If submanifold $P$ is geodesically complete then $P=M$, for connected $M$.

Question

Let $P$ be an open submanifold of a semi-Riemannian manifold $M$. If $P$ is geodesically complete and $M$ is connected then $P=M$.

My attempted proof:

Define the following metric on $M$, $d(x,y)=\inf\{L(\gamma):\text{smallest lenght}\}$

Let $t_n$ define a cauchy sequnce in $\mathbb{R}$ such that for $m,n>N$, for $\epsilon>0$

$|t_n-t_m|<\epsilon$

Then:

$$d(\gamma(t_n),\gamma(t_m))\leqslant|t_n-t_m|<\epsilon$$

so that $\gamma(t_n)$ deintes a cauchy sequnce in $P$

Since $\lim_{n\to\infty} t_n=t$, because $P$ is geodesically complete then:

$$\lim_{n\to\infty}\gamma(t_n)=\gamma(t)$$

where $\gamma(t)\in P$

Hence, $P$ is complete. Once it is complete, it must be closed in the subset topology in $M$. However, by hypothesis it is open. Since $M$ is connected either $P=\emptyset$ or $P=M$. Since $P\neq\emptyset$, then $P=M$.

Question:

Is this the correct approach? Or is there a way to do it without exploring the metric properties of the manifold?

Thanks in advance!

Answer 1

Here are the steps of the proof.

1. Assume that $P\ne M$ and let $x\in M\setminus P$ be a boundary point of $P$ in $M$.

2. Pick a point $y\in P$ such that there exists a geodesic $\gamma: [0,1]\to M$ connecting $y$ to $x$. (You should figure out why does such $y$ exists.)

3. Consider a maximal open interval $I\subset [0,1]$ such that $\gamma(I)\subset P$. Using geodesic completeness of $P$ conclude that the boundary points of $I$ are mapped by $\gamma$ to $P$.

4. Obtain a contradiction with $x\notin P$ and finish the proof.