
#1
Aug1207, 02:41 PM

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Is a function from R^n to R^m for aritrary m a considered a 0form on R^n, or does 0form refers only to functions from R^n to R ?




#2
Aug1207, 02:56 PM

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What's the definition of a pform on R^n ?




#3
Aug1207, 03:27 PM

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In particular, a 0form 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 0form would simply be a real number. But more exotic things are possible, and sometimes even fruitful. 



#4
Aug1207, 03:48 PM

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What's a 0form and what's not?
In 'Calculus on manifolds', Spivak defines a kform 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 0form we mean a function f. I was 90% sure he meant a function f:R^n>R but wanted to make sure. 



#5
Aug1207, 04:08 PM

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So a 0form 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 0form on the tangent space at P, which is the same thing as an element of R. 



#6
Aug1707, 11:52 PM

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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?




#7
Aug1807, 12:22 AM

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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 firstorder differential information  thus you want dx dx = 0. An immediate consequence of this identity is that differentials must be alternating. 



#9
Aug1807, 01:19 AM

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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 1forms are dual to tangent vectors...



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