Symplectic homology product via Legendrian surgery

Let $R$ be an algebra over a field $K$ of characteristic $0$ generated by idempotents $1_{1},\dots,1_{m}$, where $1_{k}1_{l}=\delta_{kl}1_{k}$, where $\delta_{kl}$ is the Kronecker delta. We define a graded module $V$ over $R$, which is generated by a countable set ${\mathcal{C}}$ decomposed as a disjoint union ${\mathcal{C}}=\bigcup_{1\leq i,j\leq m}{\mathcal{C}}_{ij}$. The degree of an element $a\in V$ will be denoted by $|a|$. The decomposition of ${\mathcal{C}}$ induces a decomposition $V=\bigoplus\limits_{1\leq i,j\leq m}V^{(ij)}$, where $V^{(ij)}$ is generated by ${\mathcal{C}}_{ij}$. When applied on the left, the element $1_{j}$, $j=1,\dots,m$, acts as the identity on $V^{(ij)}$ and as the $0$-map on $V^{(il)}$, $l\neq j$. Similarly, when applied on the right, the element $1_{j}$ acts as the identity on $V^{(ji)}$ and as $0$ on $V^{(li)}$, $l\neq j$.

We define an extension $\widehat{V}$ of the module $V$ as follows. The module $\widehat{V}$ is generated by $\widehat{\mathcal{C}}\cup\mathcal{X}$, where $\widehat{\mathcal{C}}=\{\widehat{c}\colon c\in\mathcal{C}\}$ and ${\mathcal{X}}=\{x_{1},\dots,x_{m}\}$, where $|\widehat{c}|=|c|+1$, for $c\in\mathcal{C}$ and $|x_{j}|=0$, $j=1,\dots,m$. We define

$$\widehat{V}:=K\langle{\mathcal{X}}\rangle\oplus V[1],$$

where $K\langle{\mathcal{X}}\rangle$ denotes the $m$-dimensional module freely generated by the set ${\mathcal{X}}$ and where $V[1]$ denotes $V$ with grading shifted up by $1$. The module $\widehat{V}$ inherits a decomposition $\widehat{V}=\bigoplus\limits_{1\leq i,j\leq m}\widehat{V}^{(ij)}$ from $V$, where

$$\widehat{V}^{(ij)}=\begin{cases}V^{(ij)}[1]&\text{if }i\neq j,\\ Kx_{j}\oplus V^{(jj)}[1]&\text{if }i=j.\end{cases}$$

Here the idempotents act as before on $V\approx V[1]$ as a subset of $\widehat{V}$, and as follows on the new generators: $1_{j}x_{i}=\delta_{ji}x_{j}$ and $x_{i}1_{j}=\delta_{ij}x_{j}$, $1\leq i,j\leq m$.

We will use the following notation. If $a\in V$ then $\widehat{a}$ denotes the element in $V[1]\subset\widehat{V}$ that corresponds to $a$. Furthermore, if $F$ and $G$ are either $V$ or $\widehat{V}$ and if $(u,v)$ is an ordered pair of elements in $F^{(ij)}\oplus G^{(kl)}$ then $(u,v)$ is composable if $j=k$.

Let

$$A:=R\oplus V\oplus V^{{\mathop{\otimes}\limits_{R}}^{2}}\oplus\dots$$

(2.1)

be the tensor algebra over $R$ generated by $V$ with its natural grading. Then $A$ is generated additively by composable monomials $w=c_{1}\dots c_{k}$, $c_{j}\in V$, $j=1,\dots,k$, where we say that $w$ is composable provided $(c_{j},c_{j+1})$ is composable for $1\leq j\leq k-1$. (Here, as often in the following, we will suppress the tensor sign in the notation for tensor products). The decomposition of $V$ induces a corresponding decomposition of $A$, $A=\bigoplus_{1\leq i,j\leq m}A^{(ij)}$, where $A^{(ij)}$ is generated by composable monomials $c_{1}\dots c_{k}$ with $c_{1}\in V^{(is)}$ for some $s\in\{1,\dots,m\}$ and $c_{k}\in V^{(tj)}$ for some $t\in\{1,\dots,m\}$.

We will assume that the algebra $A$ has a differential $d\colon A\to A$ of degree $-1$ which leaves the subspaces $A^{(ij)}$, $i,j\in\{1,\dots,m\}$ invariant (i.e., $d^{2}=0$, $d$ satisfies the Leibniz rule, and $d(A^{(ij)})\subset A^{(ij)}$). We will usually write $H(A)$ for the homology algebra $H_{*}(A,d)$.

Define

$$M=M(A):=A\mathop{\otimes}\limits_{R}\widehat{V}\mathop{\otimes}\limits_{R}A.$$

(2.2)

Then $M$ is a graded left-right module over the differential graded algebra $A$ which is additively generated by composable monomials $w=c_{1}\dots c_{k}uf_{1}\dots f_{l}$, where $c_{i}\in{\mathcal{C}}$, $i=1,\dots,k$, $u\in\widehat{\mathcal{C}}\cup{\mathcal{X}}$, and $f_{j}\in{\mathcal{C}}$, $j=1,\dots,l$. We say that a composable monomial $w$ is cyclically composable if the pair $(f_{l},c_{1})$ is composable. Let $M^{{\operatorname{diag}}}\subset M$ be the submodule generated (over $K$) by cyclically composable monomials and define $M^{{\rm cyc}}$ as the quotient space of $M^{{\operatorname{diag}}}$ in which monomials which differ by graded cyclic permutation are identified. If $w\in M^{{\operatorname{diag}}}$ then we write $\llbracket w\rrbracket$ for its equivalence class in $M^{{\rm cyc}}$, i.e. if $\pi\colon M^{{\operatorname{diag}}}\to M^{\rm cyc}$ denotes the natural projection then $\llbracket w\rrbracket:=\pi(w)\in M^{\rm cyc}$.

The differential $d\colon A\to A$ induces a differential $d_{M}\colon M\to M$ as follows. If $w_{1}u\,w_{2}\in M$, $u\in\widehat{V}$, and $w_{1},w_{2}\in A$, then

$$d_{M}(w_{1}u\,w_{2}):=(dw_{1})u\,w_{2}+(-1)^{|w_{1}|}w_{1}(d_{M}u)w_{2}+(-1)^{|w_{1}u|}w_{1}u(dw_{2}).$$

(2.3)

Here

$$d_{M}u=\begin{cases}ax_{i}-x_{j}a-S(da)&\text{if }u=\widehat{a},\,\,a\in V^{(ij)},\\ 0&\text{if }u=x_{j},\;j\in\{1,\dots,m\},\end{cases}$$

(2.4)

where the $R$-module homomorphism $S:A\to M$ is defined by

	$\displaystyle S(a_{1}\dots a_{k})$	$\displaystyle=\widehat{a}_{1}a_{2}a_{3}\dots a_{k}\,\,+\,\,(-1)^{|a_{1}|}a_{1}\widehat{a}_{2}a_{3}\dots a_{k}$
		$\displaystyle+\,\,\dots\,\,+\,\,(-1)^{|a_{1}\dots a_{k-1}|}a_{1}a_{2}\dots a_{k-1}\widehat{a}_{k},$		(2.5)

and $S(1_{j})=0$, for $j=1,\ldots,m$.

We write $H(M)$ and $H(M^{\rm cyc})$, respectively, for the homologies $H_{*}(M,d_{M})$ and $H_{*}(M^{\rm cyc},d_{M})$. Note that $H(M)$ is a left-right module over $H(A)$, while $H(M^{\rm cyc})$ is just a $K$-vector space.

Recall that the module $V$ is generated over $R$ by a set ${\mathcal{C}}$ and that the module $M$ is generated over $A$ as a left-right module by $\widehat{\mathcal{C}}\cup{\mathcal{X}}$, where $\widehat{\mathcal{C}}=\{\widehat{c}\colon c\in{\mathcal{C}}\}$ and ${\mathcal{X}}=\{x_{1},\dots,x_{m}\}$. Let $\widehat{V}^{\ast}$ denote the $K$-vector space dual to $\widehat{V}$. Then $\widehat{V}^{\ast}$ is the direct product of the 1-dimensional $K$-vector spaces which are generated by elements of $\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast}$, where $\widehat{\mathcal{C}}^{\ast}=\{\widehat{c}^{\,\ast}\colon c\in{\mathcal{C}}\}$ and ${\mathcal{X}}^{\ast}=\{x_{j}^{\ast}\colon x_{j}\in{\mathcal{X}}\}$. Furthermore, we have

$$\widehat{V}^{\ast}=\bigoplus_{1\leq i,j\leq m}\widehat{V}^{\ast\,(ij)},\quad\text{where}\quad\widehat{V}^{\ast\,(ij)}:=\bigl{(}\widehat{V}^{(ij)}\bigr{)}^{\ast}.$$

In $\widehat{V}^{\ast}$, the idempotent $1_{j}$, $j=1,\dots,m$, acts from the left as the identity on $\widehat{V}^{\ast\,(ji)}$ and as $0$ on $\widehat{V}^{\ast\,(li)}$, $l\neq j$, and acts from the right as the identity on $\widehat{V}^{\ast\,(ij)}$ and as $0$ on $\widehat{V}^{\ast\,(il)}$, $l\neq j$. Fix an integer $n\geq 2$ (it will correspond to half the dimension of the Liouville domain $X_{0}$ in Section 3). We endow $\widehat{V}^{\ast}$ with a grading defined as follows:

	$\displaystyle|{\widehat{c}}^{\,\ast}|=n-3-|\widehat{c}\,|,\text{ for }\widehat{c}\in\widehat{\mathcal{C}},$		(2.6)
	$\displaystyle|x_{j}^{\ast}|=n-3,\text{ for }j=1,\dots,m.$		(2.7)

Define

$$M^{\ast}:=A\mathop{\otimes}\limits_{R}\widehat{V}^{\ast}\mathop{\otimes}\limits_{R}A.$$

(2.8)

We define the tensor algebra

$$U(0,0):={\bigoplus}_{k\geq 0}\,(M\oplus M^{\ast})^{{\mathop{\otimes}\limits_{A}}^{k}}.$$

(2.9)

as a certain completion of the free associative algebra generated by ${\mathcal{C}}\cup\widehat{\mathcal{C}}\cup\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}\cup\mathcal{X}^{\ast}$. Its elements are infinite series in the $(\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast})$-variables with polynomial coefficients in the $({\mathcal{C}}\cup\widehat{\mathcal{C}}\cup{\mathcal{X}})$-variables (compare to the above definitions of $M$ as an infinite sum and $M^{\ast}$ as an infinite product). For simplicity of notation and since all completions we consider below are of the same type as that of (2.9), we will, as in that equation, suppress completions from notation and use the direct sum symbol in decompositions. Consider the extension $U$ of $U(0,0)$ obtained by adding multiplicative generators $\hbar,\hbar^{-1}$ and $\sigma,\sigma^{-1}$ of degrees

$$|\sigma|=1,\quad|\sigma^{-1}|=-1,\quad|\hbar|=n-3,\quad|\hbar^{-1}|=-(n-3)$$

which satisfy the following relations:

$$\sigma\sigma^{-1}=\sigma^{-1}\sigma=1,\quad\hbar\hbar^{-1}=\hbar^{-1}\hbar=1,\quad\sigma\hbar=(-1)^{n-3}\hbar\sigma,$$

and

	$\displaystyle\sigma u$	$\displaystyle=\begin{cases}(-1)^{|u|}u\sigma,&u\in{\mathcal{C}};\cr(-1)^{|u|+1}u\sigma,&u\in\widehat{\mathcal{C}}\cup\mathcal{X}\cup\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast},\end{cases}$
	$\displaystyle\hbar u$	$\displaystyle=\begin{cases}(-1)^{|u|(n-3)}u\hbar,&u\in{\mathcal{C}}\cup{\mathcal{X}}\cup\widehat{\mathcal{C}};\cr(-1)^{(|u|+n-3)(n-3)}u\hbar,&u\in\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}.\end{cases}$

The algebra $U$ can be decomposed according to tensor type: $U=\bigoplus_{p,q\geq 0}U^{p}_{q}$, where $U^{p}_{q}\subset U$ is the $K$-subspace generated by monomials with $q$ factors from $M$ and $p$ from $M^{\ast}$. We decompose further: $U^{p}_{q}=\bigoplus_{s,h\in{\mathbb{Z}}}U^{p}_{q}(s,h)$, where $U^{p}_{q}(s,h)\subset U^{p}_{q}$ is the subspace spanned by all monomials of total $\sigma$-degree $s$ and total $\hbar$-degree $h$. In particular, $U^{0}_{0}(0,0)=A$, $U^{0}_{1}(0,0)=M$, and $U^{1}_{0}(0,0)=M^{\ast}$. We also write $U^{p}:=\bigoplus_{q\geq 0}U^{p}_{q}$, $U^{+}:=\bigoplus_{p>0}U^{p}$, $U_{q}:=\bigoplus_{p\geq 0}U^{p}_{q}$, $U_{q}(s,h):=\bigoplus_{p\geq 0}U^{p}_{q}(s,h)$, and $U^{p}(s,h):=\bigoplus_{q\geq 0}U^{p}_{q}(s,h)$.

For $X\in U^{p}_{q}(s,h)$, let

$$\operatorname{st}(X):=p+h\quad\text{and}\quad{\operatorname{jp}}(X):=p+q+s.$$

In analogy with the definition of $M^{\operatorname{diag}}$ in Section 2.1.B, we define $U^{\operatorname{diag}}$. Further, define $U^{\rm cyc}$ as the quotient of $U^{{\operatorname{diag}}}$ obtained by dividing by the following graded cyclic permutation rule: if $a_{1},\dots,a_{k}\in{\mathcal{C}}\cup\widehat{\mathcal{C}}\cup{\mathcal{X}}\cup\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast}$ and if $a_{1}\dots a_{k}\in U^{{\operatorname{diag}}}$ then $a_{2}\dots a_{k}a_{1}\in U^{{\operatorname{diag}}}$, and we identify $a_{1}\dots a_{k}\in U^{{\operatorname{diag}}}$ with $\varepsilon_{1}\varepsilon_{2}\varepsilon_{3}\;a_{2}\dots a_{k}a_{1}\in U^{{\operatorname{diag}}}$, where

	$\displaystyle\varepsilon_{1}=(-1)^{|a_{1}||a_{2}\dots a_{k}|},$
	$\displaystyle\varepsilon_{2}={\begin{cases}1&\text{if }a_{1}\in{\mathcal{C}}\cup\{\hbar,\hbar^{-1}\},\cr(-1)^{{\operatorname{jp}}(a_{2}\dots a_{k})}&\text{if }a_{1}\in\widehat{\mathcal{C}}\cup{\mathcal{X}}\cup\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast}\cup\{\sigma,\sigma^{-1}\},\cr\end{cases}}$
	$\displaystyle\varepsilon_{3}={\begin{cases}1&\text{if }a_{1}\in{\mathcal{C}}\cup\widehat{\mathcal{C}}\cup{\mathcal{X}}\cup\{\sigma,\sigma^{-1}\},\cr(-1)^{\operatorname{st}(a_{2}\dots a_{k})}&\text{if }a_{1}\in\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast}\cup\{\hbar,\hbar^{-1}\}.\end{cases}}$

We will often think of cyclic monomials as written on a circle $S^{1}$ with the first letter at $1\in S^{1}$. This allows us to speak about generators in a cyclic monomial as ordered in the (counter) clockwise direction staring from some fixed generator in the monomial. In this language, the above permutations correspond to rotations of the circle.

The vector space $U^{{\rm cyc}}$ inherits the decomposition by tensor type:

$$U^{{\rm cyc}}=\bigoplus_{p,q\geq 0}(U^{p}_{q})^{{\rm cyc}}=\bigoplus_{p,q\geq 0}\bigoplus_{s,h\in{\mathbb{Z}}}\,\left(U^{p}_{q}(s,h)\right)^{{\rm cyc}}.$$

If the monomials $X$ and $X^{\prime}$ differ by a graded cyclic permutation, i.e. if $\llbracket X\rrbracket=\llbracket X^{\prime}\rrbracket$, then, for $z\in\{\sigma,\sigma^{-1},\hbar,\hbar^{-1}\}$, $\llbracket Xz\rrbracket=\llbracket X^{\prime}z\rrbracket$ and $\llbracket zX\rrbracket=\llbracket zX^{\prime}\rrbracket$. We thus define $z\llbracket X\rrbracket:=\llbracket zX\rrbracket$ and $\llbracket X\rrbracket z:=\llbracket Xz\rrbracket$, and note that this definition does not depend on choice of representative $X$.

A monomial $X\in\left(U^{p}_{q}(s,h)\right)^{{\rm cyc}}$ is balanced if $h=-1$ and $p-q-s=0$. The subspace of $U^{\rm cyc}$ generated by balanced monomials will be denoted by $U^{\mathrm{bal}}$. It contains balanced versions $M^{\mathrm{bal}}$ and $\left(M^{\ast}\right)^{{\mathrm{bal}}}$ of $M^{\rm cyc}$ and $\left(M^{\ast}\right)^{{\rm cyc}}$, respectively. More precisely,

	$\displaystyle M^{\mathrm{bal}}$	$\displaystyle:=\left(U^{0}_{1}(-1,-1)\right)^{\rm cyc}=U^{0}_{1}\cap U^{\mathrm{bal}},$
	$\displaystyle\left(M^{\ast}\right)^{{\mathrm{bal}}}$	$\displaystyle:=\left(U^{1}_{0}(1,-1)\right)^{\rm cyc}=U^{1}_{0}\cap U^{\mathrm{bal}}.$

Consider a monomial $X\in\left(U^{p}_{q}(s,h)\right)^{\rm cyc}$ and fix a factor $c\in{\mathcal{C}}$ in $X$. Then $X=\llbracket X_{c}^{\prime}cX_{c}^{\prime\prime}\rrbracket$ for monomials $X_{c}^{\prime},X_{c}^{\prime\prime}\in U$. Define

$${\mathcal{S}}(X)=\sum\limits_{c}\llbracket X_{c}^{\prime}\sigma^{-1}\widehat{c}X_{c}^{\prime\prime}\rrbracket\in\left(U^{p}_{q+1}(s-1,h)\right)^{\rm cyc},$$

where the sum is taken over all factors $c\in{\mathcal{C}}$ in $X$. It is straightforward to check that this gives a well defined operation and by definition ${\mathcal{S}}(U^{{\mathrm{bal}}})\subset U^{{\mathrm{bal}}}$. In particular, if $X=\llbracket c_{1}\dots c_{m}\widehat{a}^{\ast}\rrbracket\in\left(U^{1}_{0}(0,0)\right)^{\rm cyc}$ then ${\mathcal{S}}(X)\in\left(U^{0}_{1}(-1,0)\right)^{\rm cyc}$ satisfies

	$\displaystyle{\mathcal{S}}(c_{1}\dots c_{m}\widehat{a}^{\ast})$	$\displaystyle=\sigma^{-1}\sum_{j=1}^{m}(-1)^{|c_{1}\dots c_{j-1}|}c_{1}\dots c_{j-1}\widehat{c}_{j}c_{j+1}\dots c_{m}\widehat{a}^{\ast}$
		$\displaystyle=\sigma^{-1}S(c_{1}\dots c_{m})\widehat{a}^{\ast},$

where $S$ is the operator $A\to M$ defined in (2.5) above. Furthermore,

$${\mathcal{S}}^{k}(c_{1}\dots c_{m}\widehat{a}^{\ast})=k!\sigma^{-k}\sum_{j_{1}<\dots<j_{k}}^{m}(-1)^{\sum\limits_{l=1}^{k}|c_{1}\dots c_{j_{l}-1}|}c_{1}\dots\widehat{c}_{j_{1}}\dots\widehat{c}_{j_{k}}\dots c_{m}\widehat{a}^{\ast}.$$

We next effectively break the cyclic symmetry of $U^{\mathrm{bal}}$. An excited monomial is a cyclic monomial $X\in U^{{\mathrm{bal}}}$ with one of its factors from $\widehat{\mathcal{C}}\cup\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}\cup{\mathcal{X}}^{\ast}$ distinguished. We call the distinguished factor of an excited monomial $X$ the excited generator of $X$. The $K$-vector space generated by excited monomials from $U^{\mathrm{bal}}$ will be denoted by ${\mathbf{U}}^{\mathrm{bal}}$. Note that ${\mathbf{U}}^{{\mathrm{bal}}}$ inherits decompositions from $U^{{\mathrm{bal}}}$, ${\mathbf{U}}^{{\mathrm{bal}}}=\bigoplus_{p,q}({\mathbf{U}}^{p}_{q})^{{\mathrm{bal}}}$, etc. We set $({\mathbf{U}}^{0}_{0})^{{\mathrm{bal}}}=0$. Define the excitation operator ${\mathcal{E}}:U^{\mathrm{bal}}\to{\mathbf{U}}^{\mathrm{bal}}$ to be the linear map which maps a monomial $X$ the sum of all its excitations.

Note that ${\mathbf{U}}^{{\mathrm{bal}}}$ contain excited versions ${\mathbf{M}}^{\mathrm{bal}}=\left({\mathbf{U}}^{0}_{1}\right)^{{\mathrm{bal}}}$ and $\left({\mathbf{M}}^{\ast}\right)^{{\mathrm{bal}}}=\left({\mathbf{U}}^{1}_{0}\right)^{{\mathrm{bal}}}$ of $M^{\mathrm{bal}}$ and $\left(M^{\ast}\right)^{{\mathrm{bal}}}$, respectively, which are isomorphic to their non-excited counterparts. (In the context of ${\mathbf{U}}^{{\mathrm{bal}}}$ we think of $M^{{\mathrm{bal}}}$ and $\left(M^{\ast}\right)^{{\mathrm{bal}}}$ as generated by monomials in $\left(U^{0}_{1}\right)^{{\mathrm{bal}}}$ and $\left(U^{1}_{0}\right)^{{\mathrm{bal}}}$ without excited generator.)

We will use the following notation to identify excited generators in monomials: we underline the excited generator in a monomial in $U^{{\mathrm{bal}}}$ and write e.g. ${\underline{x}}$, ${\underline{x}}^{\ast}$ $\widehat{\underline{c}}$, and $\widehat{\underline{c}}^{\ast}$.

We next define a product on ${\mathbf{U}}^{\mathrm{bal}}$. Consider two excited cyclic monomials $X,Y\in{\mathbf{U}}^{\mathrm{bal}}$. Fix a factor $u\in\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}$ from $X$ and a factor $v\in\widehat{\mathcal{C}}\cup\mathcal{X}$ from $Y$. Write $X=\varepsilon_{1}\llbracket X_{u}^{\prime}u\rrbracket$ and $Y=\varepsilon_{2}\llbracket vY_{v}^{\prime}\rrbracket$, where $\varepsilon_{1}$ is the sign which arises from cyclically rotating $X$ so that $u$ is its last factor and where $\varepsilon_{2}$ is the sign which arises from cyclically rotating $Y$ so that $v$ is its first factor. Define the contraction operation $\mathop{\star}\limits\colon{\mathbf{U}}^{{\mathrm{bal}}}\otimes{\mathbf{U}}^{{\mathrm{bal}}}\to{\mathbf{U}}^{{\mathrm{bal}}}$,

$$X\star Y:=\sum\limits_{(u,v)}\varepsilon_{1}\varepsilon_{2}u(v)\llbracket X_{u}^{\prime}Y_{v}^{\prime}\rrbracket,$$

(2.10)

where the sum is taken over all pairs $(u,v)$ of factors $u\in\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}$ from $X$ and $v\in\widehat{\mathcal{C}}\cup\mathcal{X}$ from $Y$, and where

$$u(v)=\begin{cases}\hbar&\text{if exactly one of $u$ and $v$ is excited, $u=v^{\ast}$, and $v\notin{\mathcal{X}}$ ,}\\ (-1)^{n-1}\hbar&\text{if $u={\underline{x}}_{j}^{\ast}$ and $v=x_{j}$, for some $j=1,\dots,m$,}\\ \hbar&\text{if $u=x_{j}^{\ast}$ and $v={\underline{x}}_{j}$, for some $j=1,\dots,m$,}\\ 0&\text{otherwise}.\end{cases}$$

If $Y\in\left({\mathbf{U}}^{1}_{q}\right)^{\mathrm{bal}}$ and $X\in{\mathbf{U}}^{{\mathrm{bal}}}$ then we define, for $j=1,\dots,q$, the partial contraction $X\mathop{\star}\limits_{j}Y$ as follows:

$$X\mathop{\star}\limits\limits_{j}Y:=\sum\limits_{u}\varepsilon_{1}\varepsilon_{2}u(v_{j})\llbracket X_{u}^{\prime}Y_{v_{j}}^{\prime}\rrbracket,$$

where the sum is taken over all $u\in\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}$ from $X$ and where $v_{j}$ is the $j^{\rm th}$ factor from $\widehat{\mathcal{C}}\cup\mathcal{X}$ in $Y$ counting counter-clockwise from the unique factor from $\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}$. Similarly, if $X\in\left({\mathbf{U}}^{p}_{1}\right)^{\mathrm{bal}}$ and $Y\in{\mathbf{U}}^{{\mathrm{bal}}}$ then we define, for $i=1,\dots,p$ the partial contraction $X\mathop{\star}\limits^{i}Y$:

$$X\mathop{\star}\limits\limits^{i}Y:=\sum\limits_{v}\varepsilon_{1}\varepsilon_{2}u_{i}(v)\llbracket X_{u_{i}}^{\prime}Y_{v}^{\prime}\rrbracket,$$

where the sum is taken over all $v\in\widehat{\mathcal{C}}\cup\mathcal{X}$ from $Y$ and where $u_{i}$ is the $i$-th factor from $\widehat{\mathcal{C}}^{\ast}\cup\mathcal{X}^{\ast}$ in $X$ counting clockwise from the unique factor from $\widehat{\mathcal{C}}\cup\mathcal{X}$.

Define the bracket or commutator $[\,,]\colon{\mathbf{U}}^{{\mathrm{bal}}}\otimes{\mathbf{U}}^{{\mathrm{bal}}}\to{\mathbf{U}}^{{\mathrm{bal}}}$ as

$$[X,Y]=X\star Y-(-1)^{|X||Y|}\,Y\star X.$$

(2.11)

Let

$$\tau{\mathbf{M}}^{{\mathrm{bal}}}={\mathbf{M}}^{{\mathrm{bal}}}\oplus M^{{\mathrm{bal}}}\quad\text{and}\quad\left(\tau{\mathbf{M}}^{\ast}\right)^{{\mathrm{bal}}}=\left({\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}\oplus\left(M^{\ast}\right)^{\mathrm{bal}}.$$

We associate with $X\in\left({\mathbf{U}}^{p}_{1}\right)^{\mathrm{bal}}$ a multi-linear operation $X^{\downarrow}\colon\left(\tau{\mathbf{M}}^{{\mathrm{bal}}}\right)^{\otimes^{p}}\to\tau{\mathbf{M}}^{{\mathrm{bal}}}$ defined by

$$X^{\downarrow}(A_{1}\otimes\dots\otimes A_{p})=\left(\dots\left(\left(X\mathop{\star}\limits^{1}A_{1}\right)\mathop{\star}\limits^{1}A_{2}\right)\mathop{\star}\limits^{1}\dots\right)\mathop{\star}\limits^{1}A_{p}.$$

(2.12)

Note that if $A_{1},\dots,A_{p}\in{\mathbf{M}}^{\mathrm{bal}}$ then $X^{\downarrow}(A_{1}\otimes\dots\otimes A_{p})\in{\mathbf{M}}^{\mathrm{bal}}$, that if exactly one of the arguments lies in $M^{\mathrm{bal}}$ then the image lies in $M^{\mathrm{bal}}$, and that the operation vanishes on $p$-tuples with more than component in $M^{\mathrm{bal}}$.

Similarly, $Y\in\left({\mathbf{U}}_{q}^{1}\right)^{\mathrm{bal}}$ defines a multi-linear operation $Y^{\uparrow}\colon\left(\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}\right)^{\otimes^{q}}\to\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}$,

$$Y^{\uparrow}(B_{1}\otimes\dots\otimes B_{q})=B_{q}\mathop{\star}\limits_{1}\left(\dots\mathop{\star}\limits_{1}\left(B_{2}\mathop{\star}\limits_{1}\left(B_{1}\mathop{\star}\limits_{1}Y\right)\right)\dots\right),$$

(2.13)

if $B_{1},\dots,B_{q}\in\left({\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}$ then $Y^{\uparrow}(B_{1}\otimes\dots\otimes B_{q})\in\left({\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}$, if exactly one of the arguments lies in $\left(M^{\ast}\right)^{\mathrm{bal}}$ then the image lies in $\left(M^{\ast}\right)^{\mathrm{bal}}$, and the operation vanishes on $q$-tuples with more than one component in $\left(M^{\ast}\right)^{\mathrm{bal}}$.

In the special case $q=2$, we associate with $Y\in\left({\mathbf{U}}^{1}_{2}\right)^{\mathrm{bal}}$ two more operations,

$$Y^{{\uparrow\downarrow}}\colon\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}\otimes\tau{\mathbf{M}}^{\mathrm{bal}}\to\tau{\mathbf{M}}^{\mathrm{bal}}$$

(2.14)

and

$$Y^{{\downarrow\uparrow}}\colon\tau{\mathbf{M}}^{\mathrm{bal}}\otimes\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}\oplus\to\tau{\mathbf{M}}^{{\mathrm{bal}}},$$

(2.15)

where

	$\displaystyle Y^{{\uparrow\downarrow}}(X,Z)$	$\displaystyle=\left(X\mathop{\star}\limits_{1}Y\right)\star Z,\quad X\in\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}},\;Z\in\tau{\mathbf{M}}^{\mathrm{bal}},$
	$\displaystyle Y^{{\downarrow\uparrow}}(Z,X)$	$\displaystyle=\left(X\mathop{\star}\limits_{2}Y\right)\star Z,\quad Z\in\tau{\mathbf{M}}^{\mathrm{bal}},\;X\in\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}.$

Similarly, $X\in\left({\mathbf{U}}^{2}_{0}\right)^{\mathrm{bal}}$ induces an operation $X^{\circlearrowleft}\colon\tau{\mathbf{M}}^{\mathrm{bal}}\to\left(\tau{\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}$,

$$X^{\circlearrowleft}(Y)=X\star Y.$$

(2.16)

We observe that $X^{\circlearrowleft}({\mathbf{M}}^{\mathrm{bal}})\subset\left({\mathbf{M}}^{\ast}\right)^{\mathrm{bal}}$ and $X^{\circlearrowleft}(M^{\mathrm{bal}})\subset\left(M^{\ast}\right)^{\mathrm{bal}}$.

We define special elements $H^{1}_{q}\in{\mathbf{U}}^{\mathrm{bal}}$ which derive from the differential on $A$ and which, in the spirit of SFT, see [9], are called Hamiltonians.

We use the Hamiltonian $H^{1}_{1}$ to define a differential on $\tau{\mathbf{U}}^{\mathrm{bal}}:={\mathbf{U}}^{{\mathrm{bal}}}\oplus M^{{\mathrm{bal}}}\oplus(M^{\ast})^{{\mathrm{bal}}}$. First define the exterior differential $d\colon\tau{\mathbf{U}}^{\mathrm{bal}}\to\tau{\mathbf{U}}^{\mathrm{bal}}$ as follows. Consider the map $d\colon U\to U$ which acts as the algebra differential $d\colon A\to A$ on generators of $U$ from $A$, which maps all other generators to $0$, and which satisfies the graded Leibniz rule. Then $d$ preserves $U^{{\operatorname{diag}}}$, descends to $U^{{\rm cyc}}$ where it preserves $U^{{\mathrm{bal}}}$. It then induces a map $d\colon\tau{\mathbf{U}}^{{\mathrm{bal}}}\to\tau{\mathbf{U}}^{{\mathrm{bal}}}$ as follows. The map is already determined on $M^{{\mathrm{bal}}}\oplus(M^{\ast})^{{\mathrm{bal}}}$ and if $\llbracket\underline{u}X\rrbracket$ is a monomial in ${\mathbf{U}}^{{\mathrm{bal}}}$, where $u\in{\mathcal{C}}\cup{\mathcal{X}}\cup\widehat{\mathcal{C}}\cup\widehat{\mathcal{C}}^{\ast}\cup{\mathcal{X}}^{\ast}$ then $d(\llbracket\underline{u}X\rrbracket)=(-1)^{|u|}\llbracket\underline{u}(dX)\rrbracket$. Then $d^{2}=0$ and we observe that $H^{1}$ satisfies the following master equation

$$d{H}^{1}+{H}^{1}\star{H}^{1}=0.$$

(2.17)

Since $|H^{1}|=-1$, ${H}^{1}\star{H}^{1}=\frac{1}{2}[{H}^{1},{H}^{1}]$ and (2.17) can be rewritten in Maurer-Cartan form:

$$d{H}^{1}+\frac{1}{2}[{H}^{1},{H}^{1}]=0.$$

(2.18)