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.

Moment maps and Hamiltonian reduction), it seems that they only claim it forfinite dimensionalmanifolds. 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