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An A-infinity algebra is smooth a'la Kontsevich if it is perfect as an A-A bimodule. I am wondering about the standard tricks to show smoothness of given algebras. A relatively basic example should be the following. I have a guess that the following Z/2Z graded A-infinity algebras over C should be smooth even though their "underlying associative algebra" isn't. The algebra A_n is a vector space 1,e, where e is in even degree. The "classical" multiplications are 1 acts as a unit and e*e=0. There is only one higher multiplication, e^(tensor power)2n-->1, n>1. If I did my math right, this defines an A-infinity algebra. The basis for the guess of smoothness is that the Hochschild homology is finite dimensional, which would be a corollary of perfection, but that doesn't quite prove it without some other statement about the perfection of finite-dimensional modules over A "tensor" A-op(which would also imply the statement directly of course).

I have struggled quite a bit unsuccessfully to brute force this. I have tried to find a suitable dg-algebra equivalent to A and then to compute explicitly A-"tensor"-A-op and then write down a resolution of A. This was too gritty for me, though maybe a more insightful person could make it work. Is the fact true? Can anyone give an explanation/proof?

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Can you be more precise about the waymultiplications work in the algebra? –  Mariano Suárez-Alvarez Jul 3 '10 at 14:49
    
Hopefully my edits made it clearer... –  Daniel Pomerleano Jul 3 '10 at 15:18

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