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

Gold Member
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.

Mentor
2022 Award
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.

George Keeling
Gold Member
Even shorter: no condition, no restriction.
That was the best bit!

Mentor
2022 Award
That was the best bit!
Thanks, but it had to be prepared. Otherwise you might have felt not taken seriously.