A simple ODE on smooth manifolds

Question

For a Riemannian manifold $(M,g)$, the geodesic flow is $\phi_t:TM\to TM, (x,v)\mapsto (\gamma(t;x,v),\dot{\gamma}(t;x,v))$, where $\gamma(\cdot;x,v)$ is the geodesic started at $x$ with direction $v$. The related vector field of this flow can be formulated as $X(x,v)=(v,0)$.

My question can be viewed as the sham geodesic flow on manifolds without Riemannian metric. Namely, let consider the smooth manifold $TM$ and a vector field

$X:TM\to T(TM),(x,v)\mapsto(v,0)$.

By ODE we know this vector field can integrate to a (local) flow $\phi_t:TM\to TM$ (regardless any Riemannian metric). What is the meaning of this flow? Could we give an explicit formula of this flow?

In the simplese case $M=\mathbb{R}^d$, we know $\phi_t(x,v)=(x+t\cdot v,v)$ (indeed this is the geodesic flow with respect to the flat metric). I do not know if there is some similar form for flows on general manifolds.

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Edit: Bill pointed out that the problem is not well formulated. I realized that I misunderstood the canonical splitting into horizontal and vertical parts:

$T(TM)=H\oplus V$ where the horizontal subspace $H$ is the kernel of the connection map $K : T(TM)\to TM$ and the vertical subspace $V = \mathrm{ker}(d\pi)$ is tangent to the fibers of $\pi:TM\to M$.

So I think the problem is that the splitting does not make sense if we do not have some metric at hand. And the vector field can not be defined like that.

Answer 1

As you noticed, you need a connection $D$ on $TM$ in order to define the splitting in $T(TM)$ and the unique vector field $\xi$ with the property that $\xi$ is horizontal and satisfies $\pi_*(\xi_X)=X$ for all $X\in TM$. Then the flow $\phi_t$ of $\xi$ is the geodesic flow of the connection $D$.

To see this, recall that from the definition of the covariant derivative, a vector field $Y$ along a curve $c$ in $M$ satisfies $D_{\dot c}Y=0$ if and only if $Y$, wieved as a curve in $TM$, is horizontal (which means that $\dot Y$ is horizontal at each point wrt the above splitting).

Now, take any $X\in TM$ and denote by $\gamma_t:=\phi_t(X)$ and $c_t:=\pi(\gamma_t)$. Since $\dot\gamma=\xi$, the curve $\gamma$ is the horizontal. Moreover, $\dot c$, wieved as a curve in $TM$, is just $\gamma$: $$\dot c_t=\pi_*(\dot\gamma_t)=\pi_*(\xi_{\gamma_t})=\gamma_t.$$ By the above, $\dot c$ is $D$-parallel along $c$.