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?

  • $\begingroup$ You call "monochromatic" any coloured operad on only one colour? The construction you describe is a classical construction from operads to PROs (see Operads and PROPs of M. Markl, Example 60). $\endgroup$ Aug 12, 2013 at 22:22
  • $\begingroup$ When you say 'no group action' you mean a non-symmetric operad or a symmetric operad where the symmetric group acts trivially? These situations are orthogonal one another in some sense. What do you mean by monocromatic? Operations in an operad only have one output, so could you make precise your construction of $S(O)$? And finally, since there are many ways of regarding a category as a space, which way are you thinking of? $\endgroup$ Aug 21, 2013 at 14:57
  • $\begingroup$ I have edited the question to hopefully clarify things. $\endgroup$ Aug 21, 2013 at 15:21
  • $\begingroup$ Your conditions seem to imply that, if a morphism $m\rightarrow n$ exists at all then $m\geq n$, and $m=n$ if and only if the morphism is the identity. Am I right? or maybe I didn't get the question yet? $\endgroup$ Aug 21, 2013 at 16:13
  • 1
    $\begingroup$ There is a functor from monoidal categories to operads (with several colors), sending $C$ to the operad with colors the objects of $C$ and with operation sets $\mathrm{Hom}_C(A_1\otimes...\otimes A_n;B)$. The left adjoint of this functor is the construction considered above. $\endgroup$ Aug 21, 2013 at 18:40

2 Answers 2


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:2-chain in the nerve.

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.

  • $\begingroup$ Hi James, thank you for your answer. I see what you mean. But I also could not manage to show that the 2-cycle is not null-homologous yet. Could you expand on the trick you've mentioned? $\endgroup$ Sep 16, 2013 at 14:18

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!

  • $\begingroup$ Thanks for your answer. I know all this stuff, and thats why I'm studying the above question. It is part of a program to generalize all this to the non-free case. In the meantime, I'm on a way to prove the above statement. $\endgroup$ Sep 5, 2013 at 10:12
  • $\begingroup$ Ah I see, I misread your question! For some reason I was thinking about contractability. I wont have time today but I'll try to find a counterexample for your question, it shouldn't be hard. $\endgroup$ Sep 5, 2013 at 12:12
  • $\begingroup$ I made a mistake in the last comment, the counterexample shouldn't be hard to find if it exists. $\endgroup$ Sep 5, 2013 at 12:19
  • $\begingroup$ I would love to see a counterexample if it exists. ;) $\endgroup$ Sep 5, 2013 at 13:11

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