Background
I am trying to understand the following way of introducing differential k-forms.
We started with the exterior algebra and defined $\Lambda^{k}V:=(\Lambda_{k}V)^{*}=(V^{\otimes k}/<\otimes v>)^{*}$. I think I am more or less fine with the extiror algebra. Then there is the following:
My Question
How is $Y^{0}=(\Lambda_{k}V)^{*}$? Isn't $\Lambda_{k}V)$ a quotient of a of $V^{\otimes k}$? So how can this be a subset of $V^{\otimes k}$? My guess is that the argument goes like: first show $\{\text{functionals that vanish on} <\otimes^{k} v>\}\simeq Alt^{k}V$ , then $(\Lambda_{k}V)^{*}\simeq \{\text{functionals that vanish on} <\otimes^{k} v>\}$. So erverything is equal only up to isomorphism.
I don't get the last line. Isn't an element of $(\Lambda_{k}V)^{*}$ a map, say $\phi :\Lambda_{k}V\to\mathbb{R},v_{1}\wedge ...\wedge v_{k}\mapsto\phi(v_{1}\wedge ...\wedge v_{k})$. What I found on the internet is that $Alt^{k}V\simeq (\Lambda_{k}V)^{*}$, which would make a lot more sense to me.
According to the definition of $\Lambda_{k}V$ I would say that $\omega_{p}$ is a map that "eats" k-vectors, but then there is the following statement:
But there is also the following:
I have to say I am pretty confused about what follows right from the definition and what not.
- How is the $\omega_{p}(X_{1},...,X_{k})=\omega(X_{1}\wedge...\wedge X_{k})$ justified? If we would start with $\omega(X_{1},...,X_{k})$ I would say that by the universal property there is a linear function $\omega$ such that.... . But since we start off with $\omega(X_{1}\wedge...\wedge X_{k})$ how do I know that there is a k-linear alternating form that conicides with $\omega(X_{1}\wedge...\wedge X_{k})$? Does $\beta(X_{1},...,X_{n}):=\omega(X_{1}\wedge...\wedge X_{k})$ do the job?
I hope the questions make at least a bit sense. Thank you very much in advance.