Does a $p$-form eat $p$-vectors or $p$ number of vectors?

Question

A bilinear form is another term for a $2$-form. So does it eat $2$ distinct vectors or a single $2$-vector?

Answer 1

To elaborate on Zhen Lin's comment: either viewpoint of a bilinear form is okay due to the universal property of the tensor product.

Given a vector space $V$ with field of scalars $k$, I've usually seen a bilinear form defined as a bilinear map $B : V \times V \to k$, so a function that eats two vectors as in your first definition.

However, by the universal property of the tensor product, there is a unique linear map $\tilde{B} : V \otimes_k V \to k$ such that $B = \tilde{B} \circ \pi$ where $\pi$ is the map $\pi : V \times V \to V \otimes_k V$, $(v_1, v_2) \mapsto v_1 \otimes v_2$. So $\tilde{B}$ eats $2$-vectors, as in your second definition.

Since this association $B \leftrightarrow \tilde{B}$ is a bijection, we can in a sense identify $B$ and $\tilde{B}$ and are thus free to think about a bilinear form as a map $V \times V \to k$ or $V \otimes_k V \to k$. For more on this see section 10.4 of Dummit and Foote.

Answer 2

A bilinear form on a vector space $V$ over a field $\Bbb{F}$ is a bilinear map $B:V\times V\to \Bbb{F}$ which eats a pair of vectors to give a scalar according to $(v,w)\mapsto B(v,w)$. This can be generalized.