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> [!definition] Definition. ([[exterior derivative]])
> Suppose $M$ is a [[smooth manifold]]. Let $\Omega^{r}(M)$ denote the [[vector space|space]] of [[differential form|differential]] $r$-[[differential form|forms]] on $M$.
>
> There are unique $\mathbb{R}$-[[linear map|linear]] operators $d:\Omega^{k}(M) \to \Omega^{k+1}(M)$ for all $k$, called **exterior derivatives** or **exterior differentials**, satisfying the following:
>
> 1. **(Specialization)** If $f \in \Omega^{0}(M)=C^{\infty}(M)$, then $df$ is precisely the [[differential of smooth maps between surfaces|differential]] of $f$, given by $df(X)=Xf$
> 2. (**Cochain condition**) $d \circ d \equiv 0$
> 3. **(Graded Leibniz)** $d(\omega \wedge \eta)=d \omega \wedge \eta + (-1)^{\text{deg }\omega} \omega \wedge d \eta$ for any two [[differential form|differential forms]] $\omega, \eta$.
>
> In particular, such conditions determine the local expression in coordinates $(x^{i})$: for $\omega=\omega_{I} \ dx^{I} \in \Omega^{k}(M)$, $d \omega = d\omega_{I} \wedge dx^{I} \in \Omega^{k+1}(M),$
> where $d\omega_{I}$ is the [[differential of a smooth map between smooth manifolds|differential]] of $\omega_{I}:U \to \mathbb{R}$. Existence then reduces to showing this is in fact coordinate-independent.
> [!basicproperties]
> - $d$ is a *local operator*: if $\omega$ and $\eta$ are two $n$-forms agreeing on a [[neighborhood]] $U$, $\omega |_{U}=\eta |_{U}$, then $d\omega |_{U}=d \eta |_{U}$.
^properties
Let $\omega \in \Omega^{r}(M)$. If $d\omega \in \Omega^{r+1}(M)$ exists, then the rules $(1), (2), (3)$ determine its local expression. Indeed, let $(U, (x^{i}))$ be a [[coordinate chart]] on $M$, with $\omega |_{U}=\omega_{I} \ dx^{I}$ for $\omega_{I} \in C^{\infty}(U)=\Omega^{0}(U)$. Then[^1] $\begin{align}
d\omega &= d ( \omega_{I} \ dx^{I} ) \\
&= \overbrace{ d \omega_{I} }^{ \text{differential (rule 1)} } \wedge dx^{I} + \omega_{I} \ \cancel{ d dx^{I} } ^{=0 \ (\text{Rule 2 - cochain condition})} \ \ (\text{Rule 3 - Leibniz}).\end{align}$
This determines $d\omega$ uniquely, if it exists. To show it exists, we need to
1. Verify that this claimed expression indeed satisfies $(1)$, $(2)$, $(3)$. This is done by direct calculation. For $(3)$, recall that [[equality of mixed partials for C2 functions|mixed partials commute]] for smooth functions; this will produce cancellations.
2. Verify that none of this depends on the choice of coordinates $(U, (x^{i}))$. This will be enough because $d$ is manifestly a local operator (as it is defined in terms of partial derivatives).
**1.**
(1) is clear.
For (3), suppose $\omega=\omega _I dx^{I}$ and $\eta=\eta_{J}dx^{J}$, so that $\omega \wedge \eta=\omega_{I} \eta_{J}dx^{I} \wedge dx^{J}$. Then compute $\begin{align}
d(\omega \wedge \eta) &= d(\omega_{I} \eta_{J} dx^{I} \wedge dx^{J}) \\
&= d(\omega _{I} \eta_{J}) \wedge dx^{I} \wedge dx^{J} \\
&= \frac{ \partial (\omega_{I} \eta_{J}) }{ \partial x^{j} } dx^{j} \wedge dx^{I} \wedge dx^{J} \\
&= \left( \frac{ \partial \omega_{I} }{ \partial x^{j} } \eta_{J} + \omega_{I} \frac{ \partial \eta_{J} }{ \partial x^{j} } \right) dx^{j} \wedge dx^{I} \wedge dx^{J} \\
&= \overbrace{ \frac{ \partial \omega_{I} }{ \partial x ^{j}} \eta_{J} dx^{j} \wedge dx^{I} \wedge dx^{J} }^{ d \omega \wedge \eta } \\
&+ \omega _{I}\frac{ \partial \eta_{J} }{ \partial x^{j} } dx^{j} \wedge dx^{I} \wedge dx^{J} .
\end{align}$
In the second term, get $\eta \wedge d\omega$ once $dx^{j}$ is swapped past $|I|=\text{deg }\omega \text{-many}$ factors, incurring the coefficient $(-1)^{\text{deg }\omega }$.
For (2), start with $\omega=\omega_{I}dx^{I}$ and apply $d$: $d\omega=d\omega_{I} \wedge dx^{I}=\partial_{j}\omega_{I}dx^{j} \wedge dx^{I}.$
Apply $d$ again: $d d \omega=d(\partial_{j}\omega_{I}dx^{j} \wedge dx^{I})=TODO$ (use that mixed partials commute because everything is smooth)
**2.** Let $\tilde{d}$ be 'another $d