Cartan decomposition gives the invariant metric and curvature of G mod K

For any Lie-algebra automorphism $T$, ${\mathrm{ad}}_{TU}=T{\mathrm{ad}}_U{T}^{-1}$, so trace cyclicity gives $B(TU,TV)=B(U,V)$. Also Jacobi gives ${\mathrm{ad}}_{[U,V]}=[{\mathrm{ad}}_U,{\mathrm{ad}}_V]$; trace cyclicity then gives $B([U,V],W)=B(U,[V,W])$. For $k\in K$, differentiate $\mathrm{\Theta }\circ C_k=C_k\circ \mathrm{\Theta }$, where $C_k$ is conjugation by $k$, to obtain ${\theta }_{\ast }{\mathrm{Ad}}_k={\mathrm{Ad}}_k{\theta }_{\ast }$. Thus ${\mathrm{Ad}}_k$ preserves both summands and $B_{{\theta }_{\ast }}$. In particular its restriction to ${\mathfrak{p}}_0$ preserves $B=B_{{\theta }_{\ast }}$.

The map $j=d{q}_e{\mathrm{\mid }}_{{\mathfrak{p}}_0}$ is an isomorphism, since $\ker d{q}_e={\mathfrak{k}}_0$. For $g^{\mathrm{\prime }}=gk$, the quotient differentials satisfy $d{L}_{g^{\mathrm{\prime }-1}}=d{L}_{k^{-1}}d{L}_{g^{-1}}$ and $j^{-1}d{L}_{k^{-1}}={\mathrm{Ad}}_{k^{-1}}{j}^{-1}$. Hence the two representatives give the same pairing by step 1.1. Positive definiteness follows from the isomorphisms in the metric formula, and left invariance follows from cancellation of translations. For smoothness, [L2] supplies local smooth sections $s:U\to G$ of $q$: in submersion coordinates fix $v$ at its value at the chosen point. The displayed metric evaluated using $s(x)$ has smooth coefficients, so it is a smooth metric.

On such a section let $\eta =s^{\ast }\omega =a+u$, split into its ${\mathfrak{k}}_0$-valued part $a$ and ${\mathfrak{p}}_0$-valued part $u$. For $V\in T_{x}U$, the identity $q\circ s=\mathrm{id}$ gives $V=d{L}_{s(x)}j(u(V))$: the $a(V)$ part is vertical and is killed by $dq$. Thus $u:TU\to U\times {\mathfrak{p}}_0$ is a smooth fibrewise isomorphism and $⟨V,W⟩=B(u(V),u(W))$. Splitting the pulled-back Maurer–Cartan equation using [L1] gives $du(V,W)+[a(V),u(W)]-[a(W),u(V)]=0,$ $da(V,W)+[a(V),a(W)]=-[u(V),u(W)]\mathrm{.}$ Here $du(V,W)=V(u(W))-W(u(V))-u([V,W])$, and similarly for $da$.

Define a local connection by $u({\mathrm{\nabla }}_{V}Z)=V(u(Z))+[a(V),u(Z)]$. Its linearity in $V$ and Leibniz rule in $Z$ follow directly, so this is an affine connection. It is metric compatible: differentiating $B(u(Z),u(W))$ gives the derivative terms, and the two extra bracket terms sum to zero by the invariant Killing identity of step 1.1. Its torsion, represented by $u$, is $du(V,W)+[a(V),u(W)]-[a(W),u(V)]$, which is zero by step 3.1. Thus [L4] identifies it as Levi–Civita. These local formulas agree on overlaps by uniqueness and define the global connection; no assertion about arbitrary fundamental fields being invariant or about isometry transport being parallel is required.

Put $z=u(Z)$ and $D_{V}z=V(z)+[a(V),z]$. Expanding $D_V{D}_{W}z-D_W{D}_{V}z-D_{[V,W]}z$ cancels all derivatives of $z$ and leaves $u(R(V,W)Z)=[da(V,W)+[a(V),a(W)],z]=-[[u(V),u(W)],u(Z)]\mathrm{.}$ The first equality follows from Jacobi for the two nested $a$-brackets; the second is step 3.1. At the origin choose a section with $s(eK)=e$, so $u=j^{-1}$ there. This yields $R(X,Y)Z=-[[X,Y],Z]$ for $X,Y,Z\in {\mathfrak{p}}_0$, in the stated identification.

Let $C=[X,Y]\in {\mathfrak{k}}_0$. The numerator for sectional curvature is $B(-[C,Y],X)=-B(C,[Y,X])=B(C,C)=-B_{{\theta }_{\ast }}(C,C)$. The first equality uses that the metric is $B$ on ${\mathfrak{p}}_0$, the second uses Killing invariance, and the last uses ${\theta }_{\ast }C=C$. It is nonpositive by positive definiteness of $B_{{\theta }_{\ast }}$. For independent $X,Y$ division by the positive Gram determinant gives the claimed formula; for an orthonormal pair that determinant is one. An isometry preserves the Levi–Civita connection by uniqueness and hence its curvature, so transitivity extends this conclusion everywhere. If $\dim {\mathfrak{p}}_0<2$, there are no two-planes and the sectional assertion is vacuous; the metric and curvature formulas still apply.