Outer product of two one forms.

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The discussion focuses on determining the components of the outer product of two one-forms, f = (1,1,0,0) and g = (-1,0,1,0). Participants clarify that the "outer product" in this context refers to the tensor product rather than the wedge product. The tensor product is expressed in component notation as pij = uivj, leading to the formation of a second rank tensor. The components of the resulting tensor from the outer product are derived from the multiplication of the components of the one-forms. Overall, the conversation emphasizes the distinction between the tensor and wedge products in the context of one-forms.
ronblack2003
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Given two one forms f = (1,1,0,0,) and g=(-1,0,1,0): what are the components of f(x)g ... would appreciate any help.
 
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Outer Product of two One Forms

Given two one forms f = (1,1,0,0) and g = (-1,0,1,0):
What would be the Components of f(outer product)g be?
would appreciate any help.
 
I'm not quite sure if the "outer product" is the wedge product or the tensor product. Wikipedia seems to think it's the former. See

This Link

The wedge product of u ^ v is

u (tensor) v - v (tensor) u

The intent here is to anti-symmetrize the tensor product, The formula above works as written only for rank 1 tensors (but that's what you have).

The tensor product p = u (tensor) v in component notation for rank 1 tensors is:

pij = uivj

Thus (1,2,3) (tensor) (4,5,6) is the second rank tensor (you can think of it as an array)

1*4 ,1*5, 1*6
2*4, 2*5, 2*6
3*4, 3*5, 3*6

Hope this helps
 
pervect said:
I'm not quite sure if the "outer product" is the wedge product or the tensor product.

Yes. Its the tensor product.

Let "@" be the tensor product and let f = f_u w^v where f_u are components of f and w^u are a coordinate basis 1-forms. Same with g. Then let h = f@g. Then

h = (f_u e^u)@(g_u v e^v) = f_u g_v (w^u@w^v)

Therefore f_u g_v are the components of f@g. w^u@w^v are the covariant basis tensors for the outer product.

Pete
 
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