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.)

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    $\begingroup$ 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 $\mathbb{R}^n$. $\endgroup$ – Qiaochu Yuan Sep 21 '14 at 6:36
  • $\begingroup$ @QiaochuYuan is isomorphic? Does this result gives us an alternative proof for the following:mathoverflow.net/questions/247935/… ? $\endgroup$ – Ali Taghavi Aug 24 '16 at 11:19

Qiaochu Yuan's comment sent me in the right direction:

  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 $1$.
  • $\begingroup$ you don't need G to be semisimple for the Iwasawa decomposition? $\endgroup$ – Tim kinsella Sep 21 '14 at 22:32
  • $\begingroup$ You're right, thanks @Timkinsella. In fact, the Iwasawa decomposition can be used in the proof of the general case, it seems. $\endgroup$ – Travis Willse Sep 22 '14 at 3:23

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