# Is there a family of Riemannian manifolds with explicitly solvable geodesics for manifold M?

By this paper https://www.mat.univie.ac.at/~michor/rie-met.pdf for any given manifold $$M$$, there is a Hilbert manifold (I believe its a Hilbert manifold, if not it is still an infinite dimensional manifold for which a Riemannian metric can be endowed on) $$\mathcal{M}(M)$$ of all Riemannian metrics on $$M$$. I am wondering if there is perhaps a finite dimensional submanifold $$N$$ of $$\mathcal{M}(M)$$ for which either or both (preferably both) of the following properties apply:

• The geodesic equation in $$N$$ (which is Riemannian via the pullback metric) has an explicit solution
• For all $$g \in N$$ the geodesic equation for $$(M, g)$$ has an explicit solution

I would mostly be interested in the case for which $$M = \mathbb{R}^n$$ but other manifolds would be interesting as well. I am doing this for a computational algorithm (that needs to run in an efficient amount of time) so the main properties I am looking for is that the geodesic equation for $$N$$ and the geodesic equation for $$(M, g)$$ for all $$g \in N$$ is efficiently computable (possibly using autograd libraries) and $$N$$ is a big enough family that it can represent a very diverse set of Riemannian metrics. Also, I am not attached to the metric on $$\mathcal{M}(N)$$ used in the paper, I would be interested in a different metric if it applies to this problem well.

• What do you mean by explicit? Do you mean closed-form in terms of elementary functions? – S.Surace Aug 15 '19 at 9:23
• If you only consider smooth metrics the manifold will not be a Hilbert one, but a Frechet one. A trivial example of $N$ you want is $N$ to be a single metric, for which the geodesic equation is explicitely solvable: e.g. the standard flat metric. Outside of this it seems pretty difficult. I might see some one dimensional examples: For example by taking $N$ to be the family of metrics $e^t g$, where $g$ is the flat metric on Euclidean space. – Thomas Rot Aug 15 '19 at 11:52
• Related: mathoverflow.net/q/37651/44143. If a closed-form metric is required, then the set of Riemannian metrics will not be very diverse, but perhaps it is possible with only closed-form geodesics. – Matt F. Aug 15 '19 at 12:03
• There are results of Matveev that prevent integrable metrics on many manifolds, if I remember correctly. – Ben McKay Aug 15 '19 at 19:44
• I would like the set of metrics to be diverse in that they can approximate arbitrary metrics fairly well (Like a Fourier series or something) or at least approximate a large number of interesting metrics. And by explicit I do mean closed-form in terms of elementary functions. Although really I only need closed-form in terms of functions that there are algorithms to compute (or approximate to an acceptable degree of accuracy) in very efficient time (this is for a deep learning algorithm that has to do this for a very large number of datapoints). – Justin Dieter Aug 16 '19 at 18:45