> [!definition] > > Let $X$ be an [[n-Manifold|n-manifold]] and $\omega \in \Lambda^k(TX)$ be a [[Differential Form|differential form]]. Let $(U, \varphi = (x_1, \cdots, x_n))$ be a coordinate [[Atlas|chart]]. If > $ > \omega = \sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k}dx_{i_1} \wedge \cdots \wedge dx_{i_k} > $ > and define > $ > d\omega = \sum_{i_1 < \cdots < i_k}d\omega_{i_1, \cdots, i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} = \varphi^*d(\varphi^{-1*}\omega) > $ > where $d\omega_{i_1, \cdots, i_k}$ is the [[Differential|differential]] identified as a 1-form. Then $d\omega$ is the same independent of the choice of chart, known as the **exterior derivative**. > > *Proof*. If $(V, \psi)$ is another chart, then the local representation of $d(\varphi^{-1*}\omega)$ in the coordinates of $\psi$ is $(\varphi \circ \psi^{-1})^*d(\varphi^{-1*}\omega)$. Since the exterior derivative on the Euclidean space commutes with pullbacks, > $ > (\varphi \circ \psi^{-1})^*d(\varphi^{-1*}\omega) = d((\varphi \circ \psi^{-1})^*\varphi^{-1*}\omega) = d(\psi^{-1*}\omega) > $ > [!theorem] > > Let $X$ be an $n$-manifold, then there exists a unique family of operators > $ > d: \Omega^k(X) \to \Omega^{k+1}(X) > $ > such that: > 1. $d$ is $\real$-linear. > 2. If $\omega \in \Omega^k(X)$ and $\eta \in \Omega^\ell(X)$, then $d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^{k} \omega \wedge d\eta$. > 3. $d^2 = 0$. > 4. If $f \in \Omega^0(X) = C^\infty(X)$, $df$ is the [[Differential|differential]]. > > *Proof*. Let $d$ be a linear operator satisfying the above properties. Let $V \subset \subset U \subset X$ be open and $\omega, \eta \in \Omega^k$ be $k$-forms such that $\omega|_U = \eta|_U$. Let $\psi \in C^\infty(U)$ be a bump function such that $\psi|_V = 1$ and $\supp{\psi} \subset U$. By the product rule, > $ > 0 = d[\psi(\omega - \eta)] = d\psi \wedge \underbrace{(\omega - \eta)}_{0} + \psi d(\omega - \eta) > $ > so $d(\omega - \eta) = 0$ on $V$, so $\omega|_U = \eta|_U$. > > From here, expanding in coordinates using a bump function yields the desired result. # On Euclidean Space > [!definition] > > Let $U \subset \real^n$ be [[Open Set|open]] and $\omega \in \Omega^k(U)$ be a [[Differential Form|differential form]]. If > $ > \omega = \sum_{i_1 < \cdots < i_k}\omega_{i_1, \cdots, i_k}dx^1 \wedge \cdots \wedge dx^k = \sum_{I}\omega_{I}dx^I > $ > where $I$ represents tuples of the form $(i_1, \cdots, i_k)$ where $i_j < i_{j+1}$ for all $j < n$. Define > $ > d\omega = \sum_{I}d\omega_I \wedge dx^{I} > $ > as the **exterior derivative** of $\omega$, $d$ itself is a $\real$-[[Linear Function|linear]] mapping > $ > d: \Omega^k(U) \to \Omega^{k+1}(U) > $ > [!theorem] > > Let $\omega \in \Omega^k(U)$ and $\eta \in \Omega^\ell(U)$, then > $ > d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^k \omega \wedge d\eta > $ > > *Proof*. Suppose that $I = (i_1, \cdots, i_k)$ is **not** necessarily a strictly increasing sequence. If $I$ has duplicate entries, then $dx^I = 0$. Otherwise, if $\sigma \in S_k$ is a permutation such that $\sigma(I)$ *is* strictly increasing, then > $ > d(udx^I) = \sgn(\sigma)d(udx^{\sigma(I)}) = \sgn(\sigma)du \wedge dx^{\sigma(I)} = du \wedge dx^I > $ > thus $d(udx^I) = du \wedge dx^I$ for *any* multi-index. > > By linearity, assume without loss of generality that $\omega = udx^I$ and $\eta = vdx^J$ where $I, J$ are strictly increasing multi-indices. From here, > $ > \begin{align*} > d(\omega \wedge \eta) &= d((udx^I) \wedge (vdx^J)) \\ > &= d(uv dx^I \wedge dx^J) \\ > &= (udv + vdu) \wedge dx^I \wedge dx^J \\ > &= udv \wedge dx^I \wedge dx^J + vdu \wedge dx^I \wedge dx^J \\ > &= (-1)^{k}udx^I \wedge dv \wedge dx^J + du \wedge dx^I \wedge vdx^J \\ > &= (-1)^k \omega \wedge d\eta + d\omega \wedge \eta > \end{align*} > $ > [!theorem] > > For any $\omega \in \Omega^k(U)$, $d^2\omega = 0$. > > *Proof*. If $f \in C^2(U)$, then > $ > \begin{align*} > d^2f &= d(df) = d\braks{\part{f}{x^i}dx^i} \\ > &= \frac{\partial f}{\partial x^i \partial x^j}dx^j \wedge dx^i = 0 > \end{align*} > $ > where terms with duplicate indices $(dx^i \wedge dx^i)$ become zero, and since $\part{f}{x^i\partial x^j} = \part{f}{x^j\partial x^i}$ and $dx^j \wedge dx^i = -dx^i \wedge dx^j$, opposite indices cancel out. > > Now for any increasing multi-index $I$ of length $k$, by the product rule, > $ > \begin{align*} > d(dx^I) &= d[dx^{i_1} \wedge dx^{i_2} \wedge \cdots \wedge dx^{i_k}] \\ > &= d^2x^{i_1} \wedge (dx^{i_2} \wedge \cdots \wedge dx^{i_k}) - dx^{i_1} \wedge d(dx^{i_2} \wedge \cdots \wedge dx^{i_k}) > \end{align*} > $ > where $dx^{i_2} \wedge \cdots \wedge dx^{i_k} = dx^J$ for a multi-index of length $k - 1$. By induction, $d^2$ vanishes on all the bases. > > From here, by the product rule, if $\omega = udx^I$, then > $ > \begin{align*} > d^2\omega &= d(d(udx^I)) = d(du \wedge dx^I) \\ > &= d^2u \wedge dx^I - du \wedge d^2x^I = 0 > \end{align*} > $ > [!theorem] > > Let $V \subset \real^m$ be an open set in another space, $F: V \to U$ be a [$C^1$](Space%20of%20Continuously%20Differentiable%20Functions) map, and $\omega \in \Omega^k(U)$. Then > $ > d(F^*\omega) = F^*d\omega > $ > *Proof*. Since the exterior derivative and the [[Pullback of Sections|pullback]] are $\real$-linear, assume without loss of generality that $\omega = udx^I = udx^{i_1} \wedge \cdots \wedge dx^{i_k}$. From here, > $ > \begin{align*} > F^*d\omega &= F^*du \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_k} \\ > &= d(u \circ F) \wedge d(x^{i_1} \circ F) \wedge \cdots \wedge d(x^{i_k} \circ F) \\ > &= d((u \circ F) \cdot d(x^{i_1} \circ F) \wedge \cdots \wedge d(x^{i_k} \circ F)) \\ > &= d(F^*\omega) > \end{align*} > $