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What's a 0-form and what's not?

  1. Aug 12, 2007 #1

    quasar987

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    Is a function from R^n to R^m for aritrary m a considered a 0-form on R^n, or does 0-form refers only to functions from R^n to R ?
     
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  3. Aug 12, 2007 #2

    dextercioby

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    What's the definition of a p-form on R^n ?
     
  4. Aug 12, 2007 #3

    Hurkyl

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    If you have a finite dimensional vector space V with scalar field k, then the space of n-forms is isomorphic to the space of alternating multilinear maps Vn --> k.

    In particular, a 0-form is an element of k.



    In the typical setting of differential geometry, when analyzing a single point, your scalar field is R and your vector space is the tangent space, so a 0-form would simply be a real number. But more exotic things are possible, and sometimes even fruitful.
     
  5. Aug 12, 2007 #4

    quasar987

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    In 'Calculus on manifolds', Spivak defines a k-form on R^n as a function w sending a point p of R^n to an alternating multilinear maps (R^n)^k-->R.

    This makes sense only for k>0, so he treats the case k=0 separately by saying that by a 0-form we mean a function f.

    I was 90% sure he meant a function f:R^n-->R but wanted to make sure.
     
  6. Aug 12, 2007 #5

    Hurkyl

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    Well, it really does make sense for k=0: an A-valued function of 0 variables is the same thing as an element of A, and it's vacuously true that such a thing is alternating and 0-linear.

    So a 0-form on the tangent bundle to R^n is, indeed, a map R^n --> R. This agrees with what I said pointwise -- if f is such a thing, then f(P) is a 0-form on the tangent space at P, which is the same thing as an element of R.
     
  7. Aug 17, 2007 #6

    quasar987

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    Why are forms defined specifically as sending points to alternating tensors? What's wrong with good old arbitrary tensors? Or equivalently, what's so special about alternating ones?
     
  8. Aug 18, 2007 #7

    Hurkyl

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    Integrating along an opposite orientation should give you the opposite answer -- thus the sign change.

    From an algebraic perspective, they are trying to capture first-order differential information -- thus you want dx dx = 0. An immediate consequence of this identity is that differentials must be alternating.
     
  9. Aug 18, 2007 #8

    quasar987

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    I like your answer :)
     
  10. Aug 18, 2007 #9

    Hurkyl

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    Oh, and there's a geometric picture too -- given two vectors, you want to combine them to form a bivector that represents the area swept out by your vectors. So this product too should satisfy v v = 0. And since 1-forms are dual to tangent vectors...
     
  11. Aug 18, 2007 #10

    mathwonk

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    license to steal: salary for answering the same question infinitely many times.
     
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