My question is if like the inner product of 2-rank tensor (matrix) an a vector gives another vector, a 2-form, since it would be a 2-rank covariant antisymmetric tensor, i can make the same operation. I'm right? if i am, how can write the last inner product on the differential form aproach?
I'm autodidact on this topic and i could have some misconcepstions, feel free to correct me, and sorry for my english, thanks.
(note: in my numbers for tensor rank, I do not know the convention for ordering the numbers, so I may have them backwards)
A rank (0,2) tensor can be contracted with a vector — a rank (1,0) tensor — to produce a rank (0,1) tensor: that is, a 1-form.
A rank (1,1) tensor can be contracted with a vector to produce a vector.
In coordinates, matrices as you are thinking of them correspond specifically to (1,1) tensors — representing (0,2) tensors or (2,0) tensors as a matrix in the usual way is an abuse of notation. E.g. the matrix form of a (0,2) tensor should have two horizontal dimensions (i.e. a row of row vectors), not one horizontal and one vertical dimension.
If we write a 2-form $F$ as a bilinear functional, its contraction with a vector $v$, which I will write as $Fv$, is the 1-form given by the linear functional
$$ Fv(w) = F(v,w)$$