Proposed Counterexample to a Theorem of Differential Geometry (on Banach manifolds) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T09:15:55Z http://mathoverflow.net/feeds/question/106292 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106292/proposed-counterexample-to-a-theorem-of-differential-geometry-on-banach-manifold Proposed Counterexample to a Theorem of Differential Geometry (on Banach manifolds) Benjamin Dickman 2012-09-04T01:54:46Z 2012-09-14T05:03:00Z <p>This question stems from Jeff Rubin's earlier <a href="http://mathoverflow.net/questions/104575/under-exactly-what-extra-conditions-if-any-is-a-connected-hausdorff-manifold" rel="nofollow">MO question</a> and a <a href="http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold" rel="nofollow">follow-up</a> that I posted. </p> <p>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):</p> <p><strong>Theorem:</strong> A connected Hausdorff Banach manifold with a Riemannian metric is a metric space. </p> <p>That said, consider 27.6 (pdf pp. 262-263) in <a href="http://www.mat.univie.ac.at/~kriegl/Skripten/apbook.pdf" rel="nofollow">The convenient setting of global analysis</a> (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."</p> <p><strong>My question:</strong> Is this second countable $X$ a counterexample to the above theorem?</p> <p>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.</p> <hr> <p><b>Edit 1:</b> Here's a sketch of why one might even consider this example:</p> <p>If indeed the proposed space $X$ described in 27.6 of the link above is second-countable, then at least one <a href="http://tinyurl.com/budt9hs" rel="nofollow">source</a> I have found claims that $X$ would, as a result, admit a Riemannian metric. [<strong>NB:</strong> 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. </p> <p>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.</p> <p><strong>Sub-question 1:</strong> 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? [<strong>NB:</strong> Particularly in the context of infinite dimensional manifolds.]</p> <p><strong>Sub-question 2:</strong> can anyone prove (or sketch a proof of) the connectedness of $X$? Alternatively, can anyone show that $X$ is not connected? [<strong>NB:</strong> This has been answered: $X$ is connected.]</p> <p>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. </p> <p>Thanks!</p> <hr> <p><strong>Edit 2:</strong> My second sub-question has been answered in the affirmative by Wolfgang Loehr: $X$ is indeed a connected space.</p> <p>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. </p> <p>Nonetheless, my initial <strong>question</strong> still stands: is the space $X$ described in the AMS book on Global Analysis a counterexample to the theorem stated above?</p> <hr> <p><strong>Edit 3:</strong> 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?</p> <hr> <p><strong>Post-bounty Edit:</strong> 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.</p> http://mathoverflow.net/questions/106292/proposed-counterexample-to-a-theorem-of-differential-geometry-on-banach-manifold/106825#106825 Answer by Wolfgang Loehr for Proposed Counterexample to a Theorem of Differential Geometry (on Banach manifolds) Wolfgang Loehr 2012-09-10T15:46:16Z 2012-09-10T15:46:16Z <p>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 $x\in\ker\lambda$, <code>$x_n:=x+\frac1n y\in A$</code> and <code>$x_n\to x$</code>, hence $x\in A$ and $A=X$.</p>