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This question arises as a possible step in answering this unsolved question on MSE. Given a unit vector $v \in S^{d-1} \subset \mathbb{R}^d$, I'm looking for an explicit formula for the set $$V = \{ u \in S^{d-1} \mid u \cdot v > 0 \} $$ in spherical coordinates. For $d=2$ we can easily take $v$'s polar coordinate $\theta \in (-\pi, \pi]$ to obtain $$V = (\theta-\pi/2, \theta+\pi/2) $$ modulo $2\pi$. Unfortunately this approach runs into a problem for $d \geq 3$. In spherical coordinates $v = (\theta, \phi)$ with $\theta \in (-\pi, \pi]$ and $\phi \in (-\pi/2, \pi/2]$, consider the following examples:

\begin{align} v_1 &= (0, 0) &\implies&&& V_1 = (-\pi/2, \pi/2) \times (-\pi/2, \pi/2) \\ v_2 &= (\pi/2, 0) &\implies&&& V_2 = (0, \pi) \times (-\pi/2, \pi/2) \\ v_3 &= (0, \pi/2) &\implies&&& V_2 = (-\pi, \pi) \times (0, \pi/2) \\ \end{align}

The first two examples suggest we could do as in the two-dimensional case (subtracting and adding $\pi/2$ to each coordinate), but the last one fails in this respect. Subtracting and adding $\pi/2$ to each coordinate does give the correct set but with respect to a different (equivalent) spherical coordinate system, namely that which takes $\theta \in (-\pi/2, \pi/2]$ and $\phi \in (-\pi, \pi]$. This is inconsistent here, and I haven't found a way to remediate this problem. Any help most welcome!

smalldog
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  • The set of $(\theta,\phi)$ you want will usually not be a nice 'rectangular' region. That only works for the very specifc cases $\phi=0,-\pi/2,\pi/2$ I think. – Simon May 05 '21 at 21:25
  • You should start with finding a formula for the scalar product in spherical coordinates. (https://math.stackexchange.com/questions/243142/what-is-the-general-formula-for-calculating-dot-and-cross-products-in-spherical). But I doubt the answer will be easy. Good luck. – Simon May 05 '21 at 21:27
  • @Simon Do you have an example where the set isn't a rectangular region? Regarding the scalar product, I'm interested in general $d \geq 3$ so I don't know if that would help anyhow... Thanks for your help. – smalldog May 05 '21 at 21:53
  • For $S^2,$ consider $v = (0, \pi/4).$ If $u = (\theta,\phi)$ and $u\cdot v > 0$ we know that $\phi > -\pi/4$. If $\phi = 0$ we know that $-\pi/2 < \theta < \pi/2,$ but if $\phi > \pi/4$ then any $\theta$ is possible. And in general for $-\pi/4 < \phi < \pi/4$ there's a somewhat complicated formula to find the set of possible $\theta$ values as a function of $\phi.$ I don't think it gets any nicer for $d > 3.$ – David K May 10 '21 at 16:46

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This doesn't seem likely to work. If you think about what you're after in 3D, you want some sort of "nice" description of an arbitrary hemisphere in terms of constraints on latitude and longitude. For very special hemispheres, that's fine, but it's going to involve some messiness no matter what. This is more or less equivalent to describing the corresponding equator ("great circle") of that hemisphere in spherical; see here. You'd want a hyper-spherical generalization. Before proceeding, you should see if the $d=3$ case is of any use to you.

The trouble with spherical coordinates is that, outside of the $d=2$ case, they're not at all symmetric. Sometimes they can be useful if your problem has some recursive structure (e.g. computing the volume of hyperspheres), but otherwise it's a likely to be closer to putting a round peg in a square hole.

Joshua P. Swanson
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  • Thanks for your answer. How would one approach the problem differently, then? What kind of explicit description of the set $V$ would be likely to work? – smalldog May 20 '21 at 21:38
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    Well, the linked question doesn't actually make sense, since it supposes there is a uniform distribution on the whole real line. The comments inconsistently suggest $u$ is taken uniformity from the unit sphere instead, which is at least reasonable. In that version, you want to compute the surface area of the union of hemispheres, or equivalently the surface area of the intersection of their complementary hemispheres. That intersection will arise from an unbounded polytope which is a cone from the origin. You'd need to find the extreme rays of that cone (a computationally expensive operation) – Joshua P. Swanson May 21 '21 at 00:10
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    ...and then compute the area of the figure they trace out on the sphere. That latter computation would be a generalization of computing the area of a convex spherical polygon, which has surely been studied. The final answer will be very complicated though and will depend very non-trivially on the exact vectors chosen. With no constraints on those vector choices, there are not going to be any asymptotic regimes to focus on. It does not seem to be a good question, though variants of it could perhaps be good. – Joshua P. Swanson May 21 '21 at 00:14