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I know how to think about (curved) $A_\infty$-algebras 'geometrically', i.e. via formal non-commutative geometry in the sense of Kontsevich etc. I also know how to think about $A_\infty$-morphisms in this way. But what if two $A_\infty$-morphisms are homotopic? Does anyone know how to interpret this fact geometrically?

This is particularly important in the curved situation, because then there's no such thing as a quasi-isomorphism, so we only have homotopy-equivalence. Also I have a vague memory of reading something about this (probably written by Kontsevich), but I've searched all the papers I can think of and not found it.

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Can you say a little more about what you mean by 'geometric'? I'm not sure what kind of answer you're looking for. –  Dylan Thurston Jan 28 '12 at 17:34
    
The Kontsevich style geometric point of view on $A_\infty$-algebras is that such an algebra is a vector field on the formal completion of a point in an (infinite-dimensional) smooth noncommutative variety. The conventional $A_\infty$-algebras are the vector fields that vanish at the closed point. The curved $A_\infty$-algebras are arbitrary vector fields. –  Leonid Positselski Jan 28 '12 at 21:17
    
If the vector at the closed point is actually nonzero, the vector field can be "straightened up" (linearized) by an automorphism of the formal neighborhood, just as one would expect. This is Kontsevich's geometric reasoning underlying the algebraic assertions in my answer below (the rigorous algebraic proofs that I've recorded are also mostly due to Kontsevich). –  Leonid Positselski Jan 28 '12 at 21:25

1 Answer 1

I don't know specifically about homotopies, but the notion of a curved $A_\infty$-algebra is generally problematic. In the conventional setting of algebras over a field, it is just trivial in the following strong sense.

Let $A$ and $B$ be two curved $A_\infty$-algebras over a field $k$ with nonzero curvature elements $m_{0,A}\ne0\ne m_{0,B}$. This is a suffient condition to trivialize nonunital curved $A_\infty$-algebras; in the (strictly) unital case, assume that $m_{0,A}$ and $m_{0,B}$ do not belong to the one-dimensional vector subspaces generated by the units of $A$ and $B$ (whcich could happen in the $\mathbb Z/2$-graded case).

Then any isomorphism of graded vector spaces $f\colon A\to B$ (preserving the units, in the unital case) can be extended to an $A_\infty$-isomorphism $(f_0,f_1,f_2,\dotsc)\colon A\to B$ with $f_0=0$ and $f_1=f$. So there precisely as many curved $A_\infty$-algebras with nonzero curvature, up to $A_\infty$-isomorphism, as there are graded vector spaces; and any curved $A_\infty$-algebra with a nonzero curvature is $A_\infty$-isomorphic to a curved $A_\infty$-algebra with $m_1=m_2=m_3=\dotsb=0$.

Similarly, any curved $A_\infty$-module over a (nonunital or strictly unital) curved $A_\infty$-algebra with a nonzero curvature element is contractible.

These results are mostly due to Kontsevich; I learned them from conversations with him while visiting IHES and subsequently recorded them in what is now AMS Memoir vol.212 #996, 2011, http://arxiv.org/abs/0905.2621, Remark 7.3.

It appears that if you want to have a nontrivial theory of curved $A_\infty$-algebras, you have to do it over, say, a local ring and require the curvature elements in your algebras to be divisible by the maximal ideal of the local ring. I am presently working on this; the writeup is available from my homepage.

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This all looks plausible, but it appears to imply that the category of cdg-modules over a curved algebra is trivial? Also it doesn't answer my question :) (sorry for my delayed reply, MO failed to email me.) –  Ed Segal Feb 10 '12 at 17:31
    
No, morphisms between cdg-modules are not the same thing as morphisms between them viewed as curved A-infinity modules. So cdg-modules are not at all contractible, generally speaking, but when viewed as curved A-infinity modules, they are. –  Leonid Positselski Feb 11 '12 at 21:55
    
Similarly, the complex of morphisms between two dg-modules is different from the complex of morphisms between them viewed as A-infinity modules. In fact, the latter complex computes morphisms between these two dg-modules in the derived category of dg-modules. –  Leonid Positselski Feb 11 '12 at 22:06
    
So for cdg-modules, it does not make sense to define the derived category in this way (when working over a field). There was a paper by Pedro Nicolas suggesting this definition (arxiv.org/abs/math/0702449 ), and then in the subsequent paper of Keller-Lowen-Nicolas (arxiv.org/abs/0905.3845 ) they were proving that this "derived category" is zero. –  Leonid Positselski Feb 11 '12 at 22:06
    
this is really interesting, but this discussion is getting too involved to have in this comment thread! Let's continue by email. –  Ed Segal Feb 14 '12 at 11:09

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