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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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