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Consider the Kähler manifold in coordinates $(a,b)$ given by the complex Riemannian metric $$\begin{pmatrix} \frac{1}{1-|a|^2}&\frac{1}{1-a\bar{b}}\\\frac{1}{1-\bar{a}b}&\frac{1}{1-|b|^2}\end{pmatrix}.$$

For statistical reasons we require that $(a,b)$ be in the unit disk in the complex plane.

What are the geodesics of this manifold?

Michael Albanese
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Wintermute
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    From your notation it seems that $a,b$ are complex variables? So when you say $(a,b)$ be in the unit disk do you mean $a,b\in D\subset \mathbb{C}$ or $(a,b) \in $some subset$\subset \mathbb{C}^2$? – Willie Wong May 07 '13 at 11:38
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    I mean that $a,b \in \mathbb{C}$ and $|a|,|b|<1$. – Wintermute May 07 '13 at 12:59
  • It seems that the coordinate axes are geodesics, and that points on the diagonal can be reached faster by staying on the diagonal than by 2 paths parallel to the axes. – Brian Rushton Jan 12 '14 at 20:24
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    (statistics)? $ $ – Did May 03 '14 at 10:13
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    @Did This manifold is the statistical manifold corresponding to linear time invariant systems. Such systems are commonly used in signal processing for linear prediction. – Wintermute May 03 '14 at 13:32
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    And there is not a whiff of statistics in the question. – Did May 03 '14 at 13:33
  • Hi @Wintermute can you provide a reference for this connection? I had trouble finding results I understood on Google. Intersections between differential geometry and statistics are always interesting to me (since they are so rare), but I agree with Did inasmuch right now I fail to see right now the connection between this problem and statistics. – Chill2Macht Apr 29 '16 at 00:49
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    @William Just google information geometry, you'll see there is a huge connection between Riemannian geometry and statistics. The goto sources are by Amari. As far as Kahler geometry and it's connection to statistics, this is my personal research. I have not published it to date. – Wintermute Apr 29 '16 at 13:02
  • Oh man this is awesome! Thanks again! – Chill2Macht Apr 29 '16 at 14:28
  • It seems like the matrix is not invertible for $a=b$, so this can at best describe a Kähler metric on $\mathbb{D}\times\mathbb{D}\setminus\Delta$, $\mathbb{D}$ being the disk and $\Delta$ the diagonal of the product. Also, a quick Christoffel symbol computation seems to indicate that the axes are not geodesics, but that might be worth re-checking. – Johnny Lemmon Apr 23 '22 at 08:32

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All you need to do is find the Christoffel symbols, defined by

$$\Gamma_{\rho\tau}^\mu=\frac{1}{2}g^{\mu\nu}\left(\frac{\partial g_{\nu\rho}}{\partial x^\tau}+\frac{\partial g_{\nu\tau}}{\partial x^\rho}-\frac{\partial g_{\rho\tau}}{\partial x^\nu}\right)$$

where $g_{\mu\nu}$ is the metric tensor you have above. Then for the non-zero symbols you can plug them into the geodesic differential equation:

$$\frac{\partial^2x^\mu}{\partial\sigma^2}+\Gamma_{\rho\tau}^\mu\frac{\partial x^\rho}{\partial\sigma}\frac{\partial x^\tau}{\partial\sigma}=\gamma(\sigma)\frac{\partial x^\mu}{\partial\sigma}$$

where the geodesic is parametrized by $\sigma$ and $\gamma(\sigma)=0$ if $\sigma$ is affine (don't assume this at the outset). You can typically split the geodesic equation up into several differential equations based on what the Christoffel symbols are, and then exploit symmetries to simplify them. Then choose either the $a$ or $b$ coordinate as your parameter $\sigma$, and solve the differential equations.

William J. Cunningham
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