Lie algebra cohomology and Lie groups

Question

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\U{U}\DeclareMathOperator\Sp{Sp}$Let $G$ be a complex Lie group and $\mathfrak{g}$ its Lie algebra (which is over $\mathbb{C}$). My first question is:

(1) When do the cohomology groups $H^\bullet(G,\mathbb{C})$ (singular cohomology) and $H^\bullet(\mathfrak{g},\mathbb{C})$ (Lie algebra cohomology) coincide? (Edit: In particular, is reductive and connected enough? The OP of this MO question implies it is true without proofs.)

For example, we have $$ H^\bullet(\GL(n,\mathbb{C}),\mathbb{C})\cong H^\bullet(\U(n),\mathbb{C}) \cong \bigwedge\nolimits_\mathbb{C}(e_1,\dotsc,e_{2n-1}) \cong H^\bullet(\mathfrak{gl}(n,\mathbb{C}),\mathbb{C}), $$ where $\bigwedge_\mathbb{C}(e_1,\dotsc,e_{2n-1})$ is the free exterior algebra over $\mathbb{C}$ with odd generators $e_{2i-1}$ of degree $(2i-1)$.

In the case of real Lie groups, a well-known sufficient condition for $H^\bullet(G,\mathbb{R})\cong H^\bullet(\mathfrak{g},\mathbb{R})$ is $G$ being compact: the Chevalley–Eilenberg complex $C^\bullet(\mathfrak{g},\mathbb{R})$ is exactly the space $\Omega^\bullet_L(G)$ of left-invariant forms, and if $G$ is compact, it gives the same cohomology as the full de Rham complex $\Omega^\bullet(G)$. Of course, we can use the compactness criterion by regarding a complex Lie group as a real Lie group to say something to an extent, but I'm interested in the non-compact case as well, including $\GL(n,\mathbb{C})$ exhibited above.

Also note that neither $H^\bullet(\GL(n,\mathbb{R}),\mathbb{R})$ nor $H^\bullet(\GL^+(n,\mathbb{R}),\mathbb{R})$ is isomorphic to $H^\bullet(\mathfrak{gl}(n,\mathbb{R}),\mathbb{R})$: the latter is a free exterior algebra over $\mathbb{R}$ of the form above, while the former two are generated only by elements of degree $(4i-1)$ when $n$ is odd.

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My second question is:

(2) How can we compute $H^\bullet(\mathfrak{sp}(2n,k),k)$ for $k$ a field of characteristic $0$?

The ring structure of the cohomology above is independent (up to the change of scalars) of $k$, so we can choose a convenient one. First of all, since $\Sp(2n,\mathbb{R})$ is not compact, we cannot easily reduce to the singular cohomology in the real case. For $\Sp(2n,\mathbb{C})$, I thought (1) will help if true, but I couldn't find any useful literatures concerning the above.

Any references, proof sketch or comments are appreciated.

Answer 1

1. By the Iwasawa decomposition, a connected Lie group deformation retracts onto its maximal compact. If our connected Lie group is $G(\mathbb C)$ for reductive algebraic $G/\mathbb C$, then the Lie algebra $\mathfrak k$ of the maximal compact satisfies $\mathfrak k \otimes_{\mathbb R} \mathbb C = \mathfrak g$. Then $H^*(\mathfrak k, \mathbb R) \otimes_{\mathbb R}\mathbb C \cong H^*(\mathfrak g, \mathbb C)$, so the complex cohomology of $G(\mathbb C)$ agrees with the Lie algebra cohomology of $\mathfrak g$.

With regards to your example of $GL_n(\mathbb R)$, again the cohomology can be computed by deformation retraction to the maximal compact, only now the maximal compact is $O_n(\mathbb R)$.

2. Borel showed that if $G$ is a reductive algebraic group, then $H^*(BG,\mathbb C)$ is a polynomial ring on generators in degrees $2d_1,\ldots, 2d_r$, where $d_i$ are the exponents of $G$. By considering the spectral sequence associated to the path-loop fibration $G \to EG \to BG$, these generators transgress to show $H^*(G,\mathbb C)$ is an exterior algebra on generators in degrees $2d_1-1,\ldots, 2d_r-1$. In the case of the symplectic group $Sp_{2n}$, $r=n$ and $d_i = 2i$.