What is the largest subset of the sphere such that inner product of any two points in the set is nonnegative

Question

I'm interested in the question of finding the maximum area of $A\subset S^{d-1}$, such that, for all $x,y \in A, \left<x,y\right>\ge 0$. The portion of the sphere lying in the positive orthant seems like a reasonable guess. Can any subset of the sphere that satisfies the positive inner product property be rotated to fit in the positive orthant? I seem to have some evidence that the answer to this last question is "no" but I would like to know for certain.

Answer 1

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling factor $(\sqrt2)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt2<2$, the cap ultimately becomes exponentially larger than the positive orthant.