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Does every locally compact second countable space have a non-trivial automorphism?

The motivation for this question comes from something I'm thinking about in logic.

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    We should probably add the requirement that the space have more than one element.2011-06-11
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    Can you expand a bit on what you mean by automorphism?.. I mean $Aut(\mathbb R)$ is trivial in some contexts...2011-06-11
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    I think automorphism just means "homeomorphism to itself", since we are only talking about topological spaces.2011-06-11
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    Here are two possibly relevant papers: 1. J. de Groot, [Groups represented by homeomorphism groups I](http://dx.doi.org/10.1007%2FBF01369667), and 2. Brian M. Scott, *[On the Existence of Totally Inhomogeneous Spaces](http://www.jstor.org/stable/2040346)*. In both papers the existence of an abundance of Hausdorff spaces with trivial automorphism groups is established, but I was unable to see at a glance whether these can be made locally compact (but there exist pretty geometric examples, i.e. subspaces of the plane, see e.g. the last example in paper 2).2011-06-11

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After a bit of looking around I found an even nicer answer: The answer is no, even for compact metric spaces.

More precisely, in H. Cook, Continua which admit only the identity mapping onto non-degenerate subcontinua, Fund. Math. 60 (1967), 241–249 there is the following Theorem 3 on page 248 (it should probably be Theorem 13, never seen such a typo...).

If $n$ is a positive integer, there exists a compact metric continuum $H_n$, with an atomic mapping onto a simple closed curve, such that there exist $n$, and only $n$, mappings of $H_n$ onto $H_n$, each of them is a homeomorphism, and there exists no mapping of $H_n$ onto a proper nondegenerate subcontinuum.

While I only understand roughly half of the words used here, I am sure that this means in particular: $\# \operatorname{Aut}(H_n) = n$. Put $n = 1$ and we have what we want (I'm cheating a bit because that case is the most difficult one, and is settled in in the "slightly technical" Theorem 6 already which I didn't want to reproduce here for that reason).


Added: The paper by the same H. Cook, Upper semi-continuous continuum-valued mappings onto circle like continua, Fund. Math. 60 (1967), 233–239 is also of relevance for understanding the theorem quoted here.

As explained by Henno Brandsma in his answer, J. de Groot's paper Groups represented by homeomorphism groups I, Math. Annalen 138 (1959), 80–102, and its predecessor J. de Groot and R. J. Wille, Rigid continua and topological group-pictures, Arch. Math. 9 (1958), 441–446 started these investigations.

Added Later: Here's a freely accessible link to Brian M. Scott's paper On the existence of totally inhomogeneous spaces, Proc. Amer. Math. Soc. 51 (1975), 489–493 mentioned in a comment and in Henno's answer.

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    Or maybe the typo is due to superstition instead? :)2011-06-11
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Scott's paper, mentioned by Buehler, proves that a compact Hausdorff space $X$ is rigid (has no non-trivial auto-homeomorphism) iff for all $x \neq y$ we have that $X \setminus \{x\}$ is not homeomorphic to $X \setminus \{y\}$; however, it has an example of a rigid locally compact metric space $X$ containing points $x \neq y$ such that $X \setminus \{x\}$ is homeomorphic to $X \setminus \{y\}$.

J. de Groot constructed for every group $G$ a compact Hausdorff space $X$ such that the homeomorphism group of $X$ equals $G$. In fact, he mentions on the first page of that paper (viewable here) that he and Wille already in a previous paper constructed for a countable group $G$ a Peano curve such that its homeomorphism group equals $G$, so there is one for the trivial group as well. I think this settles the original question in the negative, as I think de Groot would call a space a Peano continuum only if it is metrisable.

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Actually I think most finite topological spaces (which are trivially both compact and second-countable) have no non-trivial automorphisms. Already the Sierpinski space $\mathbb{S} = \{ 0, 1 \}$ with topology $\{ \emptyset, \{ 1 \}, \{ 0, 1 \} \}$ is a counterexample.

For an infinite counterexample along the same lines take $\mathbb{N}$ with the topology $\{ \emptyset, \mathbb{N} - \{ 1 \}, \mathbb{N} - \{ 1, 2 \}, ... \}$. I think you want at least to assume that the space is Hausdorff. Note that any locally compact Hausdorff second-countable space is metrizable by Urysohn's metrization theorem, so we are in more familiar territory here.

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    The infinite example is the cofinite topology, right? Also +1 on the Hausdorff requirement, I was thinking that same thing myself.2011-06-11
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    @Asaf: I think Qiaochu just means the complements of the sets $\{1,\ldots,n\}$, not the entire cofinite topology. Actually, I'm not sure the cofinite topology on $\mathbb{N}$ is second countable.2011-06-11
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    @Zev: the cofinite topology has countably many open sets, so it's definitely second-countable. @Asaf: the cofinite topology has lots of automorphisms; since the closed sets are precisely the finite sets, a bijection $S \to S$ ($S$ an infinite set) is actually always a homeomorphism in the cofinite topology.2011-06-11
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    @Qiaochu Yuan: I am having a bit of trouble understanding the question. You could probably help me out (this may be a flaw in my basic understanding): both local compactness and second countability are unaffected by homeomorphisms, so I can't understand how they could affect the triviality or non-triviality of automorphisms? Thanks2011-06-11
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    @Qiaochu: Ah, of course.2011-06-11
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    @Dactyl: I don't understand what you don't understand. If you have a statement $P$ and you suspect it doesn't hold for all spaces, you might still want to find an assumption $Q$ on spaces that will make it hold. Here $P$ happens to be "has a non-trivial automorphism" and $Q$ happens to be "locally compact and second-countable."2011-06-11
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    @Qiaochu Yuan: I am sorry for being unclear. What I am trying to get at is: Is it not true that every question of the kind: 'Does every space with topological property, say P, have a non-trivial automorphism' can be answered with a counter-example?2011-06-11
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    @Dactyl: no. For example, let $P$ be "has a non-trivial automorphism." Less facetiously, let $P$ be "is an infinite manifold." You are definitely confused about something but I have no idea what it is.2011-06-11
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    @Qiaochu Yuan: Thanks.2011-06-11
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    @Qiaochu : You're right about the Hausdorff requirement, but I think there is a convention in some contexts that "compact" implies Hausdorff (otherwise we use the term "precompact"). At least I know this convention is commonly used in France (I think it's used in Bourbaki, and the EGA).2011-06-11
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    @Joel: yes, that convention exists, but I think it is silly (with apologies to Pete Clark!). Compact and Hausdorff naturally exist as two separate conditions, as shown by the following lovely pair of characterizations: a space is compact if and only if every ultrafilter on it converges to at least one point, and a space is Hausdorff if and only if every ultrafilter on it converges to at most one point. The nice properties of compact Hausdorff spaces are due to the fact that _both_ of these things are true (so ultrafilters converge to exactly one point).2011-06-11
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    @Joel: you mean *quasicompact*. Precompact is something else in Bourbaki: It's a subset of a uniform space with compact completion.2011-06-11
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    @Qiaochu : I think this debate might never be settled :)@ Theo Buehler : Yes, you're right.2011-06-11
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    @Joel: of course I think in the absence of strong reasons to do otherwise we use the conventions we were taught. I was taught topology by Munkres, from Munkres, so...2011-06-12
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    @Qiaochu: **Boo! Hiss!** Seriously, you don't have to agree with the convention or use it yourself, but it's unnecessarily dismissive it to call it silly. A lot of mathematicians have found the terminology to be natural and useful (and a lot haven't, of course). Anyway, I think this case is not such a good argument for your cause: imposing your terminology on the OP makes the question almost trivial -- we only need to consider a famous two-point space to get a counterexample -- whereas the terminology as intended by the OP made for a much more interesting problem.2011-08-05
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    P.S.: I first learned topology from Munkres as well, and I am pretty sure that "quasi-compact" was not in my vocabulary. But then I went to graduate school and started studying algebraic / arithmetic geometry and reading texts and papers in French. Watch out -- it could happen to you too. :)2011-08-05
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    I am actually considering asking "Should compact imply Hausdorff?" as a question (which should, presumably, be made CW). The one clear advantage to this would be to siphon off the back-and-forth discussion on this which inevitably takes place elsewhere on the site. But I guess I cannot really bring myself to ask it unless it seems like a worthy question in its own right. I think it could be: in the course of giving arguments people may actually give useful information, like Qiaochu's bit about ultrafilters above. (I think he knows that argument won't sway me...) What do people think?2011-08-05
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    @Pete: fair point. When I called it silly I was being a little tongue-in-cheek, which doesn't seem to have come across well. In any case, "should compact imply Hausdorff?" strikes me as a little subjective and argumentative. I can't imagine what would count as a reasonably objective answer.2011-08-06