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What is the general concept of an *operad automorphism*$?$ Is there a "standard" definition?

[added after comment] If an operad automorphism is an invertible operad endomorphism, how then is operad endomorphism defined?

I'm being drawn into these ideas for a section of a paper I'm workign on and most likely this is written-up nicely somewhere,so

references would be most welcome.

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If you remove the word "correct" from your question, I think most people would agree James Griffin's 1st paragraph is the answer. Somehow putting "correct" in there makes your question more confusing. "Correct" in what sense, i.e. what properties do you want your "automorphisms" to have, if not simply being operad maps that have inverses which are operad maps? –  Ryan Budney Sep 28 '10 at 13:30
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An endomorphism of an operad $\mathcal O$ is a collection of morphisms $f_i: \mathcal O(i) \to \mathcal O(i)$ which give commutative diagrams when you apply the structure maps for the operad (on the "domain side" and "range side" of the $f_i$'s respectively). –  Ryan Budney Sep 28 '10 at 17:02
    
Bruno Vallette's website contains many links on operads: math.unice.fr/~brunov/ODH.html The current version of the book on algebraic operads by Jean-Louis Loday and Bruno Vallette: math.unice.fr/~brunov/Operads.html –  Thomas Riepe Dec 21 '10 at 6:20

2 Answers 2

I'm not sure what the question you're trying to ask is, but the answer to the question that you have asked is that an operad automorphism is an invertible operad endomorphism.

[EDIT] (just restating Ryan's comment on the original post) An operad endomorphism is an operad morphism where the source and target operads are the same. An operad morphism is a collection of maps $\mathcal{O}(n)\rightarrow\mathcal{P}(n)$, one for each arity, such that the obvious squares involving the operad structure maps commute. [/EDIT]

Perhaps you want to know about operad automorphisms in a homotopy category. In that case you want to understand what an operad quasi-isomorphism is. This is an operad morphism which is a quasi-isomorphism on each underlying space of operations.

Would you like to refine the question?

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Re-edited in light of your comments. If you can elaborate, that would d be great; it sounds like you have the answer I'm looking for! –  Romeo Sep 28 '10 at 16:39

Suppose we only care about operads in chain complexes, although i think this can all work more generally. Then an operad is a monoid in the category of symmetric sequences with respect to a particular monoidal product. so a morphism of operads would then be a morphism of monoids. I think this framework might help clarify things. The reference I have in mind, although i am sure there are earlier ones, is Kathryn Hess's lecture notes on the cobar construction. You want to look at the second lecture, page 9 specifically.

This different monoidal product is just what you want it to be in order for a monoid to be an operad! I would explain more, but I can't do any better than Hess: http://sma.epfl.ch/~hessbell/Minicourse_Louvain_Notes.pdf

PS: a symmetric sequence is a functor from the groupoid $\Sigma$ (where the objects are sets {1}, {1,2}, ... {1,...,n},... and the morphisms are bijections) to chain complexes.

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