There are at least a couple of definitions in the literature for an $E_2$-algebra, also known as a homotopy Gerstenhaber algebra, also known as $G_{\infty}$-algebra.

Suppose $V$ is a graded vector space. Let $S(V), \text{Lie}(v)$ denote the graded symmetric algebra and the graded free Lie algebra of $V$ respectively and $S_+(V)$ be the symmetric algebra in strictly positive degree.

$\textbf{Definition (Tamarkin)}$ A $G_{\infty}$-algebra structure on $V$ is a degree $+1$ map $$ \delta: S_+(\text{Lie}(V^*[1])[1]) \rightarrow S_+(\text{Lie}(V^*[1])[1]) $$ which behaves as a derivation of both $\cdot$ and $[\,,]$ such that $\delta^2 = 0$.

Unpacking this leads to saying that we have a collection of maps $$ m_{k_1, \dots, k_n}: V^{\otimes k_1} \otimes \dots \otimes V^{\otimes k_n} \rightarrow V$$ of degree $3-(k_1 + \dots k_n + n)$ obeying appropriate symmetry and associativity relations.

The second definition pertains to algebras over the little disk operad. Let $D_2(k)$ denote the configuration space of $k$ little disks inside a big disk and let $\text{Chains}_{\bullet}(D_2(k))$ be the singular chain complex. Letting $\mathcal{P}(k) = \text{Chains}_{\bullet}(D_2(k))$, one can define an operadic structure on this collection of vector spaces. This leads one to the

$\textbf{Definition (Getzler-Jones?)}$ An $E_2$-algebra is an algebra over the operad $\text{Chains}(D_2)$.

How can one show that these two definitions are equivalent?

Short of a full proof of equivalence, it would be nice to understand a description of the cycle in $D_2(k_1 + \dots +k_n)$ which corresponds to the map $m_{k_1, \dots, k_n}$. For example: if one were working in $H_{\bullet}(D_2)$, the homology operad, one associates to the point class in configuration space of two disks the operation $\cdot = m_2$, and to the cycle involving one little disk going around the other the bracket $[\,,] = m_{1,1}$ in the Gerstenhaber algebra. Is there an explicit description of the cycle corresponding to $m_{k_1, \dots, k_n}$?


2 Answers 2


The equivalence between these two notions is nontrivial, since it amounts to a choice of formality isomorphism for the operad of little disks.

Let $D_2$ be the little disks operad. The easy part of the equivalence is to show that $G_\infty$ algebras are equivalent to algebras over the operad $H_*(D_2)$. If you take the operad of graded $\mathbb Q$ vector spaces, $H_*(D_2)$, its algebras are Gerstenhaber algebras-- this can be seen by computing the homology of the configuration spaces. Then Koszul self-duality for the Gerstenhaber operad gives the relationship between $G_\infty$ algebras and Gerstenhaber algebras. I.e. every $G_\infty$ algebra may be rectified to an algebra over $H_*(D_2)$.

Let $C_*(D_2)$ be chains on the little disk operad. To obtain an equivalence between $C_*(D_2)$ algebras and $H_*(D_2)$ algebras, one needs to choose a zig-zag of quasi-isomorphisms between the two operads. These formality isomorphisms are difficult to construct, and there are many possible choices, because these operads admit a large automorphism group. See Kontsevich's paper, https://arxiv.org/abs/math/9904055, for a discussion.

  • $\begingroup$ Thanks for the answer. Please take a look at the edits I made. $\endgroup$
    – dayar
    Commented Feb 19, 2020 at 19:50

Here are the operads that are involved in that game:

  • the operad $D_2$ of little disks, which is a topological operad.

  • its chain operad $C_{-*}(D_2,\mathbb{k})$, which is an operad in cochain complexes (of $\mathbb{k}$-modules).

  • its homology operad $H_{-*}(D_2,\mathbb{k})$, which is known to be isomorphic to the Gerstenhaber operad $G^{\mathbb{k}}$, which is itself a binary quadratic operad satisfying the Koszul property.

  • the minimal resolution of $G^{\mathbb{k}}$ is the operad $G^{\mathbb{k}}_\infty$ governing ($\mathbb{k}$-linear) $G_\infty$-algebras.

Being a resolution, $G^{\mathbb{k}}_\infty$ is obviously quasi-isomorphic to $G^{\mathbb{k}}$. As Phil Tosteson mentions in his answer, the difficult part relies on proving that $C_{-*}(D_2,\mathbb{k})$ is formal. It's only proven over a field $\mathbb{k}$ of characteristic zero, and it's actually not formal for $\mathbb{k}=\mathbb{F}_p$ (see e.g. the introduction of https://arxiv.org/pdf/1903.09191.pdf). To my knowledge, there are essentially two different proofs:

Note that the formality quasi-isomorphisms from these two proofs happen to coincide, after the correct "choice of associator" has been made. See https://arxiv.org/abs/0905.1789.

Finally, I don't think it is meaningful to ask which cycle corresponds to the map $m_{k_1,\dots,k_n}$. The reason is that $m_{k_1,\dots,k_n}$ is not closed. You may ask if they are represented by nice chains... I don't know the answer to that question (and I suspect that it is close to be as hard as proving the formality itself), but for the $m_{1,\dots,1}$ the answer is known:

  • first observe that $D_2$ is weakly homotopy equivalent to the operad of compactified configuration spaces of points in the plane.

  • $m_{1,\dots,1}$ (with $n$ "$1$"s) can be represented by the top-dimensional/fundamental cell of the compactified configuration space $\overline{C}_n\simeq D_2(n)$ of $n$-points in the plane.

  • $\begingroup$ Thanks! In $\overline{C}_n$ is the group of dilations and overall translations implicitly modded out? So that the fundamental cell of $\overline{C}_n$ is $2n-3$ dimensional? $\endgroup$
    – dayar
    Commented Feb 21, 2020 at 1:58
  • $\begingroup$ Yes, the group we use to mod out is the semi-direct product of translations of (real, positive) dilations. hence, yes, the dimension of the top dimensional cell is 2n-3. $\endgroup$
    – DamienC
    Commented Feb 21, 2020 at 13:48
  • $\begingroup$ Great. It might be worth asking this as a separate question, but what is the argument that there is a morphism of operads from the $L_{\infty}$-operad to the suboperad of top dimensional cells in the compactified configuration space? If you know any references I'd appreciate it. $\endgroup$
    – dayar
    Commented Feb 21, 2020 at 15:43
  • 1
    $\begingroup$ I don't have a reference in mind, but I think the argument is not so difficult. As a graded operad (ie without the differential), the $L_\infty$-operad is free. So, you define a morphism of graded operads sending $m_{1,\dots,1}$ to the top dimensional cell in $\overline{C}_n$. Then you check that it is compatible with the differentials (intuitively, the boundary of the top dimensional cell is a union of products of top-dimensional cells of $\overline{C}_m$'s for $m<n$. $\endgroup$
    – DamienC
    Commented Feb 22, 2020 at 9:58

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