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I would like to understand how I can endow an n-fold (complete) Segal Space with a symmetric monoidal structure. My question is basically the same as in this post: What is a symmetric monoidal $(\infty,n)$-category? but I'm still not sure how to formulate this in terms of complete Segal spaces. If I just try to copy the definition from Lurie's book "Higher Algebra" I guess I get something like this:

Let $X:\Delta^{op}\rightarrow Top$ be a complete Segal space ( or $(\infty,1)$-category) then a symmetric monoidal structure on $X$ is a morphism $F: X^{\otimes}\rightarrow N(Fin_{*})$ between two Segal spaces with $X^{\otimes}_{<1>}:=F^{-1}(<1>)=X$ (so $F$ is a natural transformation $F$= { $F_n: X_n^{\otimes} \rightarrow N(Fin_{*})_n $} and I guess $F^{-1}(<1>)$ = { $F_n^{-1}(<1>) $ }$_{n\geq 0}$ ) with the following two properties:

1)$F$ is a coCartesian fibration

2)For all $n\geq 0$ the morphisms $ p^i:< n > \rightarrow <1>;p^i(i)=1;p^i(j\neq i)=* $ induce an equivalence of Segal spaces: $X_{< n >}^{\otimes} \rightarrow X_{<1>}^{\otimes}\times...\times X^{\otimes}_{<1>}$.

$Fin_*$ is the category of pointed sets $< n >$ = { * ,1,...,n } and functions with $ f(*) = * $ as morphisms and $N(Fin_{*}):\Delta^{op} \rightarrow Set$ is the nerve of $Fin_{*}$

My question is if this is correct or is there a mistake (I'm sure there is) or some additional condition that I'm missing? How do I generalize this to n-fold Segal spaces? Just replace everything with n-fold versions? I'm pretty new at this stuff so I'm quite uncertain how to formlate this. As far as I know there is not much literature on this topic.

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I think the question title is a bit misleading. What you're really asking is whether your proposed definition is equivalent to the ones in the other thread, not if a given n-fold complete Segal space can be endowed with the structure of a symmetric monoidal $(\infty,n)$-category. I don't think the latter can be done, since a symmetric monoidal $(\infty,n)$-category is basically an $E_\infty$-algebra in the category $C$ of n-fold complete Segal spaces and there's no reason a given object of $C$ would admit such algebraic structure, even up to weak equivalence. – David White Aug 4 '14 at 9:21

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