Let $\mathbf{sSet}$ be the model category of simplicial sets and $\mathbf{Op}$ the model category of symmetric operads. Equipped with Boardman-Vogt tensor product $ \otimes\_{BV}$, the category $\mathbf{Op}$ is symmetric monoidal. 


**Questions:**


1) is $(\mathbf{Op}, \otimes\_{BV})$ a symmetric monoidal model category ?


2) let $P$ be a $E\_{n}$-operad and $Q$ be a $E\_{m}$-operad, is $P\otimes\_{BV}Q$ a $E\_{n+m}$-operad ?


**Edit**


1) $P$ and $Q$ are cofibrant symmetric operads.


2) a weak equivalence (fibration) of symmetic operads $f: A\rightarrow B$ is a level-wise weak equivalence (fibration) of simplicial sets $f\_{n}: A(n)\rightarrow B(n)$.


3) the Boardman-Vogt tensor product $A\otimes\_{BV} B$ is a (tricky) quotient of $A\sqcup B$ (the coproduct in $\mathbf{Op}$).


4) a $A\otimes\_{BV}B$-algebra is a $A$-algebra in the category of $B$-algebras, or similarly a
$B$-algebra in the category of $A$-algebras. 


5) it is natural to ask if the category of $E\_{n}$-algebras in the category of $E\_{m}$-algebras is equivalent (in homotopical sense) to $E\_{n+m}$-algebras. The question can be formulated as follows. It is true that  

$$E\_{n}\simeq E\_{1}^{\otimes\_{BV}^{n}}$$



Boardman-Vogt tensor product
The answer to 2) is yes for cofibrant operads, see <http://arxiv.org/abs/1102.1311> by Fiedorowicz and Vogt. The answer to 1) is no; the Boardman-Vogt tensor product does not interact well with cofibrations.


EDIT: As an answer to Chris' comment, here is a counterexample to 2) in the setting of colored operads, although you can adapt it to the one-color case as well. In fact, the example just concerns simplicial categories (i.e. simplicial operads with only unary operations), where the tensor product coincides with the Cartesian product of simplicial categories. Therefore it also shows why the Bergner model structure on simplicial categories is not Cartesian.


Write $I$ for the category with objects 0 and 1, with one non-trivial morphism from 0 to 1, and write $\partial I$ for the disjoint union of 0 and 1 (with just their identity morphisms). Clearly $\partial I \rightarrow I$ is a cofibration. However, the pushout-product $\partial I \times I \cup I \times \partial I \rightarrow I \times I$ is not a cofibration; it's not even a monomorphism. On the left-hand side, the (discrete) simplicial set of morphisms from $(0,0)$ to $(1,1)$ consists of two points, on the right-hand side there is only one.


If you tweak this example to apply to the one-object case (and identifying categories with one object with monoids), you run into the morphism from the free monoid on two generators to the free commutative monoid on two generators, which again is not a monomorphism.


By the way, this is not the only issue: it is also possible to cook up counterexamples to the pushout-product axiom for the Boardman-Vogt tensor product by playing around with nullary operations, using an Eckmann-Hilton style argument.


