I've been watching this series: Intro to differential forms, where the author says every now and then something like "differential forms carry their own scale", "I can eyeball a path integral with differential forms, but not with a gradient picture, as I need a scale for the latter".

Can someone explain or justify these statements? And also perhaps explain the idea of seeing differential forms (or perhaps just the exterior derivative of a 0-form $f$) as 'level sets' (of $f$)?

  • 1
    $\begingroup$ A differential form is a way to "measure" tangent vectors at each point. $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '17 at 5:06

The issue of scaling is one of the deeper but also trickier aspects of differential forms. To focus on that issue exclusively, it’s simplest to work just in $\mathbb{R}^2$ (the plane) and only consider constant differential $1$-forms, that is,$$\alpha = a \,\text{d}x + b \,\text{d}y$$where $a$, $b$ are constants. We can also note that$$\alpha = \text{d}(ax + by)$$that is, it is the exterior differential of the linear function $f(x,y) = ax + by$. To be specific, say $\alpha = 2 \,\text{d}x + 3 \,\text{d}y$. We can draw the "stack" picture of this form by drawing the integer level sets of $f$, that is, the sets $2x + 3y = k$ for all integers $k$. If $v = \langle v_1,v_2\rangle$ is a tangent vector (at a point $(x_0,y_0)$, say, although it won’t matter here) then the number $\alpha(v)$ created by letting the $1$-form $\alpha$ "eat" the vector v is$$\alpha(v) = (a \,\text{d}x + b\,\text{d}y)(\langle v_1, v_2\rangle) = av_1 + bv_2.$$But there is a much more geometric interpretation, namely, this is just how many "sheets" of the "stack" the arrow representing $v$ passes through (at least, that’s the integer part of $\alpha(v)$). To see this, let $(x_0,y_0) = (0,0)$ for simplicity, and note that the tip of the vector $v$ is at $(v_1,v_2)$, where the function $f$ has the value $f(v_1,v_2) = av_1 + bv_2$. So the arrow has passed through all of the sheets labeled $1,2,3,\ldots$ through $av_1 + bv_2$. For example, if $\alpha = 2 \,\text{d}x + 3 \,\text{d}y$ and $v = \langle5,-1\rangle$ then the vector $v$ will start on the set $f(x,y) = 0$ and end on the set $f(x,y) = f(5,-1) = 2(5) + 3(-1) = 7$, and $\alpha(v) = 7$ in this case.

So far I haven’t talked about the scaling issue. But try this: draw a coordinate system with carefully scaled $x$ and $y$-axes on a blank piece of paper, and use that to set up the example just given, with the "sheets" of alpha (with their labels $0,1,2,3,4,5,6,7$) and the arrow representing the vector $v$. Now erase the scale from the axes and see if you can still calculate $\alpha(v)$. Of course you can! It’s just the number of sheets passed through. (Or imagine handing the sheet to someone else who knows about $1$-forms and asking them to calculate $\alpha(v)$—they’ll have no trouble.)

To see it even more purely, don’t start with any axes at all. Just draw a bunch of parallel lines with labels on them (these are the level sets of a linear function) and an arrow going through the "stack" of lines. You can certainly count how many lines the arrow goes through without putting any axes or any scale of any kind on the picture.

In contrast, try this with the dot product of two vectors. Draw two vectors based at the same point, but do not draw a scale. Try to calculate the dot product of the vectors—you won’t be able to do it. The notion of the dot product depends in an essential way on the scale. (Recall that a vector is a unit vector if and only if its dot product with itself is $1$. So in particular, the dot product knows how to recognize a vector with length $1$.) The notion of a $1$-form eating a vector, on the other hand, is scale-independent.

If you want a "gotcha" version of this for the vectors, start with a scale with, say, $1\,\text{cm}$ being $100$ units on each scale, and draw the vectors $v = \langle500,-100\rangle$ and $w = \langle400,600\rangle$. (Don’t label the vectors numerically though!) Now erase the scale markings on the axes, erasing the "$00$"s very well, and erasing the leading digits less well, so it looks like the scale is $1,2,3,\ldots$ instead of $100,200,300,\ldots$. Now ask someone who knows dot products to calculate the dot product. They should say "I can’t do that", but they may assume (seeing the faint marks where you erased the scale numbers) that $1\,\text{cm}$ is $1$ unit. Then they would get $v \cdot w = 14$. You can then reveal (putting back in the scale that you really used) that $v \cdot w$ is really $140000$—quite a bit off!

Similarly, let’s stick with the same linear $\alpha = a \,\text{d}x + b\,\text{d}y$ but now integrate it over an oriented curve $C$ going from point $P$ to point $Q$ instead of having it eat a vector. The integral is$$\int_C \alpha = \int_C \text{d}f = f(Q) – f(P)$$which requires no scale, once again, to calculate. If you’re looking at the "stack" picture for $\alpha$, then the integral of $\alpha$ over $C$ is just how many sheets of $\alpha$ the curve passes through. This is another scale-invariant calculation.

In contrast, if you have a constant vector field $V$ and you want to integrate it over a curve $C$, you won’t be able to do that without a given scale. (You could run a very similar "gotcha" as above, for example.)

I hope that makes the statements about scaling more clear, and also show a simple example where the level sets of a function (here linear) are used to visually represent the data of a $1$-form (here with constant coefficients). There are two steps to a general $1$-form. First, we can still only look at exact forms, that is, those representable as $\alpha = \text{d}f$ where $f$ is a function. Second, we could look at a totally general form $\alpha = g(x,y)\,\text{d}x + h(x,y)\,\text{d}y$ that is not exact. Let’s just think about exact ones for now. I’ll also make one assumption, for simplicity: the range of $f$ is big enough (or, we could say, $f$ varies fast enough) that drawing a contour map using integer level sets of $f$ gives a precise picture with pretty dense level sets (pretty small spacing), one that doesn’t miss essential features of $f$. (If that’s not true we could always change the contour interval to $0.1$, or $0.01$, etc., but that just clutters up the conceptual understanding.) For example we could say $f(x,y) = x^2 + y^2$ and note that near the origin this is not the most precise picture (since the level sets are pretty far spaced there).

In any case, draw the contour map of $f$, using integer level sets, and let’s see if we can use it (and it alone, with no axis scaling!) to calculate

  1. $\alpha(v)$ where $\alpha = \text{d}f$ and $v$ is a tangent vector and
  2. the integral of $\alpha$ over an oriented curve $C$.

For (1), we’ll first assume that $v$ is a pretty small vector, in the sense that as we go from the tail to the tip of $v$, the level sets of $f$ look close to parallel and evenly spaced. I claim that $\alpha(v)$ is (approximately) still just the number of level sets of $f$ that the arrow for $v$ crosses. Algebraically,$$\begin{align*} \text{d}f(v) & = \text{the directional derivative of } f \text{ along the vector }v \text{ (this is the best definition of d}f\text{)} \\ & = f(\text{tip of }v) – f(\text{tail of }v).\end{align*}$$(You may be familiar only with the notion of directional derivative with respect to a unit vector; here we’re using the more general notion of the directional derivative with respect to any vector $v$. The official definition is$$\lim_{t\to 0} {{f(p+tv) – f(p)}\over{t}}$$and I’m assuming that $v$ is already small enough that setting $t=1$ gives a good approximation to the limit.)

Clearly $f(\text{tip of }v) – f(\text{tail of v})$ is just the number of "sheets" crossed by $v$.

If $v$ is not small, say $|v|=100$, then apply the above to, say, $w= v/10000$ to get $\alpha(w)$, and then put back in the scaling factor to get $\alpha(v)$. (This corresponds to putting $t = 1/10000$ into the official limit definition.)

For (2), it’s actually easier; since we’re not trying to model infinitesimal tangent vectors with actual arrows, we don’t need to worry about things being big or small or scaling them to make things work well. We again have (just as in the linear case)$$\int_C \alpha = \int_C \text{d}f = f(Q) – f(P)$$which just counts how many "sheets of alpha" the curve $C$ crosses.

So note so far how the $1$-form $\alpha = \text{d}f$ can be thought of as having almost exactly the same information content as $f$, just interpreted differently. (One subtlety is that we only ever counted how many sheets got crossed by things, i.e. differences of $f$-values; so adding a constant to $f$ will not change $\alpha(v)$ or $\int_C \alpha$. Of course that’s correct, since $\text{d}(f + C) = \text{d}f$, as the derivative of a constant is zero.) That tells us that $1$-forms and $\text{d}$ are extremely natural ideas; but it also begs the question of why we should bother.

It gets more interesting when $\alpha$ is not exact, i.e. $\alpha$ is not of the form $\text{d}f$ for some function. Then, we don’t have the global level set picture; instead, though, near any point $P$ (especially if we zoom in very tight) we can recreate the linear picture, putting a little stack near that point and using it to calculate $\alpha(v)$ for a tangent vector $v$ at $P$. If $\alpha$ secretly is exact, then we would eventually discover that all of these little stacks fit seamlessly together into the level set picture; but usually, the stacks are tighter in some places than others, or are rotated weirdly relative to each other, in a way that means they don’t join up.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.