Global iwasawa decomposition

Consequences of the global Cartan decomposition of [L1]: $K$ is connected because $G\cong K\times {\mathfrak{p}}_0$ is connected; $\exp {\mathrm{\mid }}_{{\mathfrak{p}}_0}$ is injective with inverse the second coordinate of the diffeomorphism, $Z\subseteq K$, and $K\cap \exp ({\mathfrak{p}}_0)=\{1\}$, because $k=\exp X$ forces $k\exp 0=\exp X$ and uniqueness of the factorization gives $k=1$ and $X=0$; moreover the adjoint group $\mathrm{Ad}(G)$ is a connected subgroup of $\mathrm{GL}({\mathfrak{g}}_0)$ with Lie algebra $\mathrm{ad}({\mathfrak{g}}_0)$, so by [L5] it equals the identity component $\mathrm{Aut}({\mathfrak{g}}_0{)}^0$ and is closed in $\mathrm{GL}({\mathfrak{g}}_0)$.

Adapted basis: fix a regular $H_0\in \mathfrak{a}$ with $\lambda (H_0)>0$ for all $\lambda \in {\mathrm{\Sigma }}^{+}$, and choose an orthonormal basis of ${\mathfrak{g}}_0$ for $B_{\theta }$ consisting of joint eigenvectors of the commuting family $\{\mathrm{ad}H:H\in \mathfrak{a}\}$, listed in non-increasing order of the value $\lambda (H_0)$, that is, with the root vectors of ${\mathrm{\Sigma }}^{+}$ first, then the vectors of ${\mathfrak{g}}_0^0$, then the root vectors of ${\mathrm{\Sigma }}^{-}$, each block ordered so that $\lambda (H_0)$ decreases; the Killing trace identity gives $B([X,Y],Z)=-B(Y,[X,Z])$, and therefore $\mathrm{ad}(X{)}^{\ast }=-\mathrm{ad}(\theta X)$ for $B_{\theta }$. Thus the matrices of $\mathrm{ad}{\mathfrak{k}}_0$ are skew, those of $\mathrm{ad}\mathfrak{a}$ are diagonal with real entries, and those of $\mathrm{ad}\mathfrak{n}$ are strictly upper triangular: indeed $\mathrm{ad}\mathfrak{n}$ is upper triangular because a nonzero matrix entry from ${\mathfrak{g}}_0^{{\lambda }_j}$ to ${\mathfrak{g}}_0^{{\lambda }_i}$ requires ${\lambda }_i={\lambda }_j+\lambda$ with $\lambda \in {\mathrm{\Sigma }}^{+}$, hence ${\lambda }_i(H_0)>{\lambda }_j(H_0)$ and $i<j$, and the diagonal entry vanishes since no positive restricted root is $0$.

Regularity of products: let $H$ be a Lie group with Lie algebra $\mathfrak{h}$, and let $\mathfrak{h}=\mathfrak{s}\oplus \mathfrak{t}$ be a vector-space direct sum of Lie subalgebras with associated connected subgroups $S,T$; then the multiplication map $\mathrm{\Phi }:S\times T\to H$, $\mathrm{\Phi }(s,t)=st$, is everywhere regular: identifying the tangent space of $S\times T$ at $(s_0,t_0)$ with $\mathfrak{s}\oplus \mathfrak{t}$ by left translations within $S$ and $T$, and the tangent space of $H$ at $s_0{t}_0$ with $\mathfrak{h}$ by left translation, one computes $d{\mathrm{\Phi }}_{(s_0,t_0)}(X)=\mathrm{Ad}(t_0^{-1})X$ for $X\in \mathfrak{s}$ and $d{\mathrm{\Phi }}_{(s_0,t_0)}(Y)=Y$ for $Y\in \mathfrak{t}$, which in the decomposition $\mathfrak{h}=\mathfrak{s}\oplus \mathfrak{t}$ is block triangular with invertible diagonal blocks ${\mathrm{Ad}}_{\mathfrak{h}\mathrm{/}\mathfrak{t}}(t_0^{-1})$ and the identity on $\mathfrak{t}$, hence is invertible.

Let $U$ be the full upper triangular unipotent matrix group in the dimension of ${\mathfrak{g}}_0$. Its strict upper entries give a global Euclidean coordinate system, so $U$ is connected and simply connected; its Lie algebra $\mathfrak{u}$ is strictly upper triangular and nilpotent, since a product of as many strict upper triangular matrices as the dimension is zero. By [L4], exponential is a diffeomorphism $\mathfrak{u}\to U$ and multiplication in these coordinates is the finite BCH polynomial. Consequently for every Lie subalgebra $\mathfrak{v}\subseteq \mathfrak{u}$, $\exp \mathfrak{v}$ is a subgroup: BCH stays in $\mathfrak{v}$ and inversion is $X\mapsto -X$. It is closed and diffeomorphic to the vector space $\mathfrak{v}$. It is therefore the unique connected subgroup with that Lie algebra. In dimension zero these groups are singletons.

Matrix types and closedness: with the basis of step 1.2 the matrices of $\mathrm{Ad}(K)$ are orthogonal (either exponentiate the skew matrices, since $K$ is connected by step 1.1, or use preservation of $B$ and commutation with $\theta$), those of $A_1=\mathrm{Ad}(A)=\exp (\mathrm{ad}\mathfrak{a})$ are diagonal with positive diagonal entries, and the connected subgroup $N_1=\mathrm{Ad}(N)$ has Lie algebra $\mathrm{ad}\mathfrak{n}$ and equals $\exp (\mathrm{ad}\mathfrak{n})$ inside the unipotent upper-triangular group. Moreover $A_1$ is closed in the group $D$ of diagonal matrices with positive entries, because the exponential map $\mathfrak{d}\to D$ is a diffeomorphism and $\mathrm{ad}\mathfrak{a}\subseteq \mathfrak{d}$ is a linear subspace; similarly $N_1$ is closed and simply connected by step 1.4 applied to $\mathfrak{v}=\mathrm{ad}\mathfrak{n}$. The homomorphism $\mathrm{Ad}{\mathrm{\mid }}_N:N\to N_1$ is onto and has invertible identity differential, since $\mathrm{ad}:\mathfrak{n}\to \mathrm{ad}\mathfrak{n}$ is injective for the centerless semisimple algebra ${\mathfrak{g}}_0$. Choose an identity neighbourhood $V$ on which this homomorphism is a diffeomorphism, shrunk so $V{V}^{-1}$ meets its discrete kernel only at the identity. The inverse image of its image is the disjoint union of kernel translates of $V$, each mapping diffeomorphically onto that image. Translating this construction in the target proves it is a covering homomorphism. Because $N$ is connected and $N_1$ is simply connected, this covering has one sheet, so $\mathrm{Ad}{\mathrm{\mid }}_N$ is an isomorphism. Thus $N$ is simply connected, without presupposing that every element of $N$ is exponential. The same injectivity for $A$ follows from $\ker \mathrm{Ad}{\mathrm{\mid }}_A=A\cap Z\subseteq K\cap \exp ({\mathfrak{p}}_0)=\{1\}$ by step 1.1, and $A_1\cap N_1=\{1\}$ because a positive diagonal unipotent matrix is the identity.

The subgroups $A$, $N$: first, $\exp H$ normalizes $N$ for every $H\in \mathfrak{a}$, since $\exp (\mathrm{ad}H)$ preserves $\mathfrak{n}$ rootwise and conjugation therefore gives the same connected subgroup by [L4]. Hence $AN$ is a subgroup, connected and generated by $A$ and $N$. By [L2] the subalgebra $\mathfrak{a}\oplus \mathfrak{n}$ is a Lie subalgebra, and $A$ and $N$ are the connected subgroups with Lie algebras $\mathfrak{a}$ and $\mathfrak{n}$; $A$ is abelian and isomorphic to $\mathfrak{a}$ through $\exp$ by step 1.1, and $N=\exp (\mathfrak{n})$ with $\exp \text{\hspace{0.17em}⁣}:\mathfrak{n}\to N$ a diffeomorphism by [L4], because $N$ is connected by definition, nilpotent since its Lie algebra $\mathfrak{n}$ is, and simply connected because $\mathrm{Ad}{\mathrm{\mid }}_N\text{\hspace{0.17em}⁣}:N\to N_1$ is an injective Lie-group homomorphism onto the simply connected group $N_1$ of step 2.1; by step 1.3 applied to $\mathfrak{h}=\mathfrak{a}\oplus \mathfrak{n}$, $\mathfrak{s}=\mathfrak{a}$, $\mathfrak{t}=\mathfrak{n}$, the multiplication $A\times N\to AN$ is everywhere regular; it is also injective, since $a_1{n}_1=a_2{n}_2$ gives $a_2^{-1}{a}_1=n_2{n}_1^{-1}\in A\cap N$, and $A\cap N=\{1\}$ because $\mathrm{Ad}(A\cap N)\subseteq A_1\cap N_1=\{1\}$ by step 2.1 and $\ker \mathrm{Ad}\cap A=A\cap Z=\{1\}$; hence $A\times N\to AN$ is a diffeomorphism, $AN$ is the analytic subgroup with Lie algebra $\mathfrak{a}\oplus \mathfrak{n}$, and the map $A\times N\to G$ is injective.

$K\times AN\to G$ is everywhere regular: by [L2] one has the vector-space direct sum ${\mathfrak{g}}_0={\mathfrak{k}}_0\oplus (\mathfrak{a}\oplus \mathfrak{n})$ in which both summands are Lie subalgebras, so by step 1.3 applied to $\mathfrak{h}={\mathfrak{g}}_0$, $\mathfrak{s}={\mathfrak{k}}_0$, $\mathfrak{t}=\mathfrak{a}\oplus \mathfrak{n}$ and the connected subgroups $K$ and $AN$ of step 3.1, the multiplication map $K\times AN\to G$ has invertible differential at every point.

The adjoint-group decomposition: put $G_1=\mathrm{Ad}(G)$, $K_1=\mathrm{Ad}(K)$, $A_1=\mathrm{Ad}(A)$, $N_1=\mathrm{Ad}(N)$; then the multiplication map $K_1\times A_1\times N_1\to G_1$ is bijective: it is injective because $k_1{a}_1{n}_1=k_1^{\mathrm{\prime }}{a}_1^{\mathrm{\prime }}{n}_1^{\mathrm{\prime }}$ with $k_1,k_1^{\mathrm{\prime }}\in K_1$ orthogonal, $a_1,a_1^{\mathrm{\prime }}$ diagonal with positive entries and $n_1,n_1^{\mathrm{\prime }}$ unipotent upper triangular forces $k:=k_1^{\mathrm{\prime }-1}{k}_1=a_1^{\mathrm{\prime }}{n}_1^{\mathrm{\prime }}{n}_1^{-1}{a}_1^{-1}=(a_1^{\mathrm{\prime }}{a}_1^{-1})\cdot (a_1(n_1^{\mathrm{\prime }}{n}_1^{-1})a_1^{-1})$, and both factors on the right are upper triangular with positive diagonal entries because $A_1$ is diagonal with positive entries and normalizes the unipotent group $N_1$; so the eigenvalues of $k$ are these positive diagonal entries, while $k$ is orthogonal and therefore has all eigenvalues of modulus $1$; hence every diagonal entry of $k$ is a positive real number of modulus $1$, and an orthogonal upper triangular matrix with all diagonal entries one is the identity (successive orthogonality of its columns gives every entry above the diagonal zero); thus $k=1$ and $k_1=k_1^{\mathrm{\prime }}$; then $a_1{n}_1=a_1^{\mathrm{\prime }}{n}_1^{\mathrm{\prime }}$, and since $A_1\cap N_1=\{1\}$ the parameterization $A_1\times N_1\to A_1{N}_1$ is injective, giving $a_1^{\mathrm{\prime }}=a_1$ and $n_1^{\mathrm{\prime }}=n_1$; and it is surjective because its image is open (it is everywhere regular by steps 3.1 and 4.1 transported through the local diffeomorphism $\mathrm{Ad}$) and closed (the image is the product $K_1\cdot A_1{N}_1$ of the compact set $K_1$ and the closed set $A_1{N}_1$, the latter being closed because if $a_m{n}_m\to x$ then $x$ is upper triangular and invertible, its diagonal entries are nonzero limits of positive entries and hence positive; the diagonal parts therefore converge in the positive diagonal group, so $a_m\to a\in A_1$ and then $n_m=a_m^{-1}(a_m{n}_m)\to a^{-1}x\in N_1$), and $G_1$ is connected.

Lifting to $G$: let $g\in G$; by step 5.1 write $\mathrm{Ad}(g)=k_1{a}_1{n}_1$ with $k_1\in K_1$, $a_1\in A_1$, $n_1\in N_1$; choose $k\in K$ with $\mathrm{Ad}(k)=k_1$, and let $a\in A$ and $n\in N$ be the unique elements with $\mathrm{Ad}(a)=a_1$ and $\mathrm{Ad}(n)=n_1$ (step 2.1); then $\mathrm{Ad}(g(kan{)}^{-1})=k_1{a}_1{n}_1(k_1{a}_1{n}_1{)}^{-1}=1$, so $z:=g(kan{)}^{-1}\in \ker \mathrm{Ad}=Z\subseteq K$ by [L1], and $g=(zk)an$ exhibits $g$ in $K\cdot A\cdot N$; hence $K\times A\times N\to G$ is surjective.

Injectivity on $G$: if $k_1{a}_1{n}_1=k_2{a}_2{n}_2$ with $k_i\in K$, $a_i\in A$, $n_i\in N$, then applying $\mathrm{Ad}$ and using the injectivity part of step 5.1 gives $\mathrm{Ad}(k_1)=\mathrm{Ad}(k_2)$, $\mathrm{Ad}(a_1)=\mathrm{Ad}(a_2)$ and $\mathrm{Ad}(n_1)=\mathrm{Ad}(n_2)$; by step 2.1 the maps $\mathrm{Ad}{\mathrm{\mid }}_A$ and $\mathrm{Ad}{\mathrm{\mid }}_N$ are injective, so $a_1=a_2$ and $n_1=n_2$, and cancelling $a_1{n}_1$ on the right gives $k_1=k_2$.

Completion: $K\times A\times N\to G$ is smooth, bijective by steps 6.1 and 6.2, and has invertible differential at every point, because it is the composite of $(\mathrm{i}\mathrm{d},\mu ):K\times A\times N\to K\times AN$, whose second component $A\times N\to AN$ is a diffeomorphism by step 3.1, with the multiplication map $K\times AN\to G$, which is everywhere regular by step 4.1; a bijective local diffeomorphism is a diffeomorphism, and its inverse is smooth by the inverse function theorem; moreover $A\cong \mathfrak{a}$ and $N\cong \mathfrak{n}$ are diffeomorphic to vector spaces by steps 1.1 and 3.1, hence simply connected, and $N=\exp (\mathfrak{n})$; finally, a diffeomorphism is a homeomorphism onto $G$, so the images of the closed subsets $\{1\}\times A\times \{1\}$ and $\{1\}\times \{1\}\times N$ of $K\times A\times N$ under the multiplication map are closed in $G$, that is, $A$ and $N$ are closed subgroups of $G$; this proves all the assertions.

If the restricted-root set is empty, step 1.2 uses only the zero-weight space and $\mathfrak{n}=0$. The direct sum [L2] and $[{\mathfrak{k}}_0,\mathfrak{a}]=0$ from [L3] then make $\mathfrak{a}$ central, hence zero by semisimplicity. Thus $G=K$ and $A=N=\{1\}$, consistently with the formula. This includes the zero Lie algebra, for which connected $G$ is the trivial group.