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(Note: I asked this question on math stackexchange but did not get an answer. So I decide to ask this question here and hopefully somebody would know the answer/how to approach).

Consider the $d$-dimensional $\ell_1$-ball $\mathbb B_d := \{x: |x_1|+\cdots+|x_d|\leq 1\}$ and the $d$-dimensional $\ell_1$-surface $\mathbb S_d := \{x: |x_1|+\cdots+|x_d|=1\}$. I'm interested in the following volume ratio: $$ \mathrm{vol}(\mathbb B_d) / \mathrm{vol}(\mathbb S_{d-1}). $$

It is well-known that the volume ratio for $\ell_2$-balls and surfaces is $d$. It is also known that $\mathrm{vol}(\mathbb B_d) = 2^d/d!$. But it seems difficult to find $\mathrm{vol}(\mathbb S_d)$.

j.c.
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Yining Wang
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  • The way you've defined it, $\mathbb S_{d-1}$ is $(d-1)$-dimensional, not $d$-dimensional. – j.c. Sep 20 '17 at 16:32

2 Answers2

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The $d$-dimensional $\ell_1$-ball is a cross polytope which has $2^d$ facets that in this case are all regular $(d-1)$-simplices with side length $\sqrt{2}$. Since the volume of a regular $(d-1)$-simplex with unit side lengths is $\frac{\sqrt{d}}{(d-1)!\sqrt{2}^{d-1}}$, the volume of each of the facets is $\frac{\sqrt{d}}{(d-1)!}$ and $\mathrm{vol}(\mathbb S_{d-1})=\frac{2^d\sqrt{d}}{(d-1)!}$.

Thus (if I didn't mess up my algebra) the ratio you're looking for is $\frac{1}{d\sqrt{d}}$. By the way, I wouldn't call this a "volume ratio", since you're dividing a $d$-dimensional volume by a $(d-1)$-dimensional volume. The quantity you get will carry a unit of length.

j.c.
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I am a little confused. $S_{d-1}$ is a disjoint union of $2^d$ isometric regular simplices (one of them is the set of $x_1, \dotsc, x_d,$ with $x_i \geq 0,$ and $\sum x_i = 1.$ The volume of such a simplex can be found from:

$$\frac{1}{d!} = \frac{1}{d} \sqrt{\frac{1}{d}} V.$$

So, your ratio is $d\sqrt{d}.$

Igor Rivin
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