Differential forms and wedge product

Question

In Sean Carroll's GR book, the differential $p$-form is defined as a $(0,p)$ tensor that is completely antisymmetric, which I would think is something like $$\frac{1}{2!}(t_{ab}-t_{ba})\textbf {e}^a \otimes \textbf{e}^b.$$ In Zee's GR book, part IX.7, he defined the $p$-form in a similar way, which he writes as $$\frac{1}{2!}t_{ab}dx^a dx^b.$$ Zee also wrote that $dx^a dx^b$ is the wedge product such that $dx^a dx^b=-dx^b dx^a$.

I have troubles reconciling these two definitions. Is $dx^a$ the basis vector $\textbf{e}^a$? How can we start with either one of the expression and manipulate it so that we get the other expression?

Answer 1

To answer my own question, $dx^a$ is indeed the dual basis vectors for one-forms, i.e. $dx^a \equiv \textbf{e}^a$.

The wedge product for one-forms is defined as $$\textbf{e}^a\wedge \textbf{e}^b = \textbf{e}^a\otimes\textbf{e}^b-\textbf{e}^b\otimes\textbf{e}^a.$$

Using this on Zee's definition, we get $$ \begin{align} \frac{1}{2!}t_{ab}dx^adx^b\equiv\frac{1}{2!}t_{ab}\textbf{e}^a \wedge \textbf{e}^b \\ =\frac{1}{2!}t_{ab}(\textbf{e}^a \otimes\textbf{e}^b -\textbf{e}^b\otimes\textbf{e}^a) \\ =\frac{1}{2!}(t_{ab} -t_{ba}) \textbf{e}^a \otimes\textbf{e}^b \end{align}$$ where indices are relabelled in the last step.

References: Schutz, Geometrical Methods of Mathematical Physics, pg. 117