A p-form ##\alpha## is a completely anti-symmetric tensor. This means that the form is completely determined by a set of indicies ##\left\{i_1, i_2, \ldots, i_p\right\}## where ##i_1 < i_2 <...<i_p## since the other components can be found by permutation of these. So in ##n##- dimensions we have then(adsbygoogle = window.adsbygoogle || []).push({});

$$\frac{n!}{p!(n-p)!}$$

independent components of the tensor. Then it's stated that this is the dimensionality of the space of p-forms in n-dimensions. But why is this called the dimensionality? Even though the components ##A_{\mu_1 \cdots \mu_k \cdots \mu_l \cdots \mu_p}## and ##A_{\mu_1 \cdots \mu_l \cdots \mu_k \cdots \mu_p}## are dependent, the basis tensors ##e^{\mu_1}\otimes \cdots\otimes e^{ \mu_k } \otimes \cdots \otimes e^{\mu_l }\otimes \cdots \otimes e^{\mu_p}## and ##e^{\mu_1}\otimes \cdots\otimes e^{ \mu_l } \otimes \cdots \otimes e^{\mu_k }\otimes \cdots \otimes e^{\mu_p}## are not, and thus the corresponding terms in

$$\alpha = \alpha_{\mu_1 \mu_2 \cdots \mu_p} e^{\mu_1} \otimes e^{\mu_2} \cdots \otimes e^{\mu_p} $$

can not be written as linear combinations of eachother. I would think that the dimensionality would be equal to that of the number of independent basis vectors. Why is this not the case for forms?

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# Dimensionality of the space of p-forms

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