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Let $\mathbb{S}^k \subset \mathbb{R}^{k+1}$, $k \geq 1$, be the unit sphere.

Leopardi proved that, given a natural number $n \geq 2$, there exists a partition of $\mathbb{S}^k$ into pairwise disjoint sets $\{S_j\}_{j=1}^n\subset \mathbb{S}^k$, such that $$\sigma(S_j)=\frac{\sigma(\mathbb{S}^k)}{n}\quad \text{and} \quad \text{diam}(S_j)\leq \frac{c_1(k)}{n^{\frac{1}{k}}},$$ for all $j=1,\ldots,n$, where $c_1(k)>0$ is a positive constant only depending on the dimension $k$, diam$(\cdot)$ denotes the diameter and $\sigma(\cdot)$ the surface area. These partitions are called area regular partitions.

Given a point $x\in\mathbb{S}^k$ and $-1\leq t\leq1$, we define the spherical cap $$C=C(x,t)=\{y\in\mathbb{S}^k;\,\langle x,y\rangle\leq t\},$$ where $\langle\cdot,\cdot\rangle$ is the standard scalar product.

Fixed a spherical cap $C=C(x,t)$, it can be proved that the cardinality of the set of indices $\mathcal{J}(C) \subset \{1,\ldots,n\}$ such that, if $j\in\mathcal{J}(C)$ then $C\cap S_j \neq \emptyset$ and $(S^k\setminus C)\cap S_j \neq \emptyset$, is less or equal than a constant depending only on $k$, $c_2(k)$, times $n^{1-\frac{1}{k}}$, i.e., $$\text{card}(\mathcal{J}(C)) \leq c_2(k)n^{1-\frac{1}{k}}.$$

The proof I am following gives no detail: it only says that we have to use the properties of the partition of $\mathbb{S}^k$ (which it is quite obvious), but I do not know how to apply them. Any ideas or possible proofs?

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  • $\begingroup$ Hint: Show that the union of the $S_j$'s in contained in a strip around the boundary of $C$. $\endgroup$ – Nate May 13 at 14:51
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For each such index $j$, the set $S_j$ intersect the boundary of the spherical cap, so it is contained in the $c_1(k)n^{-1/k}$-neighborhood of this boundary. The spherical area of this neighborhood is bounded above by $c(k) n^{-1/k}$ (maximized when it is a hemisphere), and since the $S_j$ are disjoint and all have the same spherical area, you get $$\textrm{card}(\mathcal{J}) \frac{\sigma(S^k)}{n} \le \frac{c(k)}{n^{1/k}},$$ which implies what you want to show.

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  • $\begingroup$ Which is the formula for the spherical area of this neighbourhood in $\mathbb{R}^{k+1}$? If we work in $\mathbb{R}^3$ I know it is $2\pi rh$ and since $r=1$ and $h\leq c_1(k)n^{-\frac{1}{k}}$ I obtain what you say. $\endgroup$ – user614222 May 14 at 9:15
  • $\begingroup$ Somehow I was only thinking of the 2-dimensional unit sphere when I wrote it, which is why I have the $4\pi$ in there. In the general case you don't really need a formula, just need to know that the $r$-neighborhood of a $k-1$-dimensional sphere sitting inside $S^k$ has a $k$-dimensional volume bounded by the $r$-neighborhood of the equator, which itself is bounded by $cr$ for some constant $c>0$. $\endgroup$ – Lukas Geyer May 14 at 14:47

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