[proofplan] We compare the coordinate formula on the overlap of two coordinate charts. The coefficient functions and the coordinate $1$-forms transform by the chain rule. After substituting these transformations into the formula for $d\omega$, the first-derivative terms give exactly the formula in the second chart, while the second-derivative terms vanish because second partial derivatives are symmetric and wedge products are antisymmetric. This proves equality on overlaps, so the local definitions glue to a global smooth $(k+1)$-form, and linearity follows directly from the formula. [/proofplan] [step:Write the coordinate formula using alternating coefficients] Let $(U,x)$ be a coordinate chart on $M$, where \begin{align*} x:U &\to x(U)\subseteq \mathbb{R}^n,\\ p &\mapsto (x_1(p),\dots,x_n(p)). \end{align*} On $U$, write $\omega$ in the equivalent alternating-index form \begin{align*} \omega\big|_U = \frac{1}{k!} \sum_{i_1,\dots,i_k=1}^n a_{i_1\cdots i_k}\, dx_{i_1}\wedge \cdots \wedge dx_{i_k}, \end{align*} where each coefficient \begin{align*} a_{i_1\cdots i_k}:U\to \mathbb{R} \end{align*} is smooth and the family $a_{i_1\cdots i_k}$ is alternating in its indices. Define the coordinate expression \begin{align*} D_x\omega = \frac{1}{k!} \sum_{j,i_1,\dots,i_k=1}^n \partial_{x_j}a_{i_1\cdots i_k}\, dx_j\wedge dx_{i_1}\wedge \cdots \wedge dx_{i_k}. \end{align*} This agrees with the increasing-index formula in the statement, because the alternating coefficients encode all permutations of each increasing multi-index. [guided] The increasing-index formula is concise, but the transformation calculation is cleaner if all indices are allowed and the coefficients are alternating. Thus, in the chart $(U,x)$, we write \begin{align*} \omega\big|_U = \frac{1}{k!} \sum_{i_1,\dots,i_k=1}^n a_{i_1\cdots i_k}\, dx_{i_1}\wedge \cdots \wedge dx_{i_k}. \end{align*} Here each coefficient is a smooth function \begin{align*} a_{i_1\cdots i_k}:U\to \mathbb{R}, \end{align*} and the coefficient family is alternating: interchanging two indices changes the sign, and repeated indices give coefficient $0$. With this notation, the [coordinate formula for the exterior derivative](/theorems/3564) becomes \begin{align*} D_x\omega = \frac{1}{k!} \sum_{j,i_1,\dots,i_k=1}^n \partial_{x_j}a_{i_1\cdots i_k}\, dx_j\wedge dx_{i_1}\wedge \cdots \wedge dx_{i_k}. \end{align*} The factor $1/k!$ corrects for summing over all permutations of each ordered increasing multi-index. This is the same expression as the statement's formula, only written in a notation better suited for checking invariance under coordinate change. [/guided] [/step] [step:Express the coefficients after changing coordinates] Let $(V,y)$ be another coordinate chart, with \begin{align*} y:V &\to y(V)\subseteq \mathbb{R}^n,\\ p &\mapsto (y_1(p),\dots,y_n(p)). \end{align*} Set $W:=U\cap V$. On $W$, write \begin{align*} \omega\big|_W = \frac{1}{k!} \sum_{\alpha_1,\dots,\alpha_k=1}^n b_{\alpha_1\cdots \alpha_k}\, dy_{\alpha_1}\wedge \cdots \wedge dy_{\alpha_k}, \end{align*} where \begin{align*} b_{\alpha_1\cdots \alpha_k}:W\to \mathbb{R} \end{align*} are smooth alternating coefficient functions. For each $\alpha\in\{1,\dots,n\}$, the coordinate function $y_\alpha$ is a smooth real-valued function on $W$, and the chain rule gives \begin{align*} dy_\alpha = \sum_{i=1}^n \partial_{x_i}y_\alpha\, dx_i. \end{align*} Substituting this into the expression for $\omega$ gives \begin{align*} a_{i_1\cdots i_k} = \sum_{\alpha_1,\dots,\alpha_k=1}^n b_{\alpha_1\cdots \alpha_k} \prod_{r=1}^k \partial_{x_{i_r}}y_{\alpha_r} \end{align*} on $W$. [guided] We now compare two charts on their overlap $W:=U\cap V$. In the $y$-chart, the same form has the expansion \begin{align*} \omega\big|_W = \frac{1}{k!} \sum_{\alpha_1,\dots,\alpha_k=1}^n b_{\alpha_1\cdots \alpha_k}\, dy_{\alpha_1}\wedge \cdots \wedge dy_{\alpha_k}, \end{align*} where each \begin{align*} b_{\alpha_1\cdots \alpha_k}:W\to \mathbb{R} \end{align*} is smooth and the coefficients are alternating. The coordinate functions $y_\alpha$ are smooth real-valued functions on $W$. Therefore their differentials can be written in the $x$-coordinate coframe as \begin{align*} dy_\alpha = \sum_{i=1}^n \partial_{x_i}y_\alpha\, dx_i. \end{align*} Substituting this formula into the $y$-coordinate expansion of $\omega$ gives \begin{align*} \omega\big|_W = \frac{1}{k!} \sum_{\alpha_1,\dots,\alpha_k=1}^n b_{\alpha_1\cdots \alpha_k} \left(\sum_{i_1=1}^n \partial_{x_{i_1}}y_{\alpha_1}\,dx_{i_1}\right) \wedge \cdots \wedge \left(\sum_{i_k=1}^n \partial_{x_{i_k}}y_{\alpha_k}\,dx_{i_k}\right). \end{align*} Expanding the wedge product yields \begin{align*} a_{i_1\cdots i_k} = \sum_{\alpha_1,\dots,\alpha_k=1}^n b_{\alpha_1\cdots \alpha_k} \prod_{r=1}^k \partial_{x_{i_r}}y_{\alpha_r}. \end{align*} This is the precise transformation rule for the alternating coefficient functions. [/guided] [/step] [step:Differentiate the transformed coefficients and separate the two kinds of terms] Differentiate the coefficient transformation with respect to $x_j$. The product rule gives \begin{align*} \partial_{x_j}a_{i_1\cdots i_k} &= \sum_{\alpha_1,\dots,\alpha_k=1}^n \sum_{\beta=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k}\, \partial_{x_j}y_\beta \prod_{r=1}^k \partial_{x_{i_r}}y_{\alpha_r} \\ &\quad+ \sum_{\alpha_1,\dots,\alpha_k=1}^n \sum_{s=1}^k b_{\alpha_1\cdots \alpha_k} \left(\partial^2_{x_jx_{i_s}}y_{\alpha_s}\right) \prod_{\substack{1\leq r\leq k\\ r\neq s}} \partial_{x_{i_r}}y_{\alpha_r}. \end{align*} The first line comes from differentiating $b_{\alpha_1\cdots \alpha_k}$ through the coordinate transition map $y\circ x^{-1}$. The second line comes from differentiating one of the factors $\partial_{x_{i_s}}y_{\alpha_s}$. [guided] We now differentiate the coefficient transformation. There are two sources of derivatives. First, the coefficient $b_{\alpha_1\cdots\alpha_k}$ is a function of the $y$-coordinates, and each $y_\beta$ is a function of the $x$-coordinates. The chain rule gives \begin{align*} \partial_{x_j} b_{\alpha_1\cdots \alpha_k} = \sum_{\beta=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k}\, \partial_{x_j}y_\beta. \end{align*} Second, the product \begin{align*} \prod_{r=1}^k \partial_{x_{i_r}}y_{\alpha_r} \end{align*} also depends on $x$, so the product rule differentiates one factor at a time. Therefore \begin{align*} \partial_{x_j}a_{i_1\cdots i_k} &= \sum_{\alpha_1,\dots,\alpha_k=1}^n \sum_{\beta=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k}\, \partial_{x_j}y_\beta \prod_{r=1}^k \partial_{x_{i_r}}y_{\alpha_r} \\ &\quad+ \sum_{\alpha_1,\dots,\alpha_k=1}^n \sum_{s=1}^k b_{\alpha_1\cdots \alpha_k} \left(\partial^2_{x_jx_{i_s}}y_{\alpha_s}\right) \prod_{\substack{1\leq r\leq k\\ r\neq s}} \partial_{x_{i_r}}y_{\alpha_r}. \end{align*} The proof now has a clear target: the first line should reconstruct the $y$-coordinate formula for $d\omega$, and the second line must vanish because it contains symmetric second derivatives multiplied by antisymmetric wedge factors. [/guided] [/step] [step:Identify the first-derivative terms with the formula in the second chart] Insert the first line from the preceding computation into $D_x\omega$. Its contribution is \begin{align*} \frac{1}{k!} \sum_{\beta,\alpha_1,\dots,\alpha_k=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k} \left(\sum_{j=1}^n \partial_{x_j}y_\beta\,dx_j\right) \wedge \bigwedge_{r=1}^k \left(\sum_{i_r=1}^n \partial_{x_{i_r}}y_{\alpha_r}\,dx_{i_r}\right). \end{align*} Using \begin{align*} dy_\beta = \sum_{j=1}^n \partial_{x_j}y_\beta\,dx_j \end{align*} and the same formula for each $dy_{\alpha_r}$, this becomes \begin{align*} \frac{1}{k!} \sum_{\beta,\alpha_1,\dots,\alpha_k=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k}\, dy_\beta\wedge dy_{\alpha_1}\wedge \cdots \wedge dy_{\alpha_k} = D_y\omega. \end{align*} [guided] We first handle the terms coming from differentiating the $b$-coefficients. Substituting the first line of the derivative formula into $D_x\omega$ gives \begin{align*} \frac{1}{k!} \sum_{\beta,\alpha_1,\dots,\alpha_k=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k} \left(\sum_{j=1}^n \partial_{x_j}y_\beta\,dx_j\right) \wedge \bigwedge_{r=1}^k \left(\sum_{i_r=1}^n \partial_{x_{i_r}}y_{\alpha_r}\,dx_{i_r}\right). \end{align*} The expression in the first parentheses is exactly $dy_\beta$, because \begin{align*} dy_\beta = \sum_{j=1}^n \partial_{x_j}y_\beta\,dx_j. \end{align*} Similarly, for every $r\in\{1,\dots,k\}$, \begin{align*} dy_{\alpha_r} = \sum_{i_r=1}^n \partial_{x_{i_r}}y_{\alpha_r}\,dx_{i_r}. \end{align*} Thus the whole first-derivative contribution is \begin{align*} \frac{1}{k!} \sum_{\beta,\alpha_1,\dots,\alpha_k=1}^n \partial_{y_\beta}b_{\alpha_1\cdots \alpha_k}\, dy_\beta\wedge dy_{\alpha_1}\wedge \cdots \wedge dy_{\alpha_k}. \end{align*} This is precisely the coordinate formula $D_y\omega$ in the $y$-chart. [/guided] [/step] [step:Cancel the second-derivative terms by symmetry and antisymmetry] It remains to show that the contribution from the second line is zero. For each smooth coordinate function \begin{align*} y_\alpha:W\to \mathbb{R}, \end{align*} the mixed second partial derivatives satisfy \begin{align*} \partial^2_{x_jx_i}y_\alpha = \partial^2_{x_ix_j}y_\alpha. \end{align*} Therefore \begin{align*} \sum_{j,i=1}^n \partial^2_{x_jx_i}y_\alpha\, dx_j\wedge dx_i = 0, \end{align*} because the coefficient is symmetric in $(j,i)$ while $dx_j\wedge dx_i$ is antisymmetric in $(j,i)$. Fix $\alpha_1,\dots,\alpha_k$ and $s\in\{1,\dots,k\}$. The corresponding second-derivative contribution contains the factor \begin{align*} \sum_{j,i_s=1}^n \partial^2_{x_jx_{i_s}}y_{\alpha_s}\,dx_j\wedge dx_{i_s}, \end{align*} wedged with the remaining one-forms \begin{align*} \sum_{i_r=1}^n \partial_{x_{i_r}}y_{\alpha_r}\,dx_{i_r} \end{align*} for $r\neq s$. Since the displayed factor is zero, every such contribution is zero. Hence all second-derivative terms vanish. [guided] The only possible obstruction to coordinate invariance is the appearance of second derivatives of the coordinate transition functions. We show that these terms vanish for a purely algebraic reason. Fix a smooth coordinate function \begin{align*} y_\alpha:W\to \mathbb{R}. \end{align*} Because $y_\alpha$ is smooth, its mixed second partial derivatives in the $x$-coordinates commute: \begin{align*} \partial^2_{x_jx_i}y_\alpha = \partial^2_{x_ix_j}y_\alpha. \end{align*} Thus the matrix with entries $\partial^2_{x_jx_i}y_\alpha$ is symmetric in the indices $(j,i)$. On the other hand, the wedge product satisfies \begin{align*} dx_j\wedge dx_i = - dx_i\wedge dx_j. \end{align*} Therefore \begin{align*} \sum_{j,i=1}^n \partial^2_{x_jx_i}y_\alpha\, dx_j\wedge dx_i = 0. \end{align*} Indeed, the term with indices $(j,i)$ cancels the term with indices $(i,j)$, and the diagonal terms vanish because $dx_i\wedge dx_i=0$. Now return to the second-derivative contribution in the formula for $D_x\omega$. Fix the coefficient indices $\alpha_1,\dots,\alpha_k$ and fix the position $s\in\{1,\dots,k\}$ where the derivative lands. The relevant part contains \begin{align*} \sum_{j,i_s=1}^n \partial^2_{x_jx_{i_s}}y_{\alpha_s}\,dx_j\wedge dx_{i_s}, \end{align*} wedged with the remaining transformed coordinate forms. The displayed factor is zero by the symmetry-antisymmetry cancellation just proved. Since wedging the zero $2$-form with any other forms still gives zero, the entire contribution for this fixed $s$ vanishes. Summing over $s$ and over all coefficient indices preserves zero. Hence every second-derivative term cancels. [/guided] [/step] [step:Glue the local formulas to obtain the global operator] On every overlap $U\cap V$, the preceding computation proves \begin{align*} D_x\omega\big|_{U\cap V} = D_y\omega\big|_{U\cap V}. \end{align*} Thus the locally defined $(k+1)$-forms agree on pairwise overlaps, so they glue to a unique smooth global $(k+1)$-form on $M$. Define \begin{align*} d\omega\in \Omega^{k+1}(M) \end{align*} to be this glued form. Finally, for $\omega,\eta\in\Omega^k(M)$ and $c\in\mathbb{R}$, the coordinate formula gives \begin{align*} D_x(\omega+\eta) = D_x\omega + D_x\eta, \qquad D_x(c\omega) = cD_x\omega \end{align*} in every chart $(U,x)$. Since the global form is determined by its local restrictions, this gives \begin{align*} d(\omega+\eta)=d\omega+d\eta, \qquad d(c\omega)=c\,d\omega. \end{align*} Therefore the coordinate formula defines a well-defined global $\mathbb{R}$-[linear map](/page/Linear%20Map) \begin{align*} d:\Omega^k(M)\to \Omega^{k+1}(M). \end{align*} [/step]