Let $V$ be a vector space over the field $\mathbb{K}$ and let $T(V)$ be it's tensor algebra. We usually define the exterior algebra $\Lambda (V)$ by the following process: we consider the bilateral ideal $I$ generated by all elements of $T(V)$ of the form $u\otimes v+v\otimes u$ for all $u,v\in V$ and define the exterior algebra as the quotient algebra $\Lambda(V) = T(V)/I$. If $\rho : T(V)\to \Lambda (V)$ is the natural projection, we define the exterior product $\wedge : \Lambda(V)\to\Lambda(V)$ by:
$$u\wedge v = \rho(u\otimes v).$$
From the algebraic point of view, in $\Lambda(V)$ the elements are skew-symmetric. That's fine, but people usually say that we can think of elements of $\Lambda(V)$ as paralelograms. In this article here the author says:
Visualizing $a\wedge b$ is easy. it's the little parallelogram that has the vectors $a$ and $b$ as its sides.
And in general I've heard that we should really think of elements $\Lambda(V)$ as "pieces of planes" in the same way we think of elements of $V$ as arrows sometimes (obviously this sometimes refers to when geometrical intuition is good).
Where this interpretation of multivectors comes from?
Thanks very much in advance!
In $\mathbb R^2$ one can visualize easily the (oriented) parallelogram and its relation to the wedge product of vectors. If
$$v=v_1e_1+v_2e_2,$$ $$w=w_1e_1+w_2e_2 $$
are two vectors in $\mathbb R^2$, with $e_1,e_2$ basis in $\mathbb R^2$, then
$$v\wedge w=v_1w_2e_1\wedge e_2+v_2w_1e_2\wedge e_1=(v_1w_2-v_2w_1)e_1\wedge e_2,$$
as $e_1\wedge e_1=e_2\wedge e_2=0$ and $e_1\wedge e_2=-e_2\wedge e_1$. But $v_1w_2-v_2w_1$ is the area of the parallelogram (up to a sign) described by $v$ and $w$.