Does anyone know how to prove the fact that $SO(n)$ is a deformation retract of $SO(n,\mathbb{C})$?
Here $SO(n,\mathbb{C}):= \{ A\in M_{n\times n}(\mathbb{C}) | A \cdot A^T = Id \} $ and $SO(n):=SO(n,\mathbb{C}) \cap M_{n\times n} (\mathbb{R})$.
I have seen an argument like this: every connected Lie group has a deformation retract to its maximal compact subgroup, and $SO(n)$ is a maximal compact subgroup of $SO(n,\mathbb{C})$. However, I don't know how to justify these steps. Could you please give me a reference on these two facts? Or, is there a easier proof for the group $SO(n,\mathbb{C})$?
Thank you very much!
You might want to consider the Iwasawa decomposition of a complex reductive group $G$. It states that $G$ is diffeomorphic to a product $K\times A\times N$, where K is a maximal compact subgroup. Here, we have chosen a maximal torus $T\subseteq K$, and a collection of positive roots. The subgroup $N$ is the unipotent radical of the Borel subgroup determined by the positive roots, and A is the real subgroup whose Lie algebra is the imaginary part of the Lie algebra of $T$. Each of $A$ and $N$ is diffeomorphic to its Lie algebra, and therefore contractible. This implies that $G$ is homotopy-equivalent to $K$.