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I'm learning about differential forms (while also reviewing my long-forgotten multivariable calculus), and in particular trying to get an inuitive understanding of the exterior derivative. One helpful suggestion has been to treat the exterior derivative of a k-form $\omega$ as (the limit of) a normalized integral of $\omega$ over the oriented boundary of a small (k+1)-parallelotope.

So for a 0-form $f$, we get:

$$df(v)\big|_p = \lim_{t \to 0}\dfrac{1}{t} \left( f(p+tv) - f(p)\right)$$

For a 1-form, intuitively we get curl (we're integrating around a parallelogram). But from reading Prof. Tao's intro to differential forms, it seems like what we should get is some kind of flux (edit: I just mean something flow-like). If we're working in Euclidean 3-space, should I view the two as something like Hodge duals of each other? And what if we're working in an arbitrary space? Should I think of an exact 2-form as curl-like or flux-like?

Any other help in thinking about all of this in an intuitive-but-formal way would be appreciated.

A_P
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  • You might want to check out some of the lectures in my YouTube videos linked in my profile. – Ted Shifrin Oct 09 '19 at 21:18
  • Thanks, @TedShifrin! I just watched 3510 day 26 and enjoyed it. It helped me with the mechanics of taking exterior derivatives. I might have to watch more to get the answer to my question. – A_P Oct 09 '19 at 22:58
  • Glad it helped. There is one specifically addressing div and curl and physical interpretations in 3-space (probably lectures 41 and 42), but lots of other ones might be helpful with developing intuition and facility with the material. – Ted Shifrin Oct 09 '19 at 22:59
  • @TedShifrin Just watched 41. I think I see where my confusion is coming from. Wikipedia has two defs of flux: flow rate per unit area and the surface integral def. It's the former sense I'm asking about. In this sense, it seems like curl $F$ can be interpreted as a kind of flux (an exact one). Do you see what I'm driving at? – A_P Oct 19 '19 at 19:52
  • No, Stokes's Theorem tells us that the flux of the curl gives us the circulation of the line integral around the boundary. "Flow rate per unit area" is a flux computation (when you interpret the vector field as a velocity field), so I think you're trying to separate two ideas that aren't separate. – Ted Shifrin Oct 19 '19 at 19:56
  • @TedShifrin We're getting closer. When we calculate flux as $\iint_S (\vec{F}\cdot \vec{s}) dS$, we are interpreting $\vec{F}$ as the direction and amount of flow of something. The associated 2-form is then something like this amount per unit area (right?), which I (perhaps wrongly) thought of as a kind of "flux." Now, that $\vec{F}$ could itself be curl $\vec{G}$. I normally think of curl as circulation density, but here we are treating it as a flow. So the 2-form associated with this circulation density seems to be what I was calling "flux," unless I got that wrong. – A_P Oct 19 '19 at 21:25
  • @Malkoun Thanks, but I don't see how it relates to my question of flux here. Also, shouldn't we say that $df$ is a form whose dual is the gradient? Calling it a gradient confuses me further (see e.g. https://math.stackexchange.com/a/1103691/161555). – A_P Oct 19 '19 at 21:28
  • Yes, the $1$-form $df$ and vector $\nabla f$ are dual objects. The language I used in my lectures is that a vector field $\vec F$ corresponds to a $1$-form $\omega$; if you take the $2$-form $\star\omega$, integrating it over an oriented surface gives the flux of $\vec F$ across that surface. I agree that you can talk about both the flux (across a surface) and the circulation (along a closed curve) for any vector field. These correspond to the $2$-form and $1$-form versions of the vector field, respectively. Having the curl in there sort of muddies the water, but it's the same principle. – Ted Shifrin Oct 19 '19 at 21:43
  • @TedShifrin Thanks, that's very clear! So a vector field $\vec{F}$ can be given a 1-form that can be thought of as circulation density, or its Hodge dual 2-form that is a "flux density." Now, the exterior derivative of a $k$-form is a $k+1$-form, and only when working with a manifold of dimension $2k+1$ might you have this funny situation where where you might confuse the two kinds of objects thanks to Hodge duality. And to answer my original question directly, an exact 2-form is a flux-like thing whose value on an infinitesimal parallelotope is equal to a circulation on its boundary. Whew. – A_P Oct 19 '19 at 22:55
  • @TedShifrin Sorry, just want to confirm: does my previous comment look okay? – A_P Oct 21 '19 at 22:40
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    Sure, it's fine. :) – Ted Shifrin Oct 21 '19 at 22:49
  • @TedShifrin Hi Ted! I came back to this question and wrote an answer that (hopefully) clears up my original confusion. Would you be kind enough to take a quick peek and let me know if it looks good to you? Thank you. – A_P Dec 19 '19 at 05:00

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Thanks to Ted Shifrin for clearing this up.

Suppose we identify a vector field $V$ with its dual, the 1-form $\omega$. This is apparently a common thing to do. Then what is $d\omega$? It is a 2-form such that $d\omega(v_1, v_2)$ gives the (normalized) circulation of $V$ around the (infinitesimal) parallelogram defined by $(v_1, v_2)$. This works in general.

But a flow only makes sense across a hypersurface. In $\mathbb{R}^n$, it must be represented by an $(n-1)$-form $\tau$, measuring the flow of the vector field corresponding to $\star\tau$. In $\mathbb{R}^3$, therefore, we can choose to interpret the 2-form $d\omega$ as measuring the flow of the field $\star d\omega$ (what we normally call curl $V$). But we cannot think of a 2-form as measuring a flow in general.

More generally, it seems useful to keep in mind that a $k$-form simply assigns a scalar to a $k$-volume. This can have more than one interpretation depending on context.

A_P
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    This is good. What generalizes is taking the flux of $V$ across a hypersurface (the integral of $\star\omega$) and then you have the divergence theorem coming out of Stokes's Theorem: $\text{div}(V)$ is the function so that $d(\star\omega)$ is that function times the volume element of $\Bbb R^n$. – Ted Shifrin Dec 19 '19 at 06:11
  • @TedShifrin Thanks! As I reflect on my motivation for posting this question, it was to understand the intuition for the given integral definition. For n=3,k=1, we see that curl is circulation density. For n=3,k=2, we see that divergence is flux density. I was trying to extrapolate, and getting confused by how a 2-form could be interpreted as either curl-like and flux-like, and what that meant for higher k. I now see that the k=2 example is really the k=n-1 example, and there probably isn't a natural interpretation for arbitrary n,k -- unless you know of one? – A_P Dec 19 '19 at 21:19
  • No, I had arguments for hours with a Ph.D. chemist 40 years ago because he insisted that curl made sense in $\Bbb R^4$. I taught him all about differential forms and he came back a week later and repeated his (wrong) assertion. Sigh. – Ted Shifrin Dec 19 '19 at 21:31