For a discrete 1-form $\omega$ we define a discrete 2-form $d\omega$, called the derivative of $\omega$:

${\displaystyle d\omega (\phi )={\int }_{\phi }d\omega :=\sum _{e\in \phi }\omega (e)}$.

The picture below illustrates this.

Recall that a 1-form $\omega$ is called exact if it is the derivative of some 0-form. $\omega$ is called closed if $d\omega =0$. It is easy to see that exact 1-forms are always closed:

$d(df)=0$.

Now that we have two maps called $d$, we sometimes distinguish them by an index. Thus

$d_0:{\mathrm{\Omega }}_0(M)\to {\mathrm{\Omega }}_1(M)$

$d_1:{\mathrm{\Omega }}_1(M)\to {\mathrm{\Omega }}_2(M)$

$d_1\circ d_0=0$.

The derivative of dual 0-forms provides us with a linear map

${\partial }_2:{\mathrm{\Omega }}_2(M{)}^{\ast }={\mathrm{\Omega }}_0(M^{\ast })\to {\mathrm{\Omega }}_1(M^{\ast })={\mathrm{\Omega }}_1(M{)}^{\ast }$.

Our definitions imply that ${\partial }_2$ is the adjoint of $d_1$:

$$\begin{array}{rl}⟨\partial f,\omega ⟩& =\sum _{e\in \hat{E}}\partial f(e)\omega (e)\\ & =\frac{1}{2}\sum _{e\in E}\partial f(e)\omega (e)\\ & =\frac{1}{2}\sum _{e\in E}(f(\text{left}(e))-f(\text{right}(e)))\omega (e)\\ & =\frac{1}{2}\sum _{e\in E}(f(\text{left}(e))-f(\text{left}(\rho (e))))\omega (e)\\ & =\frac{1}{2}(\sum _{e\in E}f(\text{left}(e))\omega (e)-\sum _{e\in E}f(\text{left}(\rho (e)))\omega (e))\\ & =\frac{1}{2}(\sum _{e\in E}f(\text{left}(e))\omega (e)-\sum _{e\in E}f(\text{left}(e)\omega (\rho (e)))\\ & =\sum _{e\in E}f(\text{left}(e))\omega (e)\\ & =\sum _{\phi \in F}\sum _{e\in \phi }f(\text{left}(e))\omega (e)\\ & =\sum _{\phi \in F}f(\phi )\sum _{e\in \phi }\omega (e)\\ & =\sum _{\phi \in F}f(\phi )d\omega (\phi )\\ & =⟨f,d\omega ⟩.\end{array}$$