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Let $M$ be a complete n dimensional Riemannian manifold. $vol$ denotes the n dimensional Hausdorff measure. Let $$ SM=\{(x,v)|x\in M, v\in T_xM, \|v\|=1\} $$ be the unit tangent bundle of $M$. Then $SM$ will be equipped with the Liouville measure $\nu$. Given a subset $A=(U,A_x)\subset SM$, where $U\subset M$ is a subset of $M$, $A_x$ is a subset of the unit sphere of the tangent space at $x\in U$, $\nu$ is defined by $$ v(A)=\int_U \int_{A_x} dS^{n-1} dvol(x) $$ where $dS^{n-1}$ is the usual Lebesgue measure on the unit sphere.

Then $\nu$ is invariant under the geodesic flow on $SM$. By the comments below, I know what it means: Let $y=(x,v)\in A$, set $\gamma_y(s)=\exp_x(sv)$, then the geodesic flow is defined by $$ \Phi_t(y)=(\gamma_y(t),\dot{\gamma}_y(t)) $$ And $\Phi_t(A)=\{\Phi_t(y)|y \in A \}$. We have $\nu(\phi_t(A))=\nu(A)$.

Can you give a direct proof without introducing cotangent bundle, 1-form, 2-form?

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  • $\begingroup$ The measure lives on the sphere bundle, the flow acts on the sphere bundle, so the meaning of invariance should be clear. $\endgroup$ – user1688 Dec 8 '15 at 11:07
  • $\begingroup$ For $y\in SM$ let $\gamma_y(t)=\exp(ty)\in M$ be the unique geodesic with initial velocity $y$. Then $\Phi_t(y)=\dot\gamma_y(t)\in SM$ defines the geodesic flow. In particular, both $\Omega\subset SM$ and $\Phi_t(\Omega)\subset SM$. You should find a proof of the invariance of the Liouville measure in any good textbook on Riemannian geometry. $\endgroup$ – Sebastian Goette Dec 8 '15 at 11:22
  • $\begingroup$ @SebastianGoette: Thank you for your comments, I correct my mistake. I have read several textbooks on Riemannian geometry, but I have not find this theorem. So can you show where I can find a direct proof? Or can you give one? $\endgroup$ – oneyear Dec 8 '15 at 13:07
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A hands down proof not using the theory of Hamiltonian systems can be done by just proving that the Jacobian determinant of the transformation is zero.

We have $$ TSM \cong \pi^* TM \oplus VSM,$$ where $VSM$ is the vertical distribution and $\pi: SM \longrightarrow M$ is the canonical projection. The isomorphism is given by the metric, which selects a horizontal subspace.

Now the geodesic flow is the flow generated by the vector field $X$ which is given by $$ X(x, v) = \begin{pmatrix} v \\ 0 \end{pmatrix}$$ in this splitting. Therefore, differentiating the defining equation $$ \dot{\Phi}_t = X(\Phi_t), ~~~~~~ \Phi_0 = \mathrm{id}$$ with respect to some metric connection gives $$ \frac{\nabla}{\mathrm{d} t} d \Phi_t = \nabla X|_{\Phi_t} \cdot d \Phi_t, ~~~~~ d\Phi_0 = \mathrm{id}.$$ For the determinant, we obtain $$ \frac{\mathrm{d}}{\mathrm{d} t} \det(d\Phi_t) = \mathrm{tr}(\nabla X)\cdot \det(d\Phi_t), ~~~~~~ \det(d\Phi_0) = 1$$ Now $VSM$ carries a natural connection, and $\pi^*TM$ carries the pullback connection. The direct sum of these connections is metric (even though it is not the Levi-Civita connection of $SM$), and $\nabla X$ is given in the splitting by $$\nabla X = \begin{pmatrix} 0 & \iota \\ 0 & 0 \end{pmatrix},$$ where $\iota$ is the inclusion of $VSM$ into $\pi^*TM$. Hence its trace is zero, and the determinant remains one for all time.

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Arnold's book "mathematical methods of classical mechanics" in section 16 proves Liouville's theorem that phase flow preserves the volume. Note that the geodesic flow is always a Hamiltonian flow but the converse is an interesting question; see When is the time evolution of a Hamiltonian system described by the geodesic flow on a Riemannian manifold?

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  • $\begingroup$ Arnold's proof is for the symplectic volume form on the cotangent bundle, not the induced volume form on the unit sphere bundle. Note that the volume form on the unit sphere bundle is not an invariant of the contact structure on the unit sphere bundle, so the result is not an immediate consequence of Arnold's proof. $\endgroup$ – Ben McKay Nov 12 '17 at 11:14
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    $\begingroup$ Thanks, @BenMcKay. I haven't thought about this recently and it would take too much time to correct my answer. If you think it is misleading I will delete it. Another possibility would be for you to correct it so as to remove inaccuracies since you seem to be on top of the material currently. $\endgroup$ – Mikhail Katz Nov 12 '17 at 11:17

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