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An A-infinity algebra $A$ 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 ${\mathbb Z}/2 {\mathbb Z}$-graded A-infinity algebras over $\mathbb C$ should be smooth even though their "underlying associative algebra" isn't. The algebra $A_n$ is a vector space with basis $\{1,e\}$, 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, $e^{\otimes 2n} \mapsto 1$, $n>1$. 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 $A \otimes A^{\mathrm {op}}$(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 $A \otimes A^{\mathrm {op}}$ 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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