Iwasawa decomposition of sl two r

Given: The Axiom of Choice; the group ${\mathrm{SL}}_2(\mathbb{R})$ of real $2\times 2$ matrices of determinant $1$; the Cartan involution $\theta (X)=-X^{\mathsf{T}}$ with ${\mathfrak{k}}_0=\mathfrak{s}\mathfrak{o}(2)$ and ${\mathfrak{p}}_0$ the symmetric traceless matrices; and a matrix $g=\left(\begin{array}{cc}p& r\\ q& s\end{array}\right)\in {\mathrm{SL}}_2(\mathbb{R})$.

The Axiom of Choice is The Axiom of Choice; it enters only through the global Iwasawa theorem of [L4] and the smooth structure of the group.

${\mathrm{SL}}_2(\mathbb{R})$ is an embedded Lie group with Lie algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ and $\mathrm{SO}(2)$ is a closed connected subgroup with Lie algebra $\mathfrak{s}\mathfrak{o}(2)=\mathbb{R}\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ (General and special linear Lie groups, Orthogonal and special orthogonal Lie groups).

The Cartan decomposition ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})={\mathfrak{k}}_0\oplus {\mathfrak{p}}_0$ has ${\mathfrak{k}}_0=\mathfrak{s}\mathfrak{o}(2)$ and ${\mathfrak{p}}_0=\{X:X^{\mathsf{T}}=X,\mathrm{tr}X=0\}$; ${\mathfrak{p}}_0$ is spanned by $H=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ and $\sigma =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and $[H,\sigma ]=\left(\begin{array}{cc}0& 2\\ -2& 0\end{array}\right)\mathrm{\ne }0$, so $\mathbb{R}H$ is a maximal abelian subspace of ${\mathfrak{p}}_0$ (Cartan involution and k plus p for sl n r, Compact and split cartan subalgebras of sl two r, Cartan decomposition of a real semisimple Lie algebra, Theta-stable Cartan subalgebras and their compact and split parts).

The matrix exponential is the Lie-group exponential of a matrix group, and $\exp (X)=I+X$ whenever $X^2=0$ (Matrix exponential as the Lie-group exponential, Exponential map of a Lie group).

In the global Cartan setup of a connected real semisimple Lie group $G$ with finite center, with $A=\exp (\mathfrak{a})$ and $N$ the connected subgroup with Lie algebra the sum $\mathfrak{n}$ of the positive restricted root spaces, the multiplication map $K\times A\times N\to G$ is a diffeomorphism (Global iwasawa decomposition).

Proof technique: direct matrix computation.

1.1 Write $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $H=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$. Then $[H,E]=2E$, so $\mathfrak{n}=\mathbb{R}E$ is the positive restricted root space for the functional $2f_1$ with $f_1(H)=1$ on the maximal abelian $\mathfrak{a}=\mathbb{R}H$ of [L2], and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})={\mathfrak{k}}_0\oplus \mathfrak{a}\oplus \mathfrak{n}$ is the Iwasawa decomposition on the Lie-algebra level, since the three spaces have dimensions $1,1,1$ and the sum is direct. [given, L2, algebra]

1.2 The subgroups generated by $\mathfrak{a}$ and $\mathfrak{n}$ are the sets displayed: $\exp (tH)=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{-t}\end{array}\right)$, so $\exp (\mathfrak{a})=A$ with $a=e^t>0$, and $E^2=0$ gives $\exp (xE)=I+xE=\left(\begin{array}{cc}1& x\\ 0& 1\end{array}\right)$ by [L3], so $\exp (\mathfrak{n})=N$. [given, L3, algebra]

2.1 Existence of the factorization. Since $g$ is invertible, its first column $(p,q)$ is nonzero, so $a:=\sqrt{p^2+q^2}>0$ is well defined. Put $k:=\frac{1}{a}\left(\begin{array}{cc}p& -q\\ q& p\end{array}\right)$ and $x:=\frac{pr+qs}{a^2}$; then $k^{\mathsf{T}}k=I$ and $\det k=\frac{p^2+q^2}{a^2}=1$, so $k\in \mathrm{SO}(2)$, and a direct multiplication gives $k^{-1}g=\frac{1}{a}\left(\begin{array}{cc}p& q\\ -q& p\end{array}\right)\left(\begin{array}{cc}p& r\\ q& s\end{array}\right)=\left(\begin{array}{cc}a& \frac{pr+qs}{a}\\ 0& \frac{ps-qr}{a}\end{array}\right)=\left(\begin{array}{cc}a& ax\\ 0& a^{-1}\end{array}\right)$, using $ps-qr=1$ and $a^2=p^2+q^2$. Hence $g=k\left(\begin{array}{cc}a& 0\\ 0& a^{-1}\end{array}\right)\left(\begin{array}{cc}1& x\\ 0& 1\end{array}\right)$ with all three factors in $K,A,N$. [given, step 1.2, algebra]

3.1 Uniqueness. Suppose $g=k_1{a}_1{n}_1=k_2{a}_2{n}_2$ with $k_i\in K$, $a_i=\mathrm{diag}(a_i,a_i^{-1})$ and $n_i\in N$. Applying both sides to the first standard basis vector and using $n_i{e}_1=e_1$ gives $k_1{a}_1{e}_1=k_2{a}_2{e}_1$, so taking norms and using that $k_1,k_2$ are orthogonal yields $a_1=a_2>0$; hence $k_1{e}_1=k_2{e}_1$ and therefore $k_1=k_2$, because a rotation of ${\mathbb{R}}^2$ fixing $e_1$ is the identity. Then $a_1{n}_1=a_2{n}_2$ forces $n_1=n_2$, so all three factors are unique. [given, step 2.1, algebra]

4.1 Define $\mathrm{\Theta }(g)=(g^{\mathsf{T}}{)}^{-1}$ on ${\mathrm{SL}}_2(\mathbb{R})$. The identities $\mathrm{\Theta }(gh)=\mathrm{\Theta }(g)\mathrm{\Theta }(h)$ and ${\mathrm{\Theta }}^2=\mathrm{i}\mathrm{d}$ make it an involutive Lie-group automorphism; its differential is $d{\mathrm{\Theta }}_I(X)=-X^{\mathsf{T}}=\theta (X)$, and its fixed group is $\{g:g^{\mathsf{T}}g=I,\det g=1\}=SO(2)=K$. It fixes the center $\{\pm I\}$ pointwise. Thus the global Cartan setup required by [L4] holds for $G={\mathrm{SL}}_2(\mathbb{R})$: its Lie algebra is semisimple, its center is finite, it is connected, and by steps 1.1 and 1.2 the data ${\mathfrak{k}}_0,\mathfrak{a},\mathfrak{n}$ are exactly those of the displayed $K,A,N$. Therefore $K\times A\times N\to {\mathrm{SL}}_2(\mathbb{R})$, $(k,a,n)\mapsto kan$, is a diffeomorphism, and by step 2.1 it is the unique factorization of each element. [step 1.1, step 1.2, step 2.1, step 3.1, L4, A1, algebra]

5.1 Endpoints and scope: for $g=I$ the factorization is $k=a=n=I$ with $a=1$ and $x=0$, which is the endpoint $t=0$ of the positive parameter domain $a>0$; for $a\to 0^{+}$ no element of $A$ is lost because $A$ is defined by positivity of $a$. The choice principle is inherited only from [L4], while the matrix computations use none. The diagonal factor is exactly $\exp (\mathfrak{a})$ and the unipotent factor is exactly $\exp (\mathfrak{n})$, so the group-level statement matches the Lie-algebra-level Iwasawa decomposition of step 1.1. [given, step 1.1, step 1.2, step 4.1, algebra] ∎