I read the beautiful (in my opinion) “Geometrical Vectors” by Gabriel Weinreich. In it, he offers a nice geometrical visualization of multivectors (elements of $\bigwedge_pV$, where $V$ is a finite dimensional vector space) and differential forms (where $p$-forms are elements of $\bigwedge^pV \equiv \bigwedge_pV^*$), though he doesn't call them that way. In summary, one could represent a $p$-vector as an oriented parallelogram (or “oriented $p$-volume”) and a form as a “bundle” of $(d - p)$-dimensional surfaces (I think this is in fact a standard way of visualizing them). The intersection of the bundle with the parallelogram is in general a certain number of points, which is the evaluation of the $p$-form on the $p$-vector.
However, I have a concern about the visualization of $1$-forms (i.e. what he calls “stacks”), and of forms in general. $1$-forms correspond to parallel plane sheets, with their density representing the “magnitude”. It appears, at first, that this picture implements in a nice way the fact that forms are “contravariant” (to avoid confusion, I know that generally forms are said to be covariant, because their components are, but the form itself is contravariant, if I understand correctly): when space is stretched, the density (i.e. the “magnitude”) decreases, whereas “arrow” vectors show the opposite behavior. However, stacks seem to behave in the same way as arrows under rotations, whereas from the algebraic point of view a form transforms with the inverse rotation.
I guess there is something I'm missing about the visualization, or maybe there is a secret metric identification going on. So, the question is:
- Why do forms seem to behave in the same way as vectors under rotations in this picture?
And, while we are at it
- What does the fact that forms algebraically transform with the inverse matrix actually mean geometrically?
Edit:
I'm not asking why forms transform like they do in the usual picture, i.e. forms are elements of $(\bigwedge_pV)^* \cong \bigwedge_pV^* = \bigwedge^pV$, and as such they assign a numerical (scalar) value to $p$-vectors. In order for this number to be a scalar, forms and vectors ought to transform in opposite ways.
What I'm asking is how this relates to the picture I described and which is found in “Geometrical Vectors” and in other sources. It appears that, in this picture, forms transform like vectors under rotations, which seems to contradict the fact that they should, according to the usual picture, transform with the inverse rotation.
I'm surely missing something.
Second edit:
There it is, the thing I missed: I completely missed the fact that the rotation matrix, when applied to the components of the form “as if” they were components of a vector, gets transposed (see the accepted answer). I should actually have been more careful with the algebra and not with the picture.
Let $$ \boldsymbol{v}=v^x\,\partial_x+v^y\,\partial_y\,,\quad\boldsymbol{\omega}=\omega_x\,dx+\omega_y\,dy $$ be a (covariant) vector and a (contravariant) one-form. Since they are dual to each other $$\tag{1} \boldsymbol{\omega}(\boldsymbol{v})=v^x\omega_x+v^y\omega_y $$ is a scalar which should not change when we rotate both, or equivalently, when we change the coordinate system by a rotation.
Equivalently to (1) the components of $\boldsymbol{v}$ can be considered a column vector, and the components of $\boldsymbol{\omega}$ a row vector.
When we rotate $\boldsymbol{v}$ we get \begin{align}\tag{2} \begin{pmatrix}u^x\\ u^y \end{pmatrix} =\underbrace{\begin{pmatrix} \cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{pmatrix}}_{\text{rotation matrix}} \begin{pmatrix}v^x\\v^y\end{pmatrix}\,. \end{align} In order that $\boldsymbol{\omega}(\boldsymbol{v})$ does not change and with the convention that the components of $\boldsymbol{\omega}$ are a row vector we must have by simple linear algebra that \begin{align} \begin{pmatrix}\eta_x & \eta_y \end{pmatrix} =\begin{pmatrix}\omega_x & \omega_y\\\end{pmatrix}\underbrace{\begin{pmatrix} \cos\phi&\sin\phi\\-\sin\phi&\cos\phi\end{pmatrix}}_{\text{ inverse rotation matrix}} \, \end{align} holds. This ensures that the scalar $\boldsymbol{\omega}(\boldsymbol{v})$ does not change: $$ u^x\eta_x+u^y\eta_y=v^x\omega_x+v^y\omega_y\,. $$
Since the transpose of a rotation matrix is its inverse this rotation can be written equivalently as \begin{align}\label{eContravariant2} \begin{pmatrix}\eta_x \\ \eta_y \end{pmatrix} =\begin{pmatrix} \cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{pmatrix}\begin{pmatrix}\omega_x \\ \omega_y\\\end{pmatrix} \end{align} which is the same as (2).