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I Differential forms as a basis for covariant antisym. tensors

  1. May 8, 2017 #1
    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 [tex] dx^{\mu_1} \otimes ... \otimes dx^{\mu_n}[/tex] with [tex] dx^{\mu_1} \wedge ... \wedge dx^{\mu_n}[/tex] of some covariant tensor we can extract its totally antisymmetric part
    [tex]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}."[/tex]

    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 [tex]T_{\mu_1 ... \mu_n}[/tex] 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?
     
  2. jcsd
  3. May 8, 2017 #2

    fresh_42

    Staff: Mentor

    ##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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