volume of "$n$-hedron"

Question

In $\mathbb{R}^n$, why does the "$n$-hedron" $|x_1|+|x_2|+\dots+|x_n| \le 1$ have volume $\cfrac{2^n}{n!}$? I came across this fact in some of Minkowski's proofs in the field of geometry of numbers.

Thank you.

Answer 1

If you take the positive orthant, all $x_i \geq 0,$ this is just Cavalieri's principle, just an integral. For example, the right triangle with legs of length $t$ has area $t^2 / 2.$ Integrate that, the right tetrahedron of legs $t$ has volume $t^3 / 6.$ And so on.

Answer 2

The $2^n$ comes because you have that many copies of the simplex $0 \le x_i \le 1$

The $n!$ comes from integrating up the volume. The area of the right triangle is $\frac 12$, the volume of the tetrahedron is $\frac 12 \cdot \frac 13$ and so on.

Answer 3

Let $X$ be a region in $\mathbb R^n\times\{0\}\subset\mathbb R^{n+1}$, with volume $V$. Now take a point in $\mathbb R^{n+1}$ and form a new solid by connecting this point by lines to all of the points of $X$. This is a "cone on X." Then it is not hard to show, for example by integration, that the $n+1$-dimensional volume of this cone is $\frac{1}{n+1} Vh$, where $h$ is the last coordinate of the cone point. Thus the orthant $x_1+x_2\leq 1,x_1,x_2\geq 0$ has area $1/2$, and in general the orthant $x_1+\cdots+x_n\leq 1, x_i\geq 0$ has volume $1/n$ times the volume of the previous orthant, which establishes the general formula of $1/n!$.