Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathfrak{F}(0)$ be the set of all bijections $\mathbb{N}\mapsto\mathbb{N}$, and let $\mathfrak{F}(n+1)$ be the set of all bijections $\mathfrak{F}(n)\mapsto\mathfrak{F}(n)$. Given $\alpha\in\mathfrak{F}(n)$, is there a lower bound on the cardinality of $C_{\mathfrak{F}(n)}(\alpha)$, the centralizer of $\alpha$? Of course, since $C_{\mathfrak{F}(n)}(I)=\mathfrak{F}(n)$, where $I$ is the identity, the cardinality of the centralizer can be the same as that of $\mathfrak{F}(n)$. So the question is, is it possible to, using the axiom of choice as necessary, exhibit $\alpha\in\mathfrak{F}(n)$ such that $|C_{\mathfrak{F}(n)}(\alpha)|<|\mathfrak{F}(n)|$? Or, making things a little more concrete, is it possible to create $f:\mathbb{R}\mapsto\mathbb{R}$, a bijection, such that $|C_{\mathfrak{F}(1)}(f)|<|2^{\mathbb{R}}|$?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

As far as I understand, the answer is negative (assuming $n>0$). It is easy to prove the following:

Claim. Let $X$ be an uncountable set. Let $G$ be the group of all bijections $X\to X$. Then for any $g\in G$, its centralizer $Z(g)\subset G$ satisfies $|Z(g)|=|G|$. (This implies the claim, since all of $\mathfrak{F}(n)$ are uncountable.)

Proof. It suffices to show that $|Z(g)|\ge|G|$. Consider the partition of $X$ into orbits under the action of $g$. Each orbit is identified with a cyclic group ${\mathbb Z}/n{\mathbb Z}$ ($n\ge 0$) with $g$ acting as the shift by one. Since there are countably many isomorphism types of orbits, and each has countably many elements, there exists a particular $n\ge 0$ such that the corresponding set of orbits $X_n$ of this type has $|X_n|=|X|$. Now it is easy to see that the map from $Z(g)$ to the bijections of $X_n$ is surjective. (And hence the cardinality of $Z(g)$ is at least the cardinality of the group of bijections of $X_n$.)

share|improve this answer
    
A side remark on this answer is the fact that if one look to the centralizer of diffeomorphisms (i.e., we add some smoothness to our bijections), then a result of Bonatti, Crovisier and Wilkinson (ams.org/mathscinet-getitem?mr=2511588) shows that the centralizer of a generic $C^1$ diffeomorphism $f$ of a compact manifold is trivial (i.e., it consists of $f$ and its iterates). –  Matheus Jul 17 '10 at 13:39
    
To know that $Z(g)$ has the same cardinality as the set of bijections, it is not sufficient just that $Z(g)$ acts transitively on $X_n$, since after all the finite support permutations also act transitively. Rather, you should observe that permutations of the orbits give rise to distinct elements of $Z(g)$. –  Joel David Hamkins Jul 18 '10 at 1:09
    
Perhaps it is worthwhile to note that this proof uses AC in several places, most plainly in the step mentioned in my previous comment, since you have to choose elements of each orbit to build those elements of $Z(g)$ out of the permutations of the orbits. –  Joel David Hamkins Jul 18 '10 at 1:28
    
@Joel: thanks, that's what I meant --- plead temporary insanity. –  t3suji Jul 18 '10 at 2:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.