4
$\begingroup$

The standard rearrangement theorem for conditionally convergent series says that the terms in a conditionally convergent series can be rearranged so that the new sum is any desired number, or $\pm\infty$.

Let me set up some notation to investigate this a little further. My coefficients will be determined by a function $\alpha:\mathbb{N}\to\mathbb{R}$ (or $\mathbb{C}$). Then I'll simply write $\sum\alpha$ for this series $\sum_{n=1}^\infty \alpha(n)$.

Now we let $\operatorname{Sym}(\mathbb{N})$ act on the set $A = \{ \alpha: \mathbb{N}\to \mathbb{R} \}$ by the rule $\alpha^\sigma (x) = \alpha(\sigma(x))$. Then we can ask simple questions

  1. is $\sum \alpha^\sigma = \sum \alpha$?
  2. if $\sum\alpha$ converges, does $\sum\alpha^\sigma$ also converge?

But these are not really the questions I want to ask.

Let $B\subseteq A$ be the subset of all sequences $\alpha$ so that the series $\sum\alpha$ is conditionally convergent.

Question 1 What is the stabilizer $G = \{ \sigma \mid \alpha^\sigma \in B\ \forall \alpha\in B\}$?

Question 2 What is the group of sum-preserving permutations $H = \{ \sigma \mid \sum \alpha^\sigma = \sum\alpha \ \forall \alpha\in B\}$?

Certainly any permutation which is the identity off a finite subset of $\mathbb{N}$ is contained in both.

There are some subgroups of $\operatorname{Sym}(\mathbb{N})$

This is related to the question here.

$\endgroup$
  • 3
    $\begingroup$ If there is a constant $C$ such that $|\sigma(n)-n| \leq C$ for all $n$, then $\sigma$ is sum-preserving. And this "boundedness" property is preserved by compositions and inverses. So one might guess that perhaps these are all of the sum-preserving permutations. But it is not the case. There are sum-preserving permutations that have "unbounded steps" (but, I suppose, very far apart). There are several known characterizations of sum-preserving permutations, see for example projecteuclid.org/euclid.pjm/1102688295 and references therein. $\endgroup$ – Zach Teitler Mar 1 '17 at 5:15
  • $\begingroup$ Maybe I'm blind, but why are G and H even groups? $\endgroup$ – Johannes Hahn Jun 20 '17 at 21:18
  • 1
    $\begingroup$ There's a finer sufficient condition for being sum-preserving given in Bourbaki, TG (=Topologie Générale), IV, §7, exercise 12 (page IV.60 in the French edition), namely that $r(n) := |\sigma(n)-n|\cdot\sup_{m\geq n}|\alpha(m)|$ tends to $0$. I seem to remember there's a slight mistake and that it should be something different (maybe replace some $m$ or $n$ by $\sigma(m)$ or $\sigma(n)$), I guess solving the exercise will unravel the correct version. $\endgroup$ – Gro-Tsen Jun 20 '17 at 22:20
3
$\begingroup$

[Making comment into answer.] If there is a constant $C$ such that $|\sigma(n)−n| \leq C$ for all $n$, then $\sigma$ is sum-preserving. And this "boundedness" property is preserved by compositions and inverses. So one might guess that perhaps these are all of the sum-preserving permutations. But it is not the case. There are sum-preserving permutations that have "unbounded steps" (but, I suppose, very far apart). There are several known characterizations of sum-preserving permutations, see for example http://projecteuclid.org/euclid.pjm/1102688295 and references therein.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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