Hopf's theorem on cohomology of compact Lie groups

Remarks

Hopf's proof uses only the Hopf algebra structure given by the multiplication of $G$, together with finite-dimensionality. Over a field of characteristic $0$, a connected graded-commutative Hopf algebra is free as a graded-commutative algebra. That is, it is a polynomial algebra on generators of even degree tensored with an exterior algebra on generators of odd degree. A polynomial generator of even degree would have infinitely many nonzero powers, which is impossible in finite dimension. So all generators have odd degree. The same argument works for any connected H-space with finite-dimensional rational cohomology.

Further 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.

Example: for $G = \mathrm{SU}(n)$, the degrees are $3, 5, \dots, 2n - 1$, with $r = n - 1$ and $3 + 5 + \cdots + (2n - 1) = n^2 - 1 = \dim \mathrm{SU}(n)$. For $G = \mathrm{SU}(2) \cong S^3$, this is $H^*(S^3; \Q) = \Lambda(x_3)$.

The characteristic $0$ hypothesis is needed. The 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 $x^2 \ne 0$.

Going deeper

For a student

The theorem explains why the rational cohomology of a compact Lie group looks like that of a product of odd spheres. For example $H^*(\mathrm{U}(n); \Q)$ is that of $S^1 \times S^3 \times \cdots \times S^{2n-1}$. The 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. With integer or mod $p$ coefficients the cohomology can have torsion, and its study is much harder.