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Trying to understand derivatives in terms of differential forms

  1. May 19, 2013 #1
    Suppose we have a curve, formed by a function f that maps real numbers to real numbers, such that f is everywhere smooth over a subset D of its domain. Let's suppose that, for all x in D, there is a vector space that contains all vectors tangent to the curve at that point, called the tangent space. A differential df is, thus, a linear functional that maps elements of the tangent space at a point x to the set of real numbers. The quantity dx denotes the differential of the identity function. As such, the derivative may be seen as a "function of proportionality" that gives, intuitively, the ratio of how much faster f grows compared to the identity function at a point x. That is, ##df = \frac{df}{dx}dx##.

    This has, roughly, been my definition of the derivative of a function. First of all, is this a correct way of thinking about it in terms of differential geometry (minus the formalisms, of course. For example, the curve might be better considered as a manifold)?

    Second, does this justify, to some degree, the idea that we can "cancel" differentials? Id est, ##\frac{dy}{dx}=\frac{dy}{\not{du}}\frac{\not{du}}{dx}##?

    Third, does this concept generalize? That is, could there be "derivatives" for higher differential forms, like ##\frac{dy\wedge dx}{dz\wedge dw}##?

    I'm a little new to the ideas of differential geometry, but feel free to use formal terms to force me to look up information on them so I can learn more. Thank you in advance for your help.
     
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  3. May 19, 2013 #2

    WannabeNewton

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    Not really because ##dx## has a well-defined, much more general role in differential geometry. For ##f\in C^{\infty}(\mathbb{R})## (the tangent spaces in ##\mathbb{R}## are all trivially identified with one another but we will forget that for the purposes of this thread) for each ##T_{p}(\mathbb{R})## the vector ##\frac{\mathrm{d} }{\mathrm{d} x}|_{p}## defined by ##\frac{\mathrm{d} }{\mathrm{d} x}|_{p}f = \frac{\mathrm{d} f}{\mathrm{d} x}(p)## forms a basis for ##T_{p}(\mathbb{R})## (here I am using the global coordinate chart ##(\mathbb{R},id_{\mathbb{R}})##). Then, there exists a natural basis for the cotangent space ##T^{*}_{p}(\mathbb{R})## made up of the covector ##dx|_p## defined by ##dx|_p(\frac{\mathrm{d} }{\mathrm{d} x}|_p) = 1##.

    The exterior derivative of ##f## is a map ##d:C^{\infty}(\mathbb{R})\rightarrow \Lambda^{1}, f \mapsto df## (here ##\Lambda^{1}## is the set of one-forms fields). Since it is nothing more than the differential of ##f##, at each ##p\in\mathbb{R}## it is defined by ##df(p):T_p(\mathbb{R})\rightarrow \mathbb{R},X \mapsto X(f)(p)##. Note that ##df(p)(\frac{\mathrm{d} }{\mathrm{d} x}|_p) = \frac{\mathrm{d} f}{\mathrm{d} x}(p)## so we can expand ##df(p)## in the coordinate basis as ##df(p) = \frac{\mathrm{d} f}{\mathrm{d} x}(p)dx|_p## and this is where it comes from.

    ##\frac{dy\wedge dx}{dz\wedge dw}## is a nonsensical expression as you are trying to divide two forms.
     
    Last edited: May 19, 2013
  4. May 19, 2013 #3

    mathwonk

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    you can divide two elements of a vector space and get a number for an answer whenever the two vectors are scalar multiples of each other. In a one dimensional space that almost always happens (except when one vector is zero). The reason you can divide df by dx, is they belong to the one dimensional space of cotangent vectors at a point of a curve, which is one dimensional.

    Now on a surface, the cotangent space is two dimensional, but the second exterior product of that space, which is where the two forms live, is one dimensional, so yes on a surface you can divide two forms. In general on an n dimensional manifold the space of n forms at a point is one dimensional. In particular you can divide the two forms you write down, if they are defined on a 2 dimensional surface, say in (x,y,z,w) space. This is analogous to dividing dx by dy on a curve in (x,y) space.
     
  5. May 19, 2013 #4

    dx

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    One should be careful though, because dX + dY can be a scalar multiple of ω, i.e.

    kω = dX + dY

    but dX is not a scalar multiple of ω, and dY is not a scalar multiple of ω, so you can't write

    k = dX/ω + dY/ω

    if you use that meaning for the quotient of two vectors.
     
    Last edited: May 19, 2013
  6. May 19, 2013 #5

    WannabeNewton

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    By the way, we already do have a way of taking derivatives of arbitrary p-forms in such a way that we end up with a p+1-form. It is called the exterior derivative: http://en.wikipedia.org/wiki/Exterior_derivative

    In more "physicsy" notation, in particular abstract index notation, if ##\nabla_{a}## is a derivative operator on your smooth manifold ##M##, the exterior derivative is given by ##(d\alpha)_{ba_1...a_n} = (p+1)\nabla_{[b}\alpha_{a_1...a_n]}## where the brackets denote anti-symmetrization. As it turns out, the exterior derivative has the property that it is independent of the derivative operator that you choose to use on your manifold so the above formula holds for the partial derivative ##\partial_{a}## associated with any coordinate chart on ##M##: ##(d\alpha)_{ba_1...a_n} = (p+1)\partial_{[b}\alpha_{a_1...a_n]}##.
     
  7. May 19, 2013 #6

    Bacle2

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    Do you mean by dividing forms, that you are using an isomorphism f with the reals, dividing
    on the reals and then pulling back the division by f?
     
  8. May 19, 2013 #7

    Bacle2

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    In your example ##df = \frac{df}{dx}dx## , ##df = \frac{df}{dx}## is the derivative
    at a given point, while the last term ##dx## is the differential. In general, in higher dimensions,
    differentials become differential forms and derivatives become Jacobian matrices, so that the proportionality does not hold beyond dimension 1. The differential in, e.g., ℝ2 ,( of a differentiable function; when your manifold is embedded in some ℝn ) at a point is the tangent plane at that point.The derivative is the Jacobian of f at the point,which ,similar to your idea, scales the differential by matrix multiplication.
     
  9. May 19, 2013 #8

    mathwonk

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    @ Bacle2: "Do you mean by dividing forms, that you are using an isomorphism f with the reals, dividing
    on the reals and then pulling back the division by f?"

    No, just use the definition of division as the inverse of multiplication. I.e. if V = cW, where V and W are vectors and c is a scalar, then c is the unique scalar which gives V when multiplied by W, so by definition of division, V/W = c.
     
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