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Does the Hopf-Rinow theorem hold if the Riemannian manifold is not necessarily connected?

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$\bf{Motivation \ and \ Minor \ Details \ About \ Question:}$

I am reading a non-standard book which covers parts of Riemannian Geometry, and in the book the author quickly mentions the Hopf-Rinow theorem without any assumptions of connectedness of the Riemannian manifold. The author has defined the Riemannian distance function in a manner that works for manifolds that are not connected, where the distance function $d(x,y)$ takes the value $+ \infty$ if there does not exist any curve segment $\gamma$ connecting $x$ and $y$. Then $d$ forms an "extended metric" in that it satisfies all the properties of a regular metric but can take the value $+\infty$. It is proven in the book that the distance function still induces the manifold topology in this case.

I have seen the Hopf-Rinow theorem mentioned in Lee's book on Riemannian Geometry as well, but only in the setting of connected manifolds. Google so far has only given revealed other people stating the theorem in the connected setting as well. For the sake of generality, I would like to know if the theorem holds without assumptions of connectedness.

Thanks!

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The main reason the Hopf-Rinow theorem is generally stated only for connected Riemannian manifolds is that there is no canonical distance function on a disconnected Riemannian manifold. It is always possible to construct a distance function that restricts to the Riemannian distance on each component; but with some distance functions the Hopf-Rinow theorem is true, while with others it is not.

Of course, the usual construction of a distance function from a Riemannian metric (infimum of lengths of curves) makes sense only for a connected manifold. If the manifold is disconnected, there are no curves connecting points in different connected components. The "distance function" you described is natural in a sense, but it is not a metric in the sense of metric spaces. (A metric, by definition, has to take values in the nonnegative real numbers.)

If $(M,g)$ is a disconnected Riemannian manifold, then an easy argument shows that $M$ is geodesically complete (in the sense that every maximal geodesic is defined for all time) if and only if each connected component is geodesically complete. However, the analogous statement for "metrically complete" depends on exactly how the distance function is defined.

One (noncanonical) way to define a distance function that restricts to the Riemannian distance function on each component is given in the proof of Corollary 13.30 in my Introduction to Smooth Manifolds, 2nd ed.: Choose a point $p_i$ in each component $M_i$, and for $x\in M_i$ and $y\in M_j$ with $i\ne j$ define $$ d(x,y) = d_g(x,p_i) + 1 + d_g(p_j,y). $$ For this distance function, the distance between points in different components is always at least $1$, so a Cauchy sequence eventually has to stay in one component. Thus with this distance function, $M$ is metrically complete if and only if each component is complete, and the Hopf-Rinow theorem is true.

However, a different way of defining distances between points in different components might yield a different result. For example, let $M$ be the following subset of $\mathbb R^2$: $$ M = \left\{(x,y): y = \frac 1 n \text{ for some } n\in \mathbb Z^+\right\}. $$ This is an embedded submanifold of $\mathbb R^2$, which is geodesically complete with the induced Riemannian metric. The ordinary Euclidean distance function restricts to the Riemannian distance function on each component. However, it is not metrically complete with this distance function, because $p_n=(0,1/n)$ is a Cauchy sequence with no convergent subsequence in $M$.

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  • $\begingroup$ Thank you for providing this insight, Professor Lee! $\endgroup$ – user137769 Mar 25 '15 at 21:25

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