[proofplan] We use the coordinate chart to produce a local frame $\partial_{x_1},\dots,\partial_{x_n}$ for the tangent bundle over $U$ and the dual coframe $dx_1,\dots,dx_n$ for the cotangent bundle. At each point, the alternating products $dx_{i_1}\wedge\cdots\wedge dx_{i_k}$ form the standard basis of the exterior power $\Lambda^k(T_p^*M)$, so the coefficients are forced by evaluating $\omega$ on the corresponding coordinate vector fields. Smoothness of the coefficients follows because a smooth differential form evaluated on smooth vector fields gives a smooth function. Uniqueness follows from pointwise [linear independence](/page/Linear%20Independence) of the exterior-basis covectors. [/proofplan]

[step:Build the coordinate frame and coframe on the chart]Let \begin{align*} \partial_{x_i}:U&\to TU\\ p&\mapsto \left.\frac{\partial}{\partial x_i}\right|_p \end{align*} denote the $i$-th coordinate vector field on $U$, for each $i\in\{1,\dots,n\}$. For every $p\in U$, the vectors \begin{align*} \left.\partial_{x_1}\right|_p,\dots,\left.\partial_{x_n}\right|_p \end{align*} form a basis of $T_pM$. For each $i\in\{1,\dots,n\}$, let \begin{align*} dx_i:U&\to T^*U\\ p&\mapsto d(x_i)_p \end{align*} be the coordinate one-form. Since $x_i$ is the $i$-th coordinate function, the duality relation is \begin{align*} dx_i|_p\left(\left.\partial_{x_j}\right|_p\right)=\delta_{ij} \end{align*} for all $p\in U$ and all $i,j\in\{1,\dots,n\}$. Hence $dx_1|_p,\dots,dx_n|_p$ is the basis of $T_p^*M$ dual to the coordinate basis of $T_pM$.[/step]

[guided]The coordinate chart gives a concrete basis of every tangent space over $U$. For each $i\in\{1,\dots,n\}$, define the smooth coordinate vector field \begin{align*} \partial_{x_i}:U&\to TU\\ p&\mapsto \left.\frac{\partial}{\partial x_i}\right|_p. \end{align*} By the definition of a coordinate chart, the vectors \begin{align*} \left.\partial_{x_1}\right|_p,\dots,\left.\partial_{x_n}\right|_p \end{align*} form a basis of $T_pM$ for each $p\in U$. The corresponding one-forms are obtained by differentiating the coordinate functions. For each $i\in\{1,\dots,n\}$, define \begin{align*} dx_i:U&\to T^*U\\ p&\mapsto d(x_i)_p. \end{align*} The defining property of coordinate vector fields gives \begin{align*} dx_i|_p\left(\left.\partial_{x_j}\right|_p\right)=\delta_{ij}. \end{align*} Thus the coordinate one-forms $dx_1|_p,\dots,dx_n|_p$ are exactly the [dual basis](/theorems/414) to the coordinate tangent basis at $p$. This is the linear-algebra input that allows us to expand any alternating covariant tensor in the wedge basis.[/guided]

[step:Define the coefficient functions by evaluating on coordinate vector fields] For every strictly increasing multi-index $I=(i_1,\dots,i_k)$, define \begin{align*} f_I:U&\to\mathbb{R}\\ p&\mapsto \omega_p\left(\left.\partial_{x_{i_1}}\right|_p,\dots,\left.\partial_{x_{i_k}}\right|_p\right). \end{align*} If $k=0$, this definition means $f_\varnothing=\omega|_U$, viewing a $0$-form as a smooth function. Because $\omega$ is a smooth $k$-form and $\partial_{x_{i_1}},\dots,\partial_{x_{i_k}}$ are smooth vector fields on $U$, the function $f_I$ is smooth on $U$. Hence $f_I\in C^\infty(U)$ for every strictly increasing multi-index $I$. [/step]

[step:Verify the pointwise exterior-basis expansion]Fix $p\in U$. For each strictly increasing multi-index $I=(i_1,\dots,i_k)$, set \begin{align*} \varepsilon_I(p):=dx_{i_1}|_p\wedge\cdots\wedge dx_{i_k}|_p\in \Lambda^k(T_p^*M). \end{align*} The family $\{\varepsilon_I(p)\}_I$ is the standard exterior basis of $\Lambda^k(T_p^*M)$ associated to the basis $dx_1|_p,\dots,dx_n|_p$ of $T_p^*M$. Define \begin{align*} \alpha_p:=\omega_p-\sum_I f_I(p)\,\varepsilon_I(p)\in \Lambda^k(T_p^*M). \end{align*} For every strictly increasing multi-index $J=(j_1,\dots,j_k)$, evaluate $\alpha_p$ on the coordinate basis vectors: \begin{align*} \alpha_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) &=\omega_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) -\sum_I f_I(p)\,\varepsilon_I(p)\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right)\\ &=f_J(p)-\sum_I f_I(p)\,\delta_{IJ}\\ &=0. \end{align*} Since an alternating $k$-linear form is determined by its values on increasing $k$-tuples from a basis, $\alpha_p=0$. Therefore \begin{align*} \omega_p=\sum_I f_I(p)\,dx_{i_1}|_p\wedge\cdots\wedge dx_{i_k}|_p. \end{align*} Because $p\in U$ was arbitrary, this gives the claimed identity in $\Omega^k(U)$.[/step]

[guided]We now prove that the coefficient functions just defined really reconstruct the form. Fix a point $p\in U$. For a strictly increasing multi-index $I=(i_1,\dots,i_k)$, define \begin{align*} \varepsilon_I(p):=dx_{i_1}|_p\wedge\cdots\wedge dx_{i_k}|_p\in \Lambda^k(T_p^*M). \end{align*} The covectors $dx_1|_p,\dots,dx_n|_p$ form a basis of $T_p^*M$, so their increasing wedge products form the standard basis of $\Lambda^k(T_p^*M)$. To prove the desired equality at $p$, subtract the proposed expression from $\omega_p$ and show that the difference is zero. Define \begin{align*} \alpha_p:=\omega_p-\sum_I f_I(p)\,\varepsilon_I(p)\in \Lambda^k(T_p^*M). \end{align*} An alternating $k$-linear form is determined by its values on increasing $k$-tuples of basis vectors. Therefore it is enough to evaluate $\alpha_p$ on \begin{align*} \left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p \end{align*} for each strictly increasing multi-index $J=(j_1,\dots,j_k)$. Using multilinearity and the definition of $f_J$, we get \begin{align*} \alpha_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) &=\omega_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) -\sum_I f_I(p)\,\varepsilon_I(p)\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right)\\ &=f_J(p)-\sum_I f_I(p)\,\varepsilon_I(p)\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right). \end{align*} The wedge basis has the Kronecker evaluation property \begin{align*} \varepsilon_I(p)\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right)=\delta_{IJ}, \end{align*} because the matrix of pairings $dx_{i_a}|_p(\partial_{x_{j_b}}|_p)$ is the identity matrix exactly when $I=J$, and otherwise has two equal rows or a zero row after restricting to the chosen increasing tuples. Hence \begin{align*} \alpha_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) =f_J(p)-\sum_I f_I(p)\delta_{IJ}=0. \end{align*} Thus $\alpha_p$ vanishes on every increasing coordinate basis tuple, so $\alpha_p=0$. Since $p$ was arbitrary, the pointwise identity assembles into the identity of smooth $k$-forms on $U$: \begin{align*} \omega|_U=\sum_I f_I\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*}[/guided]

[step:Prove uniqueness from the coordinate basis evaluations] Suppose that $(g_I)_I$ is another family of functions $g_I\in C^\infty(U)$ satisfying \begin{align*} \omega|_U=\sum_I g_I\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} Fix a strictly increasing multi-index $J=(j_1,\dots,j_k)$ and a point $p\in U$. Evaluating both sides on \begin{align*} \left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p \end{align*} gives \begin{align*} f_J(p)=\omega_p\left(\left.\partial_{x_{j_1}}\right|_p,\dots,\left.\partial_{x_{j_k}}\right|_p\right) =\sum_I g_I(p)\,\delta_{IJ} =g_J(p). \end{align*} Since $p\in U$ and $J$ were arbitrary, $g_I=f_I$ on $U$ for every strictly increasing multi-index $I$. This proves uniqueness and completes the proof. [/step]