Cohomology Isomorphism of Classifying Spaces and Equivalence of Compact Lie Groups

Question

Let $R$ be a commutative ring with unity, and let $G$ and $H$ be compact Lie groups with $BG$ and $BH$ as their respective classifying spaces. If there exists isomorphism $H^j(BH; R) \to H^j(BG; R)$ for all $j\geq 0$, can we conclude that the groups $G$ and $H$ are isomorphic or homotopy equivalent?

Answer 1

Here is a counterexample with $G, H$ connected and $R = \mathbb{Q}$. Associated to any simple Lie algebra $\mathfrak{g}$ of rank $r$ is a list of $r$ positive integers $d_1, \dots d_r$, the degrees of the fundamental invariants. If $G$ is the corresponding compact simply connected Lie group, then it's known that

• the rational cohomology $H^{\bullet}(G, \mathbb{Q})$ is an exterior algebra on generators of degrees $2d_1 - 1, \dots 2d_r - 1$, and
• the rational cohomology $H^{\bullet}(BG, \mathbb{Q})$ is a polynomial algebra on generators of degrees $2d_1, \dots 2d_r$.

The link above is a table of the degrees $d_i$, and it reveals that the simple Lie algebras of type $B_n$ and $C_n$ have the same degrees. It follows that the corresponding Lie groups $\text{Spin}(2n+1)$ and $\text{Sp}(n)$ are counterexamples as long as they aren't isomorphic. They are isomorphic for $n = 1, 2$ but not past this point, so

$$G = \text{Spin}(7), H = \text{Sp}(3)$$

is a counterexample, where $BG$ and $BH$ both have rational cohomology an exterior algebra on generators of degrees $4, 8, 12$ (the first three Pontryagin classes $p_1, p_2, p_3$ and then what I guess ought to be called "symplectic Pontryagin classes," respectively).

On the other hand, I think this is the only coincidence among the degrees of the fundamental invariants. I'm not sure what happens if you allow semisimple groups or non-semisimple groups.

Answer 2

This is already false in the finite case. For example, if $p$ is odd or $n>1$, $H^\ast(C_{p^n};\mathbb{F}_p)\cong \mathbb{F}_p[x,y]/(y^2)$ as a graded ring, where $x$ lives in degree $2$ and $y$ in degree $1$, but clearly $C_{p^n}$ is not isomorphic to $C_{p^k}$ if $k\neq n$.