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What can one say about the topology of $\text{PSL}(2, \mathbb{C})$, for example its cohomology groups?

By using the long exact sequence in homotopy one can show that the inclusion $$\text{SO}_3 \to \text{PSL}(2, \mathbb{C})$$ that sends any rotation $\phi \in \text{SO}_3$ to the moebius transformation mapping $0 \to \phi(0), 1 \to \phi(1), \infty \to \phi(\infty)$ is a homotopy equivalence.

Are there more geometric ways to see this?

Casaubon
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    Indeed as in the answers below every Lie group is diffeomorphic to $K \times \Bbb R^n$, where $K$ is its maximal compact subgroup. This is a version of the Iwasawa-Malcev theorem. So anything purely homotopical about $G$ can be reduced to working with $K$. If you're interested in how the group structure and topology play together (eg, group cohomology with continuous/smooth cochains), there's more interesting things to say. –  Sep 28 '15 at 17:36

3 Answers3

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One generic way, after identifying $PSL_{2}(\mathbb{C})$ with $SO(3,1)(\mathbb{R})$, would be to use the Iwasawa decomposition $G=NAK$, where $K=SO(n)$ and $A$ is the split Cartan, $N$ the unipotent group. As $N,A$ are contractible (isomorphic to $\mathbb{R},\mathbb{R}^{n-1}$), we see that the topology is "concentrated" in $K=SO(n)$.

Anyhow, I second Igor's reference as well.

Asaf
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See "Topology of Lie Groups" by Hans Samelson.

Igor Rivin
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Another way to see this which is more elementary, is to realise that the image of this embedding is $PSU_2(\mathbb{C})$ and because of the existence of QR decompositions, $PSL_2(\mathbb{R})$ is topologically the product of $PSU_2(\mathbb{R})$ and the subgroup $\begin{pmatrix} 1 & \mathbb{C} \\ 0 & 1\end{pmatrix}$. So the homotopy equivalence is given by contracting the triangular part.

Johannes Hahn
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