What's the sense in which A_\infty algebras are "deformable"? - MathOverflow

What's the sense in which A_\infty algebras are "deformable"?

2009-10-07T20:30:05Z

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.

Answer by Mikael Vejdemo-Johansson for What's the sense in which A_\infty algebras are "deformable"?

2009-10-09T06:08:03Z

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)

Answer by James Griffin for What's the sense in which A_\infty algebras are "deformable"?

2009-10-09T15:38:23Z

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.

The short and snappy answer:

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.