Restricted roots of sl n r

Given: An integer $n\ge 2$, the real Lie algebra ${\mathfrak{g}}_0={\mathfrak{s}\mathfrak{l}}_n(\mathbb{R})$ of real traceless matrices, the Cartan involution $\theta (X)=-X^{\mathsf{T}}$ with ${\mathfrak{p}}_0$ the symmetric traceless matrices, the diagonal subspace $\mathfrak{a}$, and the matrix units $E_{ij}$, $i\mathrm{\ne }j$.

${\mathfrak{s}\mathfrak{l}}_n(\mathbb{R})$ is a real Lie algebra under $[X,Y]=XY-YX$, with $\mathrm{tr}X=0$ for every element and $[H,E_{ij}]=(h_i-h_j)E_{ij}$ for $H=\mathrm{diag}(h_1,\dots ,h_n)$ (General and special linear Lie groups, Lie algebras over a field).

$\mathfrak{a}$ consists of diagonal symmetric traceless matrices, hence is a subspace of ${\mathfrak{p}}_0$; the Cartan decomposition of ${\mathfrak{g}}_0$ is ${\mathfrak{g}}_0={\mathfrak{k}}_0\oplus {\mathfrak{p}}_0$ with ${\mathfrak{k}}_0=\mathfrak{s}\mathfrak{o}(n)$ (Cartan involution and k plus p for sl n r, Cartan decomposition of a real semisimple Lie algebra).

A restricted root of $({\mathfrak{g}}_0,\mathfrak{a})$ for a maximal abelian $\mathfrak{a}\subseteq {\mathfrak{p}}_0$ is a nonzero real functional $\lambda$ on $\mathfrak{a}$ whose restricted root space ${\mathfrak{g}}_0^{\lambda }=\{X\in {\mathfrak{g}}_0:[H,X]=\lambda (H)X\text{ for all }H\in \mathfrak{a}\}$ is nonzero, and its multiplicity is ${\dim }_{\mathbb{R}}{\mathfrak{g}}_0^{\lambda }$; a maximal abelian subspace of ${\mathfrak{p}}_0$ has dimension equal to the real rank (Restricted root and restricted root space, Maximal split abelian subspace and real rank).

In the complex analogue, the diagonal traceless subalgebra of ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{C})$ is a Cartan subalgebra with roots ${\epsilon }_i-{\epsilon }_j$ and one-dimensional root spaces $\mathbb{C}{E}_{ij}$ (Diagonal Cartan subalgebra and roots of ${\mathfrak{s}\mathfrak{l}}_n$).

Proof technique: direct matrix computation.

1.1 The subspace $\mathfrak{a}$ is a maximal abelian subspace of ${\mathfrak{p}}_0$. It is abelian because its elements are diagonal, and it lies in ${\mathfrak{p}}_0$ by [L2]. Conversely, let $X\in {\mathfrak{p}}_0$ satisfy $[H,X]=0$ for every $H\in \mathfrak{a}$. Choosing $H=\mathrm{diag}(h_1,\dots ,h_n)$ with pairwise distinct entries $h_i$ (possible with ${\sum }_i{h}_i=0$ in dimension $n\ge 2$), the identity $[H,X]={\sum }_{i,j}(h_i-h_j)X_{ij}{E}_{ij}$ of [L1] shows $(h_i-h_j)X_{ij}=0$ for all $i,j$, so $X_{ij}=0$ whenever $i\mathrm{\ne }j$: the centralizer of $\mathfrak{a}$ in ${\mathfrak{p}}_0$ is $\mathfrak{a}$ itself, which is therefore maximal abelian. [given, L1, L2, algebra]

1.2 Every functional $f_i-f_j$ with $i\mathrm{\ne }j$ is a restricted root with $\mathbb{R}{E}_{ij}\subseteq {\mathfrak{g}}_0^{f_i-f_j}$: for $H=\mathrm{diag}(h)\in \mathfrak{a}$, [L1] gives $[H,E_{ij}]=(h_i-h_j)E_{ij}=(f_i-f_j)(H)E_{ij}$, and $E_{ij}$ is a nonzero real matrix of trace zero, while $f_i-f_j\mathrm{\ne }0$ since $i\mathrm{\ne }j$. [given, L1, L3, algebra]

2.1 There are no further restricted roots. Let $X={\sum }_{i,j}{x}_{ij}{E}_{ij}\in {\mathfrak{g}}_0$ and suppose $[H,X]=\lambda (H)X$ for every $H\in \mathfrak{a}$. Comparing the $(i,j)$-entry using [L1] gives $x_{ij}(f_i-f_j)(H)=x_{ij}\lambda (H)\text{for every }H\in \mathfrak{a}\mathrm{.}$ Thus, if an off-diagonal coefficient $x_{ij}$ is nonzero, then $\lambda (H)=(f_i-f_j)(H)$ for every $H$, so $\lambda =f_i-f_j$ as functionals. If $\lambda \mathrm{\ne }0$, choose $H$ with $\lambda (H)\mathrm{\ne }0$; the diagonal-entry equations then force every $x_{ii}=0$. Distinct functionals $f_i-f_j$ have disjoint eigenspaces, so step 1.2 now gives ${\mathfrak{g}}_0^{f_i-f_j}=\mathbb{R}{E}_{ij}$. For $\lambda =0$, choose one diagonal $H\in \mathfrak{a}$ with pairwise distinct entries. If $X\in {\mathfrak{g}}_0^0$ then $[H,X]=0$, so the same entry computation forces every off-diagonal coefficient of $X$ to vanish; as $X$ is traceless, it is a diagonal traceless matrix and hence belongs to $\mathfrak{a}$. The reverse inclusion is immediate because diagonal matrices commute, so ${\mathfrak{g}}_0^0=\mathfrak{a}$. Hence the nonzero restricted roots are exactly the $f_i-f_j$, each with multiplicity one. [step 1.1, step 1.2, L1, L3, algebra]

3.1 The decomposition is consistent dimensionally: $\dim \mathfrak{a}=n-1$ and there are $n(n-1)$ roots each of multiplicity $1$, so $\dim {\mathfrak{g}}_0=(n-1)+n(n-1)=n^2-1$, which is the dimension of ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{R})$; the restricted root system $\{f_i-f_j:i\mathrm{\ne }j\}$ is the standard realization of $A_{n-1}$ and is reduced, since for every root $f_i-f_j$ its double $2(f_i-f_j)$ is not of the form $f_k-f_l$. [step 1.1, step 2.1, L3, algebra]

3.2 The computation matches the complex root computation of [L4]: the functionals $f_i-f_j$ are the restrictions to the real diagonal traceless subspace of the root functionals ${\epsilon }_i-{\epsilon }_j$ of the complexification, and the real root space $\mathbb{R}{E}_{ij}$ is the real form of $\mathbb{C}{E}_{ij}$ fixed by complex conjugation, which is why each multiplicity is $1$. [step 2.1, L4, algebra]

4.1 Endpoints and scope: for $n=2$ there is a single pair of opposite roots $\pm (f_1-f_2)$ with one-dimensional spaces $\mathbb{R}{E}_{12}$ and $\mathbb{R}{E}_{21}$, and $\dim \mathfrak{a}=1$; the case $n=1$ is excluded because ${\mathfrak{s}\mathfrak{l}}_1(\mathbb{R})=0$ has no nonzero diagonal traceless element. The computation uses no choice principle, and the diagonal element with pairwise distinct entries exists by an explicit choice of coordinates. [given, step 2.1, step 3.1, algebra] ∎