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Prove that, for any $k$-form $\omega\in\Omega^k(M)$ the exterior derivative $\text{d}:\Omega^{\cdot}(M) \to \Omega^{\cdot} (M)$ of $\omega$ can be explicitly written as:

$$\text{d}\omega(X_1, ..., X_{k+1})=\sum_{i=1}^{k+1}(-1)^{i-1}X_i\omega(X_1, ..., \hat{X}_i, ...,X_{k+1})+\sum_{1\leq i<j\leq k+1}(-1)^{i+j}\omega([X_i, X_j], X_1, ..., \hat{X}_i, ...,\hat{X}_j, ..., X_{k+1})$$

(where $\hat{X}_i$ means that $X_i$ is ommited, e.g.: $\omega(X_1, \hat{X}_2, X_3)=\omega(X_1, X_3)$)

Tip: First show that it is enough to consider $X_i=\frac{\partial}{\partial x_i}$.

Since $\text{d}$ is uniquely defined by the following properties:

i) $\mathbb{R}(M)$-linearity

ii) $\text{d}(\Omega^k(M))\subset \Omega^{k+1}(M)$

iii) $\text{d}(\omega)(X_1)=X_1\omega$ for every $\omega\in \Omega^{0}(M)$

iv) $\text{d}(\omega\wedge\sigma)=\text{d}(\omega)\wedge\sigma+(-1)^k\omega\wedge\text{d}(\sigma)$, for all $\omega \in \Omega^{k}(M)$, $\sigma \in \Omega^{\cdot}(M)$

v) $\text{d}^2 \equiv 0$

I need to prove that the right-hand side satisfies the same properties. I've already managed to do the first three in a reasonably simple way, but the last two seem like a ridiculous amount of work, even using the tip.

I feel like there is some simplification I'm not seeing. Is there a simple way to do this? Thanks!

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rmdmc89
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1 Answers1

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Define the operator $\phi$ by:

$$\phi(X_1, ..., X_{k+1}):=\sum_{i=1}^{k+1}(-1)^{i-1}X_i\omega(X_1, ..., \hat{X}_i, ...,X_{k+1})+\sum_{1\leq i<j\leq k+1}(-1)^{i+j}\omega([X_i, X_j], X_1, ..., \hat{X}_i, ...,\hat{X}_j, ..., X_{k+1})$$

Try to prove the following: if $\omega=f\text{d}x_{i_{1}}\wedge...\wedge\text{d}x_{i_{k}}\in\Omega^k(M)$, then:

$$\phi(\omega)=\sum_{j=1}^{n}\frac{\partial f}{\partial x_{j}}\text{d}x_{j}\wedge\text{d}x_{i_{1}}\wedge...\wedge\text{d}x_{i_{k}}$$

[Hint: notice that if you already know that $\phi(\omega)$ is a $(k+1)$-form, then it is enough to prove that $\phi(\omega)(\frac{\partial}{\partial x_j}, \frac{\partial}{\partial x_{i_{1}}}, ..., \frac{\partial}{\partial x_{i_{k}}})=\frac{\partial f}{\partial x_{j}}$ for every $j\neq i_1, ..., i_{k}$.]

Once you have that, by linearity you can find $\phi$ for a general $\omega$ in terms of covectors. Then it is much simpler to prove that $\phi^2\equiv 0$ and that $\phi(\omega\wedge\sigma)=\phi(\omega)\wedge\sigma+(-1)^k\omega\wedge\phi(\sigma)$.

rmdmc89
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