[proofplan] We first use compactness of $G$ as a smooth manifold to obtain finite-dimensional real de Rham cohomology. We then check directly that wedge product descends from closed forms to cohomology classes and satisfies graded commutativity. Finally, in the abelian case, the structure theorem for compact connected abelian Lie groups identifies $G$ with a torus, and the de Rham [cohomology ring](/theorems/2271) of a torus is the exterior algebra on its degree-one cohomology. [/proofplan] [step:Use compactness of the underlying smooth manifold to obtain finite-dimensional cohomology] Let $n = \dim G$. Since $G$ is a compact Lie group, its underlying space is a compact smooth manifold of dimension $n$. By the finite-dimensionality theorem for de Rham cohomology of compact smooth manifolds (citing a result not yet in the wiki: finite-dimensionality of de Rham cohomology for compact smooth manifolds), each real [vector space](/page/Vector%20Space) $H_{\mathrm{dR}}^k(G;\mathbb{R})$ is finite-dimensional for $0 \leq k \leq n$. For $k < 0$ and $k > n$, the space of smooth differential $k$-forms on $G$ is zero, so \begin{align*} H_{\mathrm{dR}}^k(G;\mathbb{R}) = 0. \end{align*} Thus every graded component of $H_{\mathrm{dR}}^*(G;\mathbb{R})$ is finite-dimensional. [guided] The hypothesis that $G$ is compact is used here through the underlying manifold of the Lie group. A Lie group is, by definition, a smooth manifold equipped with smooth multiplication and inversion maps. Thus $G$ has a well-defined de Rham complex \begin{align*} 0 \longrightarrow \Omega^0(G) \xrightarrow{d} \Omega^1(G) \xrightarrow{d} \cdots \xrightarrow{d} \Omega^n(G) \longrightarrow 0, \end{align*} where $n = \dim G$ and $\Omega^k(G)$ denotes the real vector space of smooth differential $k$-forms on $G$. We now invoke the finite-dimensionality theorem for de Rham cohomology of compact smooth manifolds (citing a result not yet in the wiki: finite-dimensionality of de Rham cohomology for compact smooth manifolds). Its hypotheses are satisfied because the underlying manifold of $G$ is smooth and compact. Therefore each cohomology group \begin{align*} H_{\mathrm{dR}}^k(G;\mathbb{R}) = \frac{\ker(d:\Omega^k(G)\to \Omega^{k+1}(G))} {\operatorname{im}(d:\Omega^{k-1}(G)\to \Omega^k(G))} \end{align*} is finite-dimensional for $0 \leq k \leq n$. For degrees outside the range of the dimension, there are no nonzero differential forms: \begin{align*} \Omega^k(G)=0 \qquad \text{for } k<0 \text{ or } k>n. \end{align*} Hence $H_{\mathrm{dR}}^k(G;\mathbb{R})=0$ in those degrees. This proves finite-dimensionality in every degree. [/guided] [/step] [step:Descend wedge product to cohomology classes] For $p,q \in \mathbb{Z}$, define the bilinear map \begin{align*} \wedge:\Omega^p(G)\times \Omega^q(G)&\to \Omega^{p+q}(G)\\ (\alpha,\beta)&\mapsto \alpha\wedge\beta . \end{align*} If $\alpha \in \Omega^p(G)$ and $\beta \in \Omega^q(G)$ are closed, then the graded Leibniz rule gives \begin{align*} d(\alpha\wedge\beta) = d\alpha\wedge\beta+(-1)^p\alpha\wedge d\beta = 0. \end{align*} Thus the wedge product of closed forms is closed. Now suppose $\alpha'=\alpha+d\eta$ for some $\eta\in\Omega^{p-1}(G)$ and $\beta'=\beta+d\theta$ for some $\theta\in\Omega^{q-1}(G)$, where $\alpha$ and $\beta$ are closed. Then \begin{align*} \alpha'\wedge\beta'-\alpha\wedge\beta = d\eta\wedge\beta+\alpha\wedge d\theta+d\eta\wedge d\theta. \end{align*} Using $d\beta=0$ and $d\alpha=0$, we compute \begin{align*} d(\eta\wedge\beta) &= d\eta\wedge\beta,\\ d\left((-1)^p\alpha\wedge\theta\right) &= \alpha\wedge d\theta,\\ d(\eta\wedge d\theta) &= d\eta\wedge d\theta. \end{align*} Therefore \begin{align*} \alpha'\wedge\beta'-\alpha\wedge\beta = d\left(\eta\wedge\beta+(-1)^p\alpha\wedge\theta+\eta\wedge d\theta\right). \end{align*} Hence the cohomology class of $\alpha\wedge\beta$ depends only on the classes of $\alpha$ and $\beta$. This defines a product \begin{align*} H_{\mathrm{dR}}^p(G;\mathbb{R})\times H_{\mathrm{dR}}^q(G;\mathbb{R}) &\to H_{\mathrm{dR}}^{p+q}(G;\mathbb{R})\\ ([\alpha],[\beta]) &\mapsto [\alpha\wedge\beta]. \end{align*} [/step] [step:Verify graded commutativity of the induced product] Let $[\alpha]\in H_{\mathrm{dR}}^p(G;\mathbb{R})$ and $[\beta]\in H_{\mathrm{dR}}^q(G;\mathbb{R})$, with closed representatives $\alpha\in\Omega^p(G)$ and $\beta\in\Omega^q(G)$. The wedge product of differential forms satisfies \begin{align*} \alpha\wedge\beta = (-1)^{pq}\beta\wedge\alpha. \end{align*} Passing to cohomology classes gives \begin{align*} [\alpha]\,[\beta] = [\alpha\wedge\beta] = (-1)^{pq}[\beta\wedge\alpha] = (-1)^{pq}[\beta]\,[\alpha]. \end{align*} Therefore $H_{\mathrm{dR}}^*(G;\mathbb{R})$ is a graded-commutative real algebra. [/step] [step:Identify compact connected abelian Lie groups with tori] Assume now that $G$ is abelian. Let $e\in G$ be the identity element and let $\mathfrak{g}=T_eG$ be the Lie algebra of $G$. Since $G$ is abelian, the Lie bracket on $\mathfrak{g}$ is zero. By the structure theorem for compact connected abelian Lie groups (citing a result not yet in the wiki: compact connected abelian Lie groups are tori), the exponential map \begin{align*} \exp:\mathfrak{g}&\to G \end{align*} is a surjective Lie group homomorphism whose kernel $\Gamma=\ker(\exp)$ is a full lattice in the real vector space $\mathfrak{g}$. Therefore \begin{align*} G \cong \mathfrak{g}/\Gamma \end{align*} as Lie groups. If $r=\dim_{\mathbb{R}}\mathfrak{g}$, then $\mathfrak{g}/\Gamma$ is diffeomorphic to the torus $\mathbb{R}^r/\mathbb{Z}^r$. [guided] The abelian hypothesis changes the nature of the Lie group completely. Let $e\in G$ be the identity element, and define the Lie algebra \begin{align*} \mathfrak{g}=T_eG. \end{align*} Because $G$ is abelian, left-invariant vector fields commute, so the Lie bracket on $\mathfrak{g}$ is zero. We now invoke the structure theorem for compact connected abelian Lie groups (citing a result not yet in the wiki: compact connected abelian Lie groups are tori). The theorem applies because $G$ is assumed compact, connected, abelian, and a Lie group. It says that the exponential map \begin{align*} \exp:\mathfrak{g}&\to G \end{align*} is surjective and that its kernel \begin{align*} \Gamma=\ker(\exp) \end{align*} is a full lattice in the finite-dimensional real vector space $\mathfrak{g}$. A full lattice means a discrete subgroup $\Gamma\subset\mathfrak{g}$ generated by a real basis of $\mathfrak{g}$. Hence, after choosing a basis of $\mathfrak{g}$ adapted to $\Gamma$, the quotient $\mathfrak{g}/\Gamma$ is diffeomorphic to $\mathbb{R}^r/\mathbb{Z}^r$, where \begin{align*} r=\dim_{\mathbb{R}}\mathfrak{g}. \end{align*} Thus $G$ is a torus. [/guided] [/step] [step:Compute the cohomology algebra of the torus] By the de Rham cohomology computation for tori (citing a result not yet in the wiki: de Rham cohomology ring of a torus), the natural map from constant translation-invariant forms on $\mathfrak{g}/\Gamma$ to de Rham cohomology induces an isomorphism of graded real algebras \begin{align*} \Lambda^*(\mathfrak{g}^*) \longrightarrow H_{\mathrm{dR}}^*(\mathfrak{g}/\Gamma;\mathbb{R}). \end{align*} Transporting this isomorphism across the Lie group diffeomorphism $G\cong \mathfrak{g}/\Gamma$ gives an isomorphism of graded real algebras \begin{align*} H_{\mathrm{dR}}^*(G;\mathbb{R}) \cong \Lambda^*(\mathfrak{g}^*). \end{align*} Since $\mathfrak{g}^*$ is placed in degree $1$, the generators are all of odd degree. This proves the asserted exterior-algebra description in the compact connected abelian case. [/step]