This (long) paper,

Guozhen Wang, Zhouli Xu. "On the uniqueness of the smooth structure of the 61-sphere." arXiv:1601.02184 [math.AT].

proves that

the only odd dimensional spheres with a unique smooth structure are $S^1$, $S^3$, $S^5$, $S^{61}$.

The new result is for $S^{61}$.

Is it possible to give some intuition on this remarkable result, for those not steeped in algebraic and differential geometry, and so not intimately familiar with homotopy groups of spheres? Any attempt would be welcomed.

Joseph O'Rourke
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    While I like the question, I'm not sure how realistic it is to expect anyone to digest the paper and summarise the intuition behind the result in a reasonable amount of time. – Dan Rust Jan 12 '16 at 16:00
  • There are a couple of things you could mean by this. One is: "how do people reduce this question to a question about the stable homotopy groups of spheres?" Another is: "what is the contribution of this new paper, and how does it single out $S^{61}$?" I think the first question is reasonable and the second question is a bit less reasonable; I expect the answer to be fairly technical. – Qiaochu Yuan Jan 12 '16 at 16:13
  • @DanRust: Experts may already know of the result. Often a first arXiv appearance is preceded by talks. – Joseph O'Rourke Jan 12 '16 at 16:21
  • @QiaochuYuan: Good points! It certainly is intriguing to see $61$ standing alone, but I acknowledge that perhaps there is no intuition communicable to non-experts. – Joseph O'Rourke Jan 12 '16 at 16:22
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    I strongly suggest reading the first three sections of the paper. They're great. –  Jan 12 '16 at 16:29
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    I for one have trouble getting intuition in dimension 3 and above, so dimension 61... – Najib Idrissi Jan 12 '16 at 16:38
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    If you just want a general background on exotic spheres and "why" they exist, I recommend the non-technical paper by Joachim and Wraith [Exotic Spheres and Curvature](http://www.maths.ed.ac.uk/~aar/papers/joacwrai.pdf) (2008), as well as [Milnor's Abel Prize lecture](https://www.youtube.com/watch?v=kQaUCDFvbjA). – Gro-Tsen Jan 12 '16 at 17:16
  • PS (to my previous comment): The slides of the Milnor lectures [are available here](http://www.math.stonybrook.edu/~jack/MilnorOslo-print.pdf) (they are very hard to read on the video). – Gro-Tsen Jan 12 '16 at 17:46

1 Answers1


Results of this form, and my intuition from them, come from the Kervaire-Milnor paper on exotic spheres. (There was never a homotopy spheres II. The purported content of that unpublished paper appears to be summarized in these notes, though I haven't read them.) I'm going to need to jump into the algebra here; personally, I couldn't tell you the difference between $S^{57}$ and $S^{61}$ without it.

For $n \not\equiv 2 \bmod 4$, there is an exact sequence $$0 \to \Theta_n^{bp} \to \Theta_n \to \pi_n/J_n \to 0.$$ For $n=4k-2$, instead we have the exact sequence $$0 \to \Theta_n^{bp} \to \Theta_n \to \pi_n/J_n \xrightarrow{\Phi_k} \Bbb Z/2 \to \Theta_{n-1}^{bp} \to 0.$$

Let's start by introducing the cast of characters. $\Theta_n$ is the group of homotopy $n$-spheres. It's smooth manifolds, up to diffeomorphism, which are homotopy equivalent (hence by Smale's h-cobordism theorem, and in low dimensions Perelman's and Freedman's work, homeomorphic) to the $n$-sphere $S^n$. (Actually, we identify $h$-cobordant manifolds. Because $h$-cobordism is now known in all dimensions at least 5, it changes nothing for high-dimensional manifolds; but it explains why $\Theta_4=1$ is possible even though it's an open problem, suspected to be false, that the 4-sphere admits a unique smooth structure. In any case, this is not an important aside.) The group operation is connected sum. The data we're really after is $|\Theta_n|$ - the number of smooth structures.

$\Theta_n^{bp}$ is the subgroup of those $n$-spheres which bound parallelizable manifolds. This subgroup is essential, because it's usually the fellow forcing us to have exotic spheres in the other dimensions.

This group is always cyclic (Kervaire and Milnor provide an explicit generator). As a rough justification for this group: the way this goes is by taking an arbitrary element, writing down a parallelizable manifold it bounds, and using the parallelizability condition to do some simplifying algebra until this bounding manifold is particularly simple - at which point you identify it as a connected sum of standard ones, hence that $\Theta_n^{bp}$ is cyclic generated by the standard one. I (or rather, Milnor and Kervaire) can tell you its order: If $n$ is even, $|\Theta_n^{bp}| = 0$; if $n=4k-1$, $$|\Theta_n^{bp}|=2^{2k-2}(2^{2k-1}-1) \cdot \text{the numerator of }\frac{4B_k}{k}$$ is sort of nasty, but in particular always nonzero when $k>1$; and for $n=4k-3$, it is either 0 or $\Bbb Z/2$, the first precisely if $\Phi_k \neq 0$ in the above exact sequence.

$\pi_n/J$, and the map $\Theta_n \to \pi_n/J$, is a bit harder to state; $\pi_n$ is the stable-homotopy group of spheres, $J$ is the image of a certain map, and the map from $\Theta_n$ sends a homotopy 7-sphere, which is stably parallelizable, to its "framed cobordism class". The real point, though, is that this term $\pi_n/J$ is entirely the realm of stable homotopy theory. This is precisely why people now say that the exotic spheres problem is "a homotopy theory problem". (To give the slightest bit more detail: The Thom-Pontryagin construction gives that $\pi_n = \Omega_n^{fr}$, the framed cobordism group, whose elements are equivalence classes of manifolds with trivializations of the "stable tangent bunde". Every homotopy sphere is stably trivial, and the image of $J$ is precisely the difference between any two stable trivializations.) This map $\Theta_n \to \pi_n/J$ might motivate the introduction of $\Theta_n^{bp}$ - since that is, more or less obviously, the kernel. The fact that this map is not always surjective - the obstruction supplied by $\Phi_k$ - is the statement that not every framed manifold is framed cobordant to a sphere. I find it somewhat surprising that so many actually are!

The last thing you should know is about the map $\Phi_k$. It's known as the Kervaire invariant. It's known to be nonzero in dimensions $k=1,2,4,8,16$, and might be nonzero in dimension $32$, but that's open. The remarkable result of Mike Hill, Mike Hopkins, and Doug Ravenel is that $\Phi_k = 0$ for $k > 32$. I don't have much to say about this, other than that it's there. Summing up what we have so far:

For dimensions $n=4k-1>3$, there are always exotic spheres coming from $\Theta_n^{bp}$ - lots of them! For dimensions $n=4k-3$, $\Theta_n^{bp} = \Bbb Z/2$ unless $k=1,2,4,8,16,32$. So the only possible odd-dimensional spheres with a unique smooth structure are $S^1$, $S^3, S^5, S^{13}, S^{29}, S^{61}$, and $S^{125}$.

Now to deal with special cases. It is classical that $S^1$ and $S^3$ have a unique smooth structure ($S^3$ is due to Moise); $S^5$ is dealt with by 1) finding a 6-manifold of nonzero Kervaire invariant, showing that $\Phi_2 \neq 0$ and hence that $\Theta_5^{bp}=0$; and then 2) calculating that $\pi_5$, the fifth stable homotopy group of spheres, is zero. You can do this with Serre's spectral sequence calculations. (It was pointed out to me that this means that three different field's medalists work went into getting $\Theta_5 = 1$ - Milnor, Serre, Smale. It is worth noting that there is a differential topological proof, coming from the explicit classification of smooth, simply-connected 5-manifolds, but it isn't substantially easier or anything.)

For $S^{13}$ and $S^{29}$, these are disqualified by the homotopy theory calculation that $\pi_{13}/J$ and $\pi_{29}/J$ are not zero. I do not know how these calculations are done - probably the Adams spectral sequence and a lot of auxiliary spectral sequences, which seems to be how a lot of these things are done. Maybe someone else can shed some light on that.

For $S^{125}$, the paper itself sketches why: There's a spectrum known as $tmf$, and the authors are able to write down a homomorphism $\pi_n/J \to \pi_n{tmf}$ and find a class in $tmf$ that's hit when $n=125$.

So what we know now is that $\pi_{61}/J \cong \Theta_n$. The content of the paper you're talking about is precisely the calculation that $\pi_{61}/J = 0$. The authors access it through the Adams spectral sequence, as far as I can tell (I am a non-expert). Adams SS is notoriously hard to calculate anything with - mostly the entire content of the paper is identifying of a single differential in the whole spectral sequence. Once this is done, they're able to finish the calculation, but it's hard work. If you want a sketch of how this is done, I found the introduction to their paper readable - see section 3 of the paper.

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    Thank you for this lucid tutorial. You have done the community a service! – Joseph O'Rourke Jan 12 '16 at 16:31
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    @JosephO'Rourke: Another advisable read is Milnor's wonderful [Differential topology 46-years later](http://www.ams.org/notices/201106/rtx110600804p.pdf). Most of this post comes from that. –  Jan 12 '16 at 16:33
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    I stand corrected in my previous comment. This is a great overview of the state of the area leading up to this result. Thanks Mike! – Dan Rust Jan 12 '16 at 16:36
  • @MikeMiller Sorry to be late to the party, but I hit this post when reading about Milnor's 46 years paper. What confuses me is whether the uniqueness of the smooth structure on $S^{61}$ is the *new* result of the cited paper, because on the right column of P807 of Milnor's paper it is stated that $S_{61}$ (or $\Theta_{61}$ in your notation) is trivial, and it is also stated at the beginning of that paragraph that "and hence the group $S_n$ is precisely known for $n\le64$, $n\neq4$," so I am wondering what is the new contribution of the paper cited by the OP. – Fan Zheng Jul 22 '18 at 05:44
  • @FanZheng Milnor's paper cites Kochman-Mahowald, but see the bottom of page 4 of Wang-Xu: "Recently, Isaksen [23] discovered several errors in Kochman and Mahowald’s computations, and he was able to give rigorous proofs of computations through the 59-stem". In particular, their paper claims to give the first correct calculation $\pi_{61} = 0$. There is some more detail on page 7. –  Jul 22 '18 at 11:58