# How is the volume of a cross-polytope in $\mathbb{R}^n = \frac{2^n}{n!}$?

I'm fairly new to convex geometry, and I wish to prove that the volume of a cross-polytope in $$\mathbb{R}^n$$ is $$\frac{2^n}{n!}$$.

A cross-polytope is the convex-hull of $$2n$$ points in $$\mathbb{R}^n$$, namely, $$(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$$. Since this is the same as the unit ball of $$l_1$$ norm in $$\mathbb{R}^n$$, we denote it by $$B^n_1$$.

$$B^n_1$$ is made up of $$2^n$$ pieces similar to the piece whose points are in the convex hull of $$(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$$ and are all non-negative. So the total volume is $$2^n$$ times the volume of this piece (denoted by $$P_n$$ hereafter), which (from what I understand) has vertices $$(1,0,...,0), (0, 1,0,...,0),...,(0,...,0, 1)$$.

How do I find the volume of $$P_n$$? I have read an inductive proof that I don't quite understand. It says that this piece $$P_n$$ is a cone of height $$1$$ (how?) over a base, which is the analogous piece in $$\mathbb{R}^{n-1}$$, i.e. $$P_{n-1}$$ (how?).

If we take the above for granted and use that the volume of a cone in $$\mathbb{R}^n$$ is $$\frac{Bh}{n}$$ (and how do we prove this?), we'll get the desired result ($$B$$ = volume of base, $$h$$ = height of the cone) with the help of $$\operatorname{vol}(P_n) = \frac{\operatorname{vol}(P_{n-1})}{n}$$

A cone is defined as the convex hull of a single point and a convex body of dimension $$\mathbb{R}^{n-1}$$ in $$\mathbb{R}^n$$.

In general, I have found it difficult carrying forward results from $$\mathbb{R}^3$$ to $$\mathbb{R}^n$$ - a lot of the stuff is not intuitive!

Edit 1:
Based on a comment by @Arthur, I have tried out integration to find the volume of $$P_n$$ as follows: $$\int \int ... \int dx_1 dx_2 ... dx_n$$ We have $$x_1 + x_2 + ... + x_n = 1$$ and $$x_i \ge 0 \forall i = 1,...,n$$. So the limits for $$x_1$$ goes from $$0$$ to $$1 - x_2 - x_3 - ... - x_n$$ We get $$\int \int ... \int (1 - x_2 - x_3 - ... - x_n) dx_2 ... dx_n$$

How do I go ahead from here? What are the limits for $$x_2$$? Is there a pattern I should see?

• It can be done pretty easily with integration. Commented Dec 24, 2020 at 11:00
• Have you worked out manually and sketched what happens for the cases $n = 1, 2, 3$? Commented Dec 24, 2020 at 11:01
• Yes, did that. How do I get to the volume of a cone in $\mathbb{R}^n$ though? @TravisWillse Commented Dec 24, 2020 at 11:06
• Integration how? I'm not that familiar with volumes beyond $\mathbb{R}^3$. Could you give some idea? @Arthur Commented Dec 24, 2020 at 11:06

These are fairly intuitive inductions if you go from $$1$$ to $$2$$ dimensions and from $$2$$ to $$3$$ dimensions. Similar steps in higher dimensions are essentially the same.

For $$n=1$$ you are measuring the length from $$-1$$ to $$+1$$ which is $$2=\frac{2^1}{1!}$$

For the induction step you start at $$n$$ dimensions with a cross-polytope in black with hypervolume $$V_{n} = \frac{2^n}{n!}$$. You then introduce an orthogonal vertical line in red in the $$n+1$$th dimension with height $$h$$ from $$-1$$ to $$+1$$ and see that similar copies in grey of the cross-polytope have linear proportion $$1-|h|$$ and so $$n$$-dimensional hypervolume proportion $$(1-|h|)^n$$; it is easier just to consider the top half and double the result. So integrating over the slices to find the next hypervolume: $$V_{n+1} = 2\int\limits_{h=0}^1 \frac{2^n}{n!}(1-h)^n\,dh = 2\left[\frac{2^n}{n!} \times \frac{-1}{n+1}(1-h)^{n+1}\right]^1_0=\frac{2^{n+1}}{(n+1)!}$$

• Nice, that solves it directly! Could you also help me with the integral I have set up for $P_n$ (or another way to find just $P_n$) in the "Edit 1" section of my post? It should amount to $\frac{1}{n!}$. Commented Dec 24, 2020 at 11:18
• Also, induction proofs work - but are there more direct ways of approaching this? Commented Dec 24, 2020 at 11:19
• I do not understand the part about "...have linear proportion $1-|h|$ and so $n$-dimensional hypervolume proportion $(1-|h|)^n$..." - could you explain a little more here, please? Commented Dec 24, 2020 at 11:28
• You could write the integral as $\int\limits_{-1}^1\cdots\int\limits_{-1}^1 \int\limits_{-1}^1 (1-|x_1|)^0(1-|x_2|)^1\cdots(1-|x_{n}|)^{n-1}\,dx_1\,dx_2\ldots\,dx_n$ Commented Dec 24, 2020 at 11:39
• Suppose you have similar squares (or triangles or other 2D shapes): if their lengths are in proportion $k:1$ then their areas are in proportion $k^2:1$. If you have similar cubes (or pyramids or other 3D shapes): if their lengths are in proportion $k:1$ then their volumes are in proportion $k^3:1$. Similarly at higher dimensions Commented Dec 24, 2020 at 11:43

As you've described, it's enough to find the volume of the convex hull $$P_n$$ of the points $$(0, \ldots, 0), (1, 0, \ldots, 0), (0, 1, 0, \ldots, 0), (0, \ldots, 0, 1)$$, which is sometimes called the standard $$n$$-simplex.

It says that this piece $$P_n$$ is a cone of height $$1$$ (how?) over a base, which is the analogous piece in $$\Bbb R^{n−1}$$, i.e. $$P_{n−1}$$ (how?).

If we think of $$\Bbb R^{n - 1}$$ embedded in $$\Bbb R^n$$ in the usual way (as the $$(n - 1)$$-plane $$\{x_n = 0\}$$, via the usual inclusion $$\iota : \Bbb R^{n - 1} \hookrightarrow \Bbb R^n, \qquad (x_1, \ldots, x_{n - 1}) \mapsto (x_1, \ldots, x_{n - 1}, 0) ,$$ then we can see that $$P_n$$ is the cone over the $$(n - 1)$$ simplex $$\iota(P_{n - 1}) \subset \iota(\Bbb R^{n - 1})$$ with vertex $$(0, \ldots, 0, 1)$$, i.e., it is a cone (in fact, pyramid) of height $$1$$ over $$\iota(P_n)$$.

If we take the above for granted and use that the volume of a cone in $$\Bbb R^n$$ is $$\frac{B h}{n}$$ (and how do we prove this?)

To prove the cone formula, suppose a cone has height $$h$$ and base $$S$$ of $$(n - 1)$$-dimensional volume $$B$$, and consider the cross-section of the cone at height $$x_n$$. That cross section is similar to $$S$$, and by linearity any of its length dimensions is $$(1 - \frac{x_n}{h})$$ that of the corresponding length of the base. So, the $$(n - 1)$$-dimensional volume of the cross-section is $$\left(1 - \frac{x_n}{h}\right)^{n - 1} B$$. Integrating over the $$x_n$$-coordinate gives that the volume of the cone is $$\int_0^h \left(1 - \frac{x_n}{h}\right)^{n - 1} B \,dx_n = \frac{B h}{n}.$$

• Thanks a lot - very helpful! Commented Dec 24, 2020 at 12:09