35
$\begingroup$

In the Soviet times there was a famous Encyclopedia of Mathematics. I think it is still familiar to every Russian mathematician maybe except very young ones, and yours truly is in possession of all 5 volumes. Browsing it recently (with no real purpose) I came across a certain peculiarity. In the article "Kervaire invariant" by M.A.Shtan'ko there was a claim that the Kervaire invariant is nontrivial in all dimensions $2^k-2$ for $2\le k\le 7$ - yes, including 126. Which, for what I know, is still an open problem Kervaire invariant: Why dimension 126 especially difficult? . In this article, the credit for the $k=6$ and $k=7$ cases (lumped together) was given to M. Barratt, M. Mahowald, and A. Milgram, but with no actual reference.

To be fair, the absence of references is understandable (in the original article, that is) because it was written in 1978 while a complete proof for dimension 62 was only published six years later https://web.math.rochester.edu/people/faculty/doug/otherpapers/barjoma.pdf It is possible that back in 1978 the result was just announced. But, what happened to the 126? And to Milgram? The simplest possible explanation is that a proof for 126 was also announced but later retracted. However this is by no means the only possibility, so I am curious what really happened. Besides, those MO folks who know more about the subject then myself might wonder what the attempted proof was like.

After a bit of search I found a reference which may be relevant (hopefully). In "Some remarks on the Kervaire invariant problem from the homotopy point of view" by M.E.Mahowald (1971) there is Theorem 8 attributed to Milgram and after it the following Remark: "It can be shown that $\theta_4^2=0$ and thus Milgram's theorem implies $\theta_6$ exists". If I get it right this indeed means a nontrivial Kervaire invariant in dimension 126, so probably there is a mistake somewhere in this argument. (But even if it is so, damned if I have a clue who has made it: Milgram, Mahowald, or somebody else.)

I have to admit that a few things about this story look suspicious. To begin with, A. Milgram died in 1961 so it should probably be R. J. Milgram if any. In the introduction to "The Kervaire invariant of extended power manifolds " J. Jones stated explicitly that the 62 case is solved by Barratt and Mahowald but not published yet while in the higher dimensions the problem is open, in contradiction to what Shtan'ko wrote the same year. In a couple of papers between 1978 and 1981 I spotted references like [Barratt M. G., Mahowald M., The Arf invariant in dimension 62, to appear] but no traces whatsoever of 126 and Milgram. (Besides this article of Mahowald from almost a decade before.) I am at a loss what to make of all this. It would be nice if someone can set it straight - at the very least, I want to know if Shtan'ko made it up.

By the way, an English translation of this Encyclopedia article can be found here
https://www.encyclopediaofmath.org/index.php/Kervaire_invariant
Only, the year is written 1989 instead of 1978 (a second edition, apparently).

$\endgroup$
1
  • 3
    $\begingroup$ A second edition is one possibility, but a more likely scenario is that a technical typist moved fingers too far when typing the year. $\endgroup$ Jan 15, 2020 at 2:34

3 Answers 3

27
$\begingroup$

I found the following remark in Zhouli Xu's paper "The strong Kervaire invariant problem in dimension 62":

In [19], R. J. Milgram claims to show that under the same condition as in Theorem 1.1, one has $θ_{n+2}$ exists. If this were true, then we would have that $\theta_6$ exists. However, Milgram’s argument fails because of a computational mistake [8].

The paper containing the mistake is

R. J. Milgram, "Symmetries and operations in homotopy theory" Amer. Math. Soc. Proc. Symposia Pure Math., 22(1971), 203-211

and the other reference is private communication with Robert Bruner.

$\endgroup$
3
  • 2
    $\begingroup$ Thank you. So this is where the mistake was. By the way, in the Remark I mentioned Mahowald have really said "it can be shown that $\theta_4^2=0$". And yes it can, except this was actually done 40+ years later! (Theorem 1.2 in this paper of Xu.) Amazing. I am still curious though why Shtan'ko counted the problem as solved in 1970s. Was there a time when experts were optimistic about it? $\endgroup$ Jan 10, 2020 at 17:19
  • 4
    $\begingroup$ @RobertBruner It would be great if you could add more details $\endgroup$ Jan 13, 2020 at 9:41
  • $\begingroup$ There seems to be some confusion about the difference between "A. Milgram" and "R.J. Milgram". The latter used a version of his middle name "James" and was a prolific algebraic topologist at Stanford (now retired). $\endgroup$ Jan 15, 2020 at 2:39
11
$\begingroup$

I'm feeling mischievous. The history of the Kervaire invariant problem is strewn with false proofs, Jim Milgram's is only one of many. A student of mine (who I will leave nameless) had a preprint (around 1980?) that solved the problem but that Mark Mahowald quickly shot down. The most recent example I know of is that of a Russian mathematician (who I will also leave nameless). See Math Reviews MR2590025 (2010k:55031) Differentials of the Adams spectral sequence and the Kervaire invariant (Russian) Dokl. Akad. Nauk 427 (2009), no. 5, 601–604; translation in Dokl. Math. 80 (2009), no. 1, 573–576. From the text (translated from the Russian): "In this paper, we study the differentials of the Adams spectral sequence for stable homotopy groups of spheres and solve the Kervaire invariant one problem for n-dimensional manifolds when $n=2^i-2$, $i\geq 6$." That is a four page paper. Would that it were so simple!

$\endgroup$
8
$\begingroup$

I'll try to give more precise detail soon, but here's my understanding of this history. The 'proof' Peter May mentions is the 'standard mistake' in the subject.

Consider elements in a spectral sequence coming from the homotopy exact couple of a tower. If you have a geometric construction that the boundary of $x$ is $y$, and you can show that $y$ is itself null-homotopic, it is tempting to think you have shown that $x$ survives to a non-zero class. However, $x$ is one of the reasons that $y$ is null-homotopic, so you haven't really observed anything about $x$ from knowing only that $y$ is null-homotopic. What you need is that $y$ was already null-homotopic before $x$ got there to kill it. In other words, you need that $y$ is null-homotopic in an appropriately high stage of the tower, not just in the $0^{th}$ term. For an example related to this case, see pp. 38-39 of

http://www.rrb.wayne.edu/papers/fin_conj_handout.pdf

The mistake Milgram made in "Symmetries and operations" was just a miscalculation in the $\Sigma_4$ extended power of the $30$-sphere, or perhaps of $S^{30} \cup_2 e^{31}$. It did seem like this would give $\theta_6$ in the $126$-stem, given what was then known about $\theta_4$, until the mistake was noticed, and apparently Shtan'ko wrote his report during this burst of enthusiasm.

Shtan'ko's 'A. Milgram' was just a mistake. Surely he meant R. J. Milgram'.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.