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The paper Geometry of diffeomorphism groups, complete integrability and optimal transport mentions the following:

The group $\textrm{Diff}(M)$ carries a natural $L^2$-metric

$\displaystyle \langle\langle u\circ\eta, v\circ\eta\rangle\rangle_{L^2}=\int_{M}\langle u\circ \eta,v\circ \eta\rangle d\mu=\int_{M}\langle u,v\rangle \textrm{Jac}_{\mu}(\eta^{-1})d\mu \tag{2.10}$ where $u,v\in T_{e}\textrm{Diff}(M)$ and $\eta\in \textrm{Diff}(M)$.

...Its significance comes from the fact that a curve $t \to η(t)$ in $\textrm{Diff}(M)$ is a geodesic if and only if $t \to η(t)(p)$ is a geodesic in $M$ for each $p \in M$.

I would like to know why the statement about geodesics holds. The paper does not give proof, so I tried to give proof myself, with no success.

It also feels too good to be true since I feel like the mapping $p\mapsto \eta(t)(p)$ has no guarantee that it is a diffeomorphism.

For what its worth, this statement is somehow removed in the publication.

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  • $\begingroup$ Perhaps I don't understand your terminology, but I don't see how the function you've defined could be interpreted as a metric on $Diff(M)$. Presumably you need a function of the form $Diff(M) \times Diff(M) \to \mathbb R$. Aha, you are using "metric" in the sense of Riemann metric? $\endgroup$ Commented Dec 31, 2022 at 23:05
  • $\begingroup$ @RyanBudney Yes, it's about Riemannian metric. $M$ is Riemannian manifold too and $\langle\cdot,\cdot \rangle$ is Riemannian metric on $M$. I should have been clear about it. $\endgroup$
    – Kaira
    Commented Dec 31, 2022 at 23:10
  • $\begingroup$ What is the assumption going into your if and only if statement? From the way you write it, I would assume the family $\eta(t)$ are always assumed to be diffeomorphisms. $\endgroup$ Commented Jan 1, 2023 at 7:47
  • $\begingroup$ @RyanBudney The paper isn't really clear about it, but it seems like $M$ is assumed to be compact or closed. For me the case $M=\mathbb{R}^n$ is enough though. $\endgroup$
    – Kaira
    Commented Jan 1, 2023 at 15:10

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