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Let $v,w \in S^{n-1}$ be two $n$ dimensional real vectors on sphere. Consider the following integral: $$ \int_{x \in S^{n-1}} \big|\langle x,v \rangle\big|\cdot\big|\langle x,w \rangle\big|\; dx. $$ Since the integration is taking over the sphere, we have rotation invariance and the value of the integration only depends on the value of $\langle v,w \rangle$. Now the question is to show that the integral is a non-decreasing function w.r.t $\langle v,w \rangle$, for $\langle v,w \rangle > 0$. I believe that this question is connected to the problem of packing two pairs of two antipodal points on a sphere (i.e. four diametrically symmetric points), but I could not show the connection. Any help is much appreciated.

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That is technically a 2D question. We can assume that $v=e^{-it}, w=e^{it}\in\mathbb R^2$ ($0<t<\pi/4$). Then in the polar coordinates the integral becomes $$ \int_0^1 \varphi(r)dr\int_0^{2\pi}|\cos(s-t)\cos(s+t)|\,ds $$ where $\varphi(r)$ is some non-negative function which I leave to you to compute (and whose exact formula does not matter in the slightest).

Now $\cos(s-t)\cos(s+t)=\frac 12[\cos(2s)+\cos(2t)]$, so we just need to show that $\int_0^{2\pi}|\cos 2s+\cos 2t|\,ds$ is a decreasing function of $t$ on $[0,\pi/4]$, which is sort of obvious: just draw the graph of the cosine and look at the speed at which the relevant areas change when you move the constant level $-\cos 2t$ from $-1$ to $0$ (the decay will be controlled by two long intervals and the growth by two short ones).

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  • $\begingroup$ Just a minor remark. I simply get the integral of a square on the sphere, instead of the 1d with $\phi$. In any case, it is just a positive number (it is $(1/n\pi)|S^{n-1}|$) and doesn't matter...but what is your $\phi$? $\endgroup$ Jan 7, 2022 at 15:03

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