# Milnor Conjecture on Lie groups for Morava K-theory

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.

• 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 $\mathbb{Z}$! – user51223 Oct 9 '18 at 2:14
• @user51223 I hoped that the direct reference to $G$ makes it clear. Anyway Ill edit that, thanks. – S. carmeli Oct 9 '18 at 5:05
• 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. – 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 $$N. We can then choose a finite spectrum $$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.

• 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. – S. carmeli Oct 9 '18 at 18:48
• @S.carmeli Check out the Atiyah-Segal theorem, which tells you what happens when G is compact Lie. – skd Oct 11 '18 at 21:12
• @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. – S. carmeli Oct 12 '18 at 12:48