Maximal compact subgroups exist and are conjugate in a connected finite center semisimple Lie group

The map $D_R(V,k)=D(k^{-1},-V{)}^{-1}=\exp (V)k$ is a diffeomorphism by [L1] and [L5]. Its first inverse coordinate $F:G\to \mathfrak{p}$ is constant on right $K$-cosets by uniqueness. Composing with local sections of $q$ proves that its induced map on $G\mathrm{/}K$ is smooth. It is inverse to $\mathrm{\Phi }(V)=\exp (V)K$, so $\mathrm{\Phi }$ is a diffeomorphism, with global section $s(\mathrm{\Phi }(V))=\exp V$. In particular $G\mathrm{/}K$ is connected, nonempty and boundaryless. No exclusion of compact ideals is needed.

Trace cyclicity and ${\mathrm{ad}}_{[U,V]}=[{\mathrm{ad}}_U,{\mathrm{ad}}_V]$ give $B([U,V],W)=B(U,[V,W])$. For an automorphism $A$, ${\mathrm{ad}}_{AU}=A{\mathrm{ad}}_U{A}^{-1}$ gives $B(AU,AV)=B(U,V)$. For $k\in K$, differentiating $\mathrm{\Theta }{C}_k=C_k\mathrm{\Theta }$ shows that ${\mathrm{Ad}}_k$ commutes with $\theta$; hence it preserves $\mathfrak{p}$ and its positive form $B=B_{\theta }{\mathrm{\mid }}_{\mathfrak{p}}$. Let $j=d{q}_e{\mathrm{\mid }}_{\mathfrak{p}}$; its kernel is zero and it is an isomorphism by [L1]. The formula $⟨V,W{⟩}_{gK}=B(j^{-1}d{L}_{g^{-1}}V,j^{-1}d{L}_{g^{-1}}W)$ is independent of the representative by the isotropy formula in [L2]. It is positive definite, left invariant and smooth using the local sections in [L1].

Pull $\omega$ back along a local section and split $s^{\ast }\omega =a+u$ into $\mathfrak{k}$ and $\mathfrak{p}$ components. Differentiating $q\circ s=\mathrm{i}\mathrm{d}$ gives $V=d{L}_{s}j(u(V))$, so $u$ identifies the tangent bundle locally with $\mathfrak{p}$ and the metric is $B(u(V),u(W))$. Splitting [L3] gives $du(V,W)+[a(V),u(W)]-[a(W),u(V)]=0$ and $da(V,W)+[a(V),a(W)]=-[u(V),u(W)]$. Thus $u({\mathrm{\nabla }}_{V}Z)=V(u(Z))+[a(V),u(Z)]$ is a connection: the first identity makes its torsion zero, and Killing invariance in step 1.2 makes it metric compatible. Uniqueness in [L4] glues these local connections. Expanding their curvature and applying the second identity gives $u(R(V,W)Z)=-[[u(V),u(W)],u(Z)]$. Hence for $C=[U,V]\in \mathfrak{k}$, $⟨R(U,V)V,U⟩=B(C,C)\le 0$. This construction works with compact ideals still present.

Finally, $K$ is maximal compact directly from the global Cartan coordinates. If a compact subgroup $K_1$ contains $K$ and $g\in K_1$, write $g=k\exp V$ by [L1]. Then $\exp V=k^{-1}g\in K_1$ and $\exp (nV)\in K_1$ for every positive integer $n$. The continuous first inverse coordinate $F$ from step 1.1 has compact, thus bounded, image on $K_1$, whereas $F(\exp (nV))=nV$. Thus $V=0$ and $g\in K$. It follows that $K_1=K$.

Along $c(t)=\mathrm{\Phi }(tV)$ the global section of step 1.1 is $s(c(t))=\exp (tV)$, whose left Maurer–Cartan velocity is the constant $V$ by the subgroup law [L5]. Thus $a(c^{\mathrm{\prime }})=0$, $u(c^{\mathrm{\prime }})=V$, and the connection formula of step 2.1 gives $D_t{c}^{\mathrm{\prime }}=0$. Every initial velocity at $o$ is $jV$ for a unique $V$. Translating these global geodesics by $G$ and using [L5] proves that every initial datum has a global geodesic; uniqueness makes each maximal domain all of $\mathbb{R}$. By [L6], closed bounded subsets are compact and minimizing geodesics exist. Moreover the Riemannian exponential at $o$, identified by $j$, equals $\mathrm{\Phi }$, and at any other point it is its isometric translate. It is therefore a diffeomorphism at every point, and the unique radial connector to each point is minimizing.

Fix $z\in G\mathrm{/}K$ and an affinely parametrized geodesic $\gamma (s)$. By step 3.1, $\alpha (s,t)={\mathrm{Exp}}_z(t{\mathrm{Exp}}_z^{-1}\gamma (s))$, $0\le t\le 1$, is a smooth variation by the unique minimizing geodesics from $z$. Put $T={\mathrm{\partial }}_t\alpha$, $J={\mathrm{\partial }}_s\alpha$ and $h(s)=\frac{1}{2}d(z,\gamma (s){)}^2=\frac{1}{2}{\int }_0^1\mathrm{\mid }T{\mathrm{\mid }}^{2}dt$. Torsion freeness gives $D_{s}T=D_{t}J$ (in coordinates the mixed partials and the symmetric Christoffel terms coincide). The curvature definition applied to the pulled-back connection gives $(D_s{D}_t-D_t{D}_s)T=R(J,T)T$; indeed expansion of the two coordinate covariant derivatives cancels the second derivatives and leaves exactly the curvature coefficients. Since $D_{t}T=0$, we get $D_t^{2}J=-R(J,T)T$. First variation [L7] gives $h^{\mathrm{\prime }}=⟨J,T⟩{\mathrm{\mid }}_0^1$. Differentiate once more: at $t=0$, $J=0$ and $D_{s}J=0$; at $t=1$, $J={\gamma }^{\mathrm{\prime }}$ and $D_{s}J=0$ since $\gamma$ is a geodesic. Consequently integration of the metric product rule gives $h^{\mathrm{\prime }\mathrm{\prime }}=⟨J,D_{t}J⟩{\mathrm{\mid }}_0^1={\int }_0^1(\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2-⟨R(J,T)T,J⟩)dt\ge {\int }_0^1\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^{2}dt\ge \mathrm{\mid }{\gamma }^{\mathrm{\prime }}(s){\mathrm{\mid }}^2\mathrm{.}$ For the last inequality, parallel translate to one fixed tangent space; the resulting vector function starts at zero and ends with norm $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }$, so integration and Cauchy–Schwarz on $[0,1]$ give the bound. The curvature inequality is step 2.1. The formula also holds when $\gamma (s)=z$, because the inverse exponential and the displayed energy are smooth there.

For the minimizing geodesic $\gamma :[0,1]\to G\mathrm{/}K$ from $x$ to $y$, its constant speed is $d(x,y)$. Step 4.1 implies that $s\mapsto d(z,\gamma (s){)}^2-d(x,y{)}^2{s}^2$ has nonnegative second derivative. Convexity at $s=1\mathrm{/}2$ therefore yields $d(z,\gamma (1\mathrm{/}2){)}^2\le \frac{1}{2}d(z,x{)}^2+\frac{1}{2}d(z,y{)}^2-\frac{1}{4}d(x,y{)}^2\mathrm{.}$ This proves the needed inequality rather than attributing it to existence of local convex neighborhoods.

Let $L\le G$ be compact and let $O=L\cdot o$. This is nonempty compact by continuity of the action. Put $f(x)={\max }_{z\in O}d(x,z)$. Triangle inequality shows $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le d(x,y)$; since the action preserves lengths, hence distance, $f$ is $L$-invariant. Also $f(x)\ge d(x,o)$ because $o\in O$. The nonempty sublevel set $f\le f(o)$ is closed and bounded, hence compact by step 3.1. Continuity implies that $f$ attains a global minimum $D$ there: outside the sublevel set $f>f(o)$. Taking maxima over $z\in O$ in step 5.1 gives the same midpoint inequality for $f^2$. If $x,y$ both minimize, it gives $D^2\le D^2-d(x,y{)}^2\mathrm{/}4$, hence $x=y$. By $L$-invariance the unique minimizer $x_0$ is fixed by every element of $L$. Writing $x_0=hK$ gives $L\subseteq hK{h}^{-1}$ by the coset action.

Conjugation preserves compactness and inclusion, so each conjugate of the maximal compact group $K$ is maximal compact. Conversely any maximal compact subgroup equals a conjugate containing it by step 6.1. This proves all assertions. If $\mathfrak{p}=0$, step 1.1 gives a singleton quotient and $G=K$; the metric and all curve calculations use zero tangent spaces and the fixed-point argument still applies. AC supplies the global Cartan theorem and countable choice required in [L1], [L3], [L5], [L6]; no compact subgroup was assumed connected.