I'm reading the documentation of this package: Manopt, and they claim that in the hyperboloid model for $\mathbb{H}^d$ the parallel transport between tangent spaces $T_x$ and $T_y$ is given for any $u\in T_x$ by $$ P_{x\mapsto y}:\, u \mapsto u - \frac{ \langle \exp_{x}^{-1}(y) , u \rangle_x }{ d_{\mathbb{H}^d}^2(x,y) } \, \big( \exp_{x}^{-1}(y) + \exp_{y}^{-1}(x) \big) $$ where $\exp_z$ is the Riemmannian exponential map at $z\in \{x,y\}$ and $d_{\mathbb{H}^n}$ is the geodesic/length metric. What is any peer-reviewed (in the mathematics community) reference which shows this/where I can refer to?
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$\begingroup$ Parallel transport is in general path dependent; so as you wrote it this cannot be true. The linked documentation however specifies that this is the parallel transport along the unique geodesic connecting the two points. (Also, in the denominator $b$ should be $y$.) $\endgroup$– Willie WongCommented Jul 28, 2023 at 1:55
1 Answer
I doubt you will find this in a modern "peer-reviewed" work: deriving the formula is suitable as a homework exercise for a course in semi-Riemannian geometry.
Here's a sketch of proof:
The hyperboloid model realizes $\mathbb{H}^n$ as a hypersurface in Minkowski space $\mathbb{R}^{1,n}$. This hypersurface is totally umbilic.
Totally umbilic hypersurfaces $\Sigma$ of a semi-Riemannian manifold $M$ have the following nice property: if $\gamma$ is a geodesic in $\Sigma$ through a point $p$, and $v\in T_p\Sigma$ is such that $v$ is orthogonal to $\gamma$, then the parallel transport of $v$ along $\gamma$ relative to the geometry of $\Sigma$ is the same as the parallel transport of $v$ along $\gamma$ relative to the geometry of $M$.
Sketch of proof: since $\gamma$ is geodesic and $v$ is parallel transported, it is always orthogonal to $\gamma$. But as $v$ is orthogonal to $\gamma$, and $\Sigma$ is umbilical, the second fundamental form $II(\dot{\gamma},v) = 0$. Hence parallel transport in $\Sigma$ agrees with the ambient parallel transport.
In general, if you parallel transport a vector $v$ along a geodesic $\gamma$, you have that $\frac{d}{dt}\langle v,\dot{\gamma}\rangle = 0$.
Ambient parallel transport of $\mathbb{R}^{1,n}$ leaves vectors constant.
So in your case: $\exp_x^{-1}(y) / d(x,y)$ is the unit vector at $T_x$ that is pointing in the direction of $y$. The above discussion shows that after parallel transporting, the components of $v$ that are orthogonal to $\exp_x^{-1}(y)$ remains the same, and the component in the direction of $\exp_x^{-1}(y)$ gets mapped to a vector in the direction of $-\exp_y^{-1}(x)$ (this being the parallel transport of $\exp_x^{-1}(y)$ along said geodesic from $T_x$ to $T_y$) with the same length. And the result follows.
Incidentally, the same computations work for every hyperquadraic embedded in arbitrary semi-Riemannian vector spaces. So it works for the sphere too.
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5$\begingroup$ Pretty much everything you need to derive the result can be found in Chapter 4 of O'Neill's Semi-Riemannian Geometry, but I don't think this explicit formula is in the book. $\endgroup$ Commented Jul 28, 2023 at 2:28
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$\begingroup$ Amazing answer Willie, thanks very much for the super detailed rundown :) $\endgroup$– ABIMCommented Jul 31, 2023 at 12:13