Compact and split real forms of sl two c

Given: ${\mathrm{A}\mathrm{C}}_{\omega }$ and $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ with the basis $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),$ so that $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$, and the two maps ${\sigma }_c(X)=-X^{\ast }$ and ${\sigma }_s(X)=\bar{X}$.

${\mathrm{A}\mathrm{C}}_{\omega }$ is countable choice; it is used only through the matrix Lie-group examples cited in [L1] and [L2].

$\mathfrak{s}$ is the Lie algebra of traceless complex $2\times 2$ matrices with bracket $[A,B]=AB-BA$, with basis $e,f,h$ and the displayed relations, and the real traceless matrices form the real Lie subalgebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ (The special linear Lie algebra sl_2, General and special linear Lie groups).

$\mathfrak{s}\mathfrak{u}(2)=\{X\in M_2(\mathbb{C}):X^{\ast }+X=0,\mathrm{tr}X=0\}$ is a real Lie subalgebra of $\mathfrak{s}$ with the same bracket (Unitary and special unitary Lie groups).

If $\sigma$ is a conjugate-linear involutive automorphism of a finite-dimensional complex Lie algebra $\mathfrak{g}$, then its fixed locus ${\mathfrak{g}}^{\sigma }$ is a real form of $\mathfrak{g}$ (Real forms correspond to conjugate-linear involutions, Real form of a complex Lie algebra).

The Killing form of ${\mathfrak{s}\mathfrak{l}}_2$ satisfies $B(h,h)=8$, $B(e,f)=B(f,e)=4$, and all other pairings of basis vectors zero; equivalently $B(X,Y)=4\mathrm{tr}(XY)$ for all $X,Y$ (Killing form of sl_2, Killing form).

When the ambient complex Lie algebra is finite-dimensional and semisimple, a real form ${\mathfrak{g}}_0$ is a compact real form exactly when $B(X,X)<0$ for every nonzero $X\in {\mathfrak{g}}_0$; it is a split real form exactly when it contains a Cartan subalgebra ${\mathfrak{h}}_0$ such that every ${\mathrm{ad}}_H$, $H\in {\mathfrak{h}}_0$, is diagonalizable over $\mathbb{R}$ (Compact real form of a complex semisimple Lie algebra, Split real form, Cartan subalgebra).

The conjugate transpose $X^{\ast }={\bar{X}}^{\mathsf{T}}$ is additive, and $(XY{)}^{\ast }=Y^{\ast }{X}^{\ast }$ (The transpose $A^{\mathsf{T}}$ of a matrix).

Proof technique: direct matrix computation.

1.1 Both maps ${\sigma }_c$ and ${\sigma }_s$ are real-linear, additive, involutive and conjugate-linear, and they preserve brackets. By [L6] the conjugate transpose reverses products, ${\sigma }_c(AB)=-(AB{)}^{\ast }=-B^{\ast }{A}^{\ast }$, while ${\sigma }_c(A){\sigma }_c(B)=A^{\ast }{B}^{\ast }$; taking the difference gives ${\sigma }_c[A,B]=-(AB-BA{)}^{\ast }=A^{\ast }{B}^{\ast }-B^{\ast }{A}^{\ast }=[{\sigma }_{c}A,{\sigma }_{c}B]$. Entrywise conjugation is multiplicative, $\overline{AB}=\bar{A}\bar{B}$, so ${\sigma }_s[A,B]=[{\sigma }_{s}A,{\sigma }_{s}B]$. [given, L6, algebra]

1.2 The fixed algebra of ${\sigma }_c$ inside $\mathfrak{s}$ is $\mathfrak{s}\mathfrak{u}(2)$: a traceless $X$ satisfies $X=-X^{\ast }$ exactly when $X$ is skew-Hermitian, so ${\mathfrak{s}}^{{\sigma }_c}=\{X\in M_2(\mathbb{C}):X^{\ast }+X=0,\mathrm{tr}X=0\}$, which is $\mathfrak{s}\mathfrak{u}(2)$ by [L2]. [given, L2, algebra]

1.3 The fixed algebra of ${\sigma }_s$ is ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$: a matrix is fixed by entrywise conjugation exactly when it is real, and a real traceless matrix lies in ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ by [L1]. [given, L1, algebra]

1.4 The Killing form of $\mathfrak{s}$ is $B(X,Y)=4\mathrm{tr}(XY)$. Both sides are symmetric bilinear, so it suffices to compare them on basis pairs: $4\mathrm{tr}(ef)=4=B(e,f)$, $4\mathrm{tr}(h^2)=8=B(h,h)$, and $4\mathrm{tr}(e^2)=4\mathrm{tr}(f^2)=4\mathrm{tr}(eh)=4\mathrm{tr}(fh)=0$, matching the vanishing pairings of [L4]. In the basis $(e,f,h)$ its Gram matrix is $\left(\begin{array}{ccc}0& 4& 0\\ 4& 0& 0\\ 0& 0& 8\end{array}\right)$, whose determinant is $-128\mathrm{\ne }0$; hence $B$ is nondegenerate and $\mathfrak{s}$ is semisimple by Cartan's criterion over the characteristic-zero field $\mathbb{C}$. Thus the ambient hypothesis in [L5] has been established before either definition is invoked. [L4, Cartan's semisimplicity criterion, algebra]

1.5 The line $\mathbb{R}h$ is a Cartan subalgebra of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$: it is abelian, hence nilpotent, and its normalizer is itself because $[h,X]=0$ for $X=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$ forces $2b=0$ and $-2c=0$, so $X\in \mathbb{R}h$. Since $[h,h]=0$, $[h,e]=2e$ and $[h,f]=-2f$ by the given relations, ${\mathrm{ad}}_h$ is diagonal on the basis $(h,e,f)$ with real eigenvalues $0,2,-2$, so ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is a split real form by [L5]. [given, L1, L5, algebra]

2.1 Both fixed loci are real forms. Every $Z\in \mathfrak{s}$ decomposes as $Z=X+iY$ with $X=\frac{1}{2}(Z+\bar{Z})$ and $Y=\frac{1}{2i}(Z-\bar{Z})$ real traceless, so ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ spans $\mathfrak{s}$ over $\mathbb{C}$ and has real dimension $3={\dim }_{\mathbb{C}}\mathfrak{s}$; alternatively this is the general conclusion of [L3] applied to the involution ${\sigma }_s$ of step 1.1. Likewise every $Z\in \mathfrak{s}$ is $Z=X+iY$ with $X=\frac{1}{2}(Z-Z^{\ast })$ and $Y=\frac{1}{2i}(Z+Z^{\ast })$ both skew-Hermitian and traceless, so $\mathfrak{s}\mathfrak{u}(2)$ spans $\mathfrak{s}$ over $\mathbb{C}$ and has real dimension $3={\dim }_{\mathbb{C}}\mathfrak{s}$; again [L3] gives the same conclusion from ${\sigma }_c$. [L3, step 1.1, step 1.2, step 1.3, algebra]

2.2 For nonzero $X\in \mathfrak{s}\mathfrak{u}(2)$ one has $X^{\ast }=-X$, so step 1.4 gives $B(X,X)=4\mathrm{tr}(X^2)=-4\mathrm{tr}(X{X}^{\ast })=-4{\sum }_{i,j}\mathrm{\mid }{X}_{ij}{\mathrm{\mid }}^2<0$, because a nonzero matrix has a nonzero entry. Hence $B$ is negative definite on $\mathfrak{s}\mathfrak{u}(2)$, and $\mathfrak{s}\mathfrak{u}(2)$ is a compact real form by [L5]. [step 1.4, L5, algebra]

3.1 The two forms are genuinely different: $B(h,h)=8>0$ on ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ while $B(e-f,e-f)=B(e,e)-2B(e,f)+B(f,f)=-8<0$, so the Killing form of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is indefinite, as a noncompact real form must be, whereas the form on $\mathfrak{s}\mathfrak{u}(2)$ is definite by step 2.2. All computations are finite; ${\mathrm{A}\mathrm{C}}_{\omega }$ enters only through [L1] and [L2]. [A1, step 1.4, step 2.2, step 1.5, algebra] ∎