I am studying the text by Barrett O’Neill referenced below. On page 130, O'Neill states, as Proposition 7, that every point in a semi-Riemannian Manifold has a convex neighborhood. Convex is defined as an open set being a normal neighborhood of each of its points. [Which, in particular, implies that it is geodesically convex.] Perhaps I am just missing something, but the proof does not seem adequate, and I do not know of any other source for this result. Any help would be appreciated.
Barrett O’Neill, Semi-Riemannian Geometry with Applications to General Relativity, Academic Press, New York, 1983.
I have gone through the proof in Lee, Introduction to Riemannian Manifolds, Second edition, of a similar result for Riemannian manifolds, but do not see a way to adapt that to the semi-Riemannian case.
The proof in O'Neill goes as follows: (In the sketch of the proof, I have changed the names of the points for fear "$o$" could be confusing.)
Let $p$ be a point in a semi-Riemannian manifold M, and let $\xi = (x^1, \dots ,x^n)$ be a normal coordinate system on a neighborhood $\mathscr{V}$ of $p$. For $q \in \mathscr{V}$, define $N(q) = \sum_{i} (x^i)^2$. For $\delta > 0$, define $\mathscr{V}(\delta) = \{q \in \mathscr{V} : N(q) < \delta\}$.
For sufficiently small $\delta$, $\mathscr{V}(\delta)$ is a neighborhood of $p$, diffeomorphic under $\xi$ to an open ball in $\mathbb{R}^n$. [I have not made any use of this statement below, so it might be a hint as to something I have missed.]
Let $\mathscr{D} \subseteq T(M)$ be the domain of the exponental map. Define $E:\mathscr{D} \rightarrow M \times M$ by $E(v) = ( \pi(v), exp(v) ).$ One can show that by choosing $\delta$ sufficiently small, $E$ is a diffeomorphism from an open set $\mathscr{W}$ of $T(M)$, containing $0 \in T_p(M)$, onto $\mathscr{V}(\delta) \times \mathscr{V}(\delta)$. Define the symmetric (0,2) tensor $B$ on $\mathscr{V}$ to have components $$B_{ij} = \delta_{ij} - \sum_k \Gamma^k_{ij} x^k.$$ [Do not worry about the fact that it is not very "tensorial", just notice that it is a smooth tensor field.] $B$ is positive definite at $p$, and by reducing $\delta$, if necessary, one can arrange that $B$ is positive definite on $\mathscr{V}(\delta)$. Setting $\mathscr{U}=\mathscr{V}(\delta)$, the proof intends to show that $\mathscr{U}$ is a normal neighborhood of each of its points. Let $q$ be an arbitrary point in $\mathscr{U}$ and define $\mathscr{W}_q = \mathscr{W} \cap T_q(M)$. By construction $E$ is a diffeomorphism from $\mathscr{W}$ to $\mathscr{U} \times \mathscr{U}$, so $\exp_q$ is a diffeomorphism from $\mathscr{W}_q$ to $\mathscr{U}$. To show that $\mathscr{U}$ is a normal neighborhood of $q$, what is needed in addition is to show that $\mathscr{W}_q$ is star-shaped about 0.
Let $v$ be an arbitrary tangent vector in $\mathscr{W}_q$. We need to show that $tv \in \mathscr{W}_q$ for $0 \le t \le 1$. This is true for $v=0$, so assume $v$ is not zero. Because $v \in \mathscr{D}$, the geodesic $\gamma_v$ starting at $q$ with initial velocity $v$ is defined on at least [0,1]. Let $r = \gamma_v(1) = \exp_q(v)$. It now suffices to prove that $\gamma_v([0,1])$ lies in $\mathscr{U}$. For simplicity let $\sigma = \gamma_v$ restricted to $[0,1]$. We therefore assume $\sigma$ leaves $\mathscr{U}$ and try to derive a contradiction. Assume that $\sigma$ lies in $\mathscr{V}$ so that $N \circ \sigma$ is defined. [I do not see how to show that, but please bracket that issue for the moment.] Because $\sigma$ leaves $\mathscr{U}$, $N \circ \sigma$ is at least $\delta$ at some time. Since $N(q) < \delta$, and $N(r) < \delta$, $N \circ \sigma$ achieves a maximum at some $t_0 \in (0,1)$, with $N(\sigma(t_0) \ge \delta$, and hence $\sigma(t_0) \notin \mathscr{U}$. Still assuming that $\sigma$ lies in $\mathscr{V}$, one can use the geodesic equation in these coordinates to show that $$ \frac{d^2 (N\circ\sigma)}{dt^2}(t_0)= 2 B(\sigma'(t_0),\sigma'(t_0)).$$ The proof then claims that the right hand side is positive, contradicting the assumption that there was a maximum at $t_0$. But we chose $\delta$ to make $B$ positive definite on $\mathscr{U}$, and $\sigma(t_0) \notin \mathscr{U}$. So I cannot see any reason for the right hand side to be positive.
If I am missing something, please point it out. Or if you know another source or proof, please let me know.
First pick a neighborhood $\mathscr{V}(\delta)$ on which $N$ is defined and $B$ is positive definite. Then ignore the rest of the manifold. As a result, all geodesics live in $\mathscr{V}(\delta)$. Now pick a $\delta^\prime < \delta$ on which $\pi(W) \subseteq \mathscr{V}(\delta^\prime)$ is such that $$E:W \rightarrow \mathscr{V}(\delta^\prime) \times \mathscr{V}(\delta^\prime)$$ is a differmorphism. O'Neill, does not make this very clear, but we can do this by setting $$W=E^{-1}(\mathscr{V}(\delta^\prime) \times \mathscr{V}(\delta^\prime))\;.$$ Now $N \circ \sigma$ is well defined for any geodesic (in fact for any point of the truncated manifold), and the problem is resolved. Since all these are open sets, the theorem holds when we bring back the rest of the manifold.
Hope this helps.