I've been tinkering with differential forms for a while now, and I've had a few questions all rolled into one trying to understand them. The exterior derivative is quite natural to me - it looks just like a regular old derivative, and so I believe I have a good intuition about how it works.

The first question that struck me was that I did not have a similar intuition about the interior derivative. I wanted to know in what sense it was a derivative, if any, and either way I wanted to get an idea of what this guy really does. A geometric interpretation would have been ideal. At one point I was thinking about tensor contractions and how that's kind of like a trace, but that's not really what the interior derivative is anyway, so I ended up putting that aside for a while.

But I noticed something interesting that has given me renewed interest in the question when I learned about Hodge duality.

We will work in $\mathbb R^3$ with the usual coordinates. Consider an arbitrary $1-$form which we'll write down as $F_x dx + F_y dy + F_z dz$. We'll compute the exterior derivative of this, and we end up with $$ \bigg ( \frac{\partial F_x}{\partial y} - \frac{\partial F_y}{\partial x} \bigg ) dx \wedge dy + \bigg ( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \bigg ) dz \wedge dx + \bigg ( \frac{\partial F_y}{\partial z} - \frac{\partial F_z}{\partial y} \bigg ) dy \wedge dz $$

Amazingly (at least to me), this is the Hodge dual of the curl!

Now, we can do a similar computation using $d(i_X)$. If we let $X = \alpha \frac{\partial }{\partial x} + \beta \frac{\partial}{\partial y} + \gamma \frac{\partial}{\partial z}$. $X$ is a vector field so these could be functions of $x,y,z$, it doesn't really matter. Now if we let $\omega$ denote the $3-$form, $dx \wedge dy \wedge dz$. We can now compute.

The computation is a little laborious to do directly from the definition, but there's a nice little formula which we can use to shortcut the computation. What we end up with is $$ \alpha dy \wedge dz + \beta dz\wedge dx + \gamma dy \wedge dz$$

after a few little rearrangements. Now we take the exterior derivative of this, and we have $$\bigg (\frac{\partial \alpha}{\partial x} + \frac{\partial \beta}{\partial y} + \frac{\partial \gamma}{\partial z} \bigg ) (dx \wedge dy \wedge dz) $$And again in a way that totally amazes me, this is Hodge dual of the divergence of $X$!

Now this motivates the question: What is the geometric significance Hodge duality? If there was some nice way to think about what the Hodge dual of something is, it would help me to understand the interior derivative, maybe.

There is a problem though, which makes me worry that all of this is just coincedence of some sort - the above analysis depends on the fact that the Hodge dual on a 3-manifold takes 1-forms to 2-forms, which is how we able to see what's going on with the curl. In particular, curl is not even defined in dimensions different from 3. Divergence is defined in general though, so that much is a welcome sign.

So somehow, I want to say that what I have learned will give me an avenue to understanding interior derivatives, and exterior derivatives better, but I don't know how to interpret Hodge duality, and I have these concerns.

In Summary: How is the interior derivative a derivative, geometrically? What is the geometric content of Hodge duality? Can I put these together to understand interior and exterior derivatives?

  • $\begingroup$ The divergence of $X$ is a scalar function, and the Hodge dual is a $3$-form. You meant to take $d\big(\iota_X(dx\wedge dy\wedge dz)\big)$? $\endgroup$ – Ted Shifrin Jul 23 '16 at 5:49
  • $\begingroup$ @TedShifrin I think so. I don't know what I thought divergence was when I wrote this up. I'll edit the post shortly. $\endgroup$ – Alfred Yerger Jul 23 '16 at 6:34

How is the interior derivative a derivative?

I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation.

What is the geometric content of Hodge duality?

Short version: you're finding the orthogonal complement of whatever you're dualizing.

Longer version: That orthogonal complement picture doesn't generalize to non-simple forms (i.e. forms that cannot be factored into wedge products of vectors). Only simple forms correspond to subspaces.

Can I put these together to understand interior and exterior derivatives?

Perhaps, but not in the way I think you want.

I think what you really want is the codifferential $\delta$, such that $\delta \omega = \star d \star \omega$ up to a minus sign.

The interior derivative is really more of a dual analogue to the wedge product. Let $\chi$ be the 1-form corresponding to a vector field $X$ under some symmetric bilinear form $g$ that determines the Hodge dual operation $\star$. Then $\iota_X \omega = \star^{-1} (\chi \wedge (\star \omega))$.

  • $\begingroup$ This is a step in the right direction for sure. When you say orthogonal complement, how can I use that to help me understand the two things I noticed in the OP? I think of divergence as how much a vector field is like a source. $\endgroup$ – Alfred Yerger Jul 24 '16 at 18:39
  • $\begingroup$ Something you might find is that the exterior derivative and the codifferential can be used to reconstruct a differential form. In this sense, I would say that a differential form in general (excluding of course 0-forms and volume forms) have two sources. A 1-form could be considered to have a scalar source (akin to divergence) and a 2-form source (akin to curl). $\endgroup$ – Muphrid Jul 25 '16 at 17:51

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