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](https://en.wikipedia.org/wiki/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](https://arxiv.org/pdf/math/0507163.pdf), 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](https://graphics.stanford.edu/papers/permutohedral/permutohedral_techreport.pdf), 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.



How can the same polytope have three different volumes?
$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$
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!


