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If the following is true, it is probably well-known to the experts. Nevertheless, I could not find a reference for it.

Suppose $P$ and $Q$ are $A_{\infty}$-operads in topological spaces and $X$ is a $P$-algebra as well as a $Q$-algebra. Moreover, assume that the actions of $P$ and $Q$ on $X$ commute, i.e. the following diagram is strictly commutative:

$$ \begin{gather} Q(n) \times P(m) \times X^{nm} & \to & Q(n) \times X^n \\\\ \downarrow & & \downarrow \\\\ P(m) \times X^m & \to & X \end{gather} $$

where the upper horizontal arrow uses the "diagonal action" of $P$ on $X^n$ and likewise for the left vertical arrow. Now I can deloop with respect to the operad $P$ to obtain $B_PX$, similarly I can obtain $B_QX$, where I either use the delooping machine of Boardman and Vogt or the bar construction - your choice.

Is it true that $B_PX$ carries an action of the operad $Q$ in this case?

Somehow this question boils down to the problem, how well products are respected by the common delooping machines. For example, I know that $B(G_1 \times G_2)$ is homeomorphic to $BG_1 \times BG_2$ for groups (or monoids). Is this still true, if $G_i$ are just $A_{\infty}$-spaces?

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Ulrich, did you mean $X^{nm}$ instead of $X^{n+m}$? –  Todd Trimble Sep 3 '12 at 15:20
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@Todd: Yes, I did. I corrected it. I had the case $n = m = 2$ in mind :-). –  Ulrich Pennig Sep 3 '12 at 15:24
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2 Answers

Something very close to this and quite possibly relevant occurs in the theory of $E_{\infty}$ ring spaces. There is a notion of an action of an operad $Q$ on an operad $P$. When $Q$ acts on $P$, the monad on the category of spaces associated to the operad $P$ restricts to a monad on the category of $Q$-spaces. This is an instance of a general distributivity theory of one monad acting on another due to Beck. The algebras over this monad give a version of your hypotheses on $X$. Adapting to the context of $A_{\infty}$ operads, and crossing $P$ with the little $1$-cubes operad (for example) without change of notation, there is a resulting delooping $B(\Sigma,P,X)$ that is a $Q$-space. The details are an adaptation of those in Sections 1, 4, 8, 9 and Appendix $A$ of my paper "What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra''. Geometry \& Topology Monographs 16(2009), 215--282. I hope that is relevant and apologize if it is not.

Alternatively, there is a theory of tensor product of operads due to Boardman (and Vogt?) that might be relevant. I don't have the reference on the top of my head.

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First of all let me mention that the structure you describe is precisely the Boardman-Vogt tensor product of the operads $P$ and $Q$. I'll denote this by $P\otimes_{BV}Q$. A space is an algebra for the BV tensor product of $P$ and $Q$ if it is a $P$-algebra in the symmetric monoidal category of $Q$-algebras (note that you need the fact that the monoidal structure is the cartesian product to show that the category of $Q$-algebras in spaces is symmetric monoidal).

I don't have an answer to your exact question but here is something close to what you ask. Maybe you are aware of all this. If so I apologize :

The category of $P$-algebras in $Q$-algebras is Quillen equivalent to the category of $Ass$-algebras in $Q$-algebras by standard facts about category of algebras over operads. Therefore you can strictify your $P$-algebra structure. You get a space $X'$ which is homotopy equivalent to $X$ with an action of $Ass\otimes_{BV} Q$. To this you can apply the Bar construction. The important fact is that it doesn't matter if you do it in the category of spaces or in the category of $Q$-algebras because the forgetful functor $Q-Alg\to Spaces$ commutes with geometric realization of simplicial objects.

If you are careful about doing everything homotopically, you are going to end up with a space $Y$ which is weakly equivalent to $B_PX$ and which is equipped with a $Q$-action.

The following paper http://arxiv.org/abs/math.AT/0410367 does something similar to show that $THH$ of an $E_2$-algebra is an $A_{\infty}$-algebra.

Something interesting is that you can only strictify one of the two $A_{\infty}$-structure. Indeed $P\otimes_{BV} Q$ is usually equivalent to the little $2$-disk operad (in fact the derived BV tensor product is equivalent to $E_2$). However $Ass\otimes Ass$ is the commutative operad. This paper http://arxiv.org/abs/1102.1311 talks about the BV tensor product of $A_\infty$ and $E_n$.

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