# Dimensionality of the space of p-forms

1. Nov 16, 2013

### center o bass

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

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

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?

2. Nov 16, 2013

### HallsofIvy

Staff Emeritus
Because we can take $\dfrac{n!}{p!(n-p)!}$ distinct matrices, representing these tensors in some coordinate system, having "1" for each of the independent components and "0" for all others. And every such tensor can then be written as a linear combination of those tensors. The number of independent basis vector is the number of independent components. The "basis vectors" you give are NOT all "completely anti-symmetric".

3. Nov 16, 2013

### center o bass

So what you're saying is that the combinatorical symbol is the dimension -in the space of anti-symmertric covariant tensors- while it's not in the space of ordinary covariant p-tensors. That males sense! Thanks! :)