It is about a "conjecture" I heard (when I was student). There would exist an algebraic structure on the homotopy groups of spheres such that this algebraic structure would be the free algebraic structure generated by one point. In the wikipedia page "Field with one element", it is written that the algebraic $K$-theory of the field with one element is related (can be identified ?) with the stable homotopy groups of spheres. Is it the answer ?

  • $\begingroup$ Insofar as the field with one element doesn't exist, the answer is no, but it can be yes if you define it appropriately, which in my opinion is what people do when they talk about this non-existent field ;) $\endgroup$ – Fernando Muro Jul 27 '16 at 12:59
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    $\begingroup$ The sphere spectrum is the free E-infinity ring on the point (just as the integers are the free commutative ring on the point). Its homotopy groups are the stable homotopy groups of spheres. Maybe that's what you heard? $\endgroup$ – Urs Schreiber Jul 27 '16 at 16:36
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    $\begingroup$ Related: mathoverflow.net/questions/1628/kf-1-sphere-spectrum. $\endgroup$ – user62675 Jul 27 '16 at 19:20

I think the answer to the last question is no. As far as I understand, the "general linear groups over the field with one element" are supposed to be the symmetric groups. Therefore the statement that the algebraic K-theory of the field with one element can be identified with the stable homotopy groups spheres is essentially the Barratt-Priddy-Quillen theorem, which says that the group completion of $\coprod_n B\Sigma_n$ is the sphere spectrum.

As far as I can tell, the Barratt-Priddy-Quillen theorem does not yield that much explicit information about stable homotopy groups of spheres. On the positive side, I would like to mention that Smirnov apparently gave an explicit presentation of the $E_\infty$ page of the Adams spectral sequence as an $A_\infty$ algebra: "Description of stable homotopy groups of spheres in the language of A∞-algebras." (Russian) Uspekhi Mat. Nauk 51 (1996), no. 1(307), 171--172; translation in Russian Math. Surveys 51 (1996), no. 1, 171–172. But I don't know the details of this work.

  • $\begingroup$ I was not expecting any answer to my soft question about a "conjecture" which is not even a conjecture :-). So I'll mark your post as the answer. $\endgroup$ – Philippe Gaucher Aug 3 '16 at 7:01

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