First, the question; after, the motivation.

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."

Question: Does this second countable $X$ admit a Riemannian metric?

For the motivation:

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

The former recalls (but also questions) 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.

For an earlier incarnation of this question, Wolfgang Loehr gave a short argument (below) indicating that the space $X$ mentioned above is connected. In particular, $X$ is a second-countable, connected Hausdorff Banach manifold, which is separable and not regular, hence non-metrizable by Urysohn's Theorem.

If $X$ admits a Riemannian metric, then it is a counterexample to the "theorem" above. In any case, I am not sure how to prove when a manifold does or does not admit a Riemannian metric, and would appreciate assistance in this direction.

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    $\begingroup$ 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. $\endgroup$ Sep 10, 2012 at 23:05
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    $\begingroup$ 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). $\endgroup$ Sep 11, 2012 at 10:36
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    $\begingroup$ 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. $\endgroup$ Sep 12, 2012 at 23:00
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    $\begingroup$ 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. $\endgroup$
    – Jeff Rubin
    Sep 13, 2012 at 19:36
  • $\begingroup$ not sure why this unanswered 9 year old question was downvoted after I bumped it with a relevant tag, but ... constructive feedback would be welcome! $\endgroup$ Aug 11, 2021 at 5:28

1 Answer 1


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 $A=X$.


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