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Here is a question that I have had in my head for a little while and was recently reminded of.

Let $X$ be a (nonempty!) topological space. What are useful (or even nontrivial) sufficient conditions for $X$ to admit a group law compatible with its topology?

There are many necessary conditions: for instance $X$ must be completely regular. In particular, if it is a Kolmogorov space it must be a Urysohn space. It must be homogeneous in the sense that the self-homeomorphism group $\operatorname{Aut}(X)$ acts transitively. In particular $\operatorname{Aut}(X)$ must act transitively on the connected components of $X$.

It is rather clear though that these conditions are nowhere near sufficient even for very nice topological spaces. A recent question on this site calls attention to this: even for the class of smooth manifolds there are a sequence of successively more intricate necessary conditions but (apparently) no good sufficient conditions.

What about for other classes of spaces? Here are two examples: it suffices for $X$ to be discrete (i.e., every set is open) or indiscrete (i.e., only $\varnothing$ and $X$ are open). These are, of course, completely trivial. Can anyone do any better than this? For instance:

If $X$ is totally disconnected and compact (for me this implies Hausdorff), must it admit the structure of a topological group?

Added: as some have rightly pointed out, the above necessary condition of homogeneity is not implied by my assumptions. So let us assume this as well. Note that assuming second-countability as well forces the space to be (finite and discrete or) homeomorphic to the Cantor set, which is well-known to carry a topological group structure, so I don't want to assume that.

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The answer to your specific question is no, for stupid reasons: such a space need not be homogeneous. For example, the space formed by adding an extra isolated point to the Cantor set is compact, totally disconnected, and has just the one isolated point. –  Chris Eagle Jan 8 '11 at 21:52
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If you add the conditions homogeneous and second countable to your last question then the answer becomes yes, because it then either is discrete (hence finite) or it is homeomorphic to the Cantor set which is homeomorphic to a countable product $\prod\{0,1\}$ and thus a group. –  t.b. Jan 8 '11 at 22:05
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If it's a simply connected closed 3-manifold... :) –  Dylan Wilson Jan 9 '11 at 7:22
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A little better: Any simply connected manifold that covers a Lie group admits the structure of one. Still though... these aren't very good –  Dylan Wilson Jan 9 '11 at 7:24

1 Answer 1

up vote 8 down vote accepted

In this short paper we have a counterexample for an even weaker statement (it admits no left topological group structure). There are also consistent examples (under CH) that are S-spaces etc. Note that any first countable compact zero-dimensional homogeneous space that is not metrisable is a counterexample, as a first countable topological group is metrisable.

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thanks for the reference. The statement "a first countable topological group is metrizable" seems to be false for trivial reasons -- consider e.g. an indiscrete space -- but after a little internet searching I gather it is true for Hausdorff topological groups. Very interesting. –  Pete L. Clark Jan 11 '11 at 2:16
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@Pete: indeed, I was assuming all topological groups to be at least $T_0$ and thus completely regular etc. The OP also considered only completely regular spaces, so this is what I did too. –  Henno Brandsma Jan 11 '11 at 15:55
    
que? I am the OP! :) Anyway, I have accepted your answer. –  Pete L. Clark Jan 21 '11 at 11:23

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