Geodesic in plane when metric depends on single variable?

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SUMMARY

The discussion centers on the geodesics in a p+1 dimensional manifold where the Fisher metric is a function of the variable beta alone. The participants analyze the geodesic connecting two points, P=(beta1,0,0,...,0) and Q=(beta2,t1,t2,...,tp), and whether it lies in the 2D-plane spanned by the vectors (1,0,0,...,0) and the unit vector towards Q. The conclusion is that the geodesic does not lie in this plane unless the manifold itself is a flat plane, which has implications for the numerical determination of geodesics and distances between probability densities.

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rhe
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Hello,

I have a p+1 dimensional manifold describing the parameter space of a family of probability densities. The p+1 dimensions are (beta, t1, t2, ..., tp), all reals, and beta restricted to the positive reals. The (fisher) metric on this manifold is a function of beta only, hence the submanifolds for a constant beta are flat, and the geodesics are straight lines in these submanifolds.

Consider two points P=(beta1,0,0,...,0) and Q=(beta2,t1,t2,...,tp), and the geodesic connecting these two points. I suspect that this geodesic lies in the 2D-plane spanned by the vector (1,0,0,...,0) and the unit vector pointing towards Q. Is this true?

If so, this could significantly speed up the numerical determination of geodesics and geodesic distances between probability densities for this type of models.

Thank you
 
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rhe said:
Consider two points P=(beta1,0,0,...,0) and Q=(beta2,t1,t2,...,tp), and the geodesic connecting these two points. I suspect that this geodesic lies in the 2D-plane spanned by the vector (1,0,0,...,0) and the unit vector pointing towards Q. Is this true?


No, it's not, unless your manifold is a plane.
 

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