Any symmetric operad can be considered as an non-$\Sigma$ operad by throwing away permutations. Does anyone know what sort of structure one gets for algebras over $C_n$ the little n-cubes operad, or some equivalent variant, considered in this way? One thing which is strange is that there is a non-$\Sigma$ operad splitting of the map $Ass \to Com$ (where $Ass$ and $Com$ are the symmetric associative and commutative operads, respectively), and so any non-$\Sigma$ $C_n$ algebra should be a loop space, but I do not see what, if any, extra structure is present.
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2$\begingroup$ My ideal answer would be something analogous to the characterization of symmetric $C_n$ algebras as $n$-fold loop spaces. Perhaps a nice description of the free algebras. It is clear you have a loop space for all $n$, and then at $C_\infty$ you get a loop space again, but does anything interesting happen in the middle? $\endgroup$– Justin YoungCommented Feb 22, 2013 at 15:28
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$\begingroup$ Is the question settled for $Ass$? I've tried to think of this case, that I assumed to be easy, but didn't get anything appart from the splitting you mention. $\endgroup$– Fernando MuroCommented Feb 23, 2013 at 8:24
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Justin, that is a cute question, and I've never thought about it. One starting thought free guess (acting on connected spaces) is that a non-$\Sigma$ $C_n$-space has $n$ possibly inequivalent but definitely related loop space structures. Do you see examples, or is this just curiosity? As to Ass and Com, I see no interest in building in the permutations to define Ass and then throwing them away. Of course, a non-$\Sigma$ `Com'-algebra is the same thing as an Ass-algebra, but the operadic modules over these kinds of algebras are different (bimodules and left modules). David, I think Justin is asking a concrete question, not the sort that model category structures shed light on.
answered Feb 22, 2013 at 15:44
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$\begingroup$ Thank you for responding. I do not have any examples in mind other than the obvious free algebras. This question was put to me and I thought it was surprising and interesting, and I had never thought about it, either. I thought that there should be a simple answer, but I was unable to come up with one quickly myself or to find anything in the literature about it. $\endgroup$ Commented Feb 23, 2013 at 6:54