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  "body": "\n## Statement\n\nLet $G$ be a connected [[compact Lie group|9TFC6ADC1C]] and let $k$ be a field of characteristic $0$.\nThen the [[cohomology|HKZABKRPN6]] $H^*(G; k)$ is isomorphic, as a graded $k$-algebra, to an [[exterior algebra|0MDMZ965VE]] $\\Lambda_k(x_1, \\dots, x_r)$ on finitely many generators $x_i$ of odd degrees $d_1, \\dots, d_r$.\n\n## Remarks\n\nHopf's proof uses only the [[Hopf algebra|WJXK0V9NE9]] structure given by the multiplication of $G$, together with finite-dimensionality.\nOver a field of characteristic $0$, a connected graded-commutative Hopf algebra is free as a graded-commutative algebra.\nThat is, it is a polynomial algebra on generators of even degree tensored with an exterior algebra on generators of odd degree.\nA polynomial generator of even degree would have infinitely many nonzero powers, which is impossible in finite dimension.\nSo all generators have odd degree.\nThe same argument works for any connected H-space with finite-dimensional rational cohomology.\n\nFurther facts: the number $r$ of generators equals the rank of $G$, the dimension of a maximal torus, and $d_1 + \\cdots + d_r = \\dim G$, since the product $x_1 \\cdots x_r$ spans the top cohomology.\n\nExample: for $G = \\mathrm{SU}(n)$, the degrees are $3, 5, \\dots, 2n - 1$, with $r = n - 1$ and $3 + 5 + \\cdots + (2n - 1) = n^2 - 1 = \\dim \\mathrm{SU}(n)$.\nFor $G = \\mathrm{SU}(2) \\cong S^3$, this is $H^*(S^3; \\Q) = \\Lambda(x_3)$.\n\nThe characteristic $0$ hypothesis is needed.\nThe group $\\mathrm{SO}(3)$ is homeomorphic to $\\R P^3$, and $H^*(\\R P^3; \\F_2) = \\F_2[x]/(x^4)$ with $\\abs{x} = 1$, which is not an exterior algebra since $x^2 \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nThe holes of a compact connected Lie group have a very simple pattern.\nThere are a few basic holes, all of odd dimension, and every other hole is a combination of distinct basic holes.\nFor the group $\\mathrm{SU}(2)$, a $3$-sphere, there is just one basic hole of dimension $3$.\n\n### L2 human\n\nThe theorem explains why the rational cohomology of a compact Lie group looks like that of a product of odd spheres.\nFor example $H^*(\\mathrm{U}(n); \\Q)$ is that of $S^1 \\times S^3 \\times \\cdots \\times S^{2n-1}$.\nThe degrees $d_i$ are $2m_i + 1$, where the $m_i$ are the exponents of the Weyl group, and the $m_i + 1$ are the degrees of its basic invariant polynomials.\nWith integer or mod $p$ coefficients the cohomology can have torsion, and its study is much harder.\n",
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