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The homology of an $E_2$-algebra is a Gerstenhaber algebra.

How precisely is the Gerstenhaber structure related to the $E_2$-structure?

Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology.

But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right?

Thanks for any hints.

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I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).

Namely, a brace algebra satisfies the relation $$ab - (-1)^{|a||b|}ba = (-1)^{|a|} d(a\{b\}) - (-1)^{|a|}(da)\{b\} + a\{db\},$$ where $a\{b\}$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace: $$[a, b] = a\{b\} - (-1)^{(|a|+1)(|b|+1)} b\{a\}.$$ Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.

So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.

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  • $\begingroup$ Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket? $\endgroup$
    – Lukas Woike
    Commented Aug 13, 2018 at 7:46
  • $\begingroup$ Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)? $\endgroup$
    – Lukas Woike
    Commented Aug 13, 2018 at 7:46
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    $\begingroup$ For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $a\{b\}$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $b\{a\}$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie\{1\}\rightarrow H_\bullet(E_2)$ in arity 2 sends the bracket to the fundamental class. $\endgroup$
    – Pavel Safronov
    Commented Aug 13, 2018 at 8:15
  • $\begingroup$ May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_{p,q} : A_p \otimes A_q \to A_{p+q+1}$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$? $\endgroup$
    – Lukas Woike
    Commented Aug 15, 2018 at 7:57

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