A conjecture by Milnor state that if $G$ is a Lie group, then the map $B(G^{disc})\to BG$ sending the classifying space of $G$ endowed with the discrete topology to the classifying space of the topological group $G$ induces an isomorphism on homology with $\mod p$ coefficients.

In chromatic homotopy theory, there are more "fields of characteristic p" than just finite fields, namely we have the Morava $K$-theories $K(p,n)$.

Question: Do we know "more of" the Milnor conjecture for those ring-spectra then for $\mathbb{F}_p$? (ultimately, but probably too ambitious, can it be proven for those ring spectra even though it is still open for $\mathbb{F}_p)$?

By "more of" I mean any progress that is special for this case and don't work for $\mathbb{F}_p$ coefficients.

  • $\begingroup$ I think you mean $G$ when it has the discrete topology. It seems to be slightly different from a discrete group which often recall groups such as $\mathbb{Z}$! $\endgroup$ – user51223 Oct 9 '18 at 2:14
  • $\begingroup$ @user51223 I hoped that the direct reference to $G$ makes it clear. Anyway Ill edit that, thanks. $\endgroup$ – S. carmeli Oct 9 '18 at 5:05
  • 3
    $\begingroup$ I don't believe that we know any cases of Milnor's conjecture with Morava K-theories that aren't derived from the one with mod-p coefficients. That would certainly be really interesting. $\endgroup$ – Tyler Lawson Oct 9 '18 at 7:44

Consider a map $f\colon X\to Y$ of spaces (such as $B(G^{\text{disc}})\to B(G)$). Say that $f$ is a $K(n)$-equivalence if $K(n)^*(f)\colon K(n)^*(Y)\to K(n)^*(X)$ is an isomorphism. We will allow the case $n=\infty$ (corresponding to $K(\infty)^*(X)=H^*(X;\mathbb{F}_p)$) but not the case $n=0$ (corresponding to $K(0)^*(X)=H^*(X;\mathbb{Q})$.

Using the Atiyah-Hirzebruch spectral sequence $$ H^i(Cf;K(n)^j) \Longrightarrow K(n)^{i+j}(Cf), $$ where $Cf$ is the cofibre of $f$, we see that if $f$ is a $K(\infty)$-equivalence then it is a $K(n)$-equivalence for all $n$.

Conversely, suppose that $f$ is a $K(n)$-equivalence for $N<n<\infty$. We can then choose a finite spectrum $F$ of type $N$ and we see that $K(n)_*(F\wedge Cf)=0$ for all $n<\infty$ (including $n=0$, which we usually exclude). However, $Cf$ is a suspension spectrum and so is harmonic by a theorem of Hopkins and Ravenel, so we can conclude that $F\wedge Cf=0$ and thus that $f$ is a $K(\infty)$-equivalence.

I think that we actually have the same conclusion if $f$ is merely a $K(n)$-equivalence for infinitely many $n$, but I will not give the argument here.

It remains possible that $f$ could be a $K(n)$-equivalence for a finite set of integers $n$, but not for $n=\infty$. For the Lie group situation, it would be natural to think about the case $n=1$, where there is a link to representation theory.

  • $\begingroup$ Thanks for the observation. I will add this to the question if you don't mind, because it makes it much more natural to ask. For a Lie group, you know what is the precise relationship between $K_1^*BG$ and the representations of $G$? for a finite group I know that $KU$-cohomology is the completion of the representation ring, but for a Lie group or for its completion I don't know what is the relation. $\endgroup$ – S. carmeli Oct 9 '18 at 18:48
  • $\begingroup$ @S.carmeli Check out the Atiyah-Segal theorem, which tells you what happens when G is compact Lie. $\endgroup$ – skd Oct 11 '18 at 21:12
  • $\begingroup$ @Neil Strickland Thanks, Ill look. I guess the hard part of the KU-Milnor conjecture should be to show that the K-theory of G as a discrete group is topologically generated by representation classes. $\endgroup$ – S. carmeli Oct 12 '18 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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