Looking at Kervaire and Milnor's1 classification of exotic spheres, for each $n$ there are a finite number of distinct (up to diffeomorphism) smooth structures you can impose on the $n$-sphere (OEIS sequence A001676). The first few terms of this sequence, denoted $[\Theta_n]$, are

$$ \begin{array}{c|cc} n & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&\dotsb\\\hline [\Theta_n] & 1&1&1&\geq1&1&1&28&2&8&6&992&1&3&2&16256&\dotsb\\ \end{array} \;.$$

There are obvious spikes in the value of $[\Theta_n]$ for $n \equiv 3 \!\pmod 4$. Is there a nice reason why this happens? The details of the linked paper are a bit over my head.

Wikipedia gives a brief explanation: $\Theta_n$ is the group of $h$-cobordism classes of the homotopy $n$-sphere, and there's a correspondence between the elements of this group and the number of smooth structures of the $n$-sphere. Each group $\Theta_n$ contains a cyclic subgroup $bP_{n+1}$ of the "$n$-spheres that bound parallelizable manifolds." And then for some reason the size of $bP_{n+1}$ can be very large for $n \equiv 3 \!\pmod 4$ and somehow relates to Bernoulli numbers.

  1. Kervaire and Milnor, Groups of homotopy spheres: I, Ann. Math. (2) 77 (1963), 504-537
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    $\begingroup$ Did you follow the construction of 28 smooth structures on the $7$-sphere ? $\endgroup$
    – reuns
    Nov 26, 2016 at 4:24
  • $\begingroup$ @user1952009 Yeah, I'm okay with that construction, but I don't see how it relates to the algebraic side of things. Like for that manifold in $\mathbb{C}^5$, I don't see why $k=29$ wouldn't give you a new smooth structure distinct from the previous $28$. $\endgroup$ Nov 26, 2016 at 4:44
  • $\begingroup$ I don't follow it but there P.21 it defines a topological invariant specific to $4k-1$ manifolds $\endgroup$
    – reuns
    Nov 26, 2016 at 5:02
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    $\begingroup$ It is because in those dimension, the size of $bP_{n+1}$ are related to numbers that can be signature of manifolds bounding homotopy spheres/ actual spheres. You need to fight very hard to obtain the formula given in Kerviare & Milnor(no proof is given there). A proof can be found in Kosinski's book Differentiable Manifolds. And the proof does not use Brieskorn's construction. $\endgroup$ Nov 26, 2016 at 5:50
  • 1
    $\begingroup$ Duplicate of mathoverflow.net/questions/219869/… $\endgroup$ Nov 27, 2016 at 3:19

1 Answer 1


I wish I can give a more detailed explanation here.

The group $\Theta_n$ fits into an exact sequence $0 \rightarrow bP_{n+1} \rightarrow \Theta_n \rightarrow Coker J$, which is essentially Theorem 4.1 in Kervaire & Milnor. $bP_{n+1}$ is the subgroup of homotopy n-spheres(under connected sum) which are bounded by some n+1-manifolds. $CokerJ$ is the cokernel of the stable J-homomorphism in that dimension. The last map will have cokernel $\mathbb{Z}/2$ if $n = 2$ mod $4$ and Kervaire invariant in that dimension is nontrivial. Otherwise it is surjective. $bP_{n+1}$ is trivial when $n+1$ is odd. And is $\mathbb{Z/2}$ or trivial, when $n+1 = 2$ mod $4$, depending on whether we have a trivial Kervaire invariant in this dimension or not. The invariant is defined in the last section of the paper. If you google Kervaire invaraint you can see a lot of stuff about this.

The case you are interested in is when $n+1$ is a multiple of $4$. The formula I mentioned in comment is given in the paper after Corollary 7.6 with a promise of a proof in a non-exist following paper. However a proof can be found in various places. I learned a proof from Kosinski's book Differential Manifold(Theorem IX.8.7 in the book), which is a great book on everything you need for understanding Kervaire & Milnor. The number you see is so big basically because such coefficients appear in the Hirzebruch signature theorem, which you can use to compute signature of a manifold by Pontrjagin classes.


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