In this paper https://arxiv.org/pdf/math/0202260.pdf, the author proves that $BP$ is homotopic to $S^2$, using homology (In the first lemma of the paper).

Well, I tried to use the definition of geometric realization to see how $BP$ looks like. I can see that everything just collapses into five 1 simplices via the multiplication defined. But I can't seem to connect these together, to form $S^2$.

What am I missing here?

Thanks in advance.

  • $\begingroup$ What do you mean by "everything collapses into five 1-simplices"? I don't know anything about monoids, but for groups, the classifying space can be obtained by looking at the geometric realization of the nerve of its one-element groupoid. In case this construction still works for monoids, then $BP$ should certainly have a $5$-fold wedge of circles as its $1$-skeleton, but the algebraic relations should also give rise to $2$-simplices glued appropriately. $\endgroup$ – Sofie Verbeek Oct 27 '18 at 13:09
  • $\begingroup$ Because $x_{ij} x_{kl} = x_{il}$, any $n$ chain in $B_n(P)$ becomes a 1-chain. ----(1) So, it gives rise to five 1 simplices. And as you mentioned, it becomes a 5 fold wedge of circles. "...also give rise to 2-simplices glued appropriately" - I can't see how it should give rise to 2 simplices, because of (1). $\endgroup$ – wanderer Oct 27 '18 at 14:59
  • $\begingroup$ Yes, I'm considering the monoid as a category with one object, with morphisms the monoid elements. And mimicking the construction as for groups (since we don't use inverse elements in the process.) $\endgroup$ – wanderer Oct 27 '18 at 15:03

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