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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.

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  • $\begingroup$ What do you mean by explicit? Do you mean closed-form in terms of elementary functions? $\endgroup$
    – S.Surace
    Commented Aug 15, 2019 at 9:23
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    $\begingroup$ 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. $\endgroup$
    – Thomas Rot
    Commented Aug 15, 2019 at 11:52
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    $\begingroup$ 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. $\endgroup$
    – user44143
    Commented Aug 15, 2019 at 12:03
  • $\begingroup$ There are results of Matveev that prevent integrable metrics on many manifolds, if I remember correctly. $\endgroup$
    – Ben McKay
    Commented Aug 15, 2019 at 19:44
  • $\begingroup$ 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). $\endgroup$
    – Justin Dieter
    Commented Aug 16, 2019 at 18:45

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