# A simple ODE on smooth manifolds

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.

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.

• You have not defined coordinates for $T(TM)$, so the question is not well-formulated. The whole point of the covariant derivative in Riemannian geometry is to define coordinates and defining a value for the second derivative of a curve as a vector. If you just use definitions from local coordinates, they will not be consistent when you change coordinates: a straight line in one coordinate system can look like any embedded smooth curve at all in another coordinate system. – Bill Thurston Jan 23 '11 at 0:48
• You might want to try math.stackexchange.com, as you're asking for context regarding basics of ordinary differential equations. – Ryan Budney Jan 23 '11 at 2:40

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 $c$.
• Thanks! Your description of $\xi$ is exactly what I am looking for! – Pengfei Jan 24 '11 at 11:56