Consider the $k$-sphere (I'm particularly interested in $k=3$). For each positive integer $n$ let $P_{n}$ be the set of primes that divide the order of $\pi_{i}S^{k}$ for some $i\leq n$.

Does the cardinality of $P_{n}$ tend to infinity with $n$?


Yes. In fact $\bigcup_{n\geq 1} P_n $ is the set of all primes. Serre proved that, for each odd prime $p$, there is some very predictable $p$-torsion in $\pi_{k+(p-1)}(S^k)$, for example.

There is a very nice theorem of McGibbon and Neisendorfer (proving, I think, a conjecture of Serre) that implies and vastly generalizes this.

Theorem (McGibbon & Neisendorfer). Let $X$ be a simply-connected finite complex such that $\widetilde{H}_*(X; \mathbb{Z}/p) \neq 0$. Then $\pi_n X$ contains $p$-torsion for infinitely many values of $n$.

| cite | improve this answer | |
  • 6
    $\begingroup$ Taking $Q_n$ be the set of primes that divide the order of $\pi_nS^3$, do we also know whether the cardinality of $Q_n$ goes to infinity? $\endgroup$ – Lennart Meier Nov 25 '19 at 7:26
  • $\begingroup$ I guess it is assumed that $k\ge 2$. Is there any other implicit assumption on $k$? $\endgroup$ – YCor Nov 25 '19 at 10:12
  • $\begingroup$ For sure $k\neq 1$, but all other $k$ are ok. $\endgroup$ – Jeff Strom Nov 25 '19 at 15:32

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.