[proofplan] The wedge product is defined pointwise from the exterior algebra structure on each cotangent space $T_x^*M$. Bilinearity, associativity, and graded commutativity therefore follow from the corresponding identities for the wedge product on a finite-dimensional [vector space](/page/Vector%20Space), applied at every point. The only nontrivial verification at the manifold level is that the pointwise product of two smooth sections is again smooth; we check this in an arbitrary chart by writing both forms in the local coframe $\{dx_{i_1} \wedge \cdots \wedge dx_{i_k}\}$ and observing that the wedge product reduces to multiplication of smooth coefficient functions. The degree statement and the truncation $\Omega^p(M) = 0$ for $p > n$ come from $\dim \Lambda^p T_x^*M = \binom{n}{p}$. [/proofplan] [step:Recall the pointwise exterior algebra structure on each cotangent space] Fix $x \in M$. The cotangent space $T_x^*M$ is an $n$-dimensional real [vector space](/page/Vector%20Space), and its exterior algebra \begin{align*} \Lambda^* T_x^*M := \bigoplus_{p=0}^{n} \Lambda^p T_x^*M \end{align*} is a graded associative $\mathbb{R}$-algebra under the wedge product \begin{align*} \wedge_x : \Lambda^p T_x^*M \times \Lambda^q T_x^*M &\to \Lambda^{p+q} T_x^*M, \\ (\omega, \eta) &\mapsto \omega \wedge_x \eta, \end{align*} satisfying $\mathbb{R}$-bilinearity, associativity, and graded commutativity $\omega \wedge_x \eta = (-1)^{pq} \eta \wedge_x \omega$ for $\omega \in \Lambda^p T_x^*M$, $\eta \in \Lambda^q T_x^*M$. Moreover $\Lambda^p T_x^*M = 0$ for $p > n$ since $\dim \Lambda^p T_x^*M = \binom{n}{p}$. [guided] The wedge product on the exterior algebra of a single [vector space](/page/Vector%20Space) is the algebraic prerequisite for all manifold-level statements about forms. Recall that for an $n$-dimensional real [vector space](/page/Vector%20Space) $V$ — here $V = T_x^*M$ — the exterior algebra is the quotient of the tensor algebra $T(V) = \bigoplus_{k \ge 0} V^{\otimes k}$ by the two-sided ideal generated by $\{v \otimes v : v \in V\}$. The resulting graded algebra \begin{align*} \Lambda^* V = \bigoplus_{p=0}^{\dim V} \Lambda^p V \end{align*} inherits an associative product, denoted $\wedge$, that is $\mathbb{R}$-bilinear, sends $\Lambda^p V \times \Lambda^q V$ into $\Lambda^{p+q} V$, and satisfies $\omega \wedge \eta = (-1)^{pq}\eta \wedge \omega$. The dimension formula $\dim \Lambda^p V = \binom{n}{p}$ forces $\Lambda^p V = 0$ for $p > n$. We will use these pointwise facts at every $x \in M$; the only work left is to confirm that pointwise products of smooth forms are smooth. [/guided] [/step] [step:Define the wedge product of forms pointwise and reduce algebraic identities to the pointwise case] For $\alpha \in \Omega^p(M)$ and $\beta \in \Omega^q(M)$, define \begin{align*} \alpha \wedge \beta : M &\to \Lambda^{p+q} T^*M, \\ x &\mapsto \alpha_x \wedge_x \beta_x \in \Lambda^{p+q} T_x^*M. \end{align*} Provisionally, $\alpha \wedge \beta$ is a set-theoretic section of $\Lambda^{p+q}T^*M$; smoothness is verified in the next step. Properties (ii)–(iv) for $\alpha, \beta, \gamma, \alpha_1, \alpha_2$ in the relevant spaces hold by evaluating at each $x \in M$ and applying the corresponding identities in $\Lambda^* T_x^*M$ from the preceding step: \begin{align*} ((a\alpha_1 + b\alpha_2) \wedge \beta)_x &= (a\alpha_{1,x} + b\alpha_{2,x}) \wedge_x \beta_x = a(\alpha_{1,x} \wedge_x \beta_x) + b(\alpha_{2,x} \wedge_x \beta_x), \\ ((\alpha \wedge \beta) \wedge \gamma)_x &= (\alpha_x \wedge_x \beta_x) \wedge_x \gamma_x = \alpha_x \wedge_x (\beta_x \wedge_x \gamma_x) = (\alpha \wedge (\beta \wedge \gamma))_x, \\ (\alpha \wedge \beta)_x &= \alpha_x \wedge_x \beta_x = (-1)^{pq} \beta_x \wedge_x \alpha_x = (-1)^{pq}(\beta \wedge \alpha)_x. \end{align*} Since two sections of $\Lambda^{p+q}T^*M$ that agree at every point are equal, (ii), (iii), and (iv) follow. Property (i), $\alpha \wedge \beta \in \Omega^{p+q}(M)$, holds at the level of fibres since $\alpha_x \wedge_x \beta_x \in \Lambda^{p+q} T_x^*M$; what remains is to verify that $\alpha \wedge \beta$ is smooth. [/step] [step:Verify smoothness of the pointwise product in an arbitrary chart] Let $(U, \varphi)$ be a chart on $M$ with coordinates $(x_1, \dots, x_n)$ and associated coordinate coframe $(dx_1, \dots, dx_n)$ on $U$. For multi-indices $I = (i_1 < \cdots < i_p)$ and $J = (j_1 < \cdots < j_q)$ with entries in $\{1, \dots, n\}$, write $dx_I := dx_{i_1} \wedge \cdots \wedge dx_{i_p}$ and similarly $dx_J$. The forms $\{dx_I : |I| = p\}$ and $\{dx_J : |J| = q\}$ are pointwise bases of $\Lambda^p T_x^*M$ and $\Lambda^q T_x^*M$ for $x \in U$. Since $\alpha \in \Omega^p(M)$ and $\beta \in \Omega^q(M)$ are smooth sections, there exist unique smooth functions $\alpha_I, \beta_J : U \to \mathbb{R}$ with \begin{align*} \alpha|_U = \sum_{|I| = p} \alpha_I \, dx_I, \qquad \beta|_U = \sum_{|J| = q} \beta_J \, dx_J. \end{align*} By the pointwise $\mathbb{R}$-bilinearity established in the previous step, for $x \in U$, \begin{align*} (\alpha \wedge \beta)_x = \sum_{|I| = p,\, |J| = q} \alpha_I(x)\, \beta_J(x)\, (dx_I \wedge dx_J)_x. \end{align*} The pointwise wedge product $dx_I \wedge dx_J$ is a fixed smooth section of $\Lambda^{p+q} T^*U$ (it equals $\pm dx_K$ for the sorted multi-index $K = I \cup J$ when $I \cap J = \varnothing$, and $0$ otherwise — these signs and indices do not depend on $x$). Each $\alpha_I \beta_J : U \to \mathbb{R}$ is smooth as a product of smooth functions. Hence $\alpha \wedge \beta|_U$ is a finite $C^\infty(U)$-linear combination of smooth sections of $\Lambda^{p+q}T^*U$ and is therefore smooth on $U$. Since the chart $(U, \varphi)$ was arbitrary and $M$ is covered by such charts, $\alpha \wedge \beta \in \Omega^{p+q}(M)$. [guided] The smoothness check is a local computation in a chart. Pick any chart $(U, \varphi)$ on $M$ with coordinates $(x_1, \dots, x_n)$. The differentials $dx_1, \dots, dx_n$ are smooth sections of $T^*U$ and their wedge products $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_p}$, indexed by ordered multi-indices $I = (i_1 < \cdots < i_p)$, form a pointwise basis of $\Lambda^p T_x^*M$ at every $x \in U$. Smoothness of a $p$-form on $U$ is equivalent to smoothness of its coefficients in this basis: this is the defining property of the smooth structure on the bundle $\Lambda^p T^*M$. Expand \begin{align*} \alpha|_U = \sum_{|I| = p} \alpha_I\, dx_I, \qquad \beta|_U = \sum_{|J| = q} \beta_J\, dx_J, \end{align*} with $\alpha_I, \beta_J \in C^\infty(U)$. By the pointwise bilinearity from the previous step, \begin{align*} (\alpha \wedge \beta)|_U = \sum_{I,J} \alpha_I \beta_J\, (dx_I \wedge dx_J). \end{align*} The factor $dx_I \wedge dx_J$ is itself a smooth $(p+q)$-form on $U$: explicitly, if $I$ and $J$ share an index then $dx_I \wedge dx_J = 0$ (by graded commutativity, $dx_i \wedge dx_i = 0$); otherwise, $dx_I \wedge dx_J = \operatorname{sgn}(\sigma) dx_K$, where $K$ is the increasing rearrangement of the concatenated multi-index $I \mid J$ and $\sigma$ is the permutation sorting it. The sign and the multi-index $K$ are independent of $x$. Thus $(\alpha \wedge \beta)|_U$ is a $C^\infty(U)$-linear combination of smooth basis sections, hence smooth. Why does this finish the proof of smoothness on $M$? Smoothness of a section is a local property: a section of $\Lambda^{p+q}T^*M$ is smooth iff its restriction to every chart in some atlas is smooth. We checked smoothness in an arbitrary chart, so $\alpha \wedge \beta$ is smooth on $M$. [/guided] [/step] [step:Identify the multiplicative identity and conclude the algebra structure] The constant function $\mathbf{1} : M \to \mathbb{R}$, $x \mapsto 1$, is an element of $\Omega^0(M) = C^\infty(M)$. For any $\alpha \in \Omega^p(M)$ and $x \in M$, $\mathbf{1}_x \wedge_x \alpha_x = 1 \cdot \alpha_x = \alpha_x$, so $\mathbf{1} \wedge \alpha = \alpha$, and symmetrically $\alpha \wedge \mathbf{1} = \alpha$. Combined with properties (i)–(iv) established above, this shows that $(\Omega^*(M), +, \wedge)$ is a unital graded associative $\mathbb{R}$-algebra graded by the degree decomposition $\Omega^*(M) = \bigoplus_{p=0}^n \Omega^p(M)$, with graded commutativity relation $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ for $\alpha \in \Omega^p(M)$, $\beta \in \Omega^q(M)$. Finally, the grading terminates at $p = n$ because $\Lambda^p T_x^*M = 0$ for $p > n$ at every $x \in M$, hence any section of $\Lambda^p T^*M$ for $p > n$ is the zero section. This completes the proof of all four asserted properties. [/step]