# How does Berger-Moerdijk's relative Boardman-Vogt work?

In "The Boardman-Vogt resolution of operads in monoidal model categories," the authors construct factorizations of sufficiently nice operad maps $P\to Q$ into a cofibration followed by a weak equivalence $P\rightarrowtail W(H,Q)_P\stackrel{\sim}{\to}Q$. This construction depends on a choice of interval $H$. When $P$ is the unit operad $I$, this is called $W(H,Q)$ and is a cofibrant replacement for $Q$. My question is about the relative construction when $P\ne I$.

The factorization $W(H,Q)\_P$ is constructed as a sequential colimit of pushouts. For simplicity, concentrate everything in arity $1$ so that we can ignore $\Sigma_n$ actions; then these pushouts are of the form

$H(T)\otimes\underline{Q}(T)\gets {(H\otimes Q)}^-\_P(T)\to W_{k-1}(H,Q)_P.$

The pushout of this diagram (taken as $T$ ranges over all appropriate trees) is $W_{k}(H,Q)_P$. It is my understanding that the middle entry of this pushout is the colimit over a cube with the final corner omitted, and that the left entry is that final corner. The map to the right factor, which is constructed inductively,

informally puts edge-lengths to 0, whenever the vertices of the edge are both labelled by elements of P, and then applies the corresponding attaching map of the absolute W-construction.

The middle entry of the pushout is somewhat involved to describe explicitly; it occurs on p833--834 and relies on p819--821 in the published version of the paper. Let's pick an example,and I'll do a calculation, and hopefully someone can point out the mistake I'm making.

Let's look at the bivalent tree with four vertices $(v_1,v_2,v_3,v_4)$ and three internal edges $(e_{12}, e_{23}, e_{34})$. I believe that one part of the diagram making up the middle entry is:

$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes I_3\otimes Q_4$

and that this maps along the edges of the cube into both

$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes Q_3\otimes Q_4$

and

$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes P_2\otimes P_3\otimes Q_4$

In the first case, the attaching map to $W_{k-1}(H,Q)\_P$ collapses the edge $e_{34}$, getting the vertex label $Q_3\otimes Q_4$ since the edge between them is labeled by $I$ and also forgets $v_2$ because it is labeled by $I$, using the product on $H$ to get the label on the resulting edge. So we get a tree with vertices $(v_1, v_{34})$ and edges $(e_{134})$. These are labeled by $Q$, $Q$, and $H$.

In the second case, the attaching map first collapses the edge $e_{23}$ because it is between two elements of $P$, composing $P_2\otimes P_3\to P$, and then collapses the other two edges because they are of length $0$, getting $Q_1\otimes P\otimes Q_4\to Q$. In this case, we get a tree with only one vertex and no internal edges.

In order for the attaching map to be well-defined on the colimit defining the middle entry, if we begin in $I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes I_3\otimes Q_4$ then we must get the same eventual answer in $W(H,Q)_P$ whichever one of these procedures we choose. Then, shifting for concreteness' sake to, say, the topological category and the Boardman-Vogt interval, we get that the formal composition on the two vertex tree $q_1, q_2$ with edge length $r$ is equivalent to the composition $q_1\circ q_2$ on the one-vertex tree. This is a problem because it means that the entire structure collapses to $Q$, which is not in general sufficiently cofibrant.

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I'd also like to understand that construction, perhaps I'll find the time to have a go this afternoon. Could I just check what you mean by concentrating in arity one? To me this means that you're dealing with algebras (which aren't necessarily sufficiently nice). But maybe I've misunderstood because later you talk about trees. –  James Griffin Nov 5 '10 at 10:51
One other thing, if you understand the construction for P=I, then it might be worth just working out (ie guessing) the construction for general P and the topological case. That's the stage of understanding that I'm at and it's probably easier to guess the answer first and then carry out the construction. –  James Griffin Nov 5 '10 at 10:56
arity one means algebras, yes. The trees I'm talking about are bivalent. I tried guessing the construction for general P and the topological case, and my guess, which works in that case by Spitzweck, is explicitly not the same as the Berger-Moerdijk construction. Then I went to the construction myself and got the result I described above. –  Gabriel C. Drummond-Cole Nov 5 '10 at 18:11
It turns out that the construction as written in the paper is incorrect, but there is a fix known to the authors. Instead of collapsing $P\otimes H\otimes P$ to $P$, one only collapses $I\otimes P\otimes H\otimes P\otimes I$ (where the $I$ factors are included in $H$ with length 1, not 0). This gets rid of the problem above and the rest of the construction checks out. Many thanks to Gijs Heuts for communicating this to me.