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We know that in Tamarkin's proof of Kontsevich's formality theorem, he defined the $G_\infty$ structure on the Hochschild cochain complex $C^\cdot(A,A)$ and constructed a $G_\infty$ morphism from $HH^\cdot(A,A)$ to $C^\cdot(A,A)$, which makes Kontsevich's original $L_\infty$ morphism as the $L_\infty$ part of it.

So what is the precise definition of "the $L_\infty$ part of a $G_\infty$ morphism"? I know it looks like we just forget the homotopies for the cup product but how to make it precise?

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There is a map of operads from $L_\infty$ to $G_\infty$. Given a $G_\infty$, just "restrict operations" along that map, just as one restricts scalars along a map of rings. –  Mariano Suárez-Alvarez Jul 8 '12 at 3:03
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@Mariano: An $L_\infty$ morphism is NOT the same as a map of algebras over the $L_\infty$ operad. The latter is much stricted than the former. –  André Henriques Jul 8 '12 at 6:47

1 Answer 1

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A $G_\infty$-morphism $\phi$ is determined by structure maps $\phi^{k_1,\dots,k_n}$, $n\geq1$, $k_1,\dots,k_n\geq1$.

The $L_\infty$-part of $\phi$ is the $L_\infty$-morphism $\ell$ with structure maps $\ell^k=\phi^{\overbrace{1,\dots,1}^{k~times}}$.

In order to make this precise you might have to put (de)suspensions at the appropriate places.

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Thank you very much! –  Zhaoting Wei Jul 10 '12 at 2:48

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