In the book "Geometry and Topology for Physicists" by Nash and Sen, in Section 7.6, after showing that the structure group $GL(n,\mathbb{R})$ of a frame bundle $F(M)$ (for a general manifold $M$ of dimension $n$) is reducible to $O(n)$, they write
[…] we have shown that every manifold $M$ always admits a Riemannian metric.
1) I am a bit confused about the following proof, can anyone clarify it to me please ?
2) Moreover, I was thaught that one defines a manifold $M$ to be Riemannian if it admits a metric, i.e. if it has $O(n)$ as structure group.
I thought that in general manifold had $GL(n,\mathbb{R})$ as structure group (and so not necessarily Riemannian), but the statement above seems to claim that every manifold is Riemannian (by always admitting a metric).
What did I get wrong?
Best Answer
2) You are wrong in that admitting a certain structure is not equivalent to having (fixed choice of) certain structure. In other words, for any smooth manifold $M$ there is a Riemannian manifold $(M,g)$, such that if you forget the metric $g$ on $M'$, you will get just $M$. The catch is that the pair $(M,g)$ is not unique.
As a silly example, take the sphere $S^2$. You can embed it into $\mathbb{R}^3$ as the unit sphere to get one (induced) metric on it, or as a cube (smoothed), and get another induced metric. But as smooth manifolds (forgetting the metric), those are the same.
As a more interesting example, take the flat torus $\mathbb{R}^2/\mathbb{Z}^2$ with the induced metric from $\mathbb{R}^2$, and the torus which is the surface of a doughnut, embedded into $\mathbb{R}^3$. As smooth manifolds they are the same, but as Riemannian manifolds they differ; for example one can be (indeed is) embedded into $\mathbb{R}^3$, and the other one cannot.