7

Let $G$ be a semisimple Lie group and let $G = KAK$ be a Cartan decomposition.

For $\mathrm{SL}_2(\mathbb{R})$ it holds for every $g \in G$ that $KgK = Kg^{-1}K$.

Does the same hold for every semisimple Lie group?

Constantin K
  • 419

2 Answers2

9

No. For instance it fails in $\mathrm{SL}_3(\mathbf{R})$. Indeed, choose $g$ the diagonal matrix $(2,2,1/4)$. Let $V_2$ be its $2$-eigenspace.

Suppose by contradiction that $g^{-1}\in KgK$ ($K=\mathrm{SO}(3)$). Equivalently, this means that for some $k\in K$ we have $gkg\in K$. Then $V_2\cap k^{-1}V_2$ has dimension $\ge 1$. For $x\in V_2\cap k^{-1}V_2$ we have $gkg(x)=2gkx=4kx$, so $\|gkg(x)\|=4\|x\|$. For $x\neq 0$, this contradicts $gkg\in K$.

YCor
  • 60,149
  • 3
    You beat me by 3 minutes :-) Interestingly, our examples are the same, but our explanations are different. Let me vote for you. – GH from MO Mar 29 '22 at 22:58
  • 1
    @GHfromMO indeed your argument is better because in $\mathrm{SL}_n$ it says exactly when elements of $A$ are in the same double coset, while my argument is quite ad-hoc. – YCor Mar 31 '22 at 09:20
9

No, this fails already in $\mathrm{SL}_3(\mathbb{R})$. If $Kg_1K=Kg_2K$, then $g_1^Tg_1$ is conjugate (by an element of $K$) to $g_2^Tg_2$. So $g_1^Tg_1$ and $g_2^Tg_2$ have the same (positive eigenvalues). In particular, if $g_1$ and $g_2$ are positive diagonal matrices, then they have the same diagonal entries up to permutation. This rarely happens when $g_1$ and $g_2$ are inverses of each other. For example, $\mathrm{diag(2,2,1/4)}$ and $\mathrm{diag(1/2,1/2,4)}$ have very different diagonal entries, and they are inverses of each other.

GH from MO
  • 98,751