Global cartan and iwasawa decompositions hold for every nonlinear cover without modified k

The group $K$ is a circle, parametrized by $r(t)=\exp (tk)$ with period $2\pi$. The group $G$ is connected: for $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ put $s=(a^2+c^2{)}^{1\mathrm{/}2}$ and $q=s^{-1}\left(\begin{array}{cc}a& -c\\ c& a\end{array}\right)\in K$; then $q^{-1}g=\left(\begin{array}{cc}s& u\\ 0& s^{-1}\end{array}\right)$, which is joined to $I$ through upper triangular determinant-one matrices with positive diagonal, while $q$ is joined to $I$ inside the circle. A central matrix of $G$ commutes with $I+te$ and $I+tf$ for every real $t$; commuting with both matrix units forces it to be scalar. Determinant one then gives center $\{I,-I\}$. The automorphism $\mathrm{\Theta }(g)=(g^{-1}{)}^{\mathsf{T}}$ fixes both central elements, has differential $\theta$, and fixed group $K$. Thus every hypothesis in [L1] holds for the base group, while no such hypothesis has been presumed for its cover.

By [L1] and step 1.1, $K\times \mathfrak{p}\to G$, $(q,X)\mapsto q\exp X$, is a diffeomorphism. Precomposing its first factor with the universal covering map $r:\mathbb{R}\to K$ shows that $P:\mathbb{R}\times \mathfrak{p}\to G$, $P(t,X)=r(t)\exp X$, is a covering. Its domain is a vector space, hence connected and simply connected by straight-line contraction. By [L3] it identifies, preserving basepoints, with the universal cover $\tilde{G}$. The identification is smooth in covering charts, since both coverings are local diffeomorphisms. Under it the entire inverse image $\tilde{K}={\pi }^{-1}(K)$ is exactly $\mathbb{R}\times \{0\}$, a connected closed embedded submanifold.

The lift starting at the identity of $r(t)$ is $t\mapsto \stackrel{}{\exp }(tk)$ by exponential naturality and uniqueness of path lifts in a covering. In the coordinates of step 2.1 it is $t\mapsto (t,0)$. Thus it is a diffeomorphism and group isomorphism from the additive line onto $\tilde{K}$, not merely an injective immersion. The kernel of $\pi$ consists of $(2\pi m,0)$, $m\in \mathbb{Z}$, and is central by [L3]. Its element $z=\stackrel{}{\exp }(2\pi k)$ is nonidentity and has infinite order; in particular the cover has infinite center.

No compact subgroup $H$ of $\tilde{G}$ can have Lie algebra $\mathbb{R}k$. Such an $H$ would be closed, hence embedded by [L4], and its identity component $H^0$ would be closed in the compact group $H$, hence compact. Uniqueness of the connected immersed subgroup would identify $H^0$, with its intrinsic Lie-group topology, with $\tilde{K}\cong \mathbb{R}$ from step 3.1, a contradiction. This uses closedness of an identity component; it does not assert that any subgroup abstractly isomorphic to $\mathbb{R}$ inside a compact group is noncompact in the subspace topology.

The cover is genuinely nonlinear. Let $\rho :\tilde{G}\to \mathrm{GL}(V)$ be any finite-dimensional real representation and complexify its differential to a complex ${\mathfrak{s}\mathfrak{l}}_2$ representation on $V_{\mathbb{C}}$. The complex matrix $-ik$ has eigenvalues $1,-1$, hence is conjugate in ${\mathrm{GL}}_2(\mathbb{C})$ to $h$. Conjugating the other two members of the standard triple gives a triple with semisimple member $-ik$, so [L5] makes $d\rho (-ik)$ diagonalizable with integer eigenvalues. Therefore $d\rho (k)=id\rho (-ik)$ has eigenvalues in $i\mathbb{Z}$ and $\exp (2\pi d\rho (k))=I$. Naturality of the exponential gives $\rho (z)=I$ for the nonidentity $z$ of step 3.1. Every such representation has nontrivial kernel, so no faithful finite-dimensional real representation exists. The same proof applies to a complex representation without first complexifying.

A verbatim compact-$K$ factorization on this nonlinear cover would require a compact subgroup with Lie algebra $\mathbb{R}k$, impossible by step 4.1. The correct lifted Cartan factor is $\tilde{K}$. Explicitly, the lift of $\mathrm{\Theta }$ in the covering coordinates of step 2.1 is $(t,X)\mapsto (t,-X)$, because $\mathrm{\Theta }(r(t)\exp X)=r(t)\exp (-X)$. It is a group automorphism: its compositions with multiplication on either side are lifts of the same base map on connected $\tilde{G}\times \tilde{G}$ and agree at the identity, hence agree everywhere; its square is the identity by the same uniqueness argument. Its fixed group is precisely $\mathbb{R}\times \{0\}=\tilde{K}$. Thus the replacement is noncompact, and neither the finite-center nor compact-$K$ qualification may be discarded in the stated theorems. This is a counterexample to the joint claim, without asserting that every nonlinear cover has identical behavior.