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

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.

• We should probably add the requirement that the space have more than one element. – Zev Chonoles Jun 11 '11 at 19:25
• Can you expand a bit on what you mean by automorphism?.. I mean $Aut(\mathbb R)$ is trivial in some contexts... – Dactyl Jun 11 '11 at 19:30
• I think automorphism just means "homeomorphism to itself", since we are only talking about topological spaces. – Zev Chonoles Jun 11 '11 at 19:33
• Here are two possibly relevant papers: 1. J. de Groot, Groups represented by homeomorphism groups I, and 2. Brian M. Scott, On the Existence of Totally Inhomogeneous Spaces. 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). – t.b. Jun 11 '11 at 19:57

## 3 Answers

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.

• Or maybe the typo is due to superstition instead? :) – t.b. Jun 11 '11 at 23:44

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.

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.

• @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. – Qiaochu Yuan Jun 11 '11 at 20:02
• @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." – Qiaochu Yuan Jun 11 '11 at 20:04
• @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. – Qiaochu Yuan Jun 11 '11 at 20:22
• @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). – Qiaochu Yuan Jun 11 '11 at 21:05
• @Joel: you mean quasicompact. Precompact is something else in Bourbaki: It's a subset of a uniform space with compact completion. – t.b. Jun 11 '11 at 21:23