Measure of a strange set on $S^{p-1}$

Question

Let $S^{p-1}$ be the unit sphere in $\mathbb R^p$, $s \in \mathbb N$ and

\begin{align*}A_s :&= \{ t \in S^{p-1} \text{ such that } t_1 > ... > t_p > 0 \text{ and } t_1 + ... + t_s \geq t_{s+1} + ... + t_{p} \} \\ & = \{ t \in \mathbb R^p \text{ such that } \parallel t \parallel_2 = 1, ~~t_1 > ... > t_p > 0 \text{ and } \parallel t_S \parallel_1 \geq \parallel t_{S^c}\parallel_1 \} \end{align*} where $t = (t_1, ..., t_p)$, $t_S = (t_1, \dots, t_s)$.

I want to compute the Lebesgue measure of $A_s$ on $S^{p-1}$. In the three dimensional case, spherical trigonometry may give the answer but I am not able to obtain a general formula in dimension $p$. An upper bound for this quantity, which is better than $\frac{2 \pi^{p/2}}{p! \Gamma(p/2)}$ (independant of the $l^1$ condition), should be helpfull.

Answer 1

[Note: The question now asks for an upper bound that depends on $s$]

Not my area, but I'll take a shot. Although you want an upper bound independent of $s$, $A$ may depend on $s$, so call it $A_s$ and try to bound the largest $A_s$. It looks like if $s=p-1$, the condition $t_1 + \cdots + t_s \geq t_{s+1} + \cdots + t_{p}$ is implied by $t_1 > \cdots > t_p > 0$, so $\mu(A)\le\mu(A_{p-1})=\mu(\{ t \in S^{p-1} \text{ such that } t_1 > ... > t_p > 0\})$.

The area of $S^{p-1}$ is $\frac{2 \pi^{p/2}}{ \Gamma(p/2)}$, and $S^{p-1}$ can be partitioned as follows: Take as one set $\{t\in S^{p-1}\textrm{ such that }t_i= t_j \textrm{ for some }i\ne j\}$, and take $p!$ additional (disjoint) sets of equal measure, one for each permutation $\pi$ of $\{1\dots n\}$. (A point $t$ is in the one of these where the $t_{\pi(i)}$ are strictly increasing for increasing $i$.) $A_{p-1}$ is one of these latter sets - the one for the identity permutation, so its measure is at most $\frac{1}{p!}\frac{2 \pi^{p/2}}{\Gamma(p/2)}$. (If the set of points with some coordinates equal has measure 0, then this bound is the best possible.)