How does topology work when taking charts on a Psuedo-Riemannian manifold? I'll first explain why I think taking charts is sane when working with Riemannian manifolds, and then show what I believe breaks down in the Pseudo-Riemannian case with a particular choice of a Pseudo Riemannian manifold (Minkowski space). I'd like to understand where I am going wrong.
A Riemannian manifold is a differentiable manifold $M$ equipped with a positive definite inner product $d: T_p M \times T_p M \rightarrow \mathbb R$. Let us concentrate attention on some chart $(U \subseteq M, \phi : U \rightarrow \mathbb R^n)$. Here $\phi$ is a homeomorphism, hence we can "push forward" $d$ along $\phi$ to get some inner product structure on $\mathbb R^n$: $d^\star: \mathbb R^n \times \mathbb R^n \rightarrow \mathbb R$. Now since this $d^\star$ is an inner product structure, it induces a metric, which induces a topology on $\mathbb R^n$. However (and this is the saving grace), due to equivalene of norm in a finite dimensional vector space, the topology induced by $d^\star$ will match the 'usual topology' on $\mathbb R^n$. So the differential calculus that we do (which depends on having limits) cannot see the difference between $d^\star$ and the regular topology, and thus we can just do 'calculus on $\mathbb R^n$' and it transfers.
Now let us look at the contrast in the Pseudo-Riemannian case.
Let us assume we have Minkowski space, which is $\mathbb M \equiv (\mathbb R^4, d')$ where the manifold structure on $M \equiv \mathbb R^4$ is the 'stupid chart': we have a single chart $\phi: M \rightarrow \mathbb R^4; \phi(x) = x$. Now, we take the bilinear form to be $d': T_p \mathbb M \times T_p \mathbb M \rightarrow \mathbb R$ as given by $d'(\mathbf p, \mathbf q) \equiv - p_0 q_0 + p_1 q_1 + p_2 q_2 + p_3 q_3$. This is no longer positive definite! Nor is it an inner product, and this cannot even induce a norm.
However, intuitively, the way $d'$ sees space is very different from the way the usual topology sees space. For example, the distance between the points $\mathbf p =(t, x, 0, 0)$ and $ \mathbf q = (x, t, 0, 0)$ is $0$ according to $d'$ but $\sqrt{2xt}$ according to the Euclidian distance. So, how is it legal for us to do things like take limits inside minkowski space? We seem to have two choices:

*

*Claim that we treat $d'$ as simply some bilinear form, while still obeying the topology of $\mathbb R^4$. This seems really weird to me, because now the structure of the topology is no longer 'intrinsic' to the manifold + bilinear form. It is rather induced by the chart into $\mathbb R^n$

*I am going wrong somewhere in my explanation above, and I'd love to know where.

 A: First of all, even the Riemannian case is more subtle than you seem to indicate.
Suppose $M$ is a differentiable manifold equipped with a Riemannian metric. On each tangent space $T_p M$ I have a positive definite inner product which I'll denote $\langle v,w \rangle_p$, $v,w \in T_p M$. From this I obtain a norm $\|v\|_p = \sqrt{\langle v,v \rangle_p}$, for $p \in T_p M$. And using this norm, I obtain a metric on $T_p M$, $d_p(v,w) = \|v-w\|_p$, for $v,w \in T_p M$.
But, how do I obtain a metric on $M$ itself which induces the given manifold topology on $M$? Given $p,q \in M$, how do I define $d(p,q)$?
Well, first you need a hypothesis, namely that $M$ is path connected; without that, there's not canonical way to define the metric. Assuming path connectivity, you can then define the length of a smooth path $\gamma : [0,1] \to M$ from $p=\gamma(0)$ to $q=\gamma(1)$, namely
$$\text{Length}(\gamma) = \int_0^1 \|\gamma'(t)\|_{\gamma(t)} dt
$$
But there are many paths, and so now you have to take an infimum:
$$d(p,q) = \inf_\gamma \bigl(\text{Length}(\gamma) \bigr)
$$
where the infimum is taken over all smooth paths from $p$ to $q$.
With all of that, it is now possible to prove that the metric topology determined by $d$ is the same as the given manifold topology on $M$; the key concept for doing this is geodesic normal coordinates.
So, what happens in a general Lorentzian metric?
Well, you're right. It doesn't work the way it does with Riemannian metrics, there isn't any way to use the Lorentzian metric to define a metric topology equivalent to the given manifold topology. You're stuck with the topology defined by the given atlas of manifold charts.
But something else interesting happens, which a general relativist could explain better than me, but I'll give you the outline.
In general relativity, on a manifold $M$ equipped with a Lorentzian metric that is used as a mathematical model of space-time, one considers proper time as a measurement of elapsed time of a clock travelling along a timelike world line through $M$, and one considers proper length of a physical object which extends along a spacelike world line through $M$. Both of these can be regarded as different kinds of path integrals of the "norm" of the Lorentzian metric. And one can consider extrema of such path integrals, which in the time-like case defines the "quickest" world line from one space-time event to another.
But in fact there is no metric topology that is induced on $M$ by a Lorentzian metric. One way to see this is that when you integrate along a light-like world line, you always get zero. Must be interesting to be a light particle! Always going nowhere fast.
