# Differential forms and wedge product

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?

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.
• Since this is right, I'd add as a note that $\{\mathrm dx^a\}$ is specifically the dual basis to $\{\partial/\partial x^a\}$, which is induced by the coordinates $\{x^a\}$. In some contexts - especially in introductory treatments - it is common to only use bases which can be written this way, but there are choices of basis which are not induced by any coordinate system (see tetrads), which makes the $\mathbf e^a /\mathbf e_a$ notation slightly more general. Mar 5, 2021 at 12:30