How do vectors act on functions? By telling you how much the function changes in the direction of the vector. This action is given by \begin{equation} \vv(f) = \vv\cdot\grad f \label{vact} \end{equation} and involves differentiation. But we have also seen that the 1-form $df$ corresponds to $\grad f$, through the Master Formula ($df = \grad f\cdot d\rr$) and we can therefore regard (\ref{vact}) as an action of the 1-form $df$ on the vector $\vv$, given by \begin{equation} df(\vv) = \vv(f) = \grad f\cdot\vv \end{equation} This action does not involve differentiation, but only linear algebra, and therefore lends itself to a geometric interpretation in terms of stacks. 1)

Consider first the 1-form $dx$. From calculus, we know that $dx=0$ along surfaces of the form $x=\hbox{constant}$. What does $dx(\vv)$ represent? We have $dx=\grad x\cdot d\rr=\xhat\cdot d\rr$, so $dx(\vv) = \xhat\cdot\vv = v_x$ is just the $x$-component of $\vv$. How can we display this information visually?

Figure 1: The stack corresponding to $dx$.

Figure 2: The geometric evaluation of $dx(\vv)=v_x$.

We represent $dx$ as an oriented stack of surfaces $x=\hbox{constant}$. Since we have associated $dx$ with $\xhat$, we choose unit spacing between the surfaces. Thus, $dx$ corresponds to the drawing in Figure 1, where the arrow gives the orientation, namely the direction in which $x$ is increasing. To evaluate $dx(\vv)$, all we need to do is superimpose $\vv$ on this stack, as shown in Figure 2, then count how many surfaces $\vv$ goes through, which in this case is $\frac32$. If the stack had been oriented the other way (corresponding to $-dx$), then the answer would have been negative.

Figure 3: The stack corresponding to $2\,dx$.

Figure 4: Adding the stacks $dx$ and $dy$ to get the stack $dx+dy$.

Now that we know how to represent $dx$, how can we represent $2\,dx$? Clearly, the result of acting on a vector with $2\,dx$ should be twice that of acting on it with $dx$. But the vector doesn't change, so it must go through more surfaces. Thus, $2\,dx$ is represented by the same drawing as in Figure 1, but with the spacing between the surfaces reduced by a factor of two, as shown in Figure 3. This construction should remind you of a topographic map, where the contour lines are closer together the steeper the terrain is.

How do we add stacks? Remarkably, the process is simple: simply superimpose the corresponding stacks, and connect points of intersection. This construction is shown in Figure 4 for the addition of $dx$ and $dy$. It is straightforward but somewhat tedious to verify that this procedure yields the correct spacing between surfaces, which is $\frac{1}{\sqrt{a^2+b^2}}$ for $a\,dx+b\,dy$. (Why?)

Figure 5: The stacks corresponding to $r\,dr=x\,dx+y\,dy$.

Figure 6: The stacks corresponding to $r^2\,d\phi=-y\,dx+x\,dy$.

The description of stacks given above works fine for constant 1-forms, but not for 1-form fields. Just as a vector field is a vector at each point, a 1-form field is a 1-form at each point. Thus, to represent a 1-form field, we need a stack at each point.

For example, Figure 5 shows the stacks corresponding to $r\,dr=x\,dx+y\,dy$, and Figure 6 shows the stacks corresponding to $r^2\,d\phi=-y\,dx+x\,dy$. Since $r\,dr=\frac12\,d(r^2)$, the first example could have been represented by a topographic map whose contours are circles; this 1-form corresponds to the conservative vector field $r\,\rhat=x\,\xhat+y\,\yhat$. However, the vector field $-y\,\xhat+x\,\yhat$ is not conservative; the second example does not correspond to any topographic map. Stacks provide a useful geometric representation in both cases, but they do not in general combine to form a global representation in terms of contours. 2)

Figure 7: A representation of $dx\wedge dy$.

Similar methods can be used to represent higher-order forms. For example, the 2-form $dx\wedge dy$ is shown in Figure 7. The 2-form $dx\wedge dy$ acts on pairs of vectors $(\uu,\vv)$. How? Construct the parallelogram whose sides are $\uu$ and $\vv$, then superimpose it on the drawing; the number of “cells” covered by the parallelogram gives the result of $dx\wedge dy$ acting on ($\uu,\vv$), with the overall sign determined by the relative orientations. Again, this drawing shows a constant 2-form, a general 2-form (field) would be represented by a similar drawing at each point.

Similar constructions apply in higher dimensions. 3)