I'd like to understand how to evaluate a differential form on vector fields, and how to 'embed' exterior algebra into tensor algebra.
Some people define differential forms as alternating tensor fields, the others as sections of 'exterior algebra bundle'. In the former case, it is clear how such a field acts on vector fields. I am not sure how it works in the second case and what is the interplay between these two definitions. Here is what I mean.
Let $V^*$ be the dual space of the $\mathbb{F}$-vector space $V$, where $\mathbb{F}$ is either $\mathbb{R}$, or $\mathbb{C}$. Consider the tensor algebra $Tensor(V^*)$ and its quotient by the two-sided ideal $I$ generated by all elements of the form $x\otimes x$ for $x\in V^*$ together with the natural projection $\pi$, i.e. corresponding exterior algebra $\pi:Tensor(V^*)\to\Lambda(V^*)=Tensor(V^*)/I$. The exterior algebra then inherits the multiplication from the tensor product which is called the wedge product, hence $\pi(a)\wedge\pi(b):=\pi(a\otimes b)$.
The other part of the story are alternating tensors. There is an endomorphism of the tensor algebra, called the alternation map $Alt$, which is actually a projection with the kernel $Ker(Alt)=I$, and the image of $Alt$ are precisely alternating tensors $AltTen(V^*)$. Therefore, one has the vector space isomorphism $\overline{Alt}:\Lambda(V^*)\cong AltTen(V^*)$ which is given by $\pi(a)\mapsto Alt(a)$.
This construction works also in the case if we start with $TM$ instead of $V$, where $TM$ is the tangent bundle to a manifold $M$. In that situation, sections of the 'exterior algebra bundle' are differential forms.
Let $dx^1,dx^2\in T^*M$ such that $dx^i(\partial_j)=\delta^i_j$, where $x^i:M\to\mathbb{R}$ are coordinate functions and $d$ is the exterior derivative.
- People write $dx^1\wedge dx^2$. Does this mean $\pi(dx^1)\wedge\pi(dx^2)$, and hence $\pi(dx^1\otimes dx^2)$?
- How such a differential form $dx^1\wedge dx^2$ acts on $(\partial_1,\partial_2)$? Is this correct; $\pi(dx^1)\wedge\pi(dx^2) (\partial_1,\partial_2)=(dx^1\otimes dx^2+I)(\partial_1,\partial_2)=dx^1(\partial_1)dx^2(\partial_2)+I(\partial_1,\partial_2)=1+0?$
- If one uses the isomorphism $\overline{Alt}$ the form $dx^1\wedge dx^2$ corresponds to $\frac{1}{2}(dx^1\otimes dx^2-dx^2\otimes dx^1)$ which evaluates on $(\partial_1,\partial_2)$ as $\frac{1}{2}$. This seems to me as the correct way because $\frac{1}{2}(dx^1\otimes dx^2-dx^2\otimes dx^1)$ is something like 'pure' representative of the class $\pi(dx^1\otimes dx^2)$.
- If $\pi(dx^1),\pi(dx^2)\in\Lambda(T^*M)$ we may consider their product which is $\pi(dx^1\otimes dx^2)$, which in turn corresponds to $\frac{1}{2}(dx^1\otimes dx^2-dx^2\otimes dx^1)$ under the map $\overline{Alt}$. Here is the wedge product defined differently. According to that definition $dx^1\wedge dx^2=dx^1\otimes dx^2-dx^2\otimes dx^1$.
Thank you for any comments.