Milnor Conjecture on Lie groups for Morava K-theory

Question

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.

Answer 1

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.