Are differential forms defined on $\Bbb{R}^{n}$ I thought $p$-forms were linear maps from $\Bbb{R}^{n} \rightarrow \Bbb{R}$. But I read something yesterday that suggested I was mistaken to think this. It seemed to be saying that $p$-forms eat $p$-vectors and not regular vectors (unless $p=1$). It also seemed to indicate that they do this in a multilinear way not a linear way. 
Can someone clarify this? What can a $p$-form eat? Does it do this in a linear or multilinear way? And why isn't it correct to say a $p$-form is a map from $\Bbb{R}^{n} \rightarrow \Bbb{R}$? 
Thanks!
 A: In one realization of forms, on a vector space $V$ over $\mathbf{R}$, a one-form is a linear map $V \to \mathbf{R}$. A two-form is an antisymmetric bilinear map $V \times V \to \mathbf{R}$. A three-form is an antisymmetric trilinear map $V \times V \times V \to \mathbf{R}$. And so forth.
In case these words are unfamiliar, "trilinear", for example, means that for vectors $w,x,y,z$ and scalars $a$:


*

*$ f(ax, y, z) = f(x, ay, z) = f(x, y, az) = a f(x, y, z) $

*$ f(w+x,y,z) = f(w,y,z) + f(x,y,z) $

*$ f(w,x+y,z) = f(w,x,z) + f(w,y,z)$

*$ f(w,x,y+z) = f(w,x,y) + f(w,x,z)$


and "antisymmetric" means


*

*$ f(x,y,z) = -f(y,x,z) = -f(x,z,y) = -f(z,y,x) $


(and note this implies "alternating": that $f(x,x,y) = f(x,y,y) = f(x,y,x) = 0$)

In another formulation, a two-form, for example, is a linear map $V \wedge V \to \mathbf{R}$. A realization of the wedge product $V \wedge V$ is as the subgroup of all antisymmetric tensors in $V \otimes V$.

To see the correspondence between the two formulations, suppose that $f$ is an antisymmetric bilinear map $V \times V \to \mathbf{R}$. Then we can construct a new function $g$ that is a linear map $V \wedge V \to \mathbf{R}$ by defining
$$ g(x_1 \wedge y_1 + x_2 \wedge y_2 + \ldots + x_n \wedge y_n) := f(x_1,y_1) + f(x_2,y_2) + \ldots + f(x_n, y_n) $$
And conversely, if $g$ is a linear map $V \wedge V \to \mathbf{R}$, then we can construct an antisymmetric, bilinear map $f: V \times V \to \mathbf{R}$ via
$$ f(x, y) := g(x \wedge y) $$

There are a myriad of other things we can do with forms too. For example, if I have a vector $v$ and a one-form $f$, I can take their "outer product" to get a linear transformation $T : V \to V$ defined by
$$ T(w) = f(w) v $$
If I have a two-form $g$ and a vector $v$, then I can construct a new one-form $f$ by defining
$$ f(w) = g(v, w) $$
There are countless different ways to combine tensors of vectors and covectors. One important ability to acquire when studying multilinear algebra is to be able to switch between them fluently as needed, and to form the abstract idea of a tensor as just being a thing that can be used in all of these different ways, rather than focusing on their specific realizations.
A: I'll illustrate bilinearity with a simple two-form $\alpha = \sum \alpha_{ij}dx^i \wedge dx^j$ gives:
$$ \alpha(v,w) = \left(\sum \alpha_{ij}dx^i \wedge dx^j \right)(v,w) = 
\sum \alpha_{ij}(dx^i \wedge dx^j)(v,w) = \sum \alpha_{ij}(v^iw^j-v^jw^i) $$
Details aside about the wedge, you see that we have linearity in both $v$ and $w$. More generally, the wedge product generalizes the determinant and you should study the multilinearity of the determinant.
Notice, I do in fact mean that $v,w$ is a pair of vectors. Of course, this may be identified with a two-vector, but, this identification I do not make. The definition of wedge product in terms of the tensor product in the two-form case is given pragmatically from:
$$ dx^i \wedge dx^j = dx^i \otimes dx^j - dx^j \otimes dx^i$$
when we evaluate this on a pair of vectors we obtain
\begin{align} (dx^i \wedge dx^j)(v,w) &= (dx^i \otimes dx^j)(v,w) - (dx^j \otimes dx^i)(v,w) \\
&= dx^i(v)dx^j(w)-dx^j(v)dx^i(w).
\end{align}
However, as $v = \sum v^i\partial_i$ and $dx^j(\partial_i) = \delta_{ij}$ we find 
$$ dx^j(v) = dx^j(\sum v^i\partial_i) =\sum v^i dx^j(\partial_i) = \sum v^i\delta_{ij} = v^j. $$ 
In other words, $dx^j$ simply selects the $j$-th component of $v$. Moreover,
$$ (dx^i \wedge dx^j)(v,w) = dx^i(v)dx^j(w)-dx^j(v)dx^i(w) = v^iw^j-v^jw^i. $$
The bilinearity is manifestly:
$$ \alpha(Cu+v,w) = C\alpha(u,w)+\alpha(v,w)  \ \ \& \ \
\alpha(u,Cv+w) = C\alpha(u,v)+\alpha(u,w). $$
The identities above follow immediately from the formula I initially wrote. In any event, I hope the added detail is useful to the OP.
A: Just as 1-forms are naturally dual to a vector space, we can consider k-forms dual to the vector space of k-vectors. Sometimes it is the definition: linear functional on k-vectors is called a k-form.
How does that definition relate to viewing k-forms as alternating multilinear maps on $V^n$?
By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $(\wedge ^kV)^*$. Moreover, if $V$ is finite-dimensional, then the latter is naturally isomorphic to  $\wedge^k(V^*)$.
So we can write, for example, our 2-forms in $dx^i\wedge dx^j$ basis dual to $e_i\wedge e_j$ basis of 2-vectors, and write terms equivalently as bilinear forms or linear forms, $(dx^i \wedge dx^j)(v,w)$ or $(dx^i \wedge dx^j)(v\wedge w)$ with the same result ($v^iw^j-v^jw^i$) as in the answer/comments by @James and @Muphrid.
Interestingly, although any alternating bilinear forms can be expressed in the $dx^i\wedge dx^j$ basis, not all 2-forms can (unless $n=3$) be written as $\alpha\wedge\beta$ (decomposable, or simple); whereas we only consider simple bivectors $v\wedge w$.
