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I apologize in advance because this question might be a bit philosophical, but I do think it is probably a genuine question with non-vacuous content.

We know as a fact that differential forms have a much richer structure than vector fields, to name a few constructions that are built on forms but not on vectors, we have:

(1)Exterior derivative, and hence Stokes theorem, de Rham cohomology and etc.

(2)Integration.

(3)Functorality, i.e. we can always pull back a differential form but we cannot always push forward a vector field.

However, I feel somewhat paradoxical considering the fact that differential forms are defined to be the dual of vector fields, and this gives me the intuition that they should be almost "symmetric". Clearly this intuition is in fact far off the mark. But why? I mean, is there an at least heuristic argument to show, just by looking at the definition from scratch, that differential forms and vector fields must be very "asymmetric"?

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@MikeMiller: you are right I should've been more accurate. As for (3) functorality, I still would like to take it as an example of the richness of the structure of diff forms, rather than the reason for it, because I don't see a way to "see" the functorality from the definition of diff forms. I in fact don't know what kind of answer will satisfy me, but I certainly hope what I said doesn't make an answer impossible :) –  Jia Yiyang Jun 11 at 3:08
    
Um, @Mike, what is a covector field? –  Ted Shifrin Jun 11 at 3:08
    
@TedShifrin: differential 1-form I suppose. –  Jia Yiyang Jun 11 at 3:09
2  
Uh huh, which is a $1$-form.:) –  Ted Shifrin Jun 11 at 3:12

2 Answers 2

up vote 15 down vote accepted

The overwhelming preference to work with $k$-covector fields ("differential forms") stems from a few basic facts:

First, you might know of $\nabla$ from vector calculus. It is related to the exterior derivative $d$ in the sense that you can do $\nabla \wedge$ on a covector field and it is equivalent to $d$. $\nabla$ itself transforms as a covector does, and so it takes 1-covectors to 2-covectors, $k$-covectors to $k+1$-covectors (and these are all fields, of course). So there is a very convenient element of closure under the operation.

Second, integration on a manifold naturally involves the tangent $k$-vector of the manifold. This is something traditional differential forms notation tends to gloss over. When you see, for example, something like this:

$$\int f \, \mathrm dx^1 \wedge \mathrm dx^2$$

It really means this:

$$\int f \, (\mathrm dx^1 \wedge \mathrm dx^2)(e_1 \wedge e_2) \, dx^1 \, dx^2$$

For this reason, the basis covectors $\mathrm dx^i$ should not be confused with the differentials $dx^i$. Further, that we use $e_1 \wedge e_2$ here, and not $e_2 \wedge e_1$, reflects an implicit choice of orientation, which is usually picked by convention from the ordering of the basis, but this need not always be the case. The tangent $k$-vector, and especially its orientation, must necessarily be considered in these integrals.

So why does this make $k$-covector-fields preferred? Because the action of these fields on the manifolds' tangent $k$-vectors is inherently nonmetrical. So, differential forms allows you to do a lot of calculus without imposing a metric.

This point, however, is somewhat obfuscated when you introduce the Hodge star operator and interior differentials, for these are metrical. Then, you get a big problem with differential forms: by working exclusively with $k$-covector fields, and expunging all reference to $k$-vector fields, the treatment when we do have a metric is extremely ham-fisted. Yes, you can do everything with wedges, exterior derivatives, and Hodge stars. But it makes much more sense to use corresponding grade-lowering operations and derivatives instead. Geometric calculus does this, but let met get to that in a moment.

Regarding the pushforward vs. pullback, I must confess a lack of understanding. I do not see why we would want to pull covectors back from a target manifold while insisting we must push vectors forward. I'm very familiar with the mathematics: that under a smooth map, the adjoint Jacobian transforms covectors from the target cotangent space to the original, and the inverse Jacobian does the same for vectors. Perhaps it has to do with defining the pushforward as the inverse of this inverse.


Now, do all these remarks put together mean that $k$-vector fields are inherently disadvantaged, or less rich, than $k$-covector fields? I would say no. I mentioned geometric calculus earlier: it is the originator of the $\nabla \wedge$ notation that I used earlier, and it handles $k$-vector fields just fine. Geometric calculus is the calculus that goes with clifford algebra, and you may find it illuminating. Many of the theorems and results of differential forms translate to geometric calculus and to $k$-vector fields. Stokes' theorem? Used extensively. de Rham cohomology? Most of the same results apply.

My point above about differential forms integrals using tangent $k$-vectors implicitly? That comes from geometric calculus, too, where the tangent $k$-vector is not digested "trivially" and you have to look at all the metrical ways in which it might interact with the vector field you're integrating.

A grade-lowering derivative is natural to use with $k$-vector fields. In geometric calculus, this is notated as $\nabla \cdot$. You can see that successive chains of $\nabla \cdot$ continually lower the grade of a field, just as successive exterior derivatives raise it.

My ultimate point is that, when you do have a metric, it's quite nonsensical to treat everything as a differential form instead of using $k$-vector fields when appropriate. I feel the tendency to do this in physics divorces students from a lot of the vector calculus they had learned, unnecessarily so. I can't speak to mathematics courses, but I imagine some of that criticism applies, too.


Now, there are some properties of covector fields and exterior derivatives that are nicer than working with vector fields. For instance, under a map $f(x) = x'$ with adjoint Jacobian $\overline f$, it's true that $\overline f(\nabla ' \wedge A') = \nabla \wedge A$ for some covector field $A$. That's a very convenient result, and there's no correspondingly nice identity for vector fields.

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Nicely put. A rich answer. –  Arkamis Jun 11 at 4:00
    
Thanks for the informative answer, +1. To summarize your point :diff forms and multi-vector fields aren't really that "asymmetric" despite the conventional preference of diff forms, geometric calculus(whatever it means) is an alternative. So would you say the preference of diff forms is a historical incident, that it is only because the language of diff forms got popular earlier than geometric calculus? –  Jia Yiyang Jun 11 at 10:42
    
Yes, that's my position. And to be fair, vanilla tensor calculus handle vector fields on the same footing, too, but that gives up some of the cleanliness of notation in forms or GC parlance. –  Muphrid Jun 11 at 13:18

You can think of vector fields as the "Lie algebra of the diffeomorphism group"; that is, you can think of a vector field as an infinitesimal diffeomorphism. This in particular explains why you shouldn't expect vector fields to be functorial, because diffeomorphism groups are not functorial, but why you should expect vector fields to act on various other objects attached to a manifold via Lie derivative (this is some kind of "infinitesimal functoriality" for these objects).

You can think of $1$-forms as the "universal derivatives" of functions; in fact you can think of differential forms on a smooth manifold $X$ as being like the free commutative differential graded algebra generated by $C^{\infty}(X)$ (although I think this is slightly wrong as stated), which in particular explains morally why taking differential forms ought to be functorial. Because vector fields act by derivations on $C^{\infty}(X)$ this explains in some sense why vector fields pair with $1$-forms, but not why this pairing is perfect. I suspect that this is a special fact about smooth manifolds and is false in greater generality, suitably interpreted.

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Thanks and +1. This looks like good stuff but the terminologies in the 2nd paragraph are beyond my mathematical background. –  Jia Yiyang Jun 11 at 10:44
    
@Jia: for a simpler thing than what I said in the second paragraph see en.wikipedia.org/wiki/K%C3%A4hler_differential for starters (although it is slightly false that $\Omega^1(X)$ is the Kahler differentials of $C^{\infty}(X)$; one needs a notion of "smooth Kahler differential" instead). –  Qiaochu Yuan Jun 11 at 16:48
    
There's also something to say about Hochschild homology vs. cohomology but I've probably already gone too far. –  Qiaochu Yuan Jun 11 at 16:49
    
Well, certainly over my head for now but thanks for the extra information! –  Jia Yiyang Jun 12 at 2:09

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