Morava K-theories for dummies?

Question

Professor Urs Würgler passed away one year ago, and his wife engraved his tombstone with "the formula he was the most proud of" :

$B(n)_*(X)\cong P(n)_*(K(n))\square_{\Sigma_n}K(n)_*(X)$

However she doesn't understand it, and she asked me if I can. I can't. But I discovered it is very close to the one in Theorem 3.1 p. 121 of

Urs Würgler "Morava K-theories - a survey" 2006, in Lecture Notes in Mathematics Vol. 1474 DOI:10.1007/BFb0084741

So I tried to understand the wikipedia page on Morava K-theory but it is way above my level (PhD in dynamics and control)

Can anybody try to explain what the formula is about in plain english, or it is definitely too abstract to express in human language ?

Answer 1

This is a result in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later people discovered various "extraordinary homologies", which give more precise information. There are very many different extraordinary homologies, including those called $P(n)_*(X)$, $B(n)_*(X)$ and $K(n)_*(X)$. Of these, $P(0)$ (also called $BP$, or Brown-Peterson homology) is the most powerful, but it is often very hard to calculate. At the other end of the scale, $K(0)$ is the weakest and easiest to calculate. In general $K(n)$ is reasonably easy. Roughly speaking, the information in $P(n+1)$ is the information in $P(n)$ minus the information in $K(n)$, so all the $P(n)$'s are often hard to calculate.

From the definitions, the obvious guess would be that $B(n)$ is only a little easier than $P(n)$. However, this turns out to be wrong: $B(n)$ contains exactly the same information as $K(n)$ (and so is much easier than $P(n)$). If you know $K(n)_*(X)$ then Würgler's theorem allows you to calculate $B(n)_*(X)$, and a different but easier theorem lets you go in the opposite direction.