(Note: I asked [this question on math stackexchange](https://math.stackexchange.com/questions/2282504/volume-ratio-of-ell-1-balls-and-ell-1-surfaces) 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 $\mathrm{vol}(\mathbb S\_d)$.



Volume ratio of $\ell_1$ balls and $\ell_1$ surfaces
The $d$-dimensional $\ell\_1$-ball is a [cross polytope](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Simplex#Volume) 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.


