## April 23, 2009

### ‘Kervaire Invariant One Problem’ Solved

#### Posted by John Baez

Big news! It seems Mike Hopkins, Doug Ravenel and Mike Hill have cracked the Kervaire Invariant One problem.

Hopkins announced this in a maximally dramatic fashion, as explained below…

Here’s what Nick Kuhn wrote on the ALGTOP mailing list:

Yesterday, at the conference on Geometry and Physics being held in Edinburgh in honor of Sir Michael Atiyah, Harvard Professor Mike Hopkins announced a solution to the 45 year old Kervaire Invariant One problem, one of the major outstanding problems in algebraic and geometric topology. This is joint work with Rochester professor Doug Ravenel and U VA postdoctoral Whyburn Instructor Mike Hill.

The solution completes the work on ‘exotic spheres’ begun by John Milnor in the 1950’s which led to his Fields Medal. This is a central part of the classification of manifolds (= curves, surfaces, and their higher dimensional analogues). A 1962 Annals of Math paper by Milnor and Michael Kervaire classified exotic differential structures on spheres, subject to one possible ambiguity of order 2 in even dimensions. A 1969 Annals Math paper by Princeton professor William Browder resolved this question, except when the dimension was 2 less than a power of 2. In these dimensions, he translated the problem into one in algebraic topology, specifically one about the existence of certain elements in the stable homotopy groups of spheres. Over the next decade, the elements in dimensions 30, 62, and 126 were shown to exist; equivalently there exist some manifolds in those dimensions with some oddball properties. Significant work on closely related problems was done by Northwestern professor Mark Mahowald.

So yesterday’s announcement was that in all higher dimensions (254, 510, 1022, etc.), the putative elements do NOT exist. This result is ‘detected’ in a generalized homology theory that is periodic of period 256 built from the complex oriented theory associated to deformations of the universal height 4 formal group law at the prime 2. (By contrast, real K-theory is has period 8 and comes from height 1 deformations, and theories based on elliptic cohomology come from height 2.) The strategy of proof has similarity to work of Ravenel’s from the late 1970’s, but the success of the strategy now illustrates the power of newly emerging control of subtle number theoretic and group theoretic structure in algebraic topology.

(2) Technical stuff, which may or may not be accurate …

Step 1. Using results/methods from Miller, Ravenel and Wilson, one can show if $\Theta_j$ is nonzero, then it is nonzero in $\pi_*(E_4^{hZ/8})$, for some well chosen action of $Z/8$ on the 4th 2-adic Morava E theory.

Step 2. Using a spectral sequence associated to a cleverly chosen filtered equivariant model for $E_4$ (or similar ??) - and this is the very new bit, I think - one shows that

(a) $\pi_{-2}(E_4^{hZ/8}) = 0$ and

(b) $\pi_*(E_4^{hZ/8})$ is 256 periodic.

Thus the $\Theta_j$’s cannot exist beginning in dimension 254.

Café regulars should be tantalized to hear that Hopkins’ argument involves invariants that live two steps higher in the ‘chromatic filtration’ than elliptic cohomology. If you have no idea what that means, try week197 for a gentle introduction. But if even that’s too tough, here’s the idea in even more simplified form.

$K$-theory is an invariant of topological spaces that’s built using vector bundles. In quantum theory we use vector bundles to study how point particles move around under the influence of forces called ‘gauge fields’, like electromagnetism .

Elliptic cohomology is an invariant of topological spaces that’s one step higher in a sequence called the ‘chromatic filtration’ — it’s ‘height two’. It began life as a piece of very abstract homotopy theory — which Mike Hopkins was instrumental in developing. But thanks to some brilliant work by people including Ed Witten, Graeme Segal, and more recently Stephan Stolz and Peter Teichner, it’s becoming clear that this math is related to the physics of strings. The worldsheet of a string has dimension one higher than the worldline of a particle.

So, while I bet it’s not yet visible in Hopkins’ work, I bet that the invariant Hopkins uses — the 4th 2-adic Morava $E$ theory — will turn out to be at least slightly related to theories of 3-branes.

(If this guess is completely stupid, please explain why.)

Posted at April 23, 2009 5:36 PM UTC

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### Re: ‘Kervaire Invariant One Problem’ Solved

My personal spy on that conference, Danny Stevenson, mentioned, if I understood him correctly, that Hopkins also said something about the existence of “generalized Clifford algebras” in some of these dimensions. Er, maybe I don’t know in which dimensions, is it these 254, 510, 1022… and then maybe 34560?

Anyway, I am guessing this must be related to (an even higher version of) the higher Clifford algebras discussed here?

One kind of higher Clifford algebra per chromatic filtration/brane dimension?

Posted by: Urs Schreiber on April 23, 2009 7:32 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

will turn out to be at least slightly related to theories of 3-branes

Somehow it would have felt better if it had been not 3 but $4 k + 1$ for $k \in \mathbb{N}$. That’s the brane dimensions where special relations to higher Chern-Simons theories, self-dual forms etc. exist, and its the dimensions for which one expects there to be “fundamental” or “NS”-branes: the string, the fundamental 5-brane, maybe the space-filling NS 9-brane.

(Not that I claim to have more then a superficial understanding of the specialty of $4 k +1$ here. )

Posted by: Urs Schreiber on April 23, 2009 7:43 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

A ‘theory of 3-branes’ can be thought of as a 4d quantum field theory. Thanks to Donaldson, Seiberg, and Witten, we can say that 4d quantum field theory is the best way to understand exotic smooth structures in 4 dimensions. And now perhaps it’s secretly lurking behind this new understanding of smooth structures in dimensions 254, 510, 1022, etcetera!

Of course, that’s a wild extrapolation from some lower-dimensional examples. In week255 I sketched Stolz and Teichner’s audacious ideas relating deRham cohomology, $K$-theory and elliptic cohomology to the set of concordance classes of supersymmetric quantum field theories in dimensions 0, 1, and 2, respectively. I don’t know if they’ve dared write anything about dimension 4.

Posted by: John Baez on April 24, 2009 2:33 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I’m waiting for Greg Egan to set some fiction in those dimensions.

Posted by: Jonathan Vos Post on April 24, 2009 3:38 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

A few days ago, people dared to write about this very stuff in dimension 4, but with more topology in mind:

Torsten Asselmeyer-Maluga, Jerzy Król, “Gerbes, SU(2) WZW models and exotic smooth R^4”

Posted by: Daniel de França MTd2 on April 24, 2009 6:01 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

A ‘theory of 3-branes’ can be thought of as a 4d quantum field theory. Thanks to Donaldson, Seiberg, and Witten, we can say that 4d quantum field theory is the best way to understand exotic smooth structures in 4 dimensions.

Posted by: Urs Schreiber on April 24, 2009 10:39 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Besides reading that book, take at the article I cited above, written by one of the authors of the book you mention, Torsten. You should also check the book “The Wild World of Exotic Manifolds”, by Scorpan, a very exciting book.

John recommended me last year, when I was totaly confused with the issue of making sense of realistic black holes and thinking them as non differentiable spheres. Luckily, this book made things clear to me and gave me a lot of enlightment.

(What I was thinking was the concept of fuzzy spheres, the real world stringy black holes, at least according to Lubos Motl.)

Posted by: Daniel de França MTd2 on April 24, 2009 11:43 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I’m sorry, the name of the book is “The Wild World of 4 Manifolds”.

Posted by: Daniel de França MTd2 on April 24, 2009 11:46 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

fuzzy spheres

I know of the idea of modelling event horizons as fuzzy spheres, but I am not aware of a relation of fuzzy spheres to exotic spheres.

Is this just wild speculation based on the general intuition of “crumpled spheres”, or is there a precise relation?

Well, googling for these keywords tuns up one article that talks about the ” fractal (fuzzy) Milnor seven-dimensional exotic sphere”. But guess by whom.

Posted by: Urs Schreiber on April 24, 2009 3:34 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

“Is this just wild speculation based on the general intuition of “crumpled spheres”, or is there a precise relation?”

This was a wild speculation, indeed, at the time, with exactly that motivation. But I wouldn’t even try to use the El Naschie article you cited as a reliable source…

Let me say that the reason right now I like non differential spheres ( the standard example, the E8 manifold, is the ponincare or homology sphere) it is it seems to me, it is that in most cases I’ve seen, they work (my opinion) be a source for the exoctiness of 4 manifolds. But, if you want to see something formal, try this:

“Foundational Essays on Topological Manifolds, Smoothings, and Triangulations. (AM-88) Robion C. Kirby and Laurence C. Siebenmann”, look at Annex C, starting at page 307. The book is exclent, but what I want to show you is on p.311, figure 2.a. This picture summarises what is written in section 2, and if you pay attention at the explanations and deductions, you will see that, although it is not written there, that only Weistrass-like Functions can describe non differentiable manifolds.

You know that quantum mechanics has also a natural ramdomess, due its probabilistic nature. So, if you attach something to a non differentiable structure, it is likely that you get an automatic quantization, in a few cases. The speculation of the last statement is not clear anywhere, except if you consider that a non differential is really a source for any kind exoticness. If this is the case, then the article I linked above is such example of a system that is quantized automaticaly. (They actualy argue for an equivalence between exotic smoothness and wzw quantization)

Posted by: Daniel de França MTd2 on April 24, 2009 4:28 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Note that it seems to be a source for small exotic manifolds, sorry if I said “most of”. But anyway, for such equivalence to be true, there must be a failure of the generalized poincare conjecture in 4 dimensions, whose proof is sketched in the article, and in there, small and large exoticness are linked.

Note that exotic smoothness only exists in 4 dimensions.

Posted by: Daniel de França MTd2 on April 24, 2009 4:36 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I don’t get why this would solve the problem of exotic spheres because there is no mention about the existence of exotic spheres in 4 dimensions…

Posted by: Daniel de França MTd2 on April 23, 2009 8:36 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

This work doesn’t complete the classification of exotic spheres — as you note, it says nothing about dimension 4. Nick Kuhn was being a bit imprecise. This work applies only to ‘high-dimensional’ spheres, where exotic smooth structures can be classified using homotopy theory. ‘High-dimensional’ means dimension 5 or more.

I’m not an expert on this stuff, but I gather the situation is like this:

In dimensions $n \ge 5$ or more, the set of smooth structures on the $n$-sphere forms a group $\theta_n$ which also has a nice description using homotopy theory. In particular, there’s a homomorphism

$K : \theta_n \to \pi_n^S / im J_n$

where $\pi_n^S$ is the $n$th stable homotopy group of spheres and

$J_n : \pi_n(SO(\infty)) \to \pi_n^S$

is the so-called $J$-homomorphism.

In dimensions that equal 2 mod 4, the kernel of $K$ is either trivial or $\mathbb{Z}/2$.

If it’s trivial, we completely know a smooth structure on the $n$-sphere as soon as we know its image under $K$. So, we are happy.

If it’s $\mathbb{Z}/2$, we don’t: there’s a weird smooth structure on the $n$-sphere whose image under $K$ is trivial, just like the usual smooth structure. So, we are nervous.

In 1969, Browder showed that the kernel of $K$ is trivial unless $n$ is 2 less than a power of 2.

Whew!

But in the next decade, it was shown that the kernel of $K$ is not trivial if $n$ is $6, 14, 30, 62$ or $126$.

The Kervaire Invariant One problem asks about all the other numbers that are 2 less than a power of two.

Now we have been told that for all these other numbers, the kernel of $K$ is trivial.

So, until recently, for all we knew, there might have been a really weird smooth structure on a 254-dimensional sphere, or the 510-dimensional sphere, or the 1026-dimensional sphere… Now we know there’s not.

So, we can all sleep a bit better tonight.

Posted by: John Baez on April 23, 2009 9:31 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Thanks John! I understand now. But, in case 126 is non trivial, you would have another set of 5 numbers to collect in your entry “5” of favorite numbers. Might these all these be related to the perfect solids somehow?

Posted by: Daniel de França MTd2 on April 24, 2009 5:48 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Let me see if I understand the whole story:

In dimensions 1,2,3 there are no exotic spheres.

In dimension 4 the problem is still totally wide open.

Dimension >4 and 2 mod 4 we use the approach you outline and see that except for 6,14,30,62 and possibly 126 the exotic spheres are classified by the cokernel of the J-homorphism. To what extent does this solve the problem and to what extent does it kick it down the road? That is how hard is it to compute the cokernel of J? Is it computable in theory? In practice?

Dimension >4 and n not 2 mod 4 what do you do? Nick Kuhn’s email seems to suggest that that’s solved, but your comment requires n not 2 mod 4?

Posted by: Noah Snyder on April 24, 2009 7:45 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

For your last question, check out the Wikipedia page on exotic spheres. Basically there is a well understood subgroup $bP$ such that the map

$\theta / bP \to \pi^s/im \; J$

is always injective. The cokernel is the Kervaire invariant (Browder 1969). Thus for dimensions not equal to $2^k - 2$ it is an isomorphism, and now we know it is also an isomorphism except in dimensions 6, 14, 30, 62, and possibly 126.

So this reduces the problem to computing the stable homotopy groups modulo the image of $J$. The image of $J$ is completely understood.

However the computing the rest of the stable homotopy groups is probably the holy grail of algebraic topology. A lot of structure is known, but they’ve only been completely computed for small dimensions. Somewhere between 100 and 300.

I think the phrase “hopelessly difficult” comes to mind.

Posted by: Chris Schommer-Pries on April 24, 2009 8:57 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

In the heat of the moment, John got a vital detail of the Kervaire invariant setup wrong: for $n=4k+2$ the Kervaire invariant is a map $K:\pi^S_n/im J_n \to Z_2$ with kernel $\theta_n$, not the map $\theta_n \to \pi^S_n/im J_n$ in the post. In view of the renewed interest in the Kervaire invariant and exotic spheres I have written an addendum to my slides on the work of Michel Kervaire in surgery and knot theory.

Posted by: Andrew Ranicki on May 6, 2009 9:27 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Posted by: Andrew Ranicki on May 6, 2009 9:37 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Andrew, I have a few requests about articles that have to say something about these topics:

-To every Casson handle one can construct a grope and vice verse.
-How the Alexander’s horn relate to a Casson Handle.

Posted by: Daniel de França MTd2 on May 6, 2009 3:39 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Daniel: while Andrew is practically omniscient when it comes to topology, I don’t think he works on the special features of 4d topological manifolds, like Casson handles. Maybe you should find people who do.

Posted by: John Baez on May 7, 2009 12:02 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Andrew said that is a page on exotic spheres, and the of the most important cases that remains unsettled is the one about the existence of exotic 4 spheres. The answer to what I asked is basicaly a way to visualize exotic structures in 4d without rather abstract diagrams like the use of liking spheres. Unfortunantely, there are no preprints available nor in arxiv neither anywhere else.

Well, for anyone who wants the literature, this is the indicated literature, althought I myself couldn’t get them:

- J.W. Cannon. The recognition problem: What is a topological manifold? BAMS, 84:832 - 866, 1978.
- J.W. Cannon. Shrinking cell-like decompositions of manifolds: Codimension three. Ann.
Math., 110:83 – 112, 1979.
-M. Freedman and F. Quinn. Topology of 4-Manifolds. Princeton Mathematical Series.
Princeton University Press, Princeton, 1990

Posted by: Daniel de França MTd2 on May 7, 2009 2:07 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Sadly, I am not practically omniscient in topology, and not even in my specialty of high dimensional manifolds. In any case, John is right about my ignorance of Casson handles etc. – especially shameful since Casson was one of my Ph.D. supervisors! Even worse, I didn’t even know how to display TeX in my $n$-category Cafe posting, until John explained to me that if I had chosen “itex to mathml with parbreaks” in the Text Filter box, the TeX in my post would have come out nicely. I have at least learnt that. Post factum, here then is the TeX in my original post, correctly displayed: for $n=4k+2$ the Kervaire invariant is a map $K:\pi^S_n/im J_n \to Z_2$ with kernel $\theta_n$, not the map $\theta_n \to \pi^S_n/im J_n$.
Posted by: Andrew Ranicki on May 7, 2009 6:49 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Isn’t the case n = 126 still open?

Posted by: Chris Schommer-Pries on April 23, 2009 10:11 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I’m a bit puzzled about this. The all-knowing Wikipedia suggests that $n = 126$ is still open — but in the quote above, Nick Kuhn says that it’s not.

Posted by: John Baez on April 23, 2009 10:20 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Ahh… Later on, Nick sent the following to the ALGTOP mailing list:

“Regarding my message of yesterday about the Hill-Hopkins-Ravenel announcement: various folks have by now written me that the Kervaire question in dimension 126 is still open. My bad.

I’m glad we still have something to work on. :)”

So I guess, yes. It is still open.

Posted by: Chris Schommer-Pries on April 23, 2009 10:26 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

The slides containing the sketch of the proof are presented here (50MB warning)!. I found it today on wikipedia on Kervaire invariant entry. The file is hosted on Andrew Ranicki’s Webpage.

Personaly, the most mysterious part for me is the second last page, where it asks about a 4 dimensional QFT (what is that?!?), and “generalized Clifford algebras” with peridiocity of 34560.

Posted by: Daniel de França MTd2 on April 25, 2009 12:57 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Here’s part of an email from Mike Hopkins:

First of all, my apologies about the size of the file for my slides. I did the talk in apple’s keynote (and used illustrator for all of the charts and artwork). For some reason keynote always makes HUGE pdf files. I suspect it is because of the “chalkboard” background I chose. I also suspect it was kind of hard to tell what I talked about from the slides alone. The talks were recorded, so the audio (and visual) should be available somewhere.

One thing I wanted to comment on was your discussion of exotic spheres. Way back in the 1930’s when Pontryagin introduced his beautiful generalization of the notion of the “degree” of a map he introduced the basic maneuver of framed surgery and made a tiny mistake. He was trying to compute the cobordism group of stably framed 2-manifolds. He argued that if the genus is zero, the manifold bounds, since the usual 2-sphere bounds, and there is only one stable framing since $\pi_2$ of a Lie group is trivial. He then introduced the notion of framed surgery (he didn’t call it that) to reduce the genus. You have to choose a circle along which to do the surgery, and it has to be chosen so that the framing of the circle you chose actually bounds a framed disk. This defines a map

$H_1(\Sigma;\mathbb{Z}/2) \to \mathbb{Z}/2$

and Pontryagin assumed at first that it was linear. In his book on differential topology which appeared much later he noted that it is a quadratic refinement of the intersection pairing, and that the Arf invariant is a cobordism invariant.

Without going through the whole history, let me just say that this exact situation generalizes to dimension $4k+2$. The Arf (Kervaire) invariant is the obstruction to doing framed surgery and reducing your manifold to a sphere. In the language of homotopy theory, there is a map

$\pi_{4k+2} S^0 \to \mathbb{Z}/2$

and the kernel consists of (stably) framed manifolds which are framed cobordant to a sphere. This map is zero on the image of the $J$ homomorphism and the Kervaire–Milnor group in dimension $4k+2$ is (if memory serves) the kernel of

$\pi_{4k+2}/Im J \to \mathbb{Z}/2.$

We now know that that, except for 5 (or maybe 6) values of $k$, this map is zero. You can think of it as finally completing the reduction of the problem of exotic spheres to homotopy theory, or, as showing that Pontryagin’s original construction only actually fails in 5 or 6 dimensions, and in all other dimensions of the form $(4k+2)$ every framed manifold is framed cobordant to a sphere.

Of course the Kervaire invariant problem had many other forms and played many other roles in algebraic and geometric topology.

One cute formulation is in the front piece of Ioan James’ book on Stiefel Manifolds. He considers the configuration space of ordered pairs of distinct and non-antipodal points $(x,y)$ in $S^n$, and asks if there is a homotopy of the identity map of this space with the map $(x,y) \to (y,x)$. If you look at the case of $S^2$ you’ll get the idea of rotating 180 degrees in the counter-clockwise direction around $(x+y)/|x+y|$. A little thought shows that this would work on $S^n$ if you had an almost complex structure. So this gets you $n=2$ and $n=6$, which are the first two numbers of the form $2^{n-2}$. I’m not sure how one might go further with this, but it’d be nice to do it somehow. I also have to confess to never having worked out the relation of this to the Kervaire invariant, though for some reason I’ve long been under the impression that constructing such a homotopy divides a certain whitehead product by 2, and that that gives you an element of Kervaire invariant 1. Oh, and if you actually look at the book you’ll see that there’s a typo in the statement, and that what he actually states is obviously wrong.

Whew! Didn’t mean to write so much!

Mike

Posted by: John Baez on April 26, 2009 3:13 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Wow, this is a much nicer way of stating the Kervaire invariant one problem: “in sufficiently large dimensions equal to 2 mod 4, every stably framed manifold is framed cobordant to a sphere.”

It might be fun to ponder this result from the viewpoint of the Cobordism Hypothesis, where stably framed $n$-manifolds are $n$-morphisms from the identity of the identity of the identity… to itself in the free stable $(\infty,n+1)$-category on a fully dualizable object, and framed cobordisms between these are $(n+1)$-morphisms.

Posted by: John Baez on April 26, 2009 3:39 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Here is an alternate view of the Kervaire invariant 1 problem.

Consider the following two immersions of the circle into the plane: the standard embedding, and an immersion whose image is the figure 8. Now take the tangent bundle and restrict it to the image of these two immersions. For the embedded circle, observe that the normal direction e1, while pointing constantly to the east, has a winding number 1 with the tangent direction. For the figure 8, the winding number between e1 and the tangent direction is zero. As stably framed circles these two immersions are quite different.

To get the framing to be stable embed the plane into (n+2)-space for large n. The immersed figure-8 will lift to an embedding (even in one more dimension), but the framing will be a result of stabilizing the framing of the figure-8.

The embedded circle clearly bounds an embedded disk — the obvious one, and the figure-8 does not bound an immersed manifold (The cone on the figure-8 has a branch point). The cobordism group of immersed oriented n-manifolds in (n+1)-space is the nth-stable stem. The figure eight map represents the Lie framing on the circle, and the standard embedding represents the other framing. The 1st-stable stem is Z/2 with the invariant being represented by the mod-2 number of double points.

In dimension 3, when considering orientable manifolds, the 2nd stable stem is also Z/2. The trivial elements are embeddings. The non-trivial element is represented by the Lie framing on the torus. This can be obtained (by a construction of Koschorke) as follows. Lift the figure-8 to an embedding in 3-space. Then put another figure-8 transverse to this. The transverse figure-8 can be arranged symmetrically with respect to the vertical axis (so it looks like an 8 on the page — duh). Then the stable framing canonically defines the immersed torus. It can be deformed to being like a children’s pool with a full twist. This torus represents a generator of the 2nd stable stem. The full twist along the double curve keeps the immersion from being cobordant to an embedding. The stable framing on this torus is induced from the Lie group structure. So if you take the standard embedding of the torus in 2d-complex space: CxC, given by (e^{i θ}, e^{i φ}), and deform it slightly, it will project to this twisted torus.

I say that you have to deform it slightly because I have drawn this torus many times with mathematica. When I use a (3x4)-matrix to project it, I always get a torus with four branch points.

So let us consider non-orientable surfaces for a moment. The cobordism group of non-orientable surfaces immersed in 3-space is the stable homotopy group of infinite projective space. The difference between the orientable and non-orientable case has to do with the nature of the Thom space of the classifying bundle. In the orientable case the Thom space is a circle, in the non-orientable case it is the classifying space of the canonical line bundle. There is also a dimension shift. The stabilization eats one of the indices.

So Boy’s surface (or as illustrated here or better yet here ) represents a generator of the third stable homotopy group of infinite projective space. By running a frame along the double curve, you can see that the double point framing has a quarter twist in it. This means that Boy’s surface is of order 8. It is of order 8 (rather than four) because the 1 full twist is not null-cobordant. That full twist represents the generator of the fundamental group of (3x3) orthogonal matrices. The double point set of Boy’s surface represents the modulo-8 Arf invariant that was defined by Ed Brown. Four times Boy’ surface is cobordant to the twisted torus, and we recover the classical Arf invariant.

Koschorke’s figure eight construction, when applied to Boy’s surface, gives a generator of the 3rd stable stem which is isomorphic to Z/24. OK, so now the fact that the normal bundle of Boy’s surface is a non-trivial line bundle comes into play. The vertical 8 is invariant under the orientation reversing loop. Also when Boy’s surface is lifted into 4-space, it does not embed. It has a lift with exactly one double point. So the immersion that results is a circle bundle over real projective two space that turns out to be the 3-sphere. The immersion of the 3-sphere has one quadruple point, a triple point set that consists of two circles, and a double point set that is a torus disjoint union with the projective plane. At the level of Z/8 mapping into Z/24, Koschorke’s figure 8 construction is the obvious inclusion.

Peter Eccles gave the following construction of an immersed 5-manifold in 6-space with one hextuple point. Start from an immersed 3-sphere in 6-space with one double point. This can be constructed From the union of the two equatorial disks (x,y,z,0,0,0) and (0,0,0,t,u,v) where the sum of the squares of the coordinates is less than or equal to 1. Now in the 5-sphere the boundaries of these two 3-disks forms an analogue of the Hopf link in the 5-sphere. It bounds an annulus (3-sphere times interval) in the 5-sphere. The union of the disks and this annulus is a 3-sphere. The normal bundle of the 3-sphere is trivial — this is induced from the Lie group structure on the 3-sphere which is also (2x2)- special unitary matrices. Since the normal bundle is trivial we can pop a Boy’s surface into the normal bundle. The resulting 5-manifold has one 6-tuple point.

Now apply Koschoke’s construction. The resulting 3-sphere x 3-sphere represents an element of the stable stem that has a non-trivial Kervaire invariant. Eccles showed that the existence of an manifold of dimension 3 less than a power of 2 immersed with an odd number of 0-dimensional multiple points (In this case, the multiplicity will be 2,6,14) is equivalent to the existence of a frame manifold of Kervaire invariant 1.

Posted by: Scott Carter on April 26, 2009 10:07 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

“in sufficiently large dimensions equal to 2 mod 4, every stably framed manifold is framed cobordant to a sphere.”

It might be fun to ponder this result from the viewpoint of the Cobordism Hypothesis,

And noteworthy from the TQFT point of view:

it says, as far as I can see, something like that every stably framed TQFT, which is a representation of stably framed cobordisms, is what in dimension 2 would be called a “genus-0 theory” or “tree-level”: it has no interactions.

It seems.

Posted by: Urs Schreiber on April 27, 2009 9:16 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

this exact situation generalizes to dimension $4 k + 2$

Glad to meet the dimension $4 k +2$ here after all, which I was missing in my comment above!

(Noticing that the worldvolume QFT of the $4k +1$-brane is $(4 k +2)$-dimensional.)

Posted by: Urs Schreiber on April 27, 2009 8:59 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Both the slides and the video of Mike Hopkins’ April 21 Edinburgh lecture are available from
http://www.maths.ed.ac.uk/~aar/atiyah80.htm

Posted by: Andrew Ranicki on May 5, 2009 7:23 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Anyone interested in this new result who will be near Lisbon next week should go to this talk:

• Douglas Ravenel, Recent Developments in Stable Homotopy Theory from the Chromatic Point of View, Room P3.10, Department of Mathematics, Instituto Superior Técnico, Lisbon, May 5-7.

Abstract: In the past three decades, homotopy theorists have found increasingly deep connections between stable homotopy theory and algebraic geometry. This series of talks will outline the necessary background and describe some recent progress that is joint work with Mike Hill and Mike Hopkins. Here is a brief description. The best approach to the stable homotopy groups of spheres is the Adams-Novikov spectral sequence. This is the Adams spectral sequence based on complex cobordism theory ($MU$) or equivalently (after localizing at a prime $p$) Brown-Peterson theory ($BP$). The relevant algebra is controlled by the theory of 1-dimensional formal group laws. This connection was first discovered by Quillen in 1969, then explored more deeply by Morava in the early 1970s. This was followed by the discovery of the chromatic filtration of the stable homotopy category in the 1980s and the work of Hopkins and Miller in the 1990s.

A formal group law over an algebraically closed field in characteristic $p$ is determined up to isomorphism by an invariant called the height, which is a positive integer $n$. A height $n$ formal group law has an automorphism group $S_n$ known as the Morava stabilizer group. It is a pro-$p$-group with interesting arithmetic properties. We now know that it has a canonical action on a certain $E_\infty$ ring spectrum $E_n$ which is difficult to describe explicitly, but very valuable to know. Knowing the cohomology of this action would tell us a lot about stable homotopy, but the problem is prohibitively difficult for $n \gt 2$.

A more tractable problem involves finite subgroups $G$ of $S_n$, which have been classified by Hewett. The most interesting cases are ones with order divisible by $p$, which occur when $n$ is divisible by $p-1$. In these cases we can look at the homotopy fixed point set of $G$ acting on $E_n$. Studying this in depth requires some techniques from equivariant stable homotopy theory, which we will introduce as needed.
Posted by: John Baez on April 27, 2009 3:36 AM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Something confuses me about the above abstract: if the Morava stabilizer S_n is a pro-p group, then how could its finite subgroups G not have order divisible by p?

Ah, OK – from looking at some other papers, I see that S_n is just the group of units in the maximal order of a rank-n division algebra over Q_p; i.e. it’s a sort of twist of GL_n(Z_p), and isn’t quite pro-p but has a pro-p subgroup of finite index.

Posted by: Jordan Ellenberg on April 28, 2009 3:28 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Nature magazine has an article on the Kervaire invariant one problem, which is worth reading just for the silly picture. Whoops!

Posted by: John Baez on May 4, 2009 1:11 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

The picture isn’t half as silly as the last paragraph of the article, which the author seems to have based on a (mis)-reading of your blog posting…

Posted by: Peter Woit on May 4, 2009 4:16 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Peter wrote:

The picture isn’t half as silly as the last paragraph of the article, which the author seems to have based on a (mis)-reading of your blog posting…

I bet they changed the picture between when you saw it and when I first saw it.

When I first saw it, the coffee cup was a styrofoam cup without a handle — obviously not the same as a doughnut! That’s why I said “Whoops”: the science department must have asked the art department for a picture of a doughnut and a coffee cup, not specifying precisely what sort of cup.

But now the coffee cup has been upgraded to a porcelain one with a handle! Correct topology — and better for the environment, too.

However, the last paragraph is still gobbledygook.

Posted by: John Baez on May 5, 2009 7:36 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I dimly recall a comment about related work of Akhmetiev but `search’ in this blog doesn’t find it - perhaps due to the transliteration problem.

Posted by: jim stasheff on May 11, 2009 2:47 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Maybe you’re talking about the translation of the abstract of Akhmet’ev’s latest paper on the Kervaire invariant. It appeared on the category theory mailing list, but I didn’t save a copy.

Some of his earlier papers appear here.

Posted by: John Baez on May 11, 2009 4:41 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

Here is the message that Peter Landweber posted to the ALGTOP mailing list on May 9, 2009:

Dear Colleagues,

The following is a translation (to the best of my ability) of the abstract of the revised paper on the Kervaire invariant, submitted to the arXiv by Peter M. Akhmet’ev on 7 May 2009, with a few following comments. This may interest people who would like to see a more detailed English abstract, and to know the structure of the paper.

ABSTRACT We present a solution to the Kervaire invariant problem. We introduce the concepts of abelian structure on skew-framed immersions, bicyclic structure on $\mathbb{Z}/2^{[3]}$-framed immersions, and biquaternionic structure on $\mathbb{Z}/2^{[5]}$-framed immersions. Using these concepts, we show that for sufficiently large $n$, $n = 2^{\ell} - 2$, an arbitrary skew-framed immersion (mapping into $\mathbb{R}^n$) has zero Kervaire invariant. The proof makes use of a retraction theorem for skew-framed immersions in a given normal cobordism class. The paper also includes a proof of the retraction theorem.

1. The first section (6 pages) of the paper discusses the manifold of self-intersections and the Kervaire invariant of an immersion into Euclidean space; the Kervaire invariant is defined in terms of the manifold of self-intersections. Wreath products $\mathbb{Z}/2^{[d]}$ are defined in this section; e.g., for $d=2$ one obtains the dihedral group of order 8. The retractions mentioned in the abstract are defined in this section.

2. The three kinds of structures on suitable classes of immersions mentioned in the abstract are defined in the next three sections (15 pages).

3. The brief section 5 (2 1/2 pages) presents a proof of the main theorem, as stated in the abstract, based on results stated to this point, and the retraction theorem which is stated and proved later in the paper. Several results from the author’s related study of the Hopf invariant are also used. +

4. The two following sections (32 pages) give proofs of four theorems stated in sections 3 and 4.

5. The retraction theorem is stated and proved in the final section (23 pages).

Peter Landweber

Posted by: Urs Schreiber on May 11, 2009 4:49 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

I followed a link from Doug Ravenel’s page on the Arf-Kervaire invariant to a Scientific American article on the Hill-Hopkins-Ravenel result.

http://www.scientificamerican.com/article.cfm?id=hypersphere-exotica

It looks like John B’s heat of the moment mistake as corrected by Andrew R on this blog has propagated to the pages of Scientific American - that’s my guess anyway. I have tried to correct it there.

Posted by: Diarmuid Crowley on August 11, 2009 1:38 PM | Permalink | Reply to this

### Re: ‘Kervaire Invariant One Problem’ Solved

The proof was uploaded to arxiv.

Posted by: Daniel de França MTd2 on August 27, 2009 1:12 AM | Permalink | Reply to this
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