Question: I can't understand why the dimension of the space of anti-symmetric $p$-tensor of a vector space $V$ is $n \choose p$ ($dimV=n$). I thought this only count the basis like $\omega(e_{i_1}, e_{i_2}\cdots)$, but doesn't $\omega(e_{i_1}, e_{i_1}\cdots)$ need to be in the basis?
Some definition:
Tensor: a multilinear map $T:V\times V\cdots V \to \mathbb{R}$
Anti-symmetric $p$-tensor: the tensor $\omega$ such that $\omega(x_1, \cdots ,x_s, \cdots ,x_l, \cdots , x_k)=-\omega(x_1, \cdots ,x_l, \cdots ,x_s, \cdots , x_k)$, with $p$ $V$s times together as the left of the arrow.
In the context of anti-symmetric p-tensors, the dimension (np) is based on combinations rather than permutations. This means that repeating indices, as in (ei1,ei1⋯), are not considered distinct elements in the basis.
For example, in a 2-dimensional vector space V with basis (e1, e2), the basis for an anti-symmetric 2-tensor (also known as a 2-form) would include terms like e1∧e2 but not e1∧e1 since it is zero due to anti-symmetry.
The (np) dimension accounts for choosing p distinct elements from n without regard to the order, avoiding repetitions in the basis.