# How to prove Liouville measure is invariant under geodesic flow?

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?

• The measure lives on the sphere bundle, the flow acts on the sphere bundle, so the meaning of invariance should be clear. – user1688 Dec 8 '15 at 11:07
• 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. – Sebastian Goette Dec 8 '15 at 11:22
• @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? – oneyear Dec 8 '15 at 13:07

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.