# Do all Lie groups admit deformation retracts onto compact subgroups?

The noncompact Lie group $SL(n, \mathbb{R})$ admits a deformation retract onto the compact Lie group $SO(n, \mathbb{R})$ via polar decomposition. Do all noncompact Lie groups admit deformation retracts onto compact subgroups? (This is motivated by Is this a valid way to show $\chi(SL_n(\mathbb{R}))=0$? , where one solution uses the existence of the aforementioned retract.)

• IIRC an even stronger statement is true, namely that every connected Lie group is diffeomorphic to the product of a maximal compact subgroup and some $\mathbb{R}^n$. – Qiaochu Yuan Sep 21 '14 at 6:36
• @QiaochuYuan is isomorphic? Does this result gives us an alternative proof for the following:mathoverflow.net/questions/247935/… ? – Ali Taghavi Aug 24 '16 at 11:19

1. The theorem he gives is correct as stated, and is due to Cartan: Any connected Lie group is diffeomorphic to the product of a maximal compact subgroup and $$\mathbb{R}^n$$ for some $$n$$ (and in fact all maximal compact subgroups are conjugate). There's some good discussion of proofs of this fact on Math Overflow: https://mathoverflow.net/questions/53080/homotopy-type-of-connected-lie-groups . Mostow gives a compact argument in this 1949 Bulletin AMS article: http://www.ams.org/journals/bull/1949-55-10/S0002-9904-1949-09325-4/home.html .
2. Something better yet is true in the semisimple case: Any semisimple Lie group $$G$$ admits a so-called Iwasawa decomposition $$G := KAN$$ into Lie groups, where
• $$K$$ is the maximal compact subgroup,
• $$A$$ is abelian, and
• $$N$$ is nilpotent. In particular, the factors $$A$$ and $$N$$ are contractible, so $$K$$ is a deformation retract of $$G$$. In the case of the given example, $$G = SL(n, \mathbb{R})$$, $$K = SO(n, \mathbb{R})$$, $$A$$ is the group of positive diagonal $$n \times n$$ matrices, and $$N$$ is the group of upper triangular matrices with diagonal elements all $$1$$.