I am working towards understanding the wedge product of two vectors.
Here is what I have so far:
Let $V$ be an $n$-dimensional vector space over $\mathbb R$ (or $\mathbb C$, I'm not sure it makes any difference) and let $\Lambda^k (V)$ be the set of $k$-vectors.
A $k$-vector is a linear combination of wedge products of $k$ $1$-basis-vectors $e_i$ and a $1$-vector is just a vector $v \in V$.
This is what the notation means.
It is possible to define the wedge product in several different (probably equivalent) ways.
(1) One way of introducing it is by merely stating its properties and then subsequently constructing $\Lambda^2(V)$ as a quotient quite similar to the tensor product.
(2) Given $u,v \in V$ the exterior product $u \wedge v \in \Lambda^2 (V)$ is the linear map to $\mathbb R$ (or $\mathbb C$ ) which maps the alternating bilinear form $B$ to $(u \wedge v)(B) = B(u,v)$. (see e.g. here)
(3) Or we can say that if $\alpha, \beta \in \Lambda^1 (V)$ are two $1$-forms then their wedge product is defined by $$ (\alpha \wedge \beta) (u,v) := \alpha (u) \beta(v) - \alpha (v) \beta (u)$$
(see e.g. here)
The third definition is the same as the first. Or so it seems to me.
Now some of them use vectors and some of them use covectors. I just need a quick sanity check:
Please can you tell me if the following is true?
Since $\Lambda^k (V)$ is finite dimensional and since finite dimensional vector spaces are isomorphic to their dual it does not matter whether we define $\Lambda (V)$ in terms of vectors or covectors.