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I would like to calculate the exponential map in the n-sphere, however, i don't know how to get started. Someone could give me a tip or bibliographic reference?

Cézar Bezerra
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2 Answers2

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$\newcommand{\Vec}[1]{\mathbf{#1}}$If $\Vec{v}$ is a non-zero tangent vector to the $n$-sphere at a point $\Vec{p}$, the geodesic starting at $\Vec{p}$ with initial velocity $\Vec{v}$ is a circle of speed $\|\Vec{v}\|$ lying in the plane spanned by $\Vec{p}$ and $\Vec{v}$: $$ \gamma(t) = \cos(\|\Vec{v}\|t) \Vec{p} + \sin(\|\Vec{v}\|t) \frac{\Vec{v}}{\|\Vec{v}\|}. $$

Andrew D. Hwang
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    How does one derive this formula? I mean this clearly works because γ(0)=p and γ'(0)=v. But how does one arrive at this? – Bizwhiz Apr 27 '20 at 09:22
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    I feel confused because one summand has length dimension while the other is dimensionless and so for the derivative they don't match either $\gamma'(t) = -\sin(|v|t)|v|p + \cos(|v|t)v$, is there a reference for this result? – Dabed Jun 03 '21 at 18:09
  • @AbhigyanSaha: This is belated but in case you're still interested, for the unit sphere centered at the origin the geodesic lies in the plane spanned by the point and initial velocity (because this plane is precisely the fixed set of the isometries of the sphere fixing the initial position and velocity). But ${\mathbf{p}, \mathbf{v}/|\mathbf{v}|}$ is an orthonormal basis for this plane, and we have the initial position and velocity. – Andrew D. Hwang Jun 03 '21 at 18:36
  • @Dabed: Note that $\gamma'(t) = |\mathbf{v}|\bigl[-\sin(|\mathbf{v}|t) \mathbf{p} + \cos(|\mathbf{v}|t) \frac{\mathbf{v}}{|\mathbf{v}|}\bigr]$. Does that (together with my last comment) address your question about units? – Andrew D. Hwang Jun 03 '21 at 18:43
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    Thanks, maybe your way of writing it offers another point of view which I still don't get but as expressions although my latex was rawer I think we wrote the same $\gamma'(t) = -\sin(|v|t)|v|p + \cos(|v|t)v= \underbrace{|\mathbf{v}|}{\color{blue}{\text{dimension of velocity}}}\bigl[\underbrace{-\sin(|\mathbf{v}|t) \mathbf{p}}{\color{red}{\text{dimension of length}}} + \underbrace{\cos(|\mathbf{v}|t) \frac{\mathbf{v}}{|\mathbf{v}|}}_{\color{blue}{\text{dimensionless}}}\bigr]$ so the expression in red still looks weird to me – Dabed Jun 03 '21 at 20:21
  • I believe it boils down to if ${p, \frac{v}{|v|}}$ is really orthonormal, while it is orthogonal I think the orthonormal basis by Gram-Schmidt should be ${e_1,e_2}={\frac{p}{|p|},\frac{v-\langle p,v\rangle/\langle p,p\rangle}{|v-\langle p,v\rangle/\langle p,p\rangle|}}={\frac{p}{|p|},\frac{v-0}{|v-0|}}={{\frac{p}{|p|},\frac{v}{|v|}}}$ (sorry because of space couldn't fit it in one comment) – Dabed Jun 03 '21 at 20:21
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    You're right, the stated formula is for a unit sphere, i.e., $|\mathbf{p}| = 1$. If the radius is $R$, the formula needs to be multiplied by $R$, and $t$ divided by $R$, i.e., $\gamma$ replaced by$$\gamma_{R}(t) = R\gamma(t/R).$$ – Andrew D. Hwang Jun 03 '21 at 20:59
  • oh sorry my bad not sure why didn't realize you were using $|p|=1$, this answer was mentioned in a hackernews thread so I think I came with another mindset if I can say that as an excuse – Dabed Jun 03 '21 at 21:42
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    "Unit" was never mentioned explicitly, to be honest. :) – Andrew D. Hwang Jun 03 '21 at 22:08
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A good book with examples on the n-sphere is Absil: Optimization Algorithms on Matrix Manifolds. Geodesics at chapter 5 I believe.

An easy way to get geodesics on the n-sphere is to use the embedding map in R^n+1. The connection on the n-sphere is the projection of the connection in R^n+1 (that is flat!). Using this (and the characterization of the tangent vectors to S^n) you can derive the formula that Andrew says.

ricvo
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