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The right-invariant metric on $SDiff\(M\)$ is not bi-invariant, so the geodesic from the identity element of SDiff(M) with an initial velocity field v should be different with the one-parameter subgroup with the same velocity field v. But along the geodesic connecting the identity and a diffeomorphism $\phi_{0t}^v\(x\)$, the Lagrangian and Eulerian velocty fields are related by the adjoint action $Ad_{\phi}$, it seems that the Eulerian velocity field along the geodesic is in fact right-invariant. If this is true, then the geodeisc is also a flow of a right-invariant vector field on SDiff(M) so that the geodesic and the one-parameter subgroup, i.e., the Riemannian exp and the Lie exp are the same. Is this true or there is a misunderstanding somewhere? thx.

Also a geodesic is a flow of the time-depedent Eulerian velocity field, which can just be regarded as part of a vector field on $SDiff\(M\)$(at least along the geodesic). why is the geodesic not a one-parameter subgroup(of course not necessary be the Lie exp)? There must be something wrong with my understanding. Please help me.

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Any smooth curve in $SDiff$ has a right logarithmic derivative which is a smooth curve in the Lie algebra, by $\partial_t \phi(t,x) = X(t,\phi(t,x))$, or $X(t) = \partial_t\phi(t)\circ\phi(t)^{-1}$, and conversely. This $X(t,x)$ is subject to the Euler equation. If and only if $X$ is constant in $t$, time independent, you get a 1-parameter subgroup. See sections 2,3,4 of (here).

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  • $\begingroup$ Thank a lot. But I still have a question. If I understand it in the right way, by $X\(t,x\)$ you mean the right logarithmic derivative, which is the pull-back of the velocity field of the geodesic by $\phy(t)$. Is it just the Lagrangian velocity field? I think I was wrong because I thought it as a constant velocity field $\endgroup$
    – user31017
    Feb 10 '13 at 17:11
  • $\begingroup$ No, it is not the pullback! $\partial_t \phi(t,x)$ is a time-dependent vector field along $\ph(t)$, i.e., a tangent vector in $T_\phi(t) SDiff$. It is not a vector field on $M$. $X(t,x)$ is the just the right translation back to $T_{Id}SDiff$ which is the Lie algebra of divergence free vector fields. Have a look at the source I cited. It is easier to understand on an abstract group. $\endgroup$ Feb 11 '13 at 14:02

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