From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Division algebras and topological groups. Date: 11 Feb 1995 22:26:54 GMT [Permission pending] wrote: >Well, OK. Supposing you have a connected space and for each pair of points >x,y in the space there are neighborhoods M and N of x and y and a homeomorphism >f_xy: M --> N with f_xy(x) = y. Then would the space have to be homogeneous? >I.e. would it always be possible to extend some such homeomorphism to the whole >space. Still no. Take a circle and attach an interval by one of the endpoints. Clearly, any homeomorphism will have to fix the point of attachment, but more is true: the points of the interval have the property that X-{x} is not connected, but the points of the circle don't; yet all these points have the same kind of neighborhoods. So homogeneity is not simply a local property. I think at some point you will be able to postulate enough internal conditions to guarantee the space is homogeneous, but I don't know what they are. Perhaps another reader can cut short this cat and mouse game. ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Division algebras and topological groups. Date: 12 Feb 1995 06:09:26 GMT Geoffrey Mess pointed out to me that I didn't really meet the given conditions: [Permission pending] wrote: >Well, OK. Supposing you have a connected space and for each pair of points >x,y in the space there are neighborhoods M and N of x and y and a homeomorphism >f_xy: M --> N with f_xy(x) = y. Then would the space have to be homogeneous? >I.e. would it always be possible to extend some such homeomorphism to the whole >space. ...and I gave a space with a pair of "locally homeomorphic points" not conjugate under any self-homeomorphism of the whole space. But my example had one point unlike these two, forbidden in the question above. For any x in X let U_x be the set of points in the same orbit as x under the group of homeomorphisms X --> X. Clearly U_x contains x, so it's not empty. Also, U_x and U_y are disjoint or equal, that is, X is partitioned into the U_x's. In some cases, one can show each U_x is open, so if you assume X is connected, then X=U_x for one x, that is, X is homogeneous. So why should U_x be open? Consider for example X a manifold. Then find a neighborhood M of x homeomorphic to the unit ball in R^n under a homeomorphism taking x to the origin. It's easy to find a homeomorphism of the unit ball carrying the origin to any other spot, and in fact to do so without moving points on the boundary of the ball. By the glueing lemma, taking this map on M and the identity map on X-M is a continuous map X --> X. Likewise its inverse may be constructed in two parts, so it's a homeomorphism. This shows U_x contains M, and so x is in the interior of U_x. Allowing x to vary shows U_x is open. Similar arguments work with the group Diff(X), although you have to be a little more careful around the boundary; even all the way up to t C^\infty this is a standard glueing problem. dave (who promises once again to try to read questions more carefully before posting answers.) ============================================================================== Date: Sun, 26 Feb 1995 17:36:35 -0500 (EST) From: [Permission pending] Subject: Re: Division algebras and topological groups. To: rusin@math.niu.edu [deletia -- djr] I still haven't figured out if there is a counterexample of a space that would be "locally homogeneous" and not homogeneous. "Locally homogeneous" being: is connected, and for each pair of points x and y there are neighborhoods M and N and a homeomorphism f_xy: M --> N with f_xy(x) = y. The best I can figure, is maybe one could find a space X which is locally homogeneous, without a countable basis, for which X-x is disconnected for each x in X. Then maybe you could find two points x and y in X, where X-x is divided into a space with a countable basis and one with an uncountable basis, and X-y is divided into two spaces with both uncountable bases. But as you pointed out such an X could not be a manifold. I was wondering about the "long line" but I realized yes it's homogeneous. [deletia -- djr] ============================================================================== Date: Mon, 27 Feb 95 12:39:22 CST From: rusin (Dave Rusin) To: [Permission pending] Subject: Re: Division algebras and topological groups. [deletia -- djr] >I still haven't figured out if there is a counterexample of a space that >would be "locally homogeneous" and not homogeneous. "Locally homogeneous" >being: is connected, and for each pair of points x and y there are >neighborhoods M and N and a homeomorphism f_xy: M --> N with f_xy(x) = y. Well, as I pointed out X can't be a manifold, so let's start with something close. A manifold is a Hausdorff locally Euclidean space. Have you seen the example showing why Hausdorff must be explicitly required? It works for your problem. Let Y = R x {0,1} and X = Y / ~ where the equivalence relation is (y,0)~(y,1) for all real y _except_ zero. Thus X is a "real line with two origins". Consider the statement P(x): "for all y in X other than x there are disjoint open sets containing x and y". P(x) is false precisely when x is (the image in X of) either (0,0) or (0,1). Thus any homemorphism of X must permute these two points. On the other hand, the space is locally Euclidean, and so locally homogeneous. Thus this X is a locally homogeneous, non-homogeneous space. I think it may be possible to find other examples based on, say, the Zariski topology, since typically an algebraic variety has only a finite number of self-isomorphisms (and thus in particular the orbit of a point under this family of maps cannot be the whole space). However, I didn't succeed in constructing an example in which the topological homeomorphisms were all algebraic. Presumably the Hausdorff property is not the critical impediment, as your further suggestions indicate. I couldn't think of any other examples though. Clearly homogeneity requires not only local homogeneity but also "anti-local" homogeneity: given x and y there exist M and N around them for which X-M and X-N are homeomorphic. This is what allows the counter-example above: X-M has two components for most points, but three for those two special points. (Homogeneity will also require a patching condition: we will need to have homeomorphisms on X-M and _the closure of_ M, _and_ the homeomorphisms will have to agree on the boundary of M. This ought to give another, more delicate, family of counterexamples.) You may be interested in a related problem. Usually we call a space "locally compact" if it has an open neighborhood with compact closure. In particular, a compact space is locally compact. A text we use here calls a space locally compact if every neighborhood of every point contains a neighborhood with compact closure. If the spaces are Hausdorff, these conditions are equivalent, but in general the latter condition is stronger and it is not immediately clear that compactness implies local compactness! (Although some authors define compactness to presuppose Hausdorff, in which case the discussion is moot.) So the problem is: find a compact (non-Hausdorff) space in which some open set around some point contains no compact neighborhoods. I haven't managed to find an example. [deletia -- djr]