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?
 A: 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?
A: 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.
