Comass of a differential form In the wikipedia article on currents https://en.wikipedia.org/wiki/Current_%28mathematics%29 it is written that
If $\omega$ is an $m$-form, then define its comass by
$\|\omega\| = \sup\ \{|\langle \omega,\xi\rangle| : \xi$ is a unit, simple $m$-vector$\}$
I don't understand this definition. 
Can someone provide an explicit example and computation of this norm of a differential $2$-form on say $\mathbb R^2$ ?

Let's first clarify this definition.  For an $m$-form $\omega \in \Lambda^m(V^*)$ on an $n$-dimensional vector space $V$ with inner product $\langle \cdot, \cdot \rangle$, we are defining
$$
\begin{align*}
\Vert \omega \Vert & = \sup\left\{\left|\omega(v_1 \wedge \cdots \wedge v_m)\right| \colon \text{ vectors } v_1, \ldots, v_m \in V \text{ with } \left|v_1 \wedge \cdots \wedge v_m\right| = 1\right\}
 \\
& = \sup\left\{\left|\omega(v_1, \ldots, v_m)\right| \colon \text{ vectors } v_1, \ldots, v_m \in V \text{ with } \sqrt{\det(\langle v_i, v_j \rangle)} = 1\right\}.
\end{align*}$$
The quantity $\sqrt{\det(\langle v_i, v_j \rangle)}$ has a geometric interpretation: it is the volume of the parallelepiped spanned by the vectors $v_1, \ldots, v_m$.
In the first line, the quantity $|v_1 \wedge \cdots \wedge v_m|$ refers to the extension of the inner product $\langle \cdot, \cdot \rangle$ on $V$ to the vector space $\Lambda^m(V)$.
Example: In the simplest case of $m = 1$, we have
$$\Vert \omega \Vert = \sup\left\{|\omega(v)| \colon \text{ vectors } v \in V \text{ with } |v| = 1 \right\} = \sup_{|v| = 1} \left|\omega(v)\right|.$$
That is, for $1$-forms $\omega \in \Lambda^1(V) \cong V^*$, this is just the usual definition of the norm of a linear functional.
Remark: In the case of $m = 2$, we have that
$$\sqrt{\det(\langle v_i, v_j \rangle)} = \sqrt{\det \begin{pmatrix} \langle v_1, v_1 \rangle & \langle v_1, v_2 \rangle \\ \langle v_2, v_1 \rangle & \langle v_2, v_2 \rangle \end{pmatrix}} = \sqrt{|v_1|^2 |v_2|^2 - \langle v_1, v_2 \rangle^2}.$$
If also $V = \mathbb{R}^2$ or $\mathbb{R}^3$, then this quantity is exactly $\left|v_1 \times v_2\right|$, where $\times$ is the cross product.
Example: As requested, we'll look at the $2$-form $\omega = dx \wedge dy$ on $V = \mathbb{R}^2$.  Then
$$
\begin{align*}
\Vert dx \wedge dy \Vert & = \sup\left\{\left|(dx \wedge dy)(v, w)\right| \colon \text{ vectors } v, w \in \mathbb{R}^2 \text{ with } |v \times w| = 1\right\} \\
& = \sup\left\{\left|v^1 w^2 - w^1v^2 \right| \colon \text{ vectors } v, w \in \mathbb{R}^2 \text{ with } |v \times w| = 1\right\} \\
& = \sup\left\{\left|v \times w \right| \colon \text{ vectors } v, w \in \mathbb{R}^2 \text{ with } |v \times w| = 1\right\} \\
& = 1.
\end{align*}$$
