiful (and closely related) probabilistic argument:

so this probability is our integral. But it is also clearly equal to 1/(n+1) since the prob of each guy being the maximum is the same.

— Stevie Tieta - formerly Oprah Quotes (@yougetaquote) May 2, 2020

Scissors integrals

I claim this argument actually proves something more, which I think is pretty cool — the point is that because it works entirely via dissection, it actually proves something in the “scissors congruence” group. As a reminder, the scissors congruence group for $\mathbb{R}^n$, denoted $S(n)$ is the free Abelian group on the set of $n$-dimensional polytopes in $\mathbb{R}^n$, modulo the following relations:

1. $[P]=[P_1]+[P_2]$ if $P$ can be dissected into $P_1, P_2$, and

2. $[P]=[gP]$ for $g$ any isometry of $\mathbb{R}^n$ (for the usual metric).

The direct sum

$$R=\bigoplus_n S(n)$$ forms a graded ring (under Cartesian product of polytopes). Throughout the rest of this section I’ll freely identify a $d$-dimensional polytope contained in a $d$-plane inside of $\mathbb{R}^n$ with the corresponding polytope in $\mathbb{R}^d$.

I would like to think of a polytope $P$ as analogous to a monomial, and describe a way of integrating it over various regions, to get new elements of $R$. Namely, let $G_n$ be the subgroup of automorphisms (not isometries!) of $\mathbb{R}^n$ generated by scalings and translations. That is, $G_n$ is the subgroup of affine-linear transformations of the form

$$(x_1, \cdots, x_n)\mapsto (a_1x_1+b_1, \cdots, a_nx_n+b_n)$$ with the $a_i$ positive. There is an evident isomorphism $$G_n\simeq \mathbb{R}^n\rtimes \mathbb{R}^n,$$ where the transformation above maps to $$(b_1, \cdots, b_n, \log a_1, \cdots, \log a_n).$$ I’ll choose the evident identification of this with $\mathbb{R}^{2n}$.

Now let $D\subset G_n$ be any $d$-dimensional polytope, under the identification of $G_n$ with $\mathbb{R}^{2n}$ above. Then for an $n$-dimensional polytope $P$ in $\mathbb{R}^n$, I will define the scissors integral

$$\int_{D}^{SC} P:=\left[\bigcup_{d\in D} (d, dP)\subset \mathbb{R}^{2n}\times \mathbb{R}^n\right]$$ to be the class of the $n+d$-dimensional polytope defined above. We may of course also let $P$ or $D$ be linear combinations of polytopes, and extend the definition bilinearly.

Note: It’s not clear to me how this depends on $D$ and $P$; for example, there are some natural (scissor-type) relations on $D, P$ so that the integral only depends on the congruence-class of $D, P$. It would be very interesting to work out the maximal (or a large) set of relations for which this is true.

How does this relate to the stuff above? There’s a very natural 1-dimensional polytope in $G_n$ for each $n$, which I’ll denote $[a, b]$ (suppressing the dependence on $n$). This consists of maps of the form

$$(x_1, \cdots, x_n)\mapsto (cx_1, \cdots, cx_n)$$ for $a\leq c\leq b$. With some good will, I hope you’ll let me denote $$\int_a^b P dP:= \int^{SC}_{[a,b]} P.$$

There’s also an extremely natural polytope

$$\mathbb{x}\subset \mathbb{R}, \mathbb{x}:=[0,1],$$ and for any non-negative real number $a$ I’ll let $$[a]:=a\mathbb{x}=[0, a].$$

Then the argument above in fact shows

$$(n+1)\int_a^b \mathbb{x}^n d\mathbb{x}=[b]^{n+1}-[a]^{n+1}.$$

This is a scissors fundamental theorem of calculus! One obtains the usual fundamental theorem of calculus (for polynomials) by applying the function that sends a (scissors congruence class of) polytope to its volume.

You’ll note that I multiplied by $n+1$ on the left side of the equation, rather than dividing on the right as one might usually do. That’s because this is what the argument actually gives, and I’m not sure if $SC(n+1)$ (where both sides of the equation live) is uniquely divisible. For $n=0,1,2$, they are, and so one can divide both sides by $n+1$. [EDIT: Sasha Shmakov points out to me that $SC(n)$ is actually a real vector space, and hence clearly uniquely divisible, so you can divide by $n+1$ if you want.]

What other integrals can you compute? How does the integral change as one varies $P, D$? I’ll let you know when I think about this more myself! In particular, I find it very plausible that there’s a “fundamental theorem of calculus” for a much broader class of polytopes $P, D$.