The intrinsic metric $ d_i(x,y) $ on a Riemannian manifold is the infimum over all curves joining $ x $ and $ y $, of the arc length of these curves.
For a surface $ S $ embedded in $ \mathbb{R}^3 $, we also have the usual Euclidean metric $ d_e $ induced on $ S $.
It's clear that $ d_e(x,y) \leq d_i(x,y) $ so that if $ p_n $ converges to $ p $ in the intrinsic distance, we get convergence in the Euclidean distance.
It's an exercise in do Carmo to prove that these metrics induce the same topology, but I can't see how to get the reverse implication; why does convergence in Euclidean distance imply convergence in intrinsic distance?
I know that length $ \alpha = \int_0^1 {\alpha'(t)} \; dt $ but I can't seem to relate the norm of $ p_n-p $ to the norm of $ \alpha'(t) $ where $ \alpha $ is a curve joining $ p_n $ to $ p $.
Best Answer
Recall that any smooth manifold is already a topological space, even before it is endowed with a Riemannian metric. So what we actually need to show is that the metric space structure induced by a Riemannian metric is compatible with the initial topology. This can be shown using exponential coordinate charts, as follows.
Let $(M,g)$ be a Riemannian manifold, and let $p\in M$. For a small enough $\delta$, the metric open ball $B_\delta(p)$ is the image of the appropriate exponential parametrization. This parametrization is a diffeomorphism, and hence the above ball is an open set in the initial topology.
Conversely, let $U$ be an open neighborhood of $p$. We want to show that there exists a metric ball around $p$ which is contained in $U$. For that, we may assume that $U$ is contained in the image of an exponential parametrization around $p$,$$\exp_p:B_\delta(0)\subset T_pM\to M$$ (otherwise, just take the intersection of $U$ with such an image). Set$$V:=\exp_p^{-1}(U).$$Then $V$ is an open neighborhood of $0$ in a Euclidean space, and so it contains some metric ball $B$ around $0$. Since $\exp_p$ is a radial isometry, that is, for every $x\in B_\delta(0)\subset T_pM$ we have$$d(p,\exp_p(x))=|x|,$$the metric ball $B$ is carried by $\exp_p$ to a metric ball in $M$ which is contained in $U$. We're done.