10

It is known from Serre's classical result that every p-torsion occurs in the homotopy groups of every sphere. Is it known: do elements of every order occur in homotopy groups of spheres?

Arshak Aivazian
  • 6,474
  • 1
    Duplicates part of https://mathoverflow.net/questions/377545/which-finite-abelian-groups-arent-homotopy-groups-of-spheres namely the special case of a cyclic group. That question has a lot of discussion but no answers.
    The discussion suggests there might never be an element of order 5.     – Noam D. Elkies Jul 25 '22 at 14:10
  • 6
    I'm aware of this question, thanks, but I'm asking a much weaker question. There may not be a group isomorphic to $Z_5$, but it is easy to prove that every sphere has homotopy groups containing elements of order $5$. – Arshak Aivazian Jul 25 '22 at 14:12
  • 1
    What's "Serra's [Serre's?] classical result"? – Noam D. Elkies Jul 25 '22 at 14:12
  • 7
    Theorem 5.30, page 44 https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf – Arshak Aivazian Jul 25 '22 at 14:19
  • 4
    I always have a little bit of difficulty parsing "any" as a quantifier. I think "every" might be clearer here. (Unless that's not what's meant!) – LSpice Jul 25 '22 at 14:48
  • 4
    I just don't know English well yet, I'm correcting, thanks. – Arshak Aivazian Jul 25 '22 at 14:49
  • 9
    Sure. Let n be a positive integer. Let N be the product of 2n and Euler phi(2n). Then for each prime p that divides n, the number p-1 divides 2N. So for each prime divisor p of n, the denominator of the Bernoulli number B_N is div'l by p. So none of the prime factors of n cancel with factors in the numerator of B_N/N. So denom(B_N/N) is divisible by n. Now denom(B_N/N) or 2*denom(B_N/N) is the order of a cyclic summand in the image of the J-homomorphism in the (2N-1)st stable homotopy group of S^0. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order n. –  Jul 25 '22 at 15:38
  • 6
    @A.S. This seems more like an answer than a comment. – Mark Grant Jul 26 '22 at 07:57
  • 1
    @A.S. Thank you! If you post this comment as an answer, I'll accept it. – Arshak Aivazian Jul 26 '22 at 19:01

1 Answers1

17

As others suggested, I am posting my earlier comment as an answer:

$\DeclareMathOperator\denom{denom}$Sure. Let $n$ be a positive integer. Let $N$ be the product of $2n$ and Euler $\phi(2n)$. Then for each prime $p$ that divides $n$, the number $p-1$ divides $N$. So for each prime divisor $p$ of $n$, the denominator of the Bernoulli number $B_N$ is divisible by $p$. So none of the prime factors of $n$ cancel with factors in the numerator of $B_N/N$. So $\denom(B_N/N)$ is divisible by $n$. Now $\denom(B_N/N)$ or $2\denom(B_N/N)$ is the order of a cyclic summand in the image of the J-homomorphism in the $(2N-1)$st stable homotopy group of $S^0$. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order $n$.

LSpice
  • 11,423