Undergrad Differential forms as a basis for covariant antisym. tensors

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Differential forms serve as the basis for totally antisymmetric covariant tensors, allowing for the extraction of their antisymmetric parts. The tensor T can be arbitrary, and its components T_{\mu_1 ... \mu_n} are interpreted based on the chosen basis of differential forms. This transition from tensor algebra to the algebra of alternating differential forms highlights the flexibility in interpreting T's representation. The discussion emphasizes that the nature of T depends on the context and basis used, illustrating the versatility of mathematical structures. Ultimately, the decomposition into differential forms provides a framework for understanding tensors in a more nuanced way.
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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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I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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