Volume ratio of $\ell_1$ balls and $\ell_1$ surfaces 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 $\mathrm{vol}(\mathbb S_d)$.
 A: Not quite sure but by parameterizing the positive part of $\mathbb{S}_d$ namely $S^+ := \{x: x_1+\cdots+x_d=1, x_i\geq 0\}$  with positive part of $\mathbb{B}_{d-1}$ namely $B^+:=\{x: x_1+\cdots+x_{d-1}\leq 1, x_i\geq 0\}$ one could compute the surface integral. The parametrization given by
$$f: B^+\to S^+,$$ $$f: (t_1,t_2,\cdots, t_{d-1})\mapsto e_1 + t_1(e_2-e_1) + \cdots + t_{d-1}(e_d-e_1)$$ with $e_i$ standard basis vectors gives rise to $\sqrt{\det df^Tdf},$ so that $$\text{Vol}(S^+) = \int_{B^+} \sqrt{\det df^Tdf}\; dt_1dt_2\dots dt_{d-1}.$$ The volume of $\mathbb{S}_d$ should follow from symmetry of $\mathbb{S}_d$.
Alternatively, without going into precise technical details, if $\mathbb{S}_d(r)$ denotes the sphere of radius $r$, then it should be provable that
$$\text{Vol}(\mathbb{S}_d(1)) \sim \int_0^1\text{Vol}(\mathbb{S}_{d-1}(1-x))\;dx$$ since $\mathbb S_d = \{x: |x_1|+\cdots+|x_d|=1\} = \{x: |x_1|+\cdots+|x_{d-1}|=1-|x_d|\}$. This could be used for an inductive computation.
Sorry for not being very precise.
