I Why are scalars and dual vectors 0- and 1-forms?

George Keeling

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I am told: "A differential p-form is a completely antisymmetric (0,p) tensor. Thus scalars are automatically 0-forms and dual vectors (one downstairs index) are one-forms."

Since an antisymmetric tensor is one where if one swaps any pair of indices the value of the component changes sign and 1) there are no indices to swap on a scalar and 2) on a dual vector swapping something with itself is not swapping, how are they automatically 0- and 1-forms? I have no problem with higher forms.
 

fresh_42

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I am told: "A differential p-form is a completely antisymmetric (0,p) tensor. Thus scalars are automatically 0-forms and dual vectors (one downstairs index) are one-forms."

Since an antisymmetric tensor is one where if one swaps any pair of indices the value of the component changes sign and 1) there are no indices to swap on a scalar and 2) on a dual vector swapping something with itself is not swapping, how are they automatically 0- and 1-forms? I have no problem with higher forms.
If you forget the swapping process as such and define it correctly as the quotient algebra of the ##n-##fold tensor product by its ideal generated by pairs of equal vectors ##V^{\otimes n}/\langle v \otimes v \rangle##, then we get ##V^0=\mathbb{F}\; , \; V^1=V## as in both cases ## \langle v\otimes v\rangle =\langle \emptyset \rangle = \{\,0\,\}##.

In other words: The condition which you called swapping is an empty condition. The empty set generates the zero ideal, and thus the quotient algebra is the vector space itself.

Even shorter: no condition, no restriction.
 

WWGD

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In a non-algebraic sense, a p-form I just a p-linear map. So a 0-form is 0-linear, I e. linear in no arguments. These are kind of annoying special cases that must be addressed. A dual is a linear map on vector s.
 

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