An exotic sphere is a differentiable manifold M that is homeomorphic but not diffeomorphic to the standard Euclidean n-sphere.

  1. Naively, I thought that there is no algebraic topological invariant that distinguishing the exotic spheres from each others (?). For example, are there any integration over the local quantities on the differentiable manifolds that can distinguish exotic spheres?

(e.g. I dont think there are any characteristic classes, homotopy or co/homology groups, or co/bordism can distinguish or exotic spheres of the same dimension. Yes?)

  1. However, it looks that we can construct exotic spheres as the non-trivial elements of an abelian monoid under connected sum, which is a finite abelian group if the dimension is not 4.

    • The fact 1 seems to have some tension, if not contradicts, to the fact 2. Because the fact 2 of the abelian monoid / group structure seems to hint that there are some algebraic topological quantities as "topological invariants" that can distinguish exotic spheres. Yes or no?

    • Moreover, there are so called Kervaire, Kervaire-Milnor invariants, Kervaire invariant problem and the Kirby–Siebenmann invariant. Are these quantities as invariants of exotic spheres topologically? In the sense that, we can obtain the topological data (global) from integration over the local quantity? (Analogous to characteristic classes?)

    • In short, are "invariants" of Exotic spheres more of the quantities of (a) differential or (b) topological?

  • 2
    $\begingroup$ As you note, an exotic sphere is homeomorphic to any other sphere of the same dimension. There is no topological invariant that can distinguish an exotic sphere from a standard sphere (or a different exotic sphere). For that, you need to bring in the differential structure. Any invariant which distinguishes these sphere is, almost by definition, not a topological invariant. There is probably a pretty slick category theoretic way of phrasing this... $\endgroup$
    – Xander Henderson
    Jul 10, 2018 at 20:27
  • $\begingroup$ But can we obtain the invariant data (global) from integration over the local quantity? $\endgroup$ Jul 10, 2018 at 20:31
  • $\begingroup$ Can we obtain the invariant data (global) from integration over the local quantity? (even involving differential functions under the integration over the whole sphere) $\endgroup$ Jul 10, 2018 at 20:32

1 Answer 1


There are some "topological" invariants that are only preserved by diffeomorphisms, not general homeomorphisms; take the higher (integral) Pontryagin classes, for example. More generally, one can, for example, try to attach some nice invariants to the tangent bundle of a smooth manifold (and smoothness isn't even a topological invariant) and show that they differ for given manifolds. Consider, for example, a closed, simply-connected, smooth $4$-manifold $X$. If $w_2(X)$ vanishes, then Rokhlin's theorem forces $\frac{1}{16}\sigma(X)$ to be integral, where $\sigma(X)$ is the signature of $X$. The $E_8$ manifold is closed and simply-connected but has $\sigma(E_8) = 8$, so it's not smoothable.

For another example of how this sort of proof works in practice, Milnor's paper on the first exotic $S^7$ used the total space $E(\xi)$ of a bundle $S^3 \to \xi \to S^4$. In brief, the proof is:

(1) Construct such a bundle $\xi = \xi_{h, j} \to S^4$ by attaching the trivial bundles over each sphere via the map $f(u, v) = u^h v u^j$. (Note that $S^3$ is a group.)

(2) Compute the Pontryagin class $p_1(\xi) = \pm 2 (h -j)$. (This is easier than it sounds, since we have explicit transition functions.)

(3) For odd $k$, let $M_k$ denote the $7$-manifold $E(\xi_{(k+1)/2, -(k-1)/2})$. Note that $M_k$ is smooth more or less by construction, and use Morse theory to show that $M_k$ is homeomorphic to $S^7$.

(4) Define a smooth invariant $\lambda(M_k)\in \mathbb{Z}_7$. This is the tricky part, though it's mostly straightforward characteristic class theory once we know that $M_k$ bounds. The invariant is defined in terms of a certain Pontryagin class, which is a smooth invariant but not a homeomorphism invariant. (That's in short why we get an invariant defined mod 7; the rational Pontryagin classes are homeomorphism invariants.)

(5) Show that if $\lambda(M_k)\not = 0$, then $M_k$ is not diffeomorphic to $S^7$. This part is actually easy once the invariant is properly defined.

(6) Compute $\lambda(M_k) = k^2 - 1\in \mathbb{Z}_7$.

The invariant $\lambda(X)$ above sounds like the sort of thing you're looking for, but it's only invariant under diffeomorphisms (as it's a function of the tangent bundle over $X$). There are more exotic constructions, like Reidemeister torsion and the Casson invariant, that are algebraic-topology invariants of finer or more specific structure than just the usual homotopy equivalence of CW-complexes; but that's the general idea.

  • $\begingroup$ this looks nice if this is correct! +1 $\endgroup$ Jul 10, 2018 at 22:08

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