$\DeclareMathOperator\Q{\mathbf{Q}}$Jack Morava has some interesting ideas stemming from stable homotopy theory and geometric topology on the Shafarevich Conjecture.

The Shafarevich Conjecture states: $\operatorname{Gal}(\bar \Q \,/ \,\Q_{cycl})$ is free. That is, the Galois group of the algebraic closure of the rationals over the cyclotomic closure of the rationals is a free group (Added: or rather, a free profinite group).

References for Morava's thoughts are

This is exciting material, but I'm having trouble coming up with a way to summarize the gist and have some questions.

(1)What exactly is Morava's definition of a mixed Tate motive?

(2) What exactly is the connection Morava is advocating between number theory and geometric topology by invoking the appearance of the Riemann zeta function in Waldhausen's A-theory/pseudo-isotopy?

(3) Morava states that the map from the K-theory of the integers to that of the sphere spectrum, $K(\mathbb {Z}) \to K(\mathbb {S})$, is a rational equivalence as a (partial) explanation of (2). How exactly does this work??

(4) Where does Shafarevich fit in here?

Down-to-earth answers to these would be much appreciated!!

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    $\begingroup$ Just a minor comment: "free profinite" is not the same as "free and profinite". $\endgroup$
    – user6976
    Commented Sep 23, 2010 at 11:58
  • 3
    $\begingroup$ Actually, the conjecture is that the absolute Galois group in question is a "free profinite group". It is obviously profinite, but the point is that it is expected to be free in the category of profinite groups. That is different from being free in the category of all groups. See e.g. Neukirch, Schmidt and Wingberg: "Cohomology of Number Fields", page 449 or so. $\endgroup$ Commented Nov 12, 2010 at 15:29
  • $\begingroup$ Context for folks outside of homotopy theory: Morava's ideas are famously hard to make precise, but it can be famously rewarding to do so. Others here could make more precise statements, but suffice it to say that there's a reason many objects in chromatic homotopy theory are named after him. One superficial technical note -- the abstract of "the motivic Thom isomorphism" , dated 2003, suggests that a theory of rational motivic Thom isomorphisms could lead to several applications -- did not such a theory exist already at that time? $\endgroup$
    – Tim Campion
    Commented Feb 5, 2021 at 23:33
  • $\begingroup$ @Romeo, In the preface and section 3 of "Towards a fundamental groupoid ..." paper Morava sketches how a full local-global Galois-theoretic link between number fields and complex cobordism spectrum MU can be established using Rognes' Galois theory of structured ring spectra if a "reasonable topology" for (e.g., Balmer) spectrum of small tensor triangulated (TT) categories can be found and used to develop a theory of fundamental groupoids for TT categories. Is such a theory now available? Is the outlined link now established? The paper is speculatively very rich and thought-provoking. $\endgroup$
    – Nimas
    Commented Feb 6, 2021 at 0:33
  • $\begingroup$ @Nimas Romeo asked this question 10 years ago and has not been on MO since 2011 according to his profile. I don't think he can hear you. Your question sounds interesting, though. I hope somebody chimes in. $\endgroup$
    – Tim Campion
    Commented Feb 6, 2021 at 0:37

1 Answer 1


(3) The statement is that the map of ring spectra $S \to HZ$ induces a rational equivalence $K(S) \to K(Z)$. A reference is Proposition 2.2 in:

Waldhausen, Friedhelm: Algebraic $K$-theory of topological spaces. I. Algebraic and geometric topology (Proc. Sympos. Pure Math., Stanford Univ., Stanford, Calif., 1976), Part 1, pp. 35--60, Proc. Sympos. Pure Math., XXXII, Amer. Math. Soc., Providence, R.I., 1978.

The proof uses the plus-construction definition of algebraic $K$-theory. The map $BGL(S) \to BGL(Z)$ is a $\pi_1$-isomorphism and a rational equivalence, since $\pi_{n+1} BGL(S) = \pi_n GL(S)$ is the group of infinite matrices over $\pi_n(S)$ for $n\ge1$, which is torsion. Hence $BGL(S)^+ \to BGL(Z)^+$ is also a rational equivalence.

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    $\begingroup$ Ah, then (3) is relatively straightforward, thanks for the reference. Can you address how this related to (2)? This is still somewhat mysterious to me. $\endgroup$
    – Romeo
    Commented Oct 1, 2010 at 0:37

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