Proof of the uniqueness of geodesics.

Question

In my notes, there is a proof of the fact that if $c_1,c_2:I \to M$ are geodesics from an interval $I \subset \mathbb{R}$ to a smooth manifold $M$, presumably semi-riemannian, where $c_1(a) = c_2(a)$ and $c_1'(a) = c_2'(a)$, for some $a \in \mathbb{R}$, then $c_1 = c_2$.

Now, I read somewhere else (Christian Bär:s notes) that this "easily" follows from the fact that the geodesic equation is a second-order non-linear ordinary differential equations, hence we can use Picard-Lindelöf, so that by uniqueness of solution to the IVP, $c_1 = c_2$.

Here is the proof in my notes:

Proof: Let $$J := \{t \in I: c_1(t) = c_2(t), \quad c_1'(t) = c_2'(t)\}$$.

1. $J \neq \emptyset$, since $a \in J$.
2. J is closed since it is defined by a continuous equation.
3. J is open; For each $b \in J$, lemma 5.2.2 gives us that there exists an $\epsilon > 0$ such that $c_1(t) = c_2(t)$ for $t \in (b-\epsilon,b+\epsilon)$. This means that $c_1'(t) = c_2'(t)$ so that $(b-\epsilon,b+\epsilon) \subset J \subset I$ hence $J$ is open.

Now, I sort of get the general idea. We want to show that $J$ is non-empty, and both open and closed in $I$, since $I$ is an interval in $\mathbb{R}$, it follows that $I$ is connected, hence the only sets that are both open and closed are $\emptyset,I$. By implication, since $J$ is not empty, we must have $I = J$.

My question relates to $2)$. What does he mean by "J is closed since it defined by a continuous equation"? I know that a continuous function $f:X \to Y$ between topological spaces $X,Y$ is such that if $A \subset Y$ is closed, then $f^{-1}(A)$ is closed. Is that what he is referring to?

Also, I am not entirely sure how to think about the geodesic equation as a system of second-order non-linear ordinary differential equations. How shall I think of $$(c^k)''+(c^i)'(c^j)'\Gamma^k_{ij} = 0 \quad (1 \leq k \leq n)?$$

Generally, a second-order differential equation is on the form $f(t,x,x',x'')$, but here we have $(c^k)'',(c^i)',(c^j)'$ and also a variable coefficient $\Gamma^k_{ij}:U \to \mathbb{R}$ that is locally defined.

Note that $c^j = (x^1 \circ c)$ for a chart $(U, \varphi = (x^1,\ldots,x^n))$.

Answer 1

If $f,g:X\to Y$ are continuous maps between topological spaces, with $Y$ being Hausdorff, then the set of points of equality $E_{f,g}:=\{x\in X\,:f(x)=g(x)\}$ is closed.

To prove this, consider the product map $f\times g:X\to Y\times Y$, $x\mapsto (f(x),g(x))$. Also, let $\Delta_Y=\{(y,y)\,:y\in Y\}$ be the diagonal in $Y\times Y$. Then, we have $E_{f,g}=(f\times g)^{-1}(\Delta_Y)$. Now, if we equip $Y\times Y$ with the product topology, then we have that $f\times g$ is continuous, and also $\Delta_Y$ is closed if and only if $Y$ is Hausforff (a standard fact). Since we are indeed assuming $Y$ is Hausdorff, we see that $E_{f,g}$ is the preimage of a closed set under a continuous map, hence closed.

In the case where $Y$ is a normed vector space (like $\Bbb{R}$) you can offer a less abstract proof by noting that $E_{f,g}=\{x\in X\,:\, f(x)-g(x)=0\}$, i.e it is the zero-level set of the difference $f-g$ (which is continuous). Note that $\{0\}$, a singleton, is indeed a closed set. So, $E_{f,g}$ is the preimage of a closed set under a continuous map, hence continuous.

In your case, you have two equality conditions, so you can think of it as the intersection of two such sets (defined by $c_1,c_2$ and $c_1’,c_2’$). Or you can think of it as the equality set of the maps $C_1(t)=(c_1(t),c_1’(t))$ and $C_2(t)=(c_2(t),c_2’(t))$.

---

For the ODEs, you just need a sufficiently general version of the existence and uniqueness theorem. To write it in a slightly recognizable format, let me abuse notation slightly and think of ‘already applying the chart’, meaning I’ll consider the Christoffel symbols $\Gamma^i_{jk}\circ\phi^{-1}$ as a mapping $U’:=\phi[U]\subset\Bbb{R}^n\to\Bbb{R}$. But, for ease of typing, I’ll simply write $\Gamma^i_{jk}$ instead.

Consider the function $f:U’\times\Bbb{R}^n\to\Bbb{R}^n$ defined as \begin{align} f(x,v):=\left(-\Gamma^1_{jk}(x)v^jv^k, \dots, -\Gamma^n_{jk}(x)v^jv^k \right). \end{align} This is a smooth function, so the existence and uniqueness theorems (we’re dealing with the autonomous case here) tell us that for any $x_0\in U’$ and $v_0\in\Bbb{R}^n$, there is a unique smooth curve $\alpha:I\to U’$ (where $I$ is some open interval around the origin in $\Bbb{R}$) such that for all $t\in I$, we have $\alpha’’(t)=f(\alpha(t),\alpha’(t)) $ and $\alpha(0)=x_0$ and $\alpha’(0)=v_0$. This is the desired (chart-representation of) geodesic.

At this point you may object that I’m dealing with second order ODEs, while the usual existence and uniqueness theorems are phrased for first order ODEs (eg on Wikipedia). However, the second order case follows trivially (since the theorem applies generally for vector-valued functions and curves). The trick for converting a second order system into a first order one is explained in any good ODE book.