How can the same polytope have three different volumes? [closed]

Question

I'm quite new to geometry and I came across the idea that the same convex polytope can have at least three different volumes.

Consider the permutohedron, formed by the convex hull of the n! points obtained by permuting $(1, 2, ..., n)$. As the polytope lives on an $(n-1)$ dimensional hyperplane (summation is commutative), it has volume zero when $\mathbb{R}^n$ is the ambient space. Fine.

However, one paper defines the volume by projecting the vertices down and computing the volume in (n-1) space. Another volume is obtained by finding the square root of the Gram determinant. But why should these lead to different quantities?

The first is from Postnikov 2009, Theorem 3.1 and the discussion on the top of page 4. He considers the generalized permutohedron but that doesn't change my question.

The second is from Baek and Adams, Prop 2.11.

Intuitively, I am at a loss as to how the same object can hold different amounts of volume.

I mentioned that I understand how the ambient space changes things, but beyond $\mathbb{R}^n$, $\mathbb{R}^{n-1}$, I'm not sure why we'd be interested in others.

Answer 1

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$ There is nothing deep here, just a lot of small issues building up. Both papers are studying the polytope $P_n$ in $\RR^n$ which is the convex hull of the $n!$ permutations of $(0,1,2,\ldots, n-1)$. This polytope in an $n-1$ dimensional hyperplane. Let's look at the two simplest examples: $P_2$ is the line segement joining $(0,1)$ and $(1,0)$, and $P_3$ is a regular hexagon whose vertices are the $6$ permutations of $(0,1,2)$ and whose center is at $(1,1,1)$.

Let $H_n$ be the hyperplane $x_1+x_2+\cdots+x_n = \binom{n}{2}$ in $\RR^n$, so $P_n$ sits in $H_n$. The intersection $H_n \cap \ZZ^n$ is an $n-1$ dimensional lattice, with fundamental domain the box whose sides are $(1,-1,0,0,\ldots,0)$, $(0,1,-1,0,\ldots,0,0)$, $(0,0,1,-1,\ldots,0,0)$, ..., $(0,0,0,0,\ldots,1,-1)$. Let's call this box $B_n$.

Issue 1 The Postnikov's $n$ is Baek and Adams' $d+1$, so their polytope is $d$-dimensional and has symmetry group $S_{d+1}$.

Issue 2 Postnikov normalizes $B_n$ to have volume $1$. Baek and Adams use the ordinary $(n-1)$-dimensional volume from a multivariable calculus class, which makes $B_n$ have volume $\sqrt{n}$. For example, $B_2$ is a line segment from $(1,0)$ to $(0,1)$, which Postnikov says has length $1$ and Baek and Adams say have length $\sqrt{2}$. $B_3$ is a parallelogram with sides $(1,-1,0)$ and $(0,1,-1)$; again, Postnikov normalizes the volume to $1$. We can compute the volume in Baek and Adams sense by computing the length of the cross product: $(1,-1,0) \times (0,1,-1) = (1,1,1)$ and $|(1,1,1)| = \sqrt{3}$. In general, this factor of $\sqrt{n}$ is the square root of the Gram determinant.

Due to these conventions, Postnikov's volume is $n^{n-2}$ and Baek and Adams is $(d+1)^{d-1/2}$. These are the actual only disagreements, but there are two other points of confusion.

Issue 3 The point of Postnikov's paper is that he is giving three very different formulas for this volume, and the one you cite is not very close to the $n^{n-2}$. See Proposition 2.4 of Postnikov for a formula where you can see the $n^{n-2}$.

Issue 4 Postnikov is addressing the more general question of computing the volume of the convex hull of the $n!$ permutations of $(x_1, x_2, \ldots, x_n)$. For example, when $n=2$, this volume is $x_2-x_1$, which clearly specializes to $1=2^{2-2}$ when $(x_1, x_2) = (0,1)$. Postnikov gives a very different formula though: His Theorem 3.1 in the $n=2$ case is $$\tfrac{\lambda_1 x_1 + \lambda_2 x_2}{\lambda_1-\lambda_2} + \tfrac{\lambda_2 x_1 + \lambda_1 x_2}{\lambda_2-\lambda_1}.$$ It is not at all obvious that the $\lambda$'s cancel, or that the result is a polynomial in the $x$'s -- but they do and it is! Try it and see!