Twisted Multiplication on the Twisted De Rham Complex of a Graded Ring

   Let $\Bbbk$ be the characteristic zero ring. Let $A$ be the graded ring with the degree function $\deg$. Then the Kahler differential $\Omega_A$ also comes with the degree function $\deg$ extended from that of $A$, which is defined as \begin{align*} \deg fdg = \deg f + \deg g, \\ \deg (\omega \wedge \eta) = \deg \omega + \deg \eta, \end{align*} where $f,g\in A$, $\omega,\eta\in \Omega_A$ are the homogeneous elements. We let $\Omega_{A,k}$ and $\Omega_{A,k}^n$ be the homogeneous $k$-part of $\Omega_A$ and $\Omega_{A}^n$, respectively, with respect to $\deg$. We also denote $\Omega_{A,+}$ for the submodule of $\Omega_A$ generated by the homogeneous part of the positive degree, $$ \Omega_{A,+} = \bigoplus_{k=1}^\infty \Omega_{A,k}. $$

   Let $\alpha\in \Omega_{A,1}^1$ be any homogeneous element of degree $1$. Usually, $\alpha$ is taken to be the exact form, $\alpha=dS$ for some $S\in A$. Then it defines a twisted de Rham complex $(\Omega_{A},d+\alpha\wedge)$, whose differential is $d+\alpha\wedge$. It is often useful to think of the subcomplex $(\Omega_{A,+},d+\alpha\wedge)$ of the positive degree.

   The wedge product $\wedge$ no longer makes $\Omega_A$ into the differential graded associative algebra, since the differential does not commute with the wedge product: $$ (d+\alpha\wedge)(\omega\wedge\eta) \neq ((d+\alpha\wedge)\omega)\wedge\eta\pm \omega \wedge ((d+\alpha\wedge)\eta). $$ I claim that there is a newly defined multiplication $m$, which I will call the twisted multiplication, on $\Omega_{A,+}$ which makes $(\Omega_{A,+},d+\alpha\wedge,m)$ the differential graded algebra.

Proposition 1. Let $\omega\in \Omega^{n_1}_{A,+}$ and $\eta\in \Omega^{n_2}_{A,+}$ be two forms, and let \begin{align*} \omega = \omega_{1} + \cdots + \omega_{k_1}, \\ \eta = \eta_{1} + \cdots + \eta_{k_2}, \end{align*} be the decomposition into the homogeneous parts. Then we define $m(\omega,\eta)$ to be $$ m(\omega,\eta) = \sum_{\substack{1\leq j_1 \leq k_1 \\ 1\leq j_2 \leq k_2}} \frac{(j_1-1)!(j_2-1)!}{(j_1+j_2-1)!} \omega_{j_1} \wedge \eta_{j_2}. $$ We extend $m$ onto $\Omega_{A,+} \otimes_{\Bbbk} \Omega_{A,+}$ linearly. Then $(\Omega_{A,+},d+\alpha\wedge,m)$ is a commutative differential graded algebra.

   The proof is straightforward, so I will omit it.

   There are two ways from which this multiplication is derived. The first way is to consider the convolution integral. Here we only consider $\Bbbk = \mathbb{R}$ or $\mathbb{C}$. In this case, we can think of another cochain complex $\mathrm{Mor}(I,\Omega_{A,+})$, whose elements are of the form $$ p_1(t) \omega_1 + \cdots + p_n(t) \omega_n $$ where $t\in [0,1]$ is a variable, $p_i(t)$ is a polynomial, and $\omega_i\in \Omega_{A,+}$ is a differential forms. Here, the differential is defined to be $d+t\alpha\wedge$. This is a commutative differential graded algebra, together with the multiplication defined as $$ \omega_t \ast\eta_t = \int^t_0 \omega_s \wedge \eta_{t-s} ds, $$ where $\omega_t$ and $\eta_t$ are the elements of $\mathrm{Mor}(I,\Omega_{A,+})$

 

   Then the twisted multiplication is actually derived from this convolution integral. There are two morphisms between two cochain complexes, $\Omega_{A,+}$ and $\mathrm{Mor}(I,\Omega_{A,+})$ \begin{align*} \psi_t: \Omega_{A,+} \to \mathrm{Mor}(I,\Omega_{A,+}), \\ \mathrm{ev}_1 : \mathrm{Mor}(I,\Omega_{A,+}) \to \Omega_{A,+}, \end{align*} each of them defined as $$ \psi_t(\omega_k) = t^{k-1} \omega_k, \\ \mathrm{ev}_1 (\eta_t) = \eta_1, $$ for $\omega_k \in \Omega_{A,k}^n$ and $\eta_t\in \mathrm{Mor}(I,\Omega_{A,+})$.

  Then $m$ can be represented as $$ m(\omega,\eta) = \mathrm{ev}_1 \psi_t(\omega) \ast \psi_t (\eta), $$ since $$ \int^1_0 t^{j_1-1}(1-t)^{j_2-1} dt = \frac{(j_1-1)!(j_2-1)!}{(j_1+j_2-1)!}. $$

   Another way to represent $m$ is to associate it with the ordinary wedge product. We first consider the automorphism on $\Omega_{A,+}$ which maps $\omega\in\Omega_{A,k}$ to $\omega/(k-1)!$. We have a newly defined differential on $\Omega_{A,+}$ which maps $\omega \in \Omega_{A,k}$ to $$ \omega \mapsto (d+k\alpha)\omega. $$ Then the automorphism is actually the isomorphism between the cochain complexes $(\Omega_{A,+},d+\alpha)$ and $(\Omega_{A,+},d+(\deg)\alpha)$. One can easily show that the multiplication becomes the ordinary wedge product by the automorphism. Hence there is an isomorphism between two A-infinity algebra $(\Omega_{A,+},d+\alpha,m)$ and $(\Omega_{A,+},d+(\deg)\alpha,\wedge)$.