# Dual 2-forms and derivatives of dual 1-forms

A function $f\in \Omega_0(M)$ assigns a function value to each vertex $\varphi\in V$. One way to picture this is that this function value is only given at some choosen point in the face . Another interpretation (which we prefer right now) is to see $f$ as a function on the surface that is constant on each face of the dual surface $M^*$.

Think about the colors in the above picture as color-coding the function values (which are real numbers) as various colors.

We can proceed in a similar way as we did for 2-forms and dual 0-forms: If $f$ is a 0-form and $\sigma^*$ a dual 2-form we can define a pairing

$\displaystyle \langle \sigma^* , f \rangle = \int_{M^*} f \, \sigma^* = \sum_{v \in V} f(v)\, \sigma^*(v)$.

This exhibits the space $\Omega_2(M^*)$ of all dual 2-forms as the dual vector space of $\Omega_0(M)$:

$\Omega_2(M^*)=\Omega_0(M)^*$.

We define the derivative of a dual 1-form just by applying the standard definition of the derivative of a 1-form to the dual surface:

$\partial_1: \Omega_1(M^*) \rightarrow \Omega_2(M^*)$

$\displaystyle \partial_1 \eta(v) = \sum_{e \in v} \eta(e)$

We show that $\partial_1$ is the adjoint of $d_0$:

\begin{align*}
\langle \partial_1 \eta, f \rangle &=\sum_{v \in V} f(v)\,\sum_{e \in v} \eta(e)\\
&=\sum_{v \in V} \,\sum_{e \in v} f(v) \eta(e)\\
&=\sum_{e \in E} f(\textrm{end}(e)) \, \eta(e)\\
&=\frac{1}{2} \left( \sum_{e \in E} f(\textrm{end}(e)) \, \eta(e)+\sum_{e \in E} f(\textrm{end}(\rho(e))) \, \eta(\rho(e))\right)\\
&=\frac{1}{2} \sum_{e \in E}( f(\textrm{end}(e)) – \textrm{start}(e)))\, \eta(e)\\
&=\sum_{e \in \hat{E}} df(e)\, \eta(e)\\
&=\langle \eta, df\rangle.
\end{align*}

Thus indeed

$\partial_1 = d_0^*$.