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I would like explanation or clear source for some things related to $A_{\infty}$-spaces, via Stasheff's polytopes.

I tried not to think about them, because they seem too complicated for me; I thought that the small $1$-cubes operad, and abstract $A_{\infty}$-operads (each $A(n)$ is contractible), would be enough. But still, when I want to derive, at least for myself, at least heuristically, the axioms of $A_{\infty}$-algebra (in the algebraic sense), I see that I would like to understand those polytopes a bit.

There are different descriptions of $K_n$, Stasheffs polytopes. What would be a clear description, which shows all of the following three features: 1) $K_n$ embed into the small $1$-cubes (non-symmetric) operad; 2) This embedding makes $K_n$ a suboperad. 3) The boundary of $K_n$ breaks to different $K_s \times K_t$, and moreover, I can read the orientations from this, i.e. the signs which I will need to put in the dg-version.

Thank you, Sasha

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I do not know if this helps, but the way I like to think of these polytopes is as the nerves of a categories. The n^{th}-category whose nerve is the n^{th} polytope is the category of all ways of parenthesizing the string, 1,2,3,...n where the morphisms are given by a poset structure of one parenthization being more parethesized than another. For example ((12)3)(45) is more parethesized than (123)(45). – Spice the Bird Mar 27 '12 at 3:58
Strictly speaking, this gives the barycentric subdivision of the polytopes. – Spice the Bird Mar 27 '12 at 4:24
Thank you! I'll think about it. Do you have a reference for this description? – Sasha Mar 28 '12 at 8:39
up vote 5 down vote accepted

I have put some notes about operads at

They are not finished (in particular, very many references are missing), but sections 13 and 14 are in reasonable shape, and they should answer your questions. They are written in terms of symmetric operads, so my $K(n)$ consists of $n!$ copies of the Stasheff polytope.

I have also put another note at

This one shows that various things I had hoped to do with the Stasheff operad are probably not possible.

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Thank you! I'll look at both references. – Sasha Mar 28 '12 at 8:39

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