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There's a workshop at MSRI in a couple months on the Kervaire invariant problem that I'd really like to attend. I saw Hopkins speak about it a while back without understanding much of the talk, but I'm thinking that maybe by now I'm more prepared to see what it's all about. At least, I'd like to be able to get something out of the workshop, even if I don't understand everything that's going on. I don't know much more about the problem than the wikipedia article can tell me. So, I'm looking for:

(a) recommendations for reading to acquaint myself with the necessary background information, and

(b) a sketch of the story -- I can somewhat follow the intro to the paper itself, but seeing some of the details spelled out would help me start to actually understand what's going on.

(Or, unfortunately possibly: (c) a warning that it's probably too difficult to pick up all the material in just a few months and that it's not worth my time to try.)

By the way, wikipedia says that Kervaire used the invariant to create a 10-dimensional PL manifold with no differentiable structure, but I thought that the invariant is zero in dimension 10. What's the deal there?

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    $\begingroup$ I think the invariant is zero in dimension 10 for smooth manifolds... he probably found a PL manifold with nontrivial Kervaire invariant, and concluded that it had no differentiable structure. $\endgroup$
    – Dylan Wilson
    Commented Jul 25, 2010 at 6:53
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    $\begingroup$ Hopkins gave a talk on this at Atiyah's birthday conference in 2009. Perhaps you meant this talk or you already know its contents, but anyways, you can find a video and the slides of this talk here: maths.ed.ac.uk/~aar/atiyah80.htm (the title is "Applications of algebra to a problem in topology") $\endgroup$
    – user717
    Commented Jul 25, 2010 at 7:39
  • $\begingroup$ @Dylan -- I'm still a little confused, because apparently to define the Kervaire invariant you need a framing, which as I understand it definitely requires differentiability. Maybe there's a combinatorial/non-smoothness-requiring way to extend the definition? $\endgroup$
    – Aaron Mazel-Gee
    Commented Jul 25, 2010 at 16:18
  • $\begingroup$ Sure, sounds great. $\endgroup$
    – Aaron Mazel-Gee
    Commented May 20, 2011 at 20:12

3 Answers 3

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Here's a recent survey article by Victor Snaith: http://chucha.math.cinvestav.mx/morfismos/v13n2/arfsurveyMFMS.pdf

(I think there's also a copy on the arxiv)

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  • $\begingroup$ Awesome, thanks. This is what I was looking for! $\endgroup$
    – Aaron Mazel-Gee
    Commented Jul 26, 2010 at 5:08
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    $\begingroup$ That's just the beginning! Snaith has a whole book on the subject: books.google.com/… $\endgroup$
    – Dan Ramras
    Commented Jul 26, 2010 at 5:18
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Snaith's book is good place to look. Also, Hopkins lecture at Atiyah's birthday is awesome. It was caled the doomsday conjecture because of what would happen in the EHP sequence if these elements were not homotopy classes. In this vein, i should point out the "recent" note that appeared online http://www.math.northwestern.edu/~pgoerss/papers/kinoteNov16.pdf by Goerss and Mahowald. both of which understand stable homotopy theory extensively. Mahowald is a major influence on this whole subject and specifically an expert in this area.

As for dylan's comment, I believe he is exactly right. The kervaire invariant does not require a smooth structure. The cohomology of a manifold has a product, and if the manifold is even dimensional then the cup product pairing gives us a quadratic form on the middle dimensional cohomology/homology group. The kervaire invariant is the arf invariant of this quadratic form. SO there is no need for a smooth structure. What is probably the case, is that he showed that the framing of a manifold could be used to show the vanishing of the kervaire invariant. I suggest looking at the work of kervaire and kervaire-milnor, they are excellent writers.

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  • $\begingroup$ Okay, great. The wikipedia article makes it sound like even though you're just looking at the intersection pairing on Z/2 homology / cup pairing on Z/2 cohomology, the framing would be important in defining the thing (helping you distinguish (co)homology classes or something??), but the note you link to clears it up. That's cute, the "democratic invariant". What were some potential apocalyptic consequences hinging on the fate of elements of EHP, do you know? $\endgroup$
    – Aaron Mazel-Gee
    Commented Jul 26, 2010 at 16:20
  • $\begingroup$ there are a couple of apocalyptic type things, one EHP style and one Adams SS style. The EHP sequence stuff is that if these elements fail to be genuine homotopy classes then the EHP SS would not work the way that Mahowald expected/hoped/thought it would. one of those words is right, but i don't know which. So the ramifications are in the complexity of the relationship between stable and unstable homotopy theory. The other i know less about (amazing i know). It has something to do with the action of "algebraic" sq^0 on the E_2 term of the Adams SS at the prime 2. Recall that ... $\endgroup$
    – Sean Tilson
    Commented Jul 26, 2010 at 17:30
  • $\begingroup$ ...there is an action of the steenrod algebra on Ext and sq^0 is not the identity operation there. This action is not coming from topology except by analogy, but from the fact that Ext is a cohomology theory, so there is some sort of steenrod algebra lurking around. This moral reason could be BS. But the relevant article is May's article from the steenrod 60th bday conference where he sets up a unifying approach. $\endgroup$
    – Sean Tilson
    Commented Jul 26, 2010 at 17:34
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Have a look at the MIT-Harvard Seminar and the lecture notes given at this link.

For some background, see Petr. M. Akhmetev's work.

I see that you are starting with the wikipedia article. For very very introductory stuff, such as statement itself, there's the article of Teichner, article in Scientific American, a Simons foundation article with some animations and an article in Nature.

Most of the above are lifted shamelessly from Douglas Ravenel's homepage.

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    $\begingroup$ Akhmetev's work is a completely different approach to the solution more related to geometry than homotopy theory. I think that is at least worth noting. Also, I don't believe his solution to the conjecture works, Landweber has a recent preprint on the arxiv with a counterexample. While I have not read the paper, I would tend to trust Landweber seeing as he has been the one translating Akhmetev's work to english. $\endgroup$
    – Sean Tilson
    Commented Jul 26, 2010 at 20:15
  • $\begingroup$ Akhmet'ev now has a new version on the arxiv, which avoids Landweber's example by going further into iterated self-intersections and deeper into their structure groups. I'm not saying that I checked his proof. As of August, I could not even understand why he needs the 5-fold self-intersections but not the 6-fold ones. $\endgroup$
    – Sergey Melikhov
    Commented Dec 14, 2010 at 0:48

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