Geodesic flow are generated by Hamiltonian vector field

Question

Let $(M,g)$ be a Riemannian manifold and consider the Hamiltonian
\begin{equation*}
\begin{array}{rcl}
H:T^*M & \rightarrow&\mathbb{R} \\
(q,p) & \mapsto&H(q,p):=\frac{1}{2}g^{ij}(q)p_ip_j.
\end{array}
\end{equation*}
We would like to show that the Hamiltonian vector field $X_H:T^*M\rightarrow TT^*M$ generates the geodesic flow.
We can explicitely compute $X_H$, since its components satisfies Hamilton's equations:
\begin{equation*}
X_H=\dfrac{\partial H}{\partial p_j}\dfrac{\partial}{\partial q^j}-\dfrac{\partial H}{\partial q^j}\dfrac{\partial}{\partial p_k}=g^{ij}p_i\dfrac{\partial}{\partial q^j}-\frac{1}{2}\dfrac{\partial g^{ij}}{\partial q^k}p_ip_j\dfrac{\partial}{\partial p_k}.
\end{equation*}
We can see that the following vector field $Y:TM\rightarrow TTM$, over $TM$ generates the geodesic flow:
\begin{equation*}
Y(q,\dot{q}):=\dot{q}^i\dfrac{\partial}{\partial q^i}-\Gamma^j_{lk}\dot{q}^l\dot{q}^k\dfrac{\partial}{\partial\dot{q}^j}.
\end{equation*}
Indeed, we show that its integral curves are geodesics.
Let $\gamma:\mathcal{I}\rightarrow TM$, with $\gamma(t):=(\alpha(t),\beta(t))$, then
\begin{equation*}
\dot{\gamma}(t)=\dot{\alpha}^i(t)\dfrac{\partial}{\partial q^i}+\dot{\beta}^j(t)\dfrac{\partial}{\partial\dot{q}^j},
\end{equation*}
so $Y(\gamma(t))=\dot{\gamma}(t)$ $\iff$ the following hold
\begin{equation*}
\left\{\begin{array}{ll}
\dot{\alpha}^i(t)=\beta^i(t)& \\
\dot{\beta}^j(t)=-\Gamma^j_{lk}\beta^l(t)\beta^k(t)&
\end{array}\right.
\end{equation*}
This is the geodesic equation and so this proves that the integral curves of $Y$ are geodesics.
The problem is that $Y$ is a vector field over $TM$, while in the Hamiltonian formalism, $X_H$ is a vector field over $T^*M$. How are $X_H$ and $Y$ related? How can we show that $X_H$ generates the geodesic flow as $Y$ does?
Thank you all.
P.S. It is enough to show that the integral curves of $X_H$ satisfies the geodesic equation in the cotangent bundle? But then, what does it look like in the cotangent bundle?

Answer 1

Throughout I'll use $x=x^1,\dots,x^n$ as local coordinates on $M$, $(x,v)$ as the induced local coordinates on $TM$, and $(x,p)$ as the induced local coordinates on $T^*M$ (with lower indices for $p_i$ and all others upper).

Your question is essentially asking about the correspondence between the Hamiltonian and Lagrangian formulations of the geodesic equations. For a general second order ODE, the correspondence is as follows: Given a Lagrangian $L:TM\to\mathbb{R}$, there is associated to $L$ a Legendre transformation $F_L:TM\to T^*M$ given by $$ F_L(x,v)=\left(x,\frac{\partial L}{\partial v}\right) $$ Provided $F_L$ is an isomorphism, we can rewrite the entire problem on the cotangent bundle in terms of a Hamiltonian $H(x,p)=p_idx^i(F_L^{-1}(x,p))-L(F_L^{-1}(x,p))$. After the rewriting, the Euler-Lagrange equations correspond to Hamilton's equations. Writing both in terms of vector fields, the Euler-Lagrange vector field $X_L$ is $F_L$-related to the Hamiltonian vector field $X_H$, i.e. $X_H=dF_LX_L$.

In the case of the geodesic equation, the Lagrangian is $L(x,v)=\frac{1}{2}g_{ij}(x)v^iv^j$, so the Legendre transformation $F_L$ is just the musical isomorphism $\flat:TM\to T^*M$, given in coordinates by $\flat(x,v)=(x,g_{ij}(x)v^j)$. Its inverse is given by $\sharp(x,p)=(x,g^{ij}(x)p_j)$, and its differential is $$ d\flat\left(\frac{\partial}{\partial x^i}\right)=\frac{\partial}{\partial x^i}+\frac{\partial g_{jk}}{\partial x^i}v^k\frac{\partial}{\partial p_j}\ \ \ \ \ \ \ d\flat\left(\frac{\partial}{\partial v^i}\right)=g_{ij}\frac{\partial}{\partial p_j} $$ You've already written down the two vector fields in coordinates, so verifying that $d\flat X_L=X_H$ is just a computation.