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Is there any resource that might help non-experts gains some understanding of why the Kervaire invariant problem remains open now only in dimension $126$? ($126 =2^7-2=2^{j+1}-2$; whether $\theta_j=\theta_6$ exists in the "$128$-stem"), i.e., why the celebrated Hill-Hopkins-Ravenel proof technique fails in this last remaining case? I read a delightful exposition by Erica Klarreich ("Mathematicians solve 45-year-old Kevaire Invariant Puzzle," The Best Writing on Mathematics, 2010, 373ff, Simons Foundation link), which piqued my interest. But Michael Hopkins' online presentation slides announcing the result in 2009 ("Applications of algebra to a problem in topology") are beyond my ken. Perhaps this an area too abstruse for all but the experts? I'd appreciate pointers to expositions. Thanks!

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The problem is complicated and either you dive into it or you won't understand anything at all, there's no way of posting a short enough answer to your question. I'd recommendo you to read Snaith's book on the topic an then the papers by Hill-Hopkins-Ravanel and Akhmetev. –  Fernando Muro Jul 1 '12 at 0:57
There was also a Bourbaki Seminar talk by Haynes Miller, the report is available on arXiv arxiv.org/abs/1104.4523 –  quid Jul 1 '12 at 1:01
Thanks, quid, that is a great resource, exactly what I was seeking. And thanks, Fernando, for the reference to (I guess?): Snaith, Victor P. (2009), "Stable homotopy around the Arf-Kervaire invariant," Progress in Mathematics, 273, Birkhäuser Verlag. –  Joseph O'Rourke Jul 1 '12 at 1:31

2 Answers 2

I'll give a shot at an answer. The relevant dimensions are of the form $2^j-2$. For $j\leq 4$, it is easy and classical that we can construct manifolds of Kervaire invariant one. The problem was ``reduced'' from differential topology to pure stable homotopy theory by Browder in 1969. Direct calculational methods in homotopy theory were used by Barratt, Jones, and Mahowald to construct a cell complex that can be used to solve the homotopy theory problem and prove that such manifolds also exist in dimensions 30 and 62. I believe a construction of such a manifold has been worked out in dimension 30, but that has certainly not been done in dimension 62. Periodicity phenomena play a huge role in modern stable homotopy theory, and a crucial feature of the Hill, Hopkins, Ravenel proof is a periodicity of order $2^8 = 256$. That enables them to solve the stable homotopy problem and prove there is no manifold of Kervaire invariant one for $j\geq 8$. The reasons $j=7$ is so hard are several. Nobody has a really good reason for guessing which way the answer will go. There is no reason to expect a relevant periodicity of order $2^7$. Direct calculation of the Adams spectral sequence through dimension $126$ is just plain hard: the calculations blow up. There is a chance that the methodology of Barratt, Jones, and Mahowald might extend to prove existence (if that is how the answer turns out!), but it will probably be much harder to prove nonexistence (if that is the answer).

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Thank you, Peter, for this succinct and clear summary! –  Joseph O'Rourke Jul 1 '12 at 12:11
I disagree with the statement that there is no reason for guessing which way the answer will go. There is a good reason, which I learned from Atiyah, to guess that the answer in dimension 126 is the same as in the lower dimensions 62, 30, 14,... Namely, there exist remarkable manifolds in dimensions 128, 64, 32, and 16. These are the octooctonionic projective plane (see math.ucr.edu/home/baez/octonions/node19.html), the quateroctonionic projective plane (see math.ucr.edu/home/baez/octonions/node18.html), the bioctonionic projective plane, and finally $OP^2$ itself. –  André Henriques Jul 1 '12 at 19:26
I should probably add that I hope you are right, since I have a student who hopes to prove that you are (homotopically of course). –  Peter May Jul 1 '12 at 21:30
I realize that this is not a complete argument. Atiyah's hope was that there should exist a geometric construction that starts with the quateroctonionic projective plane and constructs a 62-manifold of Kervaire invariant one, and that the same construction, applied to the octooctonionic projective plane would produce a 126-manifold of Kervaire invariant one. –  André Henriques Jul 2 '12 at 12:07
I am over a year late but it seems worth mentioning that you are of course correct about the construction of the 30-dimensional manifold. According to these notes of Jones: math.rochester.edu/people/faculty/doug/otherpapers/jones3.pdf, the 30-dimensional case can be obtained as $M = Y\times S^7\times S^7\times S^7\times S^7/D_8$ where $Y$ is a genus-5 surface, $D_8$ is the dihedral group and the octonionic framing of $S^7$ determines the one used for M. –  Oliver Nash Aug 4 '13 at 10:55

More detail on Jones's 30 manifold and analogous constructions can be found in my Extended powers of manifolds and the Adams spectral sequence , Cont. Math. 271, 41--51. The basic idea in it dates from 1979 or so (immediately upon seeing Jones's construction) but didn't make it into print for another 20 years. There are also comments about the homotopy theoretic approach to the problem in the talk I gave in Edinburgh in 2011 at the workshop on the Kervaire invariant.

The main point is that there is a small set of homology classes in low dimension (dimension 2 for Jones' construction) which one wants to realize as the fundamental class of a manifold with tangent bundle realized by a permutation representation of $\pi_1$. To get the 30 dimensional Kervaire manifold from $S^7$ one only needs a two manifold and an appropriate representation in $S_4$. ($D_8$ will do: only the Sylow 2-subgroup matters.) To get from $S^7$ to the 62 dimensional Kervaire manifold requires a 6-manifold with a representation of $\pi_1$ in $S_8$. Unfortunately, no such manifold exists. In dimension 126, one would need a 14 manifold, and presumably such also fails to exist. So, one needs to find another approach that relaxes the input data needed.

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Thanks, Robert, this is very insightful concerning the $14$-manifold's (apparent) non-existence. –  Joseph O'Rourke Oct 4 at 0:05

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