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This is probably not "down-to-Earth" enough for your purposes, but it was one of the first uses of de Rham cohomology that I really enjoyed and I feel like I must share it. (I learned it from Bott's article "The geometry and representation theory of compact Lie groups" in the 1977 Proceedings of the SRC/LMS Research Symposium on Representations of Lie Groups.)

Theorem: If the $n$-sphere $S^n$ is a Lie group, then $n$ must be odd (or zero!).

Proof: $S^0$ is obviously a Lie group, so that's that. If $S^{2n}$ were a Lie group , (and $n>0$), then we'd have a smooth multiplication map $m \colon S^{2n} \times S^{2n} \to S^{2n}$ and an identity element $e \in S^{2n}$. The map $m$ induces a ring homomorphism $m^\ast \colon H^\ast(S^{2n}) \to H^\ast(S^{2n} \times S^{2n})$. Composing this with the Kunneth isomorphism $$H^\ast(S^{2n} \times S^{2n}) \stackrel{\sim}{\longrightarrow} H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$$ yields a map $f \colon H^\ast(S^{2n}) \to H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$.

Now, $$H^\ast(S^{2n}) = \begin{cases} \mathbb{R} & \text{if }\ast=0 \text{ or } 2n \\ 0 & \text{otherwise.}\end{cases}$$ So, writing $1$ and $\lambda$ for the generators of $H^0$ and $H^{2n}$, respectively, we have $f(1)=1\otimes1$ and $f(\lambda)=a(\lambda\otimes1)+b(1\otimes\lambda)$ for some $a,b\in\mathbb{R}$. To determine $a$ and $b$, first restrict the multiplication map to $S^{2n} \times \{e\}$, where it is the identity map. Consequently $m^*$ gives the identification $H^\ast(S^{2n} \times \{e\}) \cong H^\ast(S^{2n})$ and it follows that $a=1$. Similarly, $b=1$.

As $f$ is a ring homomorphism, we also have \begin{align} f(\lambda^2) = f(\lambda)^2 &= (\lambda\otimes1 + 1\otimes\lambda)^2 \\ &= \lambda^2\otimes1+ (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)+1\otimes\lambda^2.\end{align} As $\lambda^2 \in H^4 H^{4n} = 0$, this reduces to $0 = (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)$. Recall, however, that the product structure on the tensor product $A \otimes B$ of graded rings is given by $$(a \otimes b)(c \otimes d) = (-1)^{\deg a \deg c}(ac\otimes bd).$$ Since $\lambda \in H^2$H^{2n}$, this means that $$(\lambda\otimes1)(1\otimes\lambda) = (1\otimes\lambda)(\lambda\otimes1) = \lambda \otimes \lambda.$$ Consequently,$\lambda \otimes \lambda = 0$, which is a contradiction! $\blacksquare$ Of course, there are several other (perhaps even better) ways of proving this theorem, but I think this proof is fairly charming. 1 This is probably not "down-to-Earth" enough for your purposes, but it was one of the first uses of de Rham cohomology that I really enjoyed and I feel like I must share it. (I learned it from Bott's article "The geometry and representation theory of compact Lie groups" in the 1977 Proceedings of the SRC/LMS Research Symposium on Representations of Lie Groups.) Theorem: If the$n$-sphere$S^n$is a Lie group, then$n$must be odd. Proof: If$S^{2n}$were a Lie group, then we'd have a smooth multiplication map$m \colon S^{2n} \times S^{2n} \to S^{2n}$and an identity element$e \in S^{2n}$. The map$m$induces a ring homomorphism$m^\ast \colon H^\ast(S^{2n}) \to H^\ast(S^{2n} \times S^{2n})$. Composing this with the Kunneth isomorphism $$H^\ast(S^{2n} \times S^{2n}) \stackrel{\sim}{\longrightarrow} H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$$ yields a map$f \colon H^\ast(S^{2n}) \to H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$. Now, $$H^\ast(S^{2n}) = \begin{cases} \mathbb{R} & \text{if }\ast=0 \text{ or } 2n \\ 0 & \text{otherwise.}\end{cases}$$ So, writing$1$and$\lambda$for the generators of$H^0$and$H^{2n}$, respectively, we have$f(1)=1\otimes1$and$f(\lambda)=a(\lambda\otimes1)+b(1\otimes\lambda)$for some$a,b\in\mathbb{R}$. To determine$a$and$b$, first restrict the multiplication map to $S^{2n} \times \{e\}$, where it is the identity map. Consequently$m^*$gives the identification $H^\ast(S^{2n} \times \{e\}) \cong H^\ast(S^{2n})$ and it follows that$a=1$. Similarly,$b=1$. As$fis a ring homomorphism, we also have \begin{align} f(\lambda^2) = f(\lambda)^2 &= (\lambda\otimes1 + 1\otimes\lambda)^2 \\ &= \lambda^2\otimes1+ (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)+1\otimes\lambda^2.\end{align} As\lambda^2 \in H^4 = 0$, this reduces to$0 = (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)$. Recall, however, that the product structure on the tensor product$A \otimes B$of graded rings is given by $$(a \otimes b)(c \otimes d) = (-1)^{\deg a \deg c}(ac\otimes bd).$$ Since$\lambda \in H^2$, this means that $$(\lambda\otimes1)(1\otimes\lambda) = (1\otimes\lambda)(\lambda\otimes1) = \lambda \otimes \lambda.$$ Consequently,$\lambda \otimes \lambda = 0$, which is a contradiction! $\blacksquare\$

Of course, there are several other (perhaps even better) ways of proving this theorem, but I think this proof is fairly charming.