Differential forms as a basis for covariant antisym. tensors

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SUMMARY

Differential forms serve as the foundational basis for the vector space of totally antisymmetric covariant tensors. By substituting the standard basis of covariant tensors, represented as dx^{\mu_1} \otimes ... \otimes dx^{\mu_n}, with the wedge product basis dx^{\mu_1} \wedge ... \wedge dx^{\mu_n}, one can extract the totally antisymmetric part of a tensor T. The tensor T is arbitrary and does not need to be antisymmetric for this decomposition to be valid. This transformation shifts the interpretation of T from a multi-dimensional scheme of numbers to an element of the algebra of alternating differential forms.

PREREQUISITES
  • Understanding of differential forms and their properties
  • Familiarity with covariant tensors and their representations
  • Knowledge of the Graßmann algebra and its applications
  • Basic concepts of multilinear algebra
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  • Study the properties of alternating differential forms in depth
  • Explore the relationship between tensor algebra and Graßmann algebra
  • Learn about the process of extracting coordinates from alternating differential forms
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This discussion is beneficial for mathematicians, physicists, and students specializing in differential geometry, tensor analysis, and algebraic structures. It is particularly relevant for those working with covariant tensors and differential forms in theoretical contexts.

Physics_Stuff
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In a text I am reading (that I unfortunately can't find online) it says:

"[...] differential forms should be thought of as the basis of the vector space of totally antisymmetric covariant tensors. Changing the usual basis dx^{\mu_1} \otimes ... \otimes dx^{\mu_n} with dx^{\mu_1} \wedge ... \wedge dx^{\mu_n} of some covariant tensor we can extract its totally antisymmetric part
T= \frac{1}{n!}T_{\mu_1 ... \mu_n}\hspace{1pt} d x^{\mu_1} \wedge ... \wedge d x^{\mu_n}= \frac{1}{n!}T_{[\mu_1 ... \mu_n]}\hspace{1pt} d x^{\mu_1} \wedge ... \wedge d x^{\mu_n}."

What is the point here? Is T an arbitrary tensor with n covariant components, or must T already be antisymmetric in order for this expression to hold? In order to know the components T_{\mu_1 ... \mu_n} of T, so we can use the expression on the RHS above, we must already know what the tensor T looks like? Then, what is the point of such a decomposition?
 
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##T## is an arbitrary tensor. Until now, this doesn't say anything more than ##T## is a multi-dimensional scheme of numbers. In the first step, you say, these numbers represent coordinates. So the question is, according to which basis? As you answer "covariant multilinear forms", it means ##T## is interpreted according to a basis ##dx^{\mu_1} \otimes \ldots \otimes dx^{\mu_n}##. It is still the same scheme of numbers. Now you say "but my multilinear forms are alternating differential forms". This means you pass from the tensor algebra ##\mathcal{T}(V^*)## onto the homomorphic image of its Graßmann algebra ##\Lambda(V^*)##. It means, the basis vectors are now alternating differential forms and ##T_{\mu_1 \ldots \mu_n}## the coordinates of ##T## according to this basis. It is still the same scheme of numbers, however, interpreted as an element of the algebra of alternating differential forms (with a normalization factor).

Your question is as if you had asked, whether ##(1,2)## is a point, a line, a tangent, a slope, a linear mapping or a differential form. It is whatever you want it to be. The usual way to get there is of course the opposite direction: given an alternating differential form ##T##, what are its coordinates according to the basis ##dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_n}\;##?
 
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