TITLE: Why is there no natural metric on manifolds?
QUESTION [31 upvotes]: One of the things that always bothered me after learning introductory differential geometry (as a physics student) and then delving deeper into this field on my own is that, the usual construction of differential manifolds are such that $M$ is an $n$-dimensional manifold if it is 1) a Hausdorff-space, 2) it is locally homeomorphic to $\mathbb{R}^{n}$, 3) the transistion functions between charts are invertable and continuous/differentiable/analytic etc. depending on manifold type.
Now as far as I understand, norm, metric, etc. are topological properties and homeomorphisms are maps that preserve topological structure. Yet, metric is an extra structure on a manifold, not part of a manifold's given properties. But if manifolds are locally homeomorphic to a set that has a very strong topological structure, including a metric ($\mathbb{R}^{n}$) then why is that structure not preserved locally?
Secondly, which is probably related to this, if the transition functions are differentiable, as it is demanded for differential manifolds, does that mean that the coordinate maps themselves are also differentiable?
Because then they are basically diffeomorphisms from $M$ to $\mathbb{R}^{n}$. And if it is so, then given the natural metric tensor $g$ on $\mathbb{R}^{n}$, why can't that metric be induced on the manifold via the pullback?

REPLY [2 votes]: Yes, in general, there is no "god-given" (canonical, natural,...) Riemannian metric on a smooth manifold. However, sometimes, there is one! Much of the research in the modern differential geometry is about finding such canonical metrics. For instance, this is how Poincare conjecture was solved in dimension 3 by Perelman. See my answer here for details and references.