# Partitions of ${\rm Sym}(\mathbb{N})$ induced by convergent, but not absolutely convergent series

Let $(a_n) \subset \mathbb{R}$ be a sequence such that the series $\sum_{n=1}^\infty a_n$ converges, but does not converge absolutely. Then there is a partition of the symmetric group ${\rm Sym}(\mathbb{N})$ into

• sets $S_x \ (x \in \mathbb{R} \cup \{\pm \infty\})$ of all $\sigma \in {\rm Sym}(\mathbb{N})$ for which the series $\sum_{n=1}^\infty a_{n^\sigma}$ converges to $x$, and

• a set $S_{\rm NaN}$ of all $\sigma \in {\rm Sym}(\mathbb{N})$ for which the series $\sum_{n=1}^\infty a_{n^\sigma}$ does not converge.

Question: Has there any work been done on properties of this partition for particular series $\sum_{n=1}^\infty a_n$?

Basic observations are:

• all sets in the partition have the same cardinality, and

• all sets in the partition are closed under multiplication from the right by permutations $\sigma$ which satisfy the condition $\forall n \in \mathbb{N} \ |n - n^\sigma| \leq C$ for some constant $C$.

Added on Oct 11, 2014: Let the spectrum $\mathcal{S}_{\sum a_n}(G)$ of a group $G < {\rm Sym}(\mathbb{N})$ with respect to the series $\sum_{n=1}^\infty a_n$ be the set of all $x \in \mathbb{R} \cup \{-\infty,+\infty,{\rm NaN}\}$ such that $G \cap S_x \neq \emptyset$. --

• Is it true that if $G$ has full spectrum with respect to a series $\sum_{n=1}^\infty a_n$ (i.e. $\mathcal{S}_{\sum a_n}(G) = \mathbb{R} \cup \{-\infty,+\infty,{\rm NaN}\}$), then $G$ has full spectrum also with respect to any other conditionally convergent series $\sum_{n=1}^\infty b_n$?

• Are there countable groups whose spectrum with respect to a particular series is dense in $\mathbb{R}$? -- And if yes, does the choice of the series actually matter here?

• All sets in the partition indeed have the same cardinality, but don't we also know that $|S_x|=|{\rm Sym}(\mathbb{N})|=|\mathbb R|$ for all $x\in\mathbb R\cup\{\infty,-\infty,{\rm NaN}\}$? – Joonas Ilmavirta Oct 8 '14 at 11:55
• @JoonasIlmavirta: Yes, of course. – Stefan Kohl Oct 8 '14 at 13:01
• Have you see this paper? "Rearrangement of conditionally convergent series on a small set", by Rafał Filipów, and Piotr Szuca. J. Math. Anal. Appl. 362 (2010) 64–71. It is not exactly the same question, but some of the information there appears relevant. – Andrés E. Caicedo Oct 11 '14 at 17:59
• Is $S_{\sum a_n}$ a subgroup? If we generalize the definition of spectrum to apply to arbitrary subsets of $\rm{Sym}(\mathbb{N})$, then a more general question is: Is it true that for any conditionally convergent $\sum b_n$, $\left|\mathcal{S}_{\sum b_n}\left(S_{\sum a_n}\right)\right| = 1$? – Amit Kumar Gupta Oct 11 '14 at 18:18
• @StefanKohl in my second question, first I consider generalizing the notion of spectrum to apply to any subset of the symmetric group, not just subgroups, i.e. for any $X \subset \rm{Sym}(\mathbb{N})$, consider $\mathcal{S}_{\sum a_n}(X)$. I ask, if $X$ has spectrum of size 1 with respect to one conditionally convergent series, does it have spectrum of size 1 with respect to all such series? – Amit Kumar Gupta Oct 12 '14 at 0:33

Is it true that if $$G$$ has full spectrum with respect to a series $$\sum_{n=1}^\infty a_n$$ then $$G$$ has full spectrum also with respect to any other conditionally convergent series $$\sum_{n=1}^\infty b_n$$?
If I understand the definitions correctly, clearly no. Let $$G$$ be the symmetric group of $$2\mathbb{N}$$ and consider conditionally convergent series supported on the evens or odds.
Are there countable groups whose spectrum with respect to a particular series is dense in $$\mathbb{R}$$?
Sure. Let $$\sum_{n=1}^\infty a_n$$ be a conditionally convergent series. For a countable dense set of $$x \in \mathbb{R}$$, pick $$\sigma_x \in \operatorname{Sym}(\mathbb{N})$$ so that the rearrangement by $$\sigma_x$$ is $$x$$. Let $$G$$ be the group generated by the $$\sigma_x$$.