# What's the sense in which A_\infty algebras are “deformable”?

I realise this is a very vague question! I've heard people say that A algebras are the right homotopy-theoretic generalization of usual associative algebras, because you can deform them. What exactly does this mean?

This roughly makes sense -- if you "deform" an associative algebra, it's generically going to stop being associative, but it will be "associative up to homotopy" in exactly the sense A algebras are.

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Well here's my shot: (skip to the punchline at the bottom if you want)

Take an associative algebra A and a k-local ring R (the formal power series over k, or the infinitesimal ring will do nicely).

The algebra A is naturally a homotopy algebra and so may be given by a degree -1 square-zero coderivative on the free coassociative coalgebra on A[1]. We write this coalgebra BA, the bar resolution. Note that in homotopy theory it often makes life easier if we forget any unit elements; BA is non-unital.

An A-infty R-deformation of A is now a square-zero coderivative on the coalgebra BA⊗R, such that the "obvious" diagram commutes (I could post this as an image when I'm permitted). The condition could alternatively by phrased as the following: "such that it extends the original coderivative on BA".

So far this has all been definitions, my answer to your question comes next: Consider now the cobar functor applied to the morphism BA⊗R→BA,

Ω(BA⊗R) ≅ (ΩBA)⊗R → ΩBA.

This is a proper algebra deformation, nothing infinity about it! Except... ΩBA is homotopy equivalent to A.

## Infinity deformations are homotopy invariant, classical algebra deformations are not.

Edit: I should have added, if you would like me to expand on anything, I'm more than willing.

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Hmmm, perhaps I could also add that for a quasi-free associative algebra A=&Omega;C this construction also says that every infinity deformation is equivalent to a classical algebra deformation via the homotopy equivalence BA &rarr; C. But if that works it's subtle and would confuse things, so I'll leave it in a comment. –  James Griffin Oct 9 '09 at 16:20
Unfortunately coalgebras still scare me sufficiently that I don't really see what the short and snappy answer had to do with the paragraph before. –  Scott Morrison Oct 9 '09 at 20:22
I don't know who this viewpoint of A-infinity algebras is originally due to, but I learned it from reading papers of Kontsevich. It's actually not scary and quite geometric. Here's a rough sketch: think of a vector space V as a "formal manifold"; take the cofree coalgebra on V, think of this as the dual or predual of "C^infty(V)"; take a coderivation of the coalgebra, think of this as a "vector field"; the "Taylor coefficients" of the coderivation will be the multiplications m_n; the coderivation being square-zero is equivalent to saying these m_n's define an A-infinity structure on V. –  Kevin H. Lin Oct 9 '09 at 21:31
Looking back at it I can see that the short and snappy answer should probably have come first, then the rest of the answer was an illustration that we can find a quasi-isomorphic algebra where each infinity deformation is equivalent to a classical deformation. Sorry. On coalgebras: The way I think about a coalgebra is as a degree -1 derivation on the free algebra TC[-1]. Pick a coalgebra you know and try it out! Classical dg-coalgebras have image in C[-1]&oplus;C[-1]&otimes;C[-1], infinity coalgebras have image in the whole of TC. –  James Griffin Oct 12 '09 at 10:32

Further on the A-infinity operad - it is what we get if we do the "obvious" moves to introduce a homological algebra on operads, and then look for a free dg-operad quasi-isomorphic to the associative operad. In that sense, the A-infinity operad is just the "free resolution" of associative algebras, and therefore a sensible homotopy equivalent replacement for the original operad.

(reposting since something got wonky with my account and login)

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I've deleted the other post and the other account and added the other OpenID as an alternate on your account. Let me know if you have any further trouble with your account. –  Anton Geraschenko Oct 9 '09 at 6:30