20
$\begingroup$

What is the relationship between $G_\infty$ (homotopy Gerstenhaber) and $B_\infty$ algebras?

In Getzler & Jones "Operads, homotopy algebra, and iterated integrals for double loop spaces" (a paper I don't well understand) a $B_\infty$ algebra is defined to be a graded vector space $V$ together with a dg-bialgebra structure on $BV = \oplus_{i \geq 0} (V[1])^{\otimes i}$, that is a square-zero, degree one coderivation $\delta$ of the canonical coalgebra structure (stopping here, we have defined an $A_\infty$ algebra) and an associative multiplication $m:BV \otimes BV \to BV$ that is a morphism of coalgebras and such that $\delta$ is a derivation of $m$.

A $G_\infty$ algebra is more complicated. The $G_\infty$ operad is a dg-operad whose underlying graded operad is free and such that its cohomology is the operad controlling Gerstenhaber algebras. I believe that the operad of chains on the little 2-discs operad is a model for the $G_\infty$ operad. Yes?

It is now known (the famous Deligne conjecture) that the Hochschild cochain complex of an associative algebra carries the structure of a $G_\infty$ algebra. It also carries the structure of a $B_\infty$ algebra. Some articles discuss the $G_\infty$ structure while others discuss the $B_\infty$ structure. So I wonder: How are these structures related in this case? In general?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
9
$\begingroup$

There is a nice summary of the relationship between B infinity and G infinity in the first chapter of the book "Operads in Algebra, Topology and Physics" by Markl, Stasheff and Schnider. The short answer is G infinity is the minimal model for the homology of the little disks operad (the G operad). B infinity is an operad of operations on the Hochschild complex. Many of the proofs of Deligne's conjecture involve constructing a map between these two operads.

Improve this answer
$\endgroup$
1
  • 10
    $\begingroup$ More precisely, there is an opeard map (constructed by Tamarkin, and depending on the choice of an associator) $G_\infty\to B_\infty$. In other words, any $B_\infty$-algebra is a $G_\infty$-algebra. To see this, one has to remember that a $G_\infty$-structure can be expressed in terms of a DG Lie bialgebra structure on a cofree Lie coalgebra. On the other hand a $B_\infty$-algebra structure can be expressed in terms of DG bialgebra structure on a cofree coassociative algebra. the relation between the two is given by Etingof-Kazhdan dequantization Theorem. $\endgroup$
    – DamienC
    Apr 25, 2011 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.