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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?

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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

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    $\begingroup$ 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. $\endgroup$
    – J. Murray
    Mar 5, 2021 at 12:30
  • $\begingroup$ @J. Murray I am indeed learning about tetrads which led me to differential forms and wedge prodcuts haha $\endgroup$
    – TaeNyFan
    Mar 5, 2021 at 15:55

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