Is there a simple proof (short and low-tech) of the following fact:

(E. Cartan) A connected real Lie group $G$ is diffeomorphic (as a manifold) to $K\times\mathbb{R}^n$ where $K$ is a maximal compact subgroup of $G$. Moreover, all maximal compact subgroups of $G$ are conjugate to $K$.

One nice corollary of this is that the group $K$ is a deformation retract of $G$ so these two groups have the same homotopy type. In the case where $G$ is semi-simple one may deduce the first part of the result above using the existence of the Iwasawa decomposition which is not so trivial to prove! But here since I'm only asking for a diffeomorphism between two manifolds, there might be a simpler argument!

  • $\begingroup$ This theorem is due to E.Cartan only in the case of a semisimple group with finite center. According to Mostow (1955), the general case was done independently by Malcev and Iwasawa. For the conjugacy result, note that the classical nonpositive curvature argument does not work in the general case. $\endgroup$
    – YCor
    Jul 25, 2012 at 1:40

4 Answers 4


About the question of uniqueness of maximal compact subgroups, up to conjugation, I propose a Riemannian geometric approach:

Theorem (É. Cartan). A compact group of isometries of a nonpositively curved complete simply-connected Riemannian manifold has a fixed point.

Proof. (Rough sketch) Consider any orbit. It is compact. By convexity of distance functions (the curvature of the ambient is nonpositive), its center of mass can be defined and it is plainly a fixed point. QED

Now one uses the fact that the symmetric space of non-compact type $G/K$ is nonpositively curved. If $H$ is a compact subgroup of $G$ then it has a fixed point $gK$ by the theorem, so $g^{-1}Hg$ fixes the basepoint $1K$ and hence is contained in $K$.

Edit: on suggestion of Ben, I complete my answer as follows. Let $\mathfrak g = \mathfrak k + \mathfrak p$ be the decomposition of the Lie algebra of $G$ into the eigenspaces of the involution. Since $M$ is complete and nonpositively curved, the Hadamard-Cartan theorem says that the map $\varphi:\mathfrak p \to G/K$ given by $\varphi(X)=(\exp X)K$ is a smooth covering. Next one sees that $\varphi$ is injective (assume $\varphi(X)=\varphi(Y)$, apply $\mathrm{Ad}$ to both sides and use the uniqueness of polar decomposition of a matrix, and the semisimplicity of $\mathfrak g$, to deduce that $X=Y$). Hence there are diffeomorphisms $\mathbf R^n\cong\mathfrak p\cong G/K$. Now it follows rather easily that $\mathfrak p\times K\to G$, $(X,k)\mapsto(\exp X)k$ is a diffeomorhism ($G=\exp[\mathfrak p]\cdot K$ in general for a symmetric space).

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    $\begingroup$ To answer the second question, you used the fact that $G/K$ is nonpostively curved. In the cast that $G$ is semisimple, this is a fairly easy computation using the Killing form. Moreover, this result answers the first question, that $G$ is diffeo to $K\times G/K$ and $G/K$ to $R^n$. $\endgroup$ Jun 19, 2011 at 4:23
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    $\begingroup$ Well I'm confused, say that you take $X=\mathbf{C}/\Lambda$ with $\Lambda=\mathbf{Z}+i\mathbf{Z}$ and you take the isometry $z\mapsto z+\frac{1}{2}$, then there is no fixed point! Are you assuming that your space is simply connected? $\endgroup$ Jun 26, 2011 at 16:18
  • $\begingroup$ Hugo: that's right, the Riemannian manifold must be simply-connected, thanks! A symmetric space of non-compact type is automatically simply-connected, and this can be seen from the argument in the "Edit". $\endgroup$ Jul 13, 2011 at 13:20

I'm a bit skeptical as to how simple and low-tech a proof can be, since a fair amount of Lie group theory has to be in hand. Short of going into the full details of Iwasawa decomposition, you might find it worthwhile to look at two old papers in the Bulletin of the AMS. One is by Mostow:

  • A new proof of E. Cartan's theorem on the topology of semi-simple groups, Bull. Amer. Math. Soc. 55 (1949), 969-980, doi:10.1090/S0002-9904-1949-09325-4,

giving a new proof of Cartan's theorem, while the other is a longer survey on the topology of Lie groups by Samelson:

Concerning the further question about conjugacy of maximal compact subgroups, this strikes me as too deeply enmeshed in the structure theory of Lie groups to be proved easily or quickly. I'd want to start by looking carefully at how this conjugacy theorem has been approached over the years (but of course without ruling out a better approach).

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    $\begingroup$ Another nice reference for Cartan's theorem is Hoschschild's "The structure of Lie groups", see Thm 3.1 in Chapter XV (which indeed requires a fair amount of Lie theory). $\endgroup$ Jun 19, 2011 at 5:26

You might like to start with the group $G=O(n,\mathbb{C})$, for which $K$ should be $O(n,\mathbb{R})$. Put $T^{n-1}=\{z\in\mathbb{C}^n:\sum_kz_k^2=1\}$. There is an evident inclusion $S^{n-1}\to T^{n-1}$, which is a homotopy equivalence. This is quite easy when $n$ is even, and a bit more work when $n$ is odd. We can define a map $\pi:O(n,\mathbb{C})\to T^{n-1}$ by $\pi(A)=\text{ last column of } A$. This is a fibre bundle projection with fibre $O(n-1,\mathbb{C})$. It restricts to give a fibration $O(n-1,\mathbb{R})\to O(n,\mathbb{R})\to S^{n-1}$. You can use this to prove inductively that the inclusion $O(n,\mathbb{R})\to O(n,\mathbb{C})$ is a homotopy equivalence. I don't know whether this can be improved to give a diffeomorphism or if there is any more efficient approach.

  • $\begingroup$ Thanks a lot Neil for this nice argument. I'm still wondering if it is possible to use this $C^{\infty}$-structure on the topological group $G$ and prove very cheaply that $K$ is a deformation retract of $G$. $\endgroup$ Jan 25, 2011 at 16:44

Yet another traditional favorite argument, in reality a variant on the more geometric argument of Cartan's, that applies immediately to $G=GL(n,\mathbb R)$ and $K=O(n,\mathbb R)$ uses the model $G/K\approx C$, the cone of positive-definite symmetric $n$-by-$n$ matrices, via a Cartan decomposition $G=C\cdot K$, which itself follows from the spectral theorem for symmetric operators.

Then, given a compact $K'\subset G$, pick any $x_o\in C$, and consider the average $x=\int_{K'} k\cdot x_o\;dk$. Since the cone $C$ is non-degenerate, $x\not=0$, and, by design, it is $K'$-invariant. Since $G$ is transitive on $C$, all isotropy groups of points are conjugate to the isotropy group of $1_n$, namely, $O(n,\mathbb R)$.

This idea can be adapted to some other of the classical groups.


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