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.



What's the sense in which A_\infty algebras are "deformable"?
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.
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Edit: I should have added, if you would like me to expand on anything, I'm more than willing.


