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 $\mathrm{vol}(\mathbb S_d)$.

  • 1
    $\begingroup$ en.wikipedia.org/wiki/N-sphere $\endgroup$ – JJR May 15 '17 at 19:39
  • $\begingroup$ Thanks very much for the Wikipedia link. However, it appears that only surface volume for the l2 distance is derived (I.e., Euclidean distance). Is the similar relation still true for L1 case ? $\endgroup$ – Yining Wang May 15 '17 at 19:53
  • $\begingroup$ See math.stackexchange.com/q/474880/27978 for related. $\endgroup$ – copper.hat May 15 '17 at 22:42

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.

  • $\begingroup$ Thanks for your useful idea. However, using your second argument I can't seem to reproduce the $d$ factor for the $\ell_2$ case. In particular, if $\mathbb S_d$ and $\mathbb B_d$ are $\ell_2$-spheres and balls I would have $\mathrm{vol}(\mathbb B_d) = \int_0^1{\mathrm{vol}(\mathbb S_{d-1}(\sqrt{1-x^2}))dx}$. Because $\mathrm{vol}(\mathbb S_{d-1}(r)) \propto r^{d-1}$, this yields $\mathrm{vol}(\mathbb B_d)/\mathrm{vol}(\mathbb S_{d-1}) = \int_0^1{(1-x^2)^{(d-1)/2}dx} = \int_0^{\pi/2}\cos^d(\theta)d\theta$, which is incorrect. $\endgroup$ – Yining Wang May 16 '17 at 21:05
  • $\begingroup$ Yes, you are right since there is a factor missing under the integral in the second argument which I can't figure out $\endgroup$ – JJR May 16 '17 at 21:31

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