1
$\begingroup$

I'm trying to show $\text{d}(\alpha\wedge\gamma)=\text{d}\alpha\wedge\gamma+(-1)^p\alpha\wedge \text{d}\gamma$ for all $p$-forms $\alpha$ and $\gamma$, $\text{d}$ is exterior derivative. I want to know the proof using the following definition in book The Road to Reality by Penrose:

The coordinate definition of the exterior derivative of the $p$-form $\alpha$ is a $(p+1)$-form that is written $\text{d}\alpha$, which has components \begin{align*} (\text{d}\alpha)_{ab\ldots i}=\frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots i]}, \end{align*} (The notation looks a bit awkward here. The antisymmetrization---which is the key feature of the expression---extends across all $p+1$ indices, including the one on the derivative symbol.)

Here's my attempt at a proof, which is partially based off of a proof that I found online. Using the above definition for the component of a differential, we have \begin{align*} \text{d}(\alpha\wedge\gamma)_{ab\ldots i}&=\frac{\partial}{\partial x^{[a}}(\alpha\wedge\gamma)_{bc\ldots i]}\\ &=\frac{\partial}{\partial x^{[a}}\alpha_{[bc\ldots e}\gamma_{fg\ldots i]]} \end{align*} It's worth mentioning that the wedge product can very neatly be written using antisymmetrization of the components of $\alpha$ and $\gamma$ like so: \begin{align*} (\alpha\wedge\gamma)_{bc\ldots i}=\alpha_{[bc\ldots e}\gamma_{fg\ldots i]} \end{align*} This next step was the part that I borrowed off the internet: the inner brackets is redundant here because composition of permutations results in another permutation of indices. \begin{align*} \frac{\partial}{\partial x^{[a}}\alpha_{[bc\ldots e}\gamma_{fg\ldots i]]}=\frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots e}\gamma_{fg\ldots i]} \end{align*} Using the chain rule, \begin{align*} \frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots e}\gamma_{fg\ldots i]}=\frac{\partial\alpha_{bc\ldots e}}{\partial x^{[a}}\gamma_{fg\ldots i]}+\alpha_{bc\ldots e}\frac{\partial\gamma_{fg\ldots i]}}{\partial x^{[a}} \end{align*} I'm uncertain about the validity of the next part: the index $a$ been shifted to be between the indices $bc\ldots e$ and $fg\ldots i$. We have \begin{align*} \text{d}(\alpha\wedge\gamma)_{ab\ldots i}&=\frac{\partial\alpha_{bc\ldots e}}{\partial x^{[a}}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}\frac{\partial\gamma_{fg\ldots i]}}{\partial x^{a}}\\ &=\frac{\partial\alpha_{bc\ldots e]}}{\partial x^{[[a}}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}\frac{\partial\gamma_{fg\ldots i]]}}{\partial x^{[a}}\\ &=(\text{d}\alpha)_{[abc\ldots e}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}(\text{d}\gamma)_{afg\ldots i]}\\ &=(\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(\alpha\wedge \text{d}\gamma)_{bc\ldots eaf\ldots i} \end{align*} The last step is shifting the index $a$ over to the start of the subscript which results in a coefficient of $(-1)^p$ because of that is the order of permutation if there are $p$ element to the left of $a$. \begin{align*} (\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(\alpha\wedge \text{d}\gamma)_{bc\ldots eaf\ldots i}&=(\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(-1)^p(\alpha\wedge d\gamma)_{ab\ldots i}\\ &=(\text{d}\alpha\wedge\gamma+(-1)^p \alpha\wedge \text{d}\gamma)_{ab\ldots i} \end{align*} So is my proof valid, or incorrect even though the conclusion is correct?

Thank you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.