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In this question, a simple formula was provided for the geodesic (and as such the exponential map) given a starting point $p$ and an initial direction $\mathbf v$ on the surface of the real unit $n$-sphere embedded in real euclidean space $\mathbb R^{n+1}$:

$$\gamma(t)=\cos(||\mathbf v||t)\mathbf p +\sin(||\mathbf v||t)\frac{\mathbf v}{||\mathbf v||}$$

Is there any equivalent for this formula for the complex unit $n$-sphere embedded in complex euclidean space $\mathbb C^{n+1}$? Does the original formula still work? Through some quick numeric investigation, the latter does not seem to be the case.

My definition of the complex unit $n$-sphere is the set of vectors in $\mathbb C^{n+1}$ which admit to $\sqrt {x^\dagger x} = 1$ (so their "euclidian norm" should equal 1).

Background of the question: I am trying to apply techniques from optimizing on a regular $n$-sphere (which only require an exponential map) to the complex $n$-sphere. My local point, the gradient of the target property at that point and its tangent projection on the tangent bundle at the point are all given as vectors with complex entries. Just using the same geodesic (as pointed out above) does not cut it for some reason.

Maximilian Rüsch
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  • Is the complex unit $n$-sphere not identical to the real $2n$-sphere in $\mathbb R^{2n+1},?$ – Kurt G. Jul 06 '23 at 18:27
  • @KurtG. do you mean that "expanding" the complex vectors such that they result in real vectors with twice the number of entries suffices? Are these new vectors then really forming a sphere? I have a hard time wrapping my head around this. Trying to implement this yielded no difference in results too. – Maximilian Rüsch Jul 07 '23 at 09:58
  • By all means: The complex numbers $z_1,\dots, z_n$ with $|z_1|^2+\dots+|z_n|^2=1$ represent a $(2n-1)$-sphere because $|z_i|^2=x_i^2+y_i^2,.$ – Kurt G. Jul 07 '23 at 10:58
  • This does make sense. There must be some other reason why my results do not improve when using $\mathbb R^2$. – Maximilian Rüsch Jul 08 '23 at 16:49
  • @KurtG. i guess my error was in using this mapping from $\mathbb C^n$ to $\mathbb R^{2n}$ inappropriately due to other bugs in my calculations. It works now. If you want to give your suggestion again as an actual stackexchange answer i will mark it as the solution. – Maximilian Rüsch Jul 08 '23 at 18:50
  • Glad that it works The math content is too trivial for a formal answer. – Kurt G. Jul 09 '23 at 01:56

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As pointed out the in the comments, $\mathbb C^n$ is isomorph to $\mathbb R^{2n}$ which enables use of the geodesic from the question in that space.

(This answer was posted for the sole reason of marking this question as answered. If there is another way to do so without posting an official answer please let me know).

Maximilian Rüsch
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  • "isomorph" here should be understood as "isometric" (so that one can identify the respective unit spheres) – Didier Jul 09 '23 at 20:49