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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 [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?
"[...] 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?