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Elementary Differential Forms Question 
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#1
Apr1912, 07:18 PM

P: 56

Let me preface by saying I am a physics major. So I am coming at differential forms from the perspective of physics, i.e. work, flows, em fields, etc.
My question is this. My understanding is that a basic 1form dx, dy, or dz takes a vector v = (v1,v2,v3) and gives back the corresponding component of that vector, i.e. the size of that vector's projection onto the corresponding coordinate axis. The complete 1form, then, such as phi = Adx + Bdy + Cdz, is the sum of multiples of these projection sizes, where the coefficients are functions of x,y,z. My confusion is regarding which vector space these projections "occur" in: R^3, the tangent space or the dual space? In any case, how is it possible to project onto the x, y, or z axes within the tangent or dual spaces? Wouldn't the "axes" in the dual space be the "dxaxis", etc.? Or am I just confusing myself unnecessarily by assuming that the 1form and the projection (or the input vector and the projection) have to be in the same space? Thanks for your help. 


#2
Apr1912, 07:22 PM

P: 3,014



#3
Apr1912, 08:15 PM

P: 367

the "projections" occur as projections onto tangent space bases. that is, {e1,..., en} situated at a point p



#4
Apr2012, 06:26 AM

Sci Advisor
P: 1,719

Elementary Differential Forms Question
For instance dx,dy,and dz pick out the x,y, and z components of tangent vectors to a 3 dimensional coordinate system. In Euclidean space this can get confusing because the tangent space to Euclidean space can be naturally identified with Euclidean space itself and often the distinction is overlooked. Try working with 1 forms on a surface. It will be less confusing. 


#5
Apr2012, 06:50 AM

Sci Advisor
P: 2,951

One should be careful calling this kind of operation a "projection". A real projection requires a metric, and one forms naturally act on vectors without need for metric.
The easiest way to see that the one forms dx, dy, and dz don't orthogonally project vectors is to consider the case where the basis vectors are not orthonormal. Consider, if they are skewed, your components for your vectors are no longer the result of orthogonal projections, but dx, dy, and dz always pick out components. 


#6
Feb1313, 11:15 PM

P: 56

First, muchbelated thankyou's for taking the time to reply!
Thanks again for the insight! 


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