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Differential forms as antiderivatives?

  1. Jul 26, 2010 #1
    Hi, I had a silly idea that probably doesn't work, but I thought I'd ask about it anyway.

    I understand that vectors can be thought of as derivative operators, e.g. [itex]\frac{d}{d\lambda} = \frac{dx^\mu}{d\lambda} \partial_\mu[/itex], where lambda parametrizes some curve.

    I also gather that one-forms are linear maps from the tangent space to the real numbers, i.e. they 'eat' a vector and 'spit out' a real number. But surely the ideal thing to take a derivative operator and spit out a real number is an (indefinite) integral operator? Something like
    [tex]\int^x d\lambda \frac{d}{d\lambda} f(\lambda) = f(x)[/tex]

    The reason I'm not sure this works is because differential forms are supposed to be the ideal thing to integrate themselves, and I'm not sure an integral operator works as something to integrate. Then again a derivative operator isn't something you can differentiate (you can apply them multiple times but that is different), but you can still get the divergence and curl of vector fields...

    I feel like it ought to work because the integral operator is obiously the inverse of the derivative operator!
     
  2. jcsd
  3. Jul 26, 2010 #2
    I speak with no authority but...

    to get a real number in your equation, the integral operator doesn't act on a differential operator: it acts on the differential operator applied to a function.
     
  4. Jul 26, 2010 #3
    I see what you mean, but aren't vectors defined by how they act on functions as well? I think I'm trying to do something analogous to how vectors are defined, except by using integrals along curves instead of derivatives along them. Maybe I'm missing the point :P
     
  5. Jul 26, 2010 #4
    I'm not sure what you're trying to do - though I'm all for it! - but when you said

    your equation didn't do this: you're not getting a real number by operating on a derivative operator.
     
  6. Jul 26, 2010 #5
    Yes that sounds right. It's an operator that returns the value of a function at a point. I think what I'm asking is what is the correct expression, if it even exists at all.

    I'm trying to understand it geometrically, really. You can write the derivative of a function as
    [tex]lim_{b\rightarrow a} \frac{f(b)-f(a)}{b-a}[/tex]
    And I think the interval [a,b] can be viewed as an oriented line segment, which is a vector, so I think that gives you a way of looking at how vectors act on functions. But then what would be equivalent for a one-form be? I understand from MTW's Gravitation that they are sets of (n-1)-surfaces that get pierced by vectors, and I wonder if there was a way of expressing that as an integral operator? It seems like there might be but I don't know really, I might just be making stuff up.
     
  7. Jul 26, 2010 #6

    Hurkyl

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    A tangent vector is indeed a differential operator that, when paired with a function, results in a scalar.

    So, by far the most obvious gadget for turning a tangent vector into a scalar is going to be a function. :tongue:

    Of course, there's "too much information" in a function; many functions give the same result when paired with the tangent vector. When you trim away all that superfluous information, you're left precisely with the notion of a one-form.



    I confess I'm mildly against trying to make a geometric picture of one-forms -- they're more algebraic than geometric. Do you try to come up with a geometric object that represents the notion of "length", or do you think of length as a means of measuring curves?

    Anyways, the most symmetric description I know of tangent vectors and one-forms is:
    • Tangent vectors come from differentiable functions R --> M
    • One-forms come from differentiable functions M --> R



    Functions R --> M are easy to depict as curves. And from that we often view a tangent vector as a localized depiction of a curve.

    If you really like to visualize functions M --> R by their level sets, you can think of a one-form as a localized depiction of a level set.




    If you're thinking of a differential form as an operator -- something that measures localized bits of a function, I suppose there is a companion description of a one-form: something that measures localized bits of a curve.

    If you have an entire tangent vector field, you can take all of those localized bits of the function they measure, and organize the result into the "directional derivative of my function".

    If you have an entire one-form field, you can take all of those localized bits of curves it measures, and organize the result into the value of the line integral of the form along the curve.

    I'm not sure if there is a more symmetric analog.
     
  8. Jul 27, 2010 #7
    Thanks, that's a nice way of looking at it. I guess I need to learn this a bit more thoroughly. Not that I really *need* to, I'm just learning for fun. :smile:

    By the way, when we go on to k-forms, I gather they map k vectors to R. But since they are alternating and multilinear, can we say they map k-vectors to R, without losing generality? A k-vector being an oriented segment of a k-dimensional surface, so vector, bivector, trivector etc. I just thought that might have a slightly nicer geometrical interpretation.
     
  9. Jul 27, 2010 #8

    quasar987

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    Nah what you call a k-vector is just a tangent vector to a k-dimensional surface (manifold). A k-form does not map a k-vector to R, it maps k vectors to R.
     
  10. Jul 28, 2010 #9
    Why not though? I don't see what the difference is! It has to be alternating in those vectors, if any two are parallel you get zero. So what's the difference between feeding it the list [itex]v_1, \ldots, v_k[/itex] and the wedge product [itex]v_1 \wedge \ldots \wedge v_k[/itex], which is a k-vector?
     
  11. Jul 28, 2010 #10

    quasar987

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    Mmh, k, forget what I said. I misinterpreted what you meant by k-vector.

    In that case you're absolutely right in the sense that there is a natural correspondance between the set of alternating multilinear maps [itex]V\times \ldots \times V\rightarrow \mathbb{R}[/itex] and the set of linear maps [itex]V\wedge \ldots \wedge V\rightarrow \mathbb{R}[/itex]. V being here a real finite dimensional vector space.
     
  12. Aug 8, 2010 #11
    There are several types of differential forms: namely 1-form, 2-form, 3-form...
    Let us look at 1-form differentials. Suppose you have a 1 dimensional manifold: M (Line or circle) and consider the tangent space to M, denoted T(M). A 1-form is a mapping [tex]\omega :T\left(M\right)\rightarrow\mathbb{R}[/tex]. In general any 1-form can be written in local coordinates as [tex]\omega = adx [/tex], a is a real number such that for any vector [tex] \partial_{ x} [/tex] in T(M), [tex]\omega (\partial_{ x}) = a dx (\partial_{ x}) = a \partial_{ x} \left(x\right) = a [/tex]. Thus, we can regard the set of 1-forms as dual space to the tangent space T(M) in the sense of linear algebra. In general k-forms are obtained by taking a tensor product of 1-forms: for example in local coordinates of the 3 dimensional euclidean space, we may have [tex] a dx\otimes dy\otimes dz[/tex]. There are plenty of resources in literature about these things. Similarly with the previous case (1-form), they can also be identified with linear duals to some tangent space.

    Also, there is a neat relationship between differentials and measures on a manifold. The simplest case is the 1-form on the real line: dx which is nothing but the Lebesgue measure studied in Real Analysis. Obtaining a measure via differential forms allow us to be able to integrate on manifolds

    Vignon S. Oussa
     
    Last edited: Aug 8, 2010
  13. Aug 13, 2010 #12
    My original dumb idea came about by thinking about whether integral operators had a nice geometric interpretation, but I just wonder if I've found it. I found the following definition of the exterior derivative

    [tex]d\alpha (v_1 ... v_k) = \lim_{h \rightarrow 0} \frac{1}{h^k} \int_{\partial P(hv_1 ... hv_k)}\alpha[/tex]

    where alpha is a (k-1)-form, v_1 ... v_k are k vectors and P is a parallelogram spanned by those vectors. But I noticed on the left the k vectors are all contracted up with alpha, so maybe we could define a k-vector (or k-multivector, sorry for not making that clearer before :smile: ) like this

    [tex]\alpha \cdot v = \lim_{h \rightarrow 0} \frac{1}{h^k} \int_{h^k S_k } \alpha[/tex]

    Where now alpha is a k-form, S_k is a k dimensional (sub)manifold associated with the k-vector v, and we're integrating over a small volume h^k. We probably need to define it over equivalence classes of S_k as well. So it seems like there is an integral operator associated with each surface S_k, which is the "tangent k-vector" to it? It also combines the idea that a k-form is the ideal thing to turn a k-vector into a real number, with the idea that a k-form is the ideal thing to integrate. It seems to naturally pair k-forms with k-vectors, but I think it returns the idea of 1-vectors as directional derivatives, if they act on exact 1-forms, as long as you use Stokes' theorem.

    Does that make any sense?
     
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