Aspherical operads

Question

Let $O$ be an operad in $\mathtt{SETS}$. Assume that $O(0)$ is empty and $O(1)$ only consists of the identity. Assume for simplicity that $O$ is monochromatic, i.e. we have no labels on the in/outputs. Assume also for simplicity that the operad is plain, i.e. neither symmetric nor braided. So the operads in question consist of a set of $n$-ary operations $O(n)$ for each $n\in\mathbb{N}$ together with an associative composition and there is a unit element in $O(1)$ (but no more elements, as required above).

Now $O$ freely generates a monoidal category $S(O)$: The objects are natural numbers and an arrow from $m$ to $n$ consists of a sequence of operations in $O$ with a total of $m$ inputs and a total of $n$ outputs. For example if $a\in O(3)$ and $b\in O(5)$, then $(a,b)$ is an arrow from $3+5=8$ to $2$. Composition is given by composition in the operad.

I know that $S(O)$ is aspherical when $O$ is free and also in some other special cases. Here I consider categories as spaces via the usual geometric realization, i.e. the geometric realization of the nerve of the category.

Question: Is the category $S(O)$ always aspherical?

Answer 1

I've left my original answer as some people may find it of interest.

I have a candidate counterexample. The idea is to find a (non-symmetric) set operad in between the free operad $Free_2$ on a single arity 2 generator and the associative operad $As$. The example I've chosen is the operad $P$ which is isomorphic to the free operad in arities 1, 2 and 3, but trivial for arities 4 and above. There is a diagram $$ Free_2 \rightarrow P \rightarrow As $$

The monoidal category $S(P)$ defined in the question has contractible fundamental groupoid. I'll leave this as an exercise. It is quick if you are used to representing Thompson's group F via pairs of trees: the trees with k leaves are all equivalent in P for $k>4$, but any group element in F is represented (in perhaps a non-reduced way) by a pair of trees with more than 4 leaves.

To finish we can construct a non-trivial cycle in the homology of the nerve. My guess is the following:.

I hope that it is fairly clear which element this is, each tree is meant to represent two composable morphisms. You can check that it's a 2-cycle with relative ease, the tricky bit is to show that it's non-trivial. I'm not 100% sure that it is, but I'll explain why I chose it. The point is that either the first two or last two terms are chosen to kill the 1-cycle which is the difference of the two trees of arity 3. In the group F this is a representative of the 1st homology group, but for S(P) it is zero. There is more than one way to kill this 1-cycle and the 2-cycle above is the difference of these two.

To prove that it's not a boundary I guess an explicit calculation using the fact that the homology of F is easy to calculate could do the trick.

Don't hesitate to ask me to expand on any of this.

Answer 2

This needs some checking but for the free non-symmetric operad generated by a single operation in arity 2, the category (PRO) I think you get is the free monoidal category generated by a single object $X$ and a single morphism $\alpha$ from $X^2$ to $X$.

But the nerve of this category is a classifying space for Thompson's group F. That it's fundamental group is F can be seen from http://arxiv.org/abs/math/0508617. I seem to recall that the nerve is actually a locally CAT(0) complex which implies that it's a classifying space. One reference of interest is Guba and Sapir's "Diagram Groups", see also Brown's http://www.math.cornell.edu/~kbrown/papers/homology.pdf. The monoidal structure means that although F isn't abelian the homology has a (non-unital) Pontryagin product.

So these categories are perhaps better studied than you may have guessed!