[proofplan] We compute the de Rham cohomology of $T^n$ by applying the Kunneth formula for de Rham cohomology repeatedly to the product decomposition $T^n = (\mathbb{R}/\mathbb{Z})^n$. Each circle factor contributes exactly one degree-$0$ generator and one degree-$1$ generator, represented by the angular form $dt$. The exterior-algebra structure comes from the graded-commutativity of the wedge product, so the only nonzero monomial classes are indexed by subsets of $\{1,\dots,n\}$. [/proofplan] [step:Record the cohomology ring of one circle] Let $S^1$ denote $\mathbb{R}/\mathbb{Z}$, and let \begin{align*} q: \mathbb{R} &\to S^1 \\ t &\mapsto t+\mathbb{Z} \end{align*} be the quotient map. The standard angular form $dt \in \Omega^1(S^1)$ is the unique smooth $1$-form satisfying $q^*dt = d t_{\mathbb{R}}$, where $t_{\mathbb{R}}: \mathbb{R} \to \mathbb{R}$ is the identity coordinate function. Define the constant function \begin{align*} 1_{S^1}: S^1 &\to \mathbb{R} \\ x &\mapsto 1. \end{align*} Throughout the proof, $[1]$ denotes the de Rham cohomology class $[1_{S^1}] \in H^0_{\mathrm{dR}}(S^1)$. We use the standard computation of the [de Rham cohomology of the circle](/theorems/3924): (citing a result not yet in the wiki: de Rham cohomology of the circle) \begin{align*} H^0_{\mathrm{dR}}(S^1) &\cong \mathbb{R}\,[1], \\ H^1_{\mathrm{dR}}(S^1) &\cong \mathbb{R}\,[dt], \\ H^m_{\mathrm{dR}}(S^1) &= 0 \qquad \text{for all } m \ge 2. \end{align*} The class $[dt]$ has degree $1$, and since $S^1$ has no nonzero de Rham cohomology in degree $2$, the product $[dt]\wedge[dt]$ is zero in $H^2_{\mathrm{dR}}(S^1)$. [/step] [step:Apply the Kunneth formula to decompose the cohomology of the product] For each $i \in \{1,\dots,n\}$, define the projection \begin{align*} \pi_i: T^n &\to S^1 \\ (t_1,\dots,t_n) &\mapsto t_i. \end{align*} Define $dt_i := \pi_i^*dt \in \Omega^1(T^n)$. We apply the Kunneth formula for de Rham cohomology repeatedly to the finite product $T^n = (S^1)^n$: (citing a result not yet in the wiki: Kunneth formula for de Rham cohomology) \begin{align*} H^*_{\mathrm{dR}}(T^n) \cong H^*_{\mathrm{dR}}(S^1) \otimes_{\mathbb{R}} \cdots \otimes_{\mathbb{R}} H^*_{\mathrm{dR}}(S^1), \end{align*} where there are $n$ tensor factors and the isomorphism is induced by exterior products of pullbacks along the coordinate projections. Under this isomorphism, a simple tensor \begin{align*} \alpha_1 \otimes \cdots \otimes \alpha_n, \qquad \alpha_i \in H^*_{\mathrm{dR}}(S^1), \end{align*} corresponds to \begin{align*} \pi_1^*\alpha_1 \wedge \cdots \wedge \pi_n^*\alpha_n \in H^*_{\mathrm{dR}}(T^n). \end{align*} Since each factor $H^*_{\mathrm{dR}}(S^1)$ has basis $\{[1],[dt]\}$, the resulting basis elements of $H^k_{\mathrm{dR}}(T^n)$ are exactly the classes \begin{align*} [dt_{i_1} \wedge \cdots \wedge dt_{i_k}], \qquad 1 \le i_1 < \cdots < i_k \le n. \end{align*} [guided] The Kunneth formula is the mechanism that turns the cohomology of a product into the [tensor product](/page/Tensor%20Product) of the cohomologies of the factors. Here every factor is the same circle $S^1 = \mathbb{R}/\mathbb{Z}$, whose de Rham cohomology has one generator in degree $0$, namely $[1]$, and one generator in degree $1$, namely $[dt]$. For each coordinate, the projection map \begin{align*} \pi_i: T^n &\to S^1 \\ (t_1,\dots,t_n) &\mapsto t_i \end{align*} allows us to pull the angular form from the $i$-th circle factor to the whole torus. This gives the global closed $1$-form \begin{align*} dt_i := \pi_i^*dt \in \Omega^1(T^n). \end{align*} Applying the Kunneth formula for de Rham cohomology repeatedly to \begin{align*} T^n = S^1 \times \cdots \times S^1 \end{align*} gives \begin{align*} H^*_{\mathrm{dR}}(T^n) \cong H^*_{\mathrm{dR}}(S^1) \otimes_{\mathbb{R}} \cdots \otimes_{\mathbb{R}} H^*_{\mathrm{dR}}(S^1). \end{align*} The formula identifies a tensor of cohomology classes with the wedge product of their pullbacks: \begin{align*} \alpha_1 \otimes \cdots \otimes \alpha_n \mapsto \pi_1^*\alpha_1 \wedge \cdots \wedge \pi_n^*\alpha_n. \end{align*} Since each $\alpha_i$ is either $[1]$ or $[dt]$, choosing a degree-$k$ class amounts exactly to choosing $k$ of the $n$ factors in which we place $[dt]$ and placing $[1]$ in the remaining factors. If those chosen indices are \begin{align*} I = \{i_1<\cdots<i_k\} \subset \{1,\dots,n\}, \end{align*} then the corresponding class is \begin{align*} [dt_{i_1} \wedge \cdots \wedge dt_{i_k}]. \end{align*} Thus these classes span $H^k_{\mathrm{dR}}(T^n)$ and are linearly independent because they correspond to the tensor-product basis supplied by Kunneth. [/guided] [/step] [step:Identify the product structure with the exterior algebra] Let $e_1,\dots,e_n$ denote the standard basis of $\mathbb{R}^n$. Define the graded-algebra homomorphism \begin{align*} \Phi: \Lambda^*(\mathbb{R}^n) &\to H^*_{\mathrm{dR}}(T^n) \\ e_i &\mapsto [dt_i]. \end{align*} This homomorphism exists because $H^*_{\mathrm{dR}}(T^n)$ is graded commutative and the elements $[dt_i]$ have degree $1$. For $i,j \in \{1,\dots,n\}$, graded commutativity gives \begin{align*} [dt_i]\wedge[dt_j] = - [dt_j]\wedge[dt_i]. \end{align*} Taking $i=j$ gives \begin{align*} [dt_i]\wedge[dt_i] = - [dt_i]\wedge[dt_i], \end{align*} and since the coefficient field is $\mathbb{R}$, this implies \begin{align*} [dt_i]\wedge[dt_i]=0. \end{align*} Therefore every product of the degree-one generators reduces uniquely to an ordered wedge product \begin{align*} [dt_{i_1}\wedge\cdots\wedge dt_{i_k}], \qquad 1 \le i_1 < \cdots < i_k \le n. \end{align*} By the previous step, these ordered products form a basis of $H^k_{\mathrm{dR}}(T^n)$ for every $k$. Hence $\Phi$ maps the standard exterior-algebra basis \begin{align*} e_{i_1}\wedge\cdots\wedge e_{i_k}, \qquad 1 \le i_1 < \cdots < i_k \le n, \end{align*} bijectively onto a basis of $H^k_{\mathrm{dR}}(T^n)$ in each degree. Thus $\Phi$ is an isomorphism of graded $\mathbb{R}$-algebras. [/step] [step:Count the basis elements in each degree] Fix $k \in \{0,\dots,n\}$. The basis of $H^k_{\mathrm{dR}}(T^n)$ consists of one element for each $k$-element subset of $\{1,\dots,n\}$. The number of such subsets is $\binom{n}{k}$, so \begin{align*} \dim_{\mathbb{R}} H^k_{\mathrm{dR}}(T^n) = \binom{n}{k}. \end{align*} This proves both the exterior-algebra description and the stated dimension formula. [/step]