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I'm trying to understand what my teacher wrote.

We build a vector field $\cal{X}$ on $TM$ for which the flow, called geodesic flow, consists in the curves $(\gamma(t),\dot{\gamma}(t))$ where $\gamma$ is a geodesic $${\cal{X}}_X=(X^1,\cdots,X^n,-\Gamma^1_{ij}X^iX^j,\cdots,-\Gamma^n_{ij}X^iX^j)$$ Then $$\frac{d\gamma^k}{dt} = X^k\\ \frac{dX^k}{dt} = -\Gamma^k_{ij}X^iX^j$$ is the system of equations of the flow of $\cal{X}$.

Why is that system the set of equations defining the flow of $\cal{X}$?

it is clear that the curves $(\gamma(t),\dot{\gamma}(t))$, with $\gamma$ a geodesic, satisfy the equations but how do we know that no other curve on $TM$ satisfies these equations

user405156
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1 Answers1

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Let $x^i$ and $v^j$ form a coordinate basis for $TM$, where $M$ is a Riemannian manifold with metric tensor $g$. Consider the vector field on $TM$ given by: $$\mathcal{X} = (v^1,\ldots,v^n,- \Gamma_{jk}^1v^jv^k,\ldots,-\Gamma_{jk}^nv^jv^k) = v^i \frac{\partial}{\partial x^i} - \Gamma_{jk}^iv^jv^k\frac{\partial}{\partial v^i} $$

(1) What is the flow of this field?

Well, the "flow" of any field is simply given by the dynamical system defined by following (or integrating along that field) for some time $t$. Consider the part of the flow on $M$ (i.e. look at the first $n$ components of $\mathcal{X}$, which determine how a point on the manifold $M$ moves. Consider a curve $\gamma:\mathbb{R}\rightarrow M$, following this part of the field: at every point, the tangent vector to the curve is given by $v$. Thus: $$ \frac{\partial \gamma^i}{\partial t} = v^i $$ But then the tangent vector of the curve also needs to follow the flow prescribed by the second half of the field (i.e. the part in $T_pM$). This constrains the rate of change of the tangent vector to $\gamma$, i.e. $\partial_t \gamma^i$. But we just saw this is equal to $v^i$. So we are constraining the tangent vector to $v^i$ over time. The rate of change of $\partial_t \gamma^i$ must follow the last $n$ components of $\mathcal{X}$. Thus: $$ \frac{\partial^2 \gamma^i}{\partial t^2} = \frac{\partial v^i}{\partial t} = - \Gamma_{jk}^iv^jv^k $$

(2) How does show the geodesic is unique?

As the comment suggests, use the Picard-Lindelof theorem. You have an ODE $\dot{v}^i=- \Gamma_{jk}^iv^jv^k$, which assuming a little smoothness can be uniquely solved. Then you have another ODE $\dot{\gamma}^i=v^i$ that may be uniquely solved for the same reason.

Related: here and here. (Also, sort of related: here)

Aside: we can see why this generates geodesics by recalling the classical geodesic equation: $$ \underbrace{ \frac{\partial^2\gamma^i}{\partial t^2} }_{\displaystyle \dot{v}^i} + \Gamma_{\mu\eta}^i \underbrace{ \frac{\partial \gamma^\mu}{\partial t} }_{\displaystyle {v}^\mu} \underbrace{ \frac{\partial \gamma^\eta}{\partial t}}_{\displaystyle {v}^\eta} = 0 $$ $$ \dot{v}^i = \frac{\partial v^i}{\partial t}= -\Gamma_{\mu\eta}^i{v}^\eta{v}^\mu $$

user3658307
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