Editing Riemannian manifold (section)

== Metric space structure ==
An ''admissible curve'' is a piecewise smooth curve <math>\gamma : [0,1] \to M</math> whose velocity <math>\gamma'(t) \in T_{\gamma(t)}M</math> is nonzero everywhere it is defined. The nonnegative function <math>t\mapsto\|\gamma'(t)\|_{\gamma(t)}</math> is defined on the interval <math>[0,1]</math> except for at finitely many points. The length <math>L(\gamma)</math> of an admissible curve <math>\gamma : [0,1] \to M</math> is defined as
: <math>L(\gamma)=\int_0^1 \|\gamma'(t)\|_{\gamma(t)} \, dt.</math>
The integrand is bounded and continuous except at finitely many points, so it is integrable. For ''<math>(M,g)</math>'' a connected Riemannian manifold, define <math>d_g:M\times M\to[0,\infty)</math> by
: <math>d_g(p,q) = \inf \{ L(\gamma) : \gamma \text{ an admissible curve with } \gamma(0) = p, \gamma(1) = q \}.</math>
'''Theorem:''' <math>(M,d_g)</math> is a [[metric space]], and the [[metric topology]] on <math>(M,d_g)</math> coincides with the topology on <math>M</math>.{{sfn|Lee|2018|p=39}}

{{Collapse top|title=Proof sketch that <math>(M,d_g)</math> is a metric space, and the metric topology on <math>(M,d_g)</math> agrees with the topology on <math>M</math>}}
In verifying that <math>(M,d_g)</math> satisfies all of the [[Metric space#Definition|axioms of a metric space]], the most difficult part is checking that <math>p\neq q</math> implies <math>d_g(p,q)>0</math>. Verification of the other metric space axioms is omitted.

There must be some precompact open set around ''p'' which every curve from ''p'' to ''q'' must escape. By selecting this open set to be contained in a coordinate chart, one can reduce the claim to the well-known fact that, in Euclidean geometry, the shortest curve between two points is a line. In particular, as seen by the Euclidean geometry of a coordinate chart around ''p'', any curve from ''p'' to ''q'' must first pass though a certain "inner radius." The assumed continuity of the Riemannian metric ''g'' only allows this "coordinate chart geometry" to distort the "true geometry" by some bounded factor.

To be precise, let <math>(U,x)</math> be a smooth coordinate chart with <math>x(p)=0</math> and <math>q\notin U.</math> Let <math>V\ni x</math> be an open subset of <math>U</math> with <math>\overline{V}\subset U.</math> By continuity of <math>g</math> and compactness of <math>\overline{V},</math> there is a positive number <math>\lambda</math> such that <math>g(X,X)\geq\lambda\|X\|^2</math> for any <math>r\in V</math> and any <math>X\in T_rM,</math> where <math>\|\cdot\|</math> denotes the Euclidean norm induced by the local coordinates. Let ''R'' denote <math>\sup\{r>0:B_r(0)\subset x(V)\}</math>.
Now, given any admissible curve <math>\gamma:[0,1]\to M</math> from ''p'' to ''q'', there must be some minimal <math>\delta>0</math> such that <math>\gamma(\delta)\notin V;</math> clearly <math>\gamma(\delta)\in\partial V.</math>

The length of <math>\gamma</math> is at least as large as the restriction of <math>\gamma</math> to <math>[0,\delta].</math> So
: <math>L(\gamma)\geq\sqrt{\lambda}\int_0^\delta\|\gamma'(t)\|\,dt.</math>
The integral which appears here represents the Euclidean length of a curve from 0 to <math>x(\partial V)\subset\mathbb{R}^n</math>, and so it is greater than or equal to ''R''. So we conclude <math>L(\gamma)\geq\sqrt{\lambda}R.</math>

The observation about comparison between lengths measured by ''g'' and Euclidean lengths measured in a smooth coordinate chart, also verifies that the metric space topology of <math>(M,d_g)</math> coincides with the original topological space structure of <math>M</math>. 
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Although the length of a curve is given by an explicit formula, it is generally impossible to write out the distance function <math>d_g</math> by any explicit means. In fact, if <math>M</math> is compact, there always exist points where <math>d_g:M\times M\to\mathbb{R}</math> is non-differentiable, and it can be remarkably difficult to even determine the location or nature of these points, even in seemingly simple cases such as when <math>(M,g)</math> is an ellipsoid.{{citation needed|date=July 2024}}

If one works with Riemannian metrics that are merely continuous but possibly not smooth, the length of an admissible curve and the Riemannian distance function are defined exactly the same, and, as before, <math>(M,d_g)</math> is a [[metric space]] and the [[metric topology]] on <math>(M,d_g)</math> coincides with the topology on <math>M</math>.{{sfn|Burtscher|2015|p=276}}

=== Diameter ===
The [[Diameter of a metric space|''diameter'']] of the metric space <math>(M,d_g)</math> is
: <math>\operatorname{diam}(M,d_g)=\sup\{d_g(p,q):p,q\in M\}.</math>
The [[Hopf–Rinow theorem]] shows that if <math>(M,d_g)</math> is [[Complete metric space|complete]] and has finite diameter, it is compact. Conversely, if <math>(M,d_g)</math> is compact, then the function <math>d_g:M\times M\to\mathbb{R}</math> has a maximum, since it is a continuous function on a compact metric space. This proves the following.
: If <math>(M,d_g)</math> is complete, then it is compact if and only if it has finite diameter.
This is not the case without the completeness assumption; for counterexamples one could consider any open bounded subset of a Euclidean space with the standard Riemannian metric. It is also not true that ''any'' complete metric space of finite diameter must be compact; it matters that the metric space came from a Riemannian manifold.