# Under exactly what (extra) conditions (if any) is a connected Hausdorff manifold with a Riemannian metric a metric space?

The setting is that manifolds are Banach manifolds, not necessarily finite dimensional. No other assumption is made about the topology of the manifold. In particular, it is not assumed to be regular or normal. Of course, this means that the manifold is not assumed to be paracompact. There is no assumption made about the manifold admitting partitions of unity. In this setting, both Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag) prove that a connected Hausdorff manifold with a Riemannian metric is a metric space. Both proofs, it seems to me, suffer from the same flaw. The proofs are similar enough that I'll just refer to Lang's. We start with a connected Hausdorff manifold $X$ and a Riemannian metric, $g$ on $X$. No special assumptions are made about the Hilbert space, $E$, on which the manifold is modeled. For example, $E$ may or may not be separable. Start by defining a length function, $L_g$ which assigns a real number $L_g(\gamma)$ to each piecewise $C^1$ path, $\gamma$, from $J=[a,b]$ into $X$, based on the metric, $g$. The distance function, $d_g:X\times X\to\mathbb{R}$ is then defined by $d_g(x,y)=\inf\{L_g(\gamma)\}$ over all piecewise $C^1$ paths, $\gamma$, defined on $J$ with $\gamma(a)=x$ and $\gamma(b)=y$.

Without any difficulty, $d_g$ is a pseudo-metric. The first main point of the proof is to show that $d_g$ is actually a metric. So we start with distinct points $x$ and $y$ of $X$ and set out to show $d_g(x,y) > 0$. We have a chart $(U, \phi)$ at $x$ for $X$ with $\phi(U)$ open in $E$, and we can arrange $U$ to be small enough that $y$ is not in $U$, since the manifold is assumed to be Hausdorff. Working in $\phi(U)$ we find an $r>0$ such that the closed ball $D(\phi(x),r)$ is contained in $\phi(U)$ and such that certain other properties hold. Let $S(\phi(x),r)$ be the boundary of $D(\phi(x),r)$. Then we define $D(x,r)=\phi^{-1}(D(\phi(x),r))$ and $S(x,r)=\phi^{-1}(S(x,r))$, both subsets of $U$.

Since $\phi$ is a homeomorphism, $D(x,r)$ and $S(x,r)$ are closed in $U$ (not necessarily closed in $X$). To me, this is a key stumbling point, as I'll explain. We next let $\gamma:J \to X$ be any piecewise $C^1$ path in $X$ from $x$ to $y$. Both proofs make the following assumption: since $x$ is in $D(x,r)$ and since $y$ is not in $U$, the path $\gamma$ must cross $S(x,r)$. Neither author explicitly proves this assumption (and AMR doesn't even state it).

When I set out to prove this, using the continuity of $\gamma$ and the connectedness of $J$, I quickly run into the need to show that $D(x,r)$ is closed in $X$, not just in $U$. If $X$ were known to be regular, it would not be a problem to take $r$ small enough that $D(x,r)$ was closed in $X$. But as I mentioned at the beginning, I don't know that $X$ is regular. If I could show that the pseudo-metric topology for $X$ induced by $d_g$ was the same as the original manifold topology, I would also get that $X$ was regular. But I don't see how to do that without first completing the first part of the proof.

The whole question seems to be, can I make $r$ small enough that $D(x,r)$ stays away from the topological (in the original manifold topology of $X$) boundary of $U$? But this does not seem to be a local issue, since it depends on what is closed in $X$ which in turn, depends on what is open everywhere in $X$ including outside of $U$.

So, the question is, are other assumptions necessary, or is it possible to fix the proof so that no other assumptions need be made?

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Is there a Hausdorff Banach manifold which is not a regular space? maik.ru/full/rusmath/97/10/rusmath10_97p49full.pdf nonchalantly states (p.53): "Note that a Hausdorff Banach manifold X is a regular space." Separately, in 1.3.1 of "Momentum Maps and Hamiltonian Reduction" (tinyurl.com/budt9hs): "it can proved that a connected Hausdorff manifold admits a Riemannian metric if and only if it is second countable." Unfortunately, neither note is sourced; otherwise you'd have regularity and second-countability, and Urysohn would finish things off without any extra conditions. – Benjamin Dickman Aug 16 '12 at 22:43
I also found a note in Abraham and Marsden, "Foundations of Mechanics", Updated 1985 Printing, p128: "Recall that we include second countable in our definition of a manifold. It is interesting that a manifold which admits a Riemannian metric (or a connection) must be second countable (see Abraham [1963])." Unfortunately, the only entry in the references that could possibly match that is Abraham, R. 1963.a Transversality in manifolds of mappings. Bull. Am. Math. Soc. 69 (4):470-474. and that has nothing to do with Riemannian metrics or second countability. – Jeff Rubin Aug 19 '12 at 21:09
An answer to my question (see comment above) posed last August: mathoverflow.net/questions/104965/… – Benjamin Dickman May 29 '13 at 7:39

There are infinite dimensional weak Riemannian manifolds with vanishing geodesic distance (in the sense as defined in the question). These are modeled on nuclear Frechet spaces, but the results extend to Sobolev completions of high enough order ($>\dim/2 +2$). They are still weak Riemannian manifolds (i.e., the Riemann metric does not generate the topology on the tangent spaces).

The first example was the $L^2$ metric on $\text{Emb}(S^1,\mathbb R^2)/\text{Diff}(S^1)$, as shown in the first paper below. Then it turned out that the right invariant $L^2$-metric on each full diffeomorphism group also has this property, also Sobolev metrics for Sobolev order $<1/2$ ($\le 1/2$ on $\text{Diff}(S^1)$). In particular, Burgers' equation and KdV are nonlinear PDE's corresponding to geodesic equations for metrics with vanishing geodesic distance.

All the papers are in the arXiv or on my homepage.

• Peter W. Michor; David Mumford: Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. (JEMS) 8 (2006), 1-48.

• Peter W. Michor; David Mumford. Vanishing geodesic distance on spaces of submanifolds and diffeomorphisms. Documenta Math. 10 (2005), 217--245 (written later)

• Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Vanishing geodesic distance for the Riemannian metric with geodesic equation the KdV-equation. Ann. Glob. Anal. Geom. 41, 4 (2012) 461-472.

• Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. Ann. Glob. Anal. Geom. 44, 1 (2013), 5-21.

• Martin Bauer, Martins Bruveris, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. II. 7 pages. To appear in: Ann. Glob. Anal. Geom.

Edit. Some more remarks:

If you have a strong Riemannian metric (such that $g_x$ induces the topology on $T_xM$ for each $x$), then you have a Hilbert manifold, the Riemannian exponential mapping is a local diffeomorphism, and by the Gauss lemma geodesic distance describes the manifold topology.

If the Riemannian metric is weak (so $g_x: T_xM\to T_x^*M$ is injective only), Then:
(1) one has to prove that the connection exists and is smooth.
(2) Even for a Banach manifold where the Riemannanian exponential mapping is automatically a local diffeomorphism, the Gauss lemma is not true in general, since the exponential mapping is a diffeomorphism on a neighborhood of $0\in T_xM$, but this need not be a $\|\cdot\|_{g_x}$-ball.
This is what happens all the examples described in the papers above.

The first paper has an example (concentric spheres, towards the end) of an incomplete geodesic where the conjugate points are dense.

We believe, that vanishing geodesic distance is tied to the fact, that sectional curvature is locally positive unbounded: behind mountain you always find a shorter geodesic.

For weak Riemannian metrics of low Sobolev order we do not know whether the geodesic equation is well posed. Of course KdV and Burgers are, but Burgers' close relative, the $L^2$-metric on $\text{Imm}(S^1,\mathbb R^2)$ or $\text{Imm}(S^1,\mathbb R^2)/\text{Diff}(S^1)$, are not known to have well posed geodesic equation.
In the paper

• arXiv:1202.5122 Right-invariant Sobolev metrics ${H}^{s}$ on the diffeomorphisms group of the circle. Joachim Escher (IFAM), Boris Kolev (LATP),

the geodesic equation for the Sobolev $1/2$-metric on $\text{Diff}(S^1)$ is shown to be well posed, but the third paper above shows that it has vanishing geodesic distance.

The OP question stats from a (Banach)-manifolds and asks for Riemannian metric on it. But even if you find one, there are many more obstacles until one ends up with a metric space described by geodesic distance.

In my view, the metric (or even the geodesic equation) is more important than the manifold, which you can adapt to the metric somewhat.

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