Compact and split cartan subalgebras of sl two r

Given: ${\mathfrak{g}}_0={\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ with the basis $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, $H=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ satisfying $[H,e]=2e$, $[H,f]=-2f$, $[e,f]=H$, and the matrices $K_0=f-e=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $\theta$.

${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ consists of the real traceless $2\times 2$ matrices, with basis $e,f,H$ and the displayed bracket relations (The special linear Lie algebra sl_2, General and special linear Lie groups).

The Cartan involution $\theta (X)=-X^{\mathsf{T}}$ of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ has ${\mathfrak{k}}_0=\{X:X^{\mathsf{T}}+X=0\}=\mathfrak{s}\mathfrak{o}(2)$ and ${\mathfrak{p}}_0=\{X\in {\mathfrak{s}\mathfrak{l}}_2(\mathbb{R}):X^{\mathsf{T}}=X\}$ (Cartan involution and k plus p for sl n r, Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra, Orthogonal and special orthogonal Lie groups).

A Cartan subalgebra of a Lie algebra is a nilpotent self-normalizing subalgebra, and a $\theta$-stable Cartan subalgebra ${\mathfrak{h}}_0$ decomposes as ${\mathfrak{h}}_0={\mathfrak{t}}_0\oplus {\mathfrak{a}}_0$ with ${\mathfrak{t}}_0={\mathfrak{h}}_0\cap {\mathfrak{k}}_0$ the compact part and ${\mathfrak{a}}_0={\mathfrak{h}}_0\cap {\mathfrak{p}}_0$ the split part; it is maximally compact when $\dim {\mathfrak{t}}_0$ is maximal and maximally noncompact when $\dim {\mathfrak{a}}_0$ is maximal (Cartan subalgebra, Theta-stable Cartan subalgebras and their compact and split parts).

Proof technique: direct matrix computation.

1.1 The elements $K_0$ and $H$ lie where claimed: $K_0^{\mathsf{T}}=-K_0$, so $\theta K_0=K_0$ and $K_0\in {\mathfrak{k}}_0$, while $H^{\mathsf{T}}=H$ and $\mathrm{tr}H=0$, so $\theta H=-H$ and $H\in {\mathfrak{p}}_0$. In particular both lines $\mathbb{R}{K}_0$ and $\mathbb{R}H$ are $\theta$-stable. [given, L2, algebra]

1.2 The normalizer of $\mathbb{R}{K}_0$ in ${\mathfrak{g}}_0$ is $\mathbb{R}{K}_0$: for $X=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$ one computes $[K_0,X]=\left(\begin{array}{cc}-b-c& 2a\\ 2a& b+c\end{array}\right)$. For this matrix to equal $r{K}_0$ its diagonal entries force $c=-b$, while its two off-diagonal entries give $2a=-r$ and $2a=r$, hence $a=r=0$. Thus $X\in \mathbb{R}{K}_0$, and conversely every such $X$ normalizes the line. Hence $\mathbb{R}{K}_0$ is abelian, nilpotent and self-normalizing, so it is a Cartan subalgebra by [L3]. [given, L1, L3, algebra]

1.3 The normalizer of $\mathbb{R}H$ in ${\mathfrak{g}}_0$ is $\mathbb{R}H$: for $X=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$ one computes $[H,X]=\left(\begin{array}{cc}0& 2b\\ -2c& 0\end{array}\right)$. If this equals $rH$, comparison of diagonal and off-diagonal entries gives $r=0$ and $b=c=0$, so $X=\mathrm{diag}(a,-a)\in \mathbb{R}H$; conversely every such $X$ normalizes the line. Hence $\mathbb{R}H$ is abelian, nilpotent and self-normalizing, so it is a Cartan subalgebra by [L3]. [given, L1, L3, algebra]

1.4 The adjoint spectra distinguish the two: by [L1], $[K_0,H]=[f-e,H]=[f,H]-[e,H]=2f+2e=2(e+f)$ and $[K_0,e+f]=[f,e]+[f,f]-[e,e]-[e,f]=[f,e]-[e,f]=-2H$, so ${\mathrm{ad}}_{K_0}$ is zero on $K_0$ and has the matrix $\left(\begin{array}{cc}0& -2\\ 2& 0\end{array}\right)$ in the basis $(H,e+f)$ of ${\mathfrak{p}}_0$, with eigenvalues $0,\pm 2i$; whereas ${\mathrm{ad}}_H$ is diagonal on $e,f,H$ with real eigenvalues $2,-2,0$. [given, L1, algebra]

2.1 Both Cartan subalgebras are $\theta$-stable by step 1.1, so the decomposition of [L3] applies. For ${\mathfrak{h}}_0=\mathbb{R}{K}_0$ one has ${\mathfrak{h}}_0\cap {\mathfrak{k}}_0=\mathbb{R}{K}_0$ and ${\mathfrak{h}}_0\cap {\mathfrak{p}}_0=0$, so ${\mathfrak{t}}_0={\mathfrak{h}}_0$ and ${\mathfrak{a}}_0=0$: every element of ${\mathfrak{h}}_0$ is compact, and the compact dimension $1$ equals $\dim {\mathfrak{k}}_0$, so ${\mathfrak{h}}_0$ is maximally compact. For ${\mathfrak{h}}_0^{\mathrm{\prime }}=\mathbb{R}H$ one has ${\mathfrak{h}}_0^{\mathrm{\prime }}\cap {\mathfrak{k}}_0=0$ and ${\mathfrak{h}}_0^{\mathrm{\prime }}\cap {\mathfrak{p}}_0=\mathbb{R}H$, so ${\mathfrak{t}}_0=0$ and ${\mathfrak{a}}_0={\mathfrak{h}}_0^{\mathrm{\prime }}$. Its noncompact dimension is maximal: if a $\theta$-stable Cartan subalgebra $\mathfrak{q}$ had $\dim (\mathfrak{q}\cap {\mathfrak{p}}_0)=2$, then ${\mathfrak{p}}_0=\mathbb{R}H\oplus \mathbb{R}(e+f)$ would lie in $\mathfrak{q}$. Closure under brackets and $[H,e+f]=2(e-f)=-2K_0$ would then put $K_0$ in $\mathfrak{q}$, hence $\mathfrak{q}={\mathfrak{g}}_0$. But ${\mathfrak{g}}_0$ is not nilpotent (indeed $[{\mathfrak{g}}_0,{\mathfrak{g}}_0]={\mathfrak{g}}_0$ from the displayed basis relations), whereas every Cartan subalgebra is nilpotent by [L3]. Thus every split part has dimension at most $1$, and $\mathbb{R}H$ attains that bound. [step 1.1, step 1.2, step 1.3, L1, L2, L3, algebra]

3.1 Consistency of brackets and signs with the general theory: $[{\mathfrak{h}}_0,{\mathfrak{h}}_0]=0$ lies in both ${\mathfrak{t}}_0$ and ${\mathfrak{a}}_0$ cases because the Cartan subalgebras are abelian, and the compact line lies in ${\mathfrak{k}}_0$ where the Killing form is negative definite while the split line lies in ${\mathfrak{p}}_0$ where it is positive definite, so the Killing form is negative definite on ${\mathfrak{t}}_0={\mathfrak{h}}_0$ in the first case and positive definite on ${\mathfrak{a}}_0={\mathfrak{h}}_0^{\mathrm{\prime }}$ in the second. Both subalgebras are one-dimensional, and the two eigenvalue computations of step 1.4 match the compact/purely imaginary and split/real terminology. [step 1.1, step 1.4, step 2.1, L2, algebra]

4.1 Endpoints and scope: the algebra is three-dimensional and both Cartan subalgebras are one-dimensional, which equals its rank; ${\mathfrak{h}}_0\mathrm{\ne }{\mathfrak{h}}_0^{\mathrm{\prime }}$, and the compact one is the compact line $\mathbb{R}{K}_0=i\mathbb{R}{h}_B$ in the notation of the source and the split one is the diagonal line $\mathbb{R}H$. No choice principle enters, and the computations are finite. [given, step 1.2, step 1.3, step 2.1, algebra] ∎