5
$\begingroup$

McDuff showed that every connected homotopy type can be realized as the classifying space of a discrete monoid, but the monoid she constructs has lots of idempotents.

Question: Which homotopy types are realized as the classifying space of a right-cancellative discrete monoid?

In the commutative case, my guess would be that $BM \simeq B(M[M^{-1}])$, so that the classifying space is aspherical. But I'm less confident that this happens in the noncommutative case.

$\endgroup$
2
  • $\begingroup$ The category of elements of a semi-simplicial sets is left cancelative. So I don't know for monoid, but at least one can represent any homotopy type as the realization of a left (or right) cancelative category by taking the category of elements (or its opposite) of any semi-simplicial sets representing it. $\endgroup$ Aug 22, 2019 at 21:51
  • $\begingroup$ Indeed, by subdivision one can represent any homotopy type as the classifying space of a poset, which is both left and right cancellative. But far from being a monoid. $\endgroup$
    – Tim Campion
    Aug 23, 2019 at 0:26

1 Answer 1

5
$\begingroup$

By universal properties, we have that $BM$ is the classifying space of the homotopy localization $B(M[M^{-1}]^h)$. Thus $BM$ is aspherical if and only if the homotopy localization is discrete.

Further, Dwyer-Kan showed that if $(M,W)$ admits a calculus of fractions, then the homotopy localization agrees with the ordinary localization. When $M$ is cancellative, $(M,M)$ admits a calculus of fractions if and only if it satisfies the Ore condition: $$\forall m_1, m_2 \in M, \exists n_1, n_2 \in M, ~ n_1 m_1 = n_2 m_2.$$ So in this case $M[M^{-1}]^h \simeq M[M^{-1}]$ and the homotopy localization is aspherical. In general, I do not know what happens.

$\endgroup$
2
  • $\begingroup$ Thanks! I believe that every cancellative commutative monoid admits a calculus of fractions, but I'm not sure I believe this in the noncommutative case; do you have a reference? I don't see why the Ore condition should be satisfied. And for instance, having a calculus of fractions should probably imply that the monoid is embeddable in a group, which is not the case for all cancellative monoids, as I learned here. $\endgroup$
    – Tim Campion
    Aug 22, 2019 at 15:00
  • $\begingroup$ @TimCampion I think you're right, I was misreading the Ore condition on the nLab page, in such a way that I thought it was vacuous for the whole monoid. $\endgroup$ Aug 22, 2019 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.