# Proposed Counterexample to a Theorem of Differential Geometry (on Banach manifolds)

This question stems from Jeff Rubin's earlier MO question and a follow-up that I posted.

The former recalls the following result proved by both Serge Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag):

Theorem: A connected Hausdorff Banach manifold with a Riemannian metric is a metric space.

That said, consider 27.6 (pdf pp. 262-263) in The convenient setting of global analysis (AMS, 1997), and in particular the example given at the end of it, which concludes with: "Then the same results are valid, but $X$ is now even second countable."

My question: Is this second countable $X$ a counterexample to the above theorem?

I'm hoping someone can shed some light on this matter, either by explaining why it fails as a counterexample (offhand, I'd deem this the more likely scenario) or by proving/sketching why it might actually suffice.

Edit 1: Here's a sketch of why one might even consider this example:

If indeed the proposed space $X$ described in 27.6 of the link above is second-countable, then at least one source I have found claims that $X$ would, as a result, admit a Riemannian metric. [NB: It has been pointed out that this source states its claim strictly in the context of finite-dimensional manifolds.] Furthermore, $X$ is described as a modification (where "the same results are valid") of a space that is a connected Hausdorff Banach manifold that is separable and not regular.

To summarize, we might have $X$ as a connected Hausdorff Banach manifold with a Riemannian metric, which is separable and not regular (hence non-metrizable by Urysohn's Theorem), in which case, $X$ would be a counterexample to the above-stated theorem.

Sub-question 1: can anyone find other sources (preferably with proof) that a second-countable connected Hausdorff manifold necessarily admits a Riemannian metric? Alternatively, can anyone find a counterexample to this? [NB: Particularly in the context of infinite dimensional manifolds.]

Sub-question 2: can anyone prove (or sketch a proof of) the connectedness of $X$? Alternatively, can anyone show that $X$ is not connected? [NB: This has been answered: $X$ is connected.]

I'd appreciate even a partial answer to my original question or either of my sub-questions. Also, if you should know (of) anyone who is doing work in this area of mathematics, perhaps you could direct them to my query.

Thanks!

Edit 2: My second sub-question has been answered in the affirmative by Wolfgang Loehr: $X$ is indeed a connected space.

I see numerous mentions of the result mentioned in my first sub-question (that second-countability alone implies a connected Hausdorff manifold admits a Riemannian metric) but I'm wondering whether this is in fact only a theorem for finite dimensional manifolds.

Nonetheless, my initial question still stands: is the space $X$ described in the AMS book on Global Analysis a counterexample to the theorem stated above?

Edit 3: As time winds down on the question's bounty, I wonder whether anyone has helpful thoughts with regard to non-regular manifolds that admit Riemannian metrics. More precisely, how could one prove that $X$ does or does not admit a Riemannian metric?

Post-bounty Edit: I awarded the bounty since my sub-question 2 was answered entirely. There is still no conclusion as to whether or not the space referenced above is a counterexample to the aforementioned theorem, but it is increasingly clear that there is a fair bit of confusion surrounding when theorems about Banach manifolds do or do not extend from the finite dimensional case to the infinite dimensional one.

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I'm a bit skeptical that you'd be able to construct a Riemannian metric on your counter example when the space is not regular - the only general method I know of uses a smooth partition of unity to go from local existence to global existence. A non regular space may not even have a continuous partition of unity. –  Brett Parker Sep 10 '12 at 23:05
For your link to "Riemannian iff second countable", if you go back to page 14 of the book you cited (Moment maps and Hamiltonian reduction), it seems that they only claim it for finite dimensional manifolds. Finite dimensional is used by noting that FD manifolds are locally Euclidean and hence locally compact. Thus if Hausdorff and second countable the manifold is paracompact (and then you can build partitions of unity and hence Riemannian metric). –  Willie Wong Sep 11 '12 at 10:36
Yes, I think that is probably a theorem of FD manifolds; however, one might still ask whether this particular example admits a Riemannian metric. My hunch is no (as Brett Parker suggested above) but I'm not sure how to show this rigorously. –  Benjamin Dickman Sep 11 '12 at 17:39
At least something seems to be wrong in [Abraham, Marsden, and Ratiu], for after the disputed theorem they claim in Corollary 6.5.13 (3rd edition, 2007 pdf-version): Paracompact (or second countable) Hausdorff manifolds modeled on separable real Hilbert spaces admit Riemannian metrics.'' This would make our $X$ a counter example to their theorem. However, they prove the corollary only in the paracompact case and $X$ is second countable but of course not paracompact. –  Wolfgang Loehr Sep 12 '12 at 23:00
I think where second-countable comes into play may be that a second-countable locally compact Hausdorff space is paracompact (Abraham, Marsden, Ratiu, 2nd Ed, Prop 5.5.5, or Kelley: locally compact Hausdorff implies regular; second-countable implies Lindelof; regular Lindelof implies paracompact (Exercise Y(a))). So maybe AMR just gave the proof for paracompact Hausdorff manifolds thinking that in the second-countable Hausdorff case you would have to assume local compactness (for example if the manifold is finite dimensional). Of course this doesn't help since $X$ is not paracompact. –  Jeff Rubin Sep 13 '12 at 19:36
Sub-question 2 is easy: $X$ is connected. Assume $A\subseteq X$ is open and closed. $Y\setminus \ker\lambda$ carries the topology inherited from $\ell^2$, hence is connected, hence we may assume w.l.o.g. that it is contained in $A$ (otherwise we take the complement of $A$). Now fix $y$ with $\lambda(y)=1$. For any $x\in\ker\lambda$, $x_n:=x+\frac1n y\in A$ and $x_n\to x$, hence $x\in A$ and $A=X$.