How does the First Fundamental Form define a metric in $M$

differential-geometrymanifoldsmetric-spacesriemannian-geometrysurfaces

I am studying differential geometry and many professors I know say that the first fundamental form is a metric on the differential manifold (or regular surface) M, while the second fundamental form is a way to measure curvature. This two statments seem to be really natural for them but not for me, here is why:

The first fundamental form is a function that for each $p \in M$ associates a symmetric bilinear form $<,>_p$ on $T_pM$. So there is an induced metric on the space $T_pM$ but how does this help me find distances of to points $p,q$ in $M$? First I though that I could find the arc lenght of a curve $\alpha: I \rightarrow M$ that starts on $p$ and ends on $q$. But that would depend on the curve $\alpha$ (maybe I could choose it as a geodesic, although M is not necessary geodesically complete, is it?) and $<\alpha,\alpha>_p$ is not necessary equal to $<\alpha,\alpha>_q$.

I would really appreciate help here. Thanks in advance!

Best Answer

These are just two totally different meanings of the term "metric". In the context of "metric spaces" the word "metric" means one thing, and in the context of Riemannian manifolds it means a different thing (namely, a smoothly varying inner product on the tangent space at each point).

That said, a Riemannian metric on a connected manifold does actually induce a metric in the sense of a distance function. You define the distance between two points as the infimum of the lengths of all smooth curves between them. For a geodesically complete manifold, this infimum is realized by some geodesic, but in general it is not realized. (For instance, in $\mathbb{R}^2\setminus\{0\}$, if you have two points such that $0$ is on the straight line between them, you can connect them by paths that get arbitrarily close to the straight line distance but you cannot realize the straight line distance itself.)

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