Is a topological space with a Minkowsi metric a topological manifold? The definition of a topological manifold from Wikipedia: tm
defines it as a topological space which locally looks like Euclidean space. But what about a topological space that uses the Minkowski metric from special relativity? It doesn't appear to be Euclidean locally, so does that mean it's not a topological manifold?
 A: You have to be careful about your definitions here.  In particular, "locally looks like Euclidean space" means locally homeomorphic to Euclidean space, not locally isometric.  In other words, the metric doesn't serve any role here except to induce a topological structure.
I'm not too familiar with the Minkowski "metric" (it's not actually a metric by the standard definition), but I don't think it's usually used to induce a topology.
If it was, it would be a really weird topology, since (as I understand it) there are non-zero vectors in Minkowski space that have a norm of zero.  This means your space would be non-Hausdorff, among other nasty properties, so sequences could converge to more than one limit, etc.  I think the Euclidean topology is usually used for these objects.
A: Let's see how Minkowski space is defined:
We take $\mathbb{R}^4$ as a topological manifold with the identity as a global chart. Every point has a tangent space with a canonical basis with respect to the identity map which we may denote as $(\partial_t, \partial_x, \partial_y, \partial_z)_p$ at the point $p$. We define a quadratic form $g$ on every tangent space that is represented with respect to this basis on every point by the matrix diag(-1, 1, 1, 1).
You see, in order to define the quadradic form $g$ we need the notion of tangent space, for which we need to know that the base space at hand is (at least) a finite dimensional real manifold. If we do not know this about the space at hand, we cannot define any $g$ as in the definition of the Minkonwski space above.
If you define a quadratic form $Q$ that is positive definite, you may use this form to define a function $d(x, y)$ for points $x, y$ by defining that this is the minimum of
$$
\int_{\gamma} \sqrt{Q(\gamma'(t), \gamma'(t))} d t
$$
over all differential paths $\gamma$ from x to y.
In order for this to make sense, we have to impose that our base space is actually a differential manifold, not just a topological manifold. If $Q$ is positive definite, you can prove that $d$ is a metric. If $Q$ is not positive definite, the function $d$ will in general not be a metric. 
An important theorem says that the metric defined above induces a topology on the differentiable manifold that agrees with the original topology. If the function $d$ is not a metric, you can still define the induced topology in the same way, by saying that all epsilon balls with respect to d are open, but this topology will in general not be the same as the original topology on the base space that we started with.
In the case of special and general relativity, the really important topology is the original one. I have never seen anyone talking about the topology that is induced by the "Minkowski metric", so if you are interested in relativity theory, you may shrug and move on :-)
Summary: The very definition of Minkowski space or - in more generality - a "Minkowski metric" has to presume that the topological space at hand is at least a finite dimenional real manifold. 
