Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


The so-called "recognition principle" of Boardman-Vogt and May leaves me unsatisfied. My problem is the following:

The "recognition principle" says that every "group-like" algebra over the little $k$-disks operad is equivalent (as an algebra over the little $k$-disks operad) to a $k$-fold loop space. However, if I start with a space homotopy equivalent to a $k$-fold loop space, then it is not a priori equipped with an action of the little disks. So:

1) First, can someone give me an explicit example of a space homotopy equivalent, say, to a double loop space and such that it admits no (compatible) actions of the little $2$-disks operad?

2) Can you characterize among the spaces homotopy equivalent to double loop spaces those which are algebras over the little disks operad?

3)I heard that the problem was related to the fact that the little disks operads are not cofibrant (in the homotopy category of operads), and that the cofibrant replacements would be the so-called Fulton-McPherson operads. These are "compactifications" of configuration spaces of points in $\mathbf{R}^k$ (modulo the action of the affine group maybe) defined using a variant of a construction due to Fulton and McPherson.

Is it obvious from the definition that this operad indeed acts on iterated loop spaces?

Many thanks,


share|improve this question
Concerning 1):<br> It might be helpful to first think about the same question for a simpler non-cofibrant operad. Let's take the operad <i>Ass</i>. What's an example of a space that is homotopy equivalent to a topological group, but that isn't a topological monoid? –  André Henriques May 9 '10 at 14:50
add comment

3 Answers

I'll give an answer from the old days. The $W$ construction of Boardman and Vogt corresponds to the modern notion of cofibrant replacement of operads. If an operad $\mathcal{C}$ acts on $X$ and $Y$ is homotopy equivalent to $X$, then $W\mathcal{C}$ acts on $Y$. By the way, there is an inherent flaw in the little discs operads $\mathcal{D}_n$, namely there is no map of operads $\mathcal{D}_n\longrightarrow \mathcal{D}_{n+1}$ that is compatible with suspension (in the obvious sense: consider $\Omega^n \longrightarrow \Omega^{n+1}\Sigma$). The little $n$-cubes operads do not have this problem, but have others not shared by the $\mathcal{D}_n$. The Steiner operads have all the good properties of both the $\mathcal{C}_n$ and the $\mathcal{D}_n$. In practice, that is in actual applications, such geometric differences are far more important than the questions of cofibrancy and homotopy invariance.

share|improve this answer
add comment

Here's a nice example for André's simplified question. Consider the interval $X=[-1,1]$ with $0$ as the basepoint. Suppose there was a continuous monoid structure $m:X^2\to X$ with $0$ as the identity element. Suppose that $u<0$ and $v>0$; then we can apply $m$ to the piecewise-linear path $(0,u)\to(v,u)\to(v,0)$ to get a path from $u$ to $v$, which must pass through $0$ by the intermediate value theorem. It follows that either $u$ or $v$ must be invertible. From this we deduce that either all positive numbers are invertible, or all negative numbers are invertible; wlog the former. Now for $0\leq t\leq 1$ the maps $m(t,-):X\to X$ are homeomorphisms, so they preserve the boundary $\{-1,1\}$. As $m(0,-)$ is the identity on the boundary, the same must be true of $m(t,-)$ (for $0\leq t\leq 1$). In particular we have $m(t,1)=1=m(0,1)$, but $1$ is invertible so $t=0$, which is a contradiction.

Now let $X$ be an arbitrary tree with finitely many edges. If $a\in X$ is not a boundary point then $X\setminus\{a\}$ will be disconnected and we can run the same kind of argument with the different connected components to see that there is no monoid structure with $a$ as the identity element. If $a$ is a boundary point then there may be a monoid structure; for example, the set $X=\{z\in\mathbb{C}:z^n\in [0,1]\}$ is a tree which is a submonoid of $\mathbb{C}$ under multiplication. My guess is that for more general trees there is no monoid structure but I do not see a proof at the moment.

share|improve this answer
add comment

This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,\ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i \lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid $\{z\in \mathbb{C}: z^{r_i}\in [0,1]\}$. Consider the monoid $A_1\times\cdots \times A_n$, and consider the subsets $M_i=\{0\}\times\cdots\times\{0\}\times A_i\times\{1\}\times \cdots \times\{1\}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $i\lt j$, then $M_i\times M_j\subset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(\underbrace{0,\ldots, 0}_i, 1,\ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.


My answer to mathoverflow.net/questions/91327 shows that the universal cover of the theta graph (and many similar trees) cannot carry a monoid structure.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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