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Is a function contravariant?

  1. Aug 16, 2010 #1
    My understanding of differential forms is that, given the map
    [tex] \phi : V \to V^{*} [/tex]
    If [tex] f \in V^{*} [/tex] is a 0-form (a real-valued function on [tex] V^{*} [/tex]), then [tex] \phi [/tex] defines the function [tex] \phi^{*} f [/tex] on [tex] V [/tex]
    where [tex] \phi^{*}: F^{0}(V^{*}) \to F^{0}(V) [/tex]
    as the function such that [tex] \phi^{*}f(p)=f(\phi(p)) [/tex].

    So, since differential forms transform contravariantly, the 0-form (function) f transforms contravariantly (and therefore is contravariant.....).

    Because [tex] f [/tex] doesn't have any indices, I decided to look online and see if that fact makes a difference in requiring the label "contravariant."
    One of the first links was to a thread on physicsforums
    https://www.physicsforums.com/showthread.php?t=311174

    The debate is between "dx" and "MarkParis." It doesn't seem to reach an agreement in that thread, so I was wondering if anyone else could shed some light on it.

    *IMO, the posts by "dx" are correct and meet my understanding of the topic. However, I'm still learning differential forms and there is certainly going to be nuances of the subject that I haven't understood yet....so I figured I'd see if anyone here has any input on this and/or the posts made in the linked thread.
     
  2. jcsd
  3. Aug 16, 2010 #2
    The function f belongs to V*, so it can be expanded in a linear combination of a basis of V*. The coefficients of this expansion form a vector, that transforms contravariantly.
     
  4. Aug 16, 2010 #3

    atyy

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    I think MarkParis's point is that strictly speaking it is df belongs to V*, not f itself.

    Similarly it is d/dl that belong to V, not the curve itself.

    And Hurkyl and dx's points are that df comes from f, just as d/dl comes from the curve, so in that sense f is "contravariant" while a curve is "covariant" (or is it the other way round, I never remember which is which).
     
  5. Aug 16, 2010 #4

    Hurkyl

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    The unifying notion of covariant and contravaraint comes from category theory.

    There are lots of things we might be inclined to associate to any particular manifold -- things like its tangent bundle, the cotangent bundle, the set of curves defined on the manifold, its homology groups, et cetera.

    Each of the things I mentioned above is a "functor", which here boils down to saying that if you have a differentiable function between manifolds, then you get a homomorphism between the kinds of things we are associating to the manifold.

    The functor is covariant if the associated homomorphism goes in the same direction as the differentiable function, and contravariant otherwise. e.g. "Cotangent bundle" is contravariant, because for any differentiable function f:M-->N, the map f*:T*N-->T*M goes in the same direction.

    Similarly "the set of curves" is covariant, because given f:M-->N, the associated function on the set of curves (i.e. "composition with f") turns a curve on M into a curve on N.


    In this sense, "the sheaf of scalar fields" is contravariant, as is the "set of differentiable functions to R".
     
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