Relation of basis of vector space $\mathfrak{sl}_2(\mathbb{R})$ - to subgroup generators in the Iwasawa decomposition?

Question

Is it possible to choose a basis for $\mathfrak{sl}_2(\mathbb{R})$ in such a way that it corresponds to the infinitesimal element's generators for the subgroups KAN (of the Iwasawa decomposition)?

The classical basis (the "split" basis) for $\mathfrak{sl}_2(\mathbb{R})$ is usually defined as $$h=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] \quad n_+=\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \quad n_{-}=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right] $$ so, the Lie algebra is determined by the known commutation relations $$[h, n_{\pm}] = \pm 2n_{\pm}\quad [n_+, n_{-}] = h$$ The basis does not provide much intuition on KAN subgroup element's generators (only $A$ and $N$, $\overline{N}$). However, in the "Representations of linear Lie groups..", p.9 (Example 1.13) Salem Ben Said introduces $$X_1=\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \quad X_2=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] \quad X_{3}=\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] $$ with $$[X_1, X_2 ] = − 2X_1\quad [X_2, X_3 ] = − 2X_2 \quad [X_1, X_3 ] = − X_2$$ At this point, the basis clearly corresponds to the generators for the elements of subgroups in the Iwasawa decomposition ($X_1 \to N$, $X_2 \to A$, $X_3 \to K$). However, it seems that the second equation $[X_2, X_3 ] = − 2X_2$ is not correct, so I doubt that such approach works. Please advice.

Answer 1

$\newcommand{\k}{\mathfrak k}$ $\newcommand{\a}{\mathfrak a}$ $\newcommand{\g}{\mathfrak g}$ $\newcommand{\p}{\mathfrak p}$ $\newcommand{\n}{\mathfrak n}$

I'm quite sure that several sources prove Iwasawa decomposition for a connected semisimple real Lie group $G$ via a corresponding decomposition of its (real) Lie algebra $\mathfrak g_0$, where one first takes the Cartan decomposition

$$\mathfrak g_0 \simeq \k_0 \oplus \p_0$$

(with $\k_0$ being a maximal compact subalgebra), then chooses a maximal abelian subalgebra (kind of a maximal split torus) $\a_0 \subset \p_0$; the "eigenvalues" of $\a_0$ form a (not necessarily reduced) root system, the "system of restricted roots" $\Sigma$, of which we finally choose a positive part $\Sigma^+$ and define $\n_0$ as the sum of the root spaces for those positive roots, leading to a decomposition

$$\g_0 \simeq \k_0 \oplus \a_0 \oplus \n_0.$$

It's worth pointing out that all three components are subalgebras, in particular that is a vector space decomposition, but it's not a direct decomposition of Lie algebras, as none of those three are (in general) ideals. (Rather, $[\a_0, \n_0] = \n_0$ and in general the best we have is $[\k_0, \a_0] \subseteq \k_0 \oplus \n_0$ (note that we have $[\k_0, \a_0] \subseteq \p_0$ but $\n_0 \not \subseteq \p_0$ in general).)

However, if you are just interested in a vector space basis, of course now you can choose bases for each of the summands, and their union gives you a basis of the Lie algebra.

In the example at hand, $\g_0 = \mathfrak{sl}_2(\mathbb R)$, a standard choice is $\k_0 = \{\pmatrix{0&c\\-c&0}:c \in \mathbb R\}$, $\a_0 = \{\pmatrix{a&0\\0&-a}:a \in \mathbb R\}$, $\n_0 = \{\pmatrix{0&b\\0&0}:b \in \mathbb R\}$, and of course if you choose one non-zero element of each, then together they form a basis.

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This generalises first to general $\mathfrak{sl}_n(\mathbb R)$, where $\k_0 =$ skew-symmetric matrices a.k.a. $\mathfrak{so}_n$, $\a_0 =$ diagonal matrices and $\n_0 =$ strictly upper triangular matrices, for which obvious standard choices of bases would be

$\{E_{i,j}-E_{j,i}$ ($j>i$)$\}$ for $\k_0$,

$\{E_{i,i}-E_{i+1,i+1}$ $\}$ for $\a_0$,

$\{E_{i,j}$ ($j>i$) $\}$ for $\n_0$.

Even more generally, if $\g_0$ is any split real semisimple Lie algebra with a chosen Chevalley / Cartan-Weyl basis $(H_\alpha, E_\alpha)_{\alpha \in \text{roots}}$, then I would bet that (signs in $\color{red}{\text{red}}$ might flip depending on chosen normalisation)

$\{E_\gamma \color{red}{\pm} E_{-\gamma} : \gamma \in \text{ positive roots} \}$ is a basis for $\k_0$

$\{H_\alpha : \alpha \in \text{ simple roots} \}$ is a basis for $\a_0$

$\{E_\gamma : \gamma \in \text{ positive roots} \}$ is a basis for $\n_0$.

This all matches your special case where $X_3 = K = n_+ - n_-$, $X_2 = A = h$, $X_1=N = n_+$.

On the other extreme, for compact forms of course we just have $\g_0 = \k_0$. Choosing a Chevalley basis for the complexification $\g := (\g_0)_{\mathbb C}$ one can transform that into a basis of $\g_0 = \k_0$ which is made up of

$\{E_\gamma \color{red}{\pm} E_{-\gamma} : \gamma \in \text{ positive roots} \} \cup \{i(E_\gamma \color{red}{\mp} E_{-\gamma}) : \gamma \in \text{ positive roots} \} \cup \{ i H_\alpha : \alpha \in \text{ simple roots} \}$.

For the forms "in between" split and compact, this seems more complicated. I fumbled with related computations in What changes in the representation theory of real Lie algebras?.