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

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Discussion Overview

The discussion centers on the classification of scalars and dual vectors as 0- and 1-forms within the context of differential forms and antisymmetric tensors. Participants explore the definitions and implications of these classifications, addressing both theoretical and conceptual aspects.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that scalars are automatically 0-forms and dual vectors are 1-forms due to their nature as completely antisymmetric tensors.
  • Others question this classification by highlighting that scalars have no indices to swap, and thus challenge the reasoning behind defining them as 0-forms.
  • A participant proposes an alternative definition of antisymmetric tensors using the quotient algebra of the tensor product, suggesting that the condition of swapping is effectively empty for scalars and dual vectors.
  • One participant mentions that a 0-form can be viewed as a 0-linear map, which raises further questions about the nature of these special cases.

Areas of Agreement / Disagreement

Participants express differing views on the classification of scalars and dual vectors as forms, with no consensus reached on the validity of the arguments presented.

Contextual Notes

The discussion involves complex definitions and assumptions related to antisymmetric tensors and the properties of forms, which may not be fully resolved within the thread.

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.
 
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George Keeling said:
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.
 
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fresh_42 said:
Even shorter: no condition, no restriction.
That was the best bit!
 
George Keeling said:
That was the best bit!
Thanks, but it had to be prepared. Otherwise you might have felt not taken seriously. :wink:
 
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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