User james griffin - MathOverflow

Answer by James Griffin for Deformation theory and differential graded Lie algebras

2009-10-13T12:21:10Z

I can offer an algebraic example generalising that of Hochschild cohomology. Let O → P be a morphism of operads, assume that O has O(1)=k (although augmented should be strong enough as well I think). Then we can form a cofibrant resolution, O' of O, this has the underlying structure of the free operad on a set of generators C, and this C has a cooperad structure.

We want to deform f. Well Hom(O',P) < Hom(FC,P) = Hom_S(C,P), where the first two hom's are in the category of operads and the final is in the category of collections. Recall that collections underlie operads, they can be given as functors from the category of finite sets and bijections into vector spaces.

But C is a cooperad and P is an operad, and this homset looks a lot like linear maps between them, so shouldn't we have a convolution operad structure. Well we do, but it's only a non-symmetric operad.

Non-symmetric operads have the natural structure of a pre-Lie algebra, the composition is defined by taking the sum of all possible ways of plugging one operation into another. If you haven't met pre-Lie algebras then don't worry as the anti-symmetrisation of the pre-Lie product is a Lie bracket. So our non-symmetric operad Hom(FC,P) naturally has the structure of a dg-Lie algebra.

The nature of the inclusion of Hom(O',P) into Hom(FC,P) should come as no surprise. They're precisely the Maurer-Cartan elements. So our morphism f corresponds to a MC element in a dg-Lie algebra.

Given a MC element in a graded-Lie algebra the deformations are MC elements in the dg-Lie algebra twisted by the original MC element.

Examples

Let P be an endomorphism operad ⊕Hom(A⊗...⊗A,A), then the theory above is the deformation theory for O algebras.
Let O be the associative operad, let P be the operad for associative algebras with an action of a comm. alg R by central elements and a Lie algebra g by derivations with some compatability conditions (in fact by a Lie-Rinehart algebra, or a Lie-algebroid). Then the deformation theory of the inclusion morphism is the study of deformations induced by Lie-algebroid actions. In fact the dg-Lie algebras involved are very often formal.

Answer by James Griffin for What is Koszul duality?

2009-10-12T14:01:11Z

I'd really like to hear a complete answer to this question. I have my own idea and although I don't think it's a perfect answer to your question, perhaps it will help someone.

There's a way of defining cofibrant replacements in certain categories call the Boardman-Vogt W-construction. I know it from the work of Berger and Moerdijk on operads. I believe it was first defined topologically, but I only know it algebraically and for algebras it goes something like this:

The unit interval could be described by the ring I=kt ⊕ ks ⊕ ke, where e is concentrated in degree 1, and d(e)=t-s. t is the identity and s is idempotent, also se=e. There's a unique augmentation. So this is a chain complex modelling the unit interval. The multiplication could be described as taking the maximal value, this is how it's defined topologically. t corresponds to the maximal element 1, s to the element 0.

Now let A be an associative algebra, the W-construction W(A) looks very much like the free product A*I (perhaps it's the same I don't remember), with the induced differential. W(A) is a cofibrant replacement for A.

This contains a subalgebra quasi-isomorphic to W(A), which may be given by the bar-cobar construction. Remember that the bar complex BA is given by taking the free coalgebra on A[1], with a square-zero coderivative induced by the multiplication map. The cobar complex ΩC on C is the free algebra on C[-1]. This carries the square-zero derivative defined by the coalgebra structure.

That's a mouthful, but all you need to remember is that the bar and cobar functors define an adjunction, actually a Quillen equivalence of model categories. And the unit and counit of this look very much like the W-construction.

In Berger and Moerdijk they do this not for associative algebras but for operads. And there are similar constructions for any operad (generalising the case for the associative operad above). In one example the bar complex is based on the free colie algebra and the cobar complex is based on the free commutative algebra.

Onto the categories of modules: there is an adjunction between the category of modules of A and the category of comodules for BA. This is actually a Quillen equivalence.

Now back to the Koszul property. An algebra is Koszul when the bar complex is quasi-isomorphic as coalgebras to something very small, in many cases quasi-isomorphic to its homology. Call this nice coalgebra C.

Then there is an adjunction between the category of modules for A and the category of comodules for C.

So to me at least the Koszul property concerns situations when large resolutions of things, say ΩBA, may be replaced by much smaller things. And this means a lot for the categories of (co)modules for the objects concerned. But that's just my personal impression.

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.

Comment by James Griffin

2009-10-14T09:39:14Z

That's certainly the notion of formality I'm familiar with. Kontsevich formality concerns just one particular class of examples: the Hochschild cohomology of a "smooth algebra" has a dg-Lie algebra structure. Kontsevich formality says that it's formal.

Comment by James Griffin

2009-10-13T12:27:23Z

I'm looking forward to a proper answer to this question, I've not come across a theorem like that before and it sounds very interesting indeed. My experience is very much weighted to the algebraic side, I should read up on geometric examples.

Comment by James Griffin

2009-10-12T10:32:39Z

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.

Comment by James Griffin

2009-10-09T16:20:14Z

Hmmm, perhaps I could also add that for a quasi-free associative algebra A=ΩC this construction also says that every infinity deformation is equivalent to a classical algebra deformation via the homotopy equivalence BA → C. But if that works it's subtle and would confuse things, so I'll leave it in a comment.