Existence of a compact real form

Given: The Axiom of Choice and a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$ with Killing form $B$.

The Axiom of Choice is The Axiom of Choice; it enters through the Cartan and root data of [L1] and through the normalization statement [L3], whose statements carry the assumption.

A Cartan subalgebra $\mathfrak{h}$ exists; with such a choice, the root system $\mathrm{\Phi }$ is finite, $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\alpha \in \mathrm{\Phi }}{\mathfrak{g}}_{\alpha }$ with one-dimensional root spaces, and for every root $\alpha$ there are $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ and $H_{\alpha }\in \mathfrak{h}$ with $[e_{\alpha },f_{\alpha }]=H_{\alpha }$, $\alpha (H_{\alpha })=2$, $[H_{\alpha },e_{\alpha }]=2e_{\alpha }$, $[H_{\alpha },f_{\alpha }]=-2f_{\alpha }$ and $\beta (H_{\alpha })\in \mathbb{Z}$ for every root $\beta$; also $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$ and the Cartan integers are rational (Existence of Cartan subalgebras, Root-space decomposition, Root spaces of a complex semisimple Lie algebra are one-dimensional, The root sl_2 triple, Serre presentation theorem).

$B$ is symmetric, invariant and nondegenerate, the center of $\mathfrak{g}$ is zero, the pairing ${\mathfrak{g}}_{\alpha }\times {\mathfrak{g}}_{-\alpha }$ induced by $B$ is nondegenerate, and in the triple above the trace formula gives $B(H_{\alpha },H_{\alpha })=2B(e_{\alpha },f_{\alpha })$ (Cartan's semisimplicity criterion, Killing form, Opposite root spaces pair nondegenerately).

The root-vector basis of [L1] can be rescaled so that, in the notation of [L1], $[e_{\alpha },f_{\alpha }]=H_{\alpha }$ for every root $\alpha$ and the structure constants $N_{\alpha \beta }$, defined by $[e_{\alpha },e_{\beta }]=N_{\alpha \beta }{e}_{\alpha +\beta }$ when $\alpha +\beta \in \mathrm{\Phi }$ and $N_{\alpha \beta }:=0$ when $\alpha +\beta \mathrm{\notin }\mathrm{\Phi }\cup \{0\}$, are integers satisfying $N_{\alpha \beta }=-N_{-\alpha ,-\beta }$ for all $\alpha ,\beta \in \mathrm{\Phi }$ with $\alpha +\beta \mathrm{\ne }0$; in this normalization the coroots $H_{{\alpha }_1},\dots ,H_{{\alpha }_r}$ attached to any base $\mathrm{\Delta }=\{{\alpha }_1,\dots ,{\alpha }_r\}$ form a basis of $\mathfrak{h}$ (Chevalley basis and real structure constants).

Proof technique: direct.

1.1 Fix the data of [L1]. For every root $\alpha$, invariance of $B$ and $[e_{\alpha },f_{\alpha }]=H_{\alpha }$ give $B(H_{\alpha },H_{\alpha })=B([e_{\alpha },f_{\alpha }],H_{\alpha })=B(e_{\alpha },[f_{\alpha },H_{\alpha }])=2B(e_{\alpha },f_{\alpha })$, so $B(e_{\alpha },f_{\alpha })=\frac{1}{2}B(H_{\alpha },H_{\alpha })$. The trace formula computes $B$ on $\mathfrak{h}$: in a basis consisting of a basis $H_1,\dots ,H_r$ of $\mathfrak{h}$ together with one nonzero vector $e_{\gamma }\in {\mathfrak{g}}_{\gamma }$ for each root $\gamma$, the operator ${\mathrm{ad}}_H$ has eigenvalues $0$ on $\mathfrak{h}$ and $\gamma (H)$ on the one-dimensional space ${\mathfrak{g}}_{\gamma }$, so the trace of ${\mathrm{ad}}_H{\mathrm{ad}}_{H^{\mathrm{\prime }}}$ is ${\sum }_{\gamma \in \mathrm{\Phi }}\gamma (H)\gamma (H^{\mathrm{\prime }})$ and $B(H,H^{\mathrm{\prime }})$ equals that sum for $H,H^{\mathrm{\prime }}\in \mathfrak{h}$. For $H={\sum }_{\alpha }{t}_{\alpha }{H}_{\alpha }$ with real $t_{\alpha }$ every $\gamma (H)\in \mathbb{R}$ by the integrality $\gamma (H_{\alpha })\in \mathbb{Z}$ recorded in [L1], and the two terms $\gamma =\pm \alpha$ give $B(H_{\alpha },H_{\alpha })={\sum }_{\gamma \in \mathrm{\Phi }}\gamma (H_{\alpha }{)}^2\ge 8>0$. Hence $B(e_{\alpha },f_{\alpha })=\frac{1}{2}B(H_{\alpha },H_{\alpha })>0\text{for every root }\alpha ;$ no rescaling is needed for this, and the inverse rescaling $({\lambda }_{\alpha }{e}_{\alpha },{\lambda }_{\alpha }^{-1}{f}_{\alpha })$, ${\lambda }_{\alpha }>0$, preserves $[e_{\alpha },f_{\alpha }]$ and leaves $B(e_{\alpha },f_{\alpha })$ unchanged. [A1, L1, L2, algebra]

2.1 By the trace formula of step 1.1, $B(H,H^{\mathrm{\prime }})={\sum }_{\gamma \in \mathrm{\Phi }}\gamma (H)\gamma (H^{\mathrm{\prime }})$ for $H,H^{\mathrm{\prime }}\in \mathfrak{h}$. For $H={\sum }_{\alpha }{t}_{\alpha }{H}_{\alpha }$ with real $t_{\alpha }$ all values $\gamma (H)$ are real by [L1]; if $H\mathrm{\ne }0$ then $\gamma (H)\mathrm{\ne }0$ for some root $\gamma$, because an element of $\mathfrak{h}$ commuting with every root vector commutes with all of $\mathfrak{g}$ and so lies in the zero center recorded in [L2]. Hence $B(H,H)={\sum }_{\gamma }\gamma (H{)}^2>0$ for $H\mathrm{\ne }0$, that is, the restriction of $B$ to ${\mathfrak{h}}_{\mathbb{R}}:={\sum }_{\alpha }\mathbb{R}{H}_{\alpha }$ is positive definite. [L1, L2, step 1.1, algebra]

2.2 (Normalized basis and closure of ${\mathfrak{k}}_0$) Choose the root vectors as in [L3], so that $[e_{\alpha },f_{\alpha }]=H_{\alpha }$, the structure constants $N_{\alpha \beta }$ are real and $N_{\alpha \beta }=-N_{-\alpha ,-\beta }$ whenever $\alpha +\beta \mathrm{\ne }0$; write $f_{\gamma }=e_{-\gamma }$ and abbreviate $A_{\alpha }:=e_{\alpha }-f_{\alpha }$, $B_{\alpha }:=i(e_{\alpha }+f_{\alpha })$, $T_{\alpha }:=i{H}_{\alpha }$. Define ${\mathfrak{k}}_0:={\mathrm{span}}_{\mathbb{R}}(\{T_{\alpha }:\alpha \in \mathrm{\Phi }\}\cup \{A_{\alpha }:\alpha \in \mathrm{\Phi }\}\cup \{B_{\alpha }:\alpha \in \mathrm{\Phi }\})\mathrm{.}$ Since these vectors span ${\mathfrak{k}}_0$ and the bracket is bilinear, it suffices to show that the bracket of any two of them again lies in ${\mathfrak{k}}_0$. For the Cartan brackets $[T_{\alpha },T_{\beta }]=0$, while $[H_{\alpha },e_{\beta }-f_{\beta }]=\beta (H_{\alpha })(e_{\beta }+f_{\beta })$ and $[H_{\alpha },e_{\beta }+f_{\beta }]=\beta (H_{\alpha })(e_{\beta }-f_{\beta })$ give $[T_{\alpha },A_{\beta }]=\beta (H_{\alpha })B_{\beta }\in {\mathfrak{k}}_0,[T_{\alpha },B_{\beta }]=-\beta (H_{\alpha })A_{\beta }\in {\mathfrak{k}}_0,$ with real coefficients because $\beta (H_{\alpha })\in \mathbb{Z}\subseteq \mathbb{R}$ by [L1]. For the root-root brackets let $\alpha \mathrm{\ne }\pm \beta$. Expanding and using $N_{-\alpha ,\beta }=-N_{\alpha ,-\beta }$ (the relation of [L3] with $(\alpha ,\beta )$ replaced by $(-\alpha ,\beta )$, legitimate here because $\beta \mathrm{\ne }\alpha$) together with $N_{-\alpha ,-\beta }=-N_{\alpha \beta }$ gives $[A_{\alpha },A_{\beta }]=N_{\alpha \beta }{A}_{\alpha +\beta }-N_{\alpha ,-\beta }{A}_{\alpha -\beta },[A_{\alpha },B_{\beta }]=N_{\alpha \beta }{B}_{\alpha +\beta }+N_{\alpha ,-\beta }{B}_{\alpha -\beta },$ $[B_{\alpha },B_{\beta }]=-N_{\alpha \beta }{A}_{\alpha +\beta }-N_{\alpha ,-\beta }{A}_{\alpha -\beta },$ where a term with vanishing structure constant is absent; all three are real linear combinations of generators. If $\beta =\pm \alpha$ then $A_{-\alpha }=-A_{\alpha }$ and $B_{-\alpha }=B_{\alpha }$ give $[A_{\alpha },A_{-\alpha }]=0=[B_{\alpha },B_{-\alpha }]$. Finally $[A_{\alpha },B_{\alpha }]=[e_{\alpha }-f_{\alpha },i(e_{\alpha }+f_{\alpha })]=i([e_{\alpha },f_{\alpha }]-[f_{\alpha },e_{\alpha }])=2i{H}_{\alpha }=2T_{\alpha }$, and $B_{-\alpha }=B_{\alpha }$ makes this also the case $\beta =-\alpha$. Hence all brackets of generators lie in ${\mathfrak{k}}_0$, so ${\mathfrak{k}}_0$ is a real Lie subalgebra of $\mathfrak{g}$. [A1, L1, L3, step 1.1, algebra]

3.1 The complex span of ${\mathfrak{k}}_0$ is $\mathfrak{g}$: from $A_{\alpha }-i{B}_{\alpha }=2e_{\alpha }$ and $A_{\alpha }+i{B}_{\alpha }=-2f_{\alpha }$ one has $e_{\alpha },f_{\alpha }\in \mathbb{C}{\mathfrak{k}}_0$ for every root, and the $H_{\alpha }=-i{T}_{\alpha }$ span $\mathfrak{h}$ over $\mathbb{C}$ by [L3]. Hence ${\mathfrak{k}}_0$ is a real form of $\mathfrak{g}$. [L1, L3, step 2.2, algebra]

4.1 The form is negative definite on ${\mathfrak{k}}_0$. On the Cartan part $B(T_{\alpha },T_{\beta })=-B(H_{\alpha },H_{\beta })$, so the restriction to $i{\mathfrak{h}}_{\mathbb{R}}$, ${\mathfrak{h}}_{\mathbb{R}}={\sum }_{\alpha }\mathbb{R}{H}_{\alpha }$, is minus $B{\mathrm{\mid }}_{{\mathfrak{h}}_{\mathbb{R}}}$, which is negative definite by step 2.1. On each root direction, using step 1.1 and the weight decomposition, $B(e_{\alpha },e_{\alpha })=0=B(f_{\alpha },f_{\alpha })$ because the two weight components do not pair, hence $B(A_{\alpha },A_{\alpha })=-2B(e_{\alpha },f_{\alpha })<0$ and $B(B_{\alpha },B_{\alpha })=-2B(e_{\alpha },f_{\alpha })<0$; also the mixed term $B(A_{\alpha },B_{\alpha })=i[B(e_{\alpha },e_{\alpha })-B(f_{\alpha },f_{\alpha })]=0$. For $X=a{A}_{\alpha }+b{B}_{\alpha }$ with real $(a,b)\mathrm{\ne }(0,0)$ this gives $B(X,X)=-2B(e_{\alpha },f_{\alpha })(a^2+b^2)<0$. Generators of distinct weights pair to zero, so the planes $W_{\alpha }=\mathbb{R}{A}_{\alpha }\oplus \mathbb{R}{B}_{\alpha }$ are pairwise orthogonal and orthogonal to $i{\mathfrak{h}}_{\mathbb{R}}$; for a set ${\mathrm{\Phi }}^{+}$ of representatives of the pairs $\{\alpha ,-\alpha \}$ the sum ${\mathfrak{k}}_0=i{\mathfrak{h}}_{\mathbb{R}}\oplus {⨁}_{\alpha \in {\mathrm{\Phi }}^{+}}{W}_{\alpha }$ is direct, and $B{\mathrm{\mid }}_{{\mathfrak{k}}_0}$ is negative definite. [L1, step 1.1, step 2.1, step 3.1, algebra]

5.1 By steps 3.1 and 4.1, ${\mathfrak{k}}_0$ is a real form of $\mathfrak{g}$ whose Killing form is negative definite, that is, a compact real form in the sense of Compact real form of a complex semisimple Lie algebra. The theorem follows. [step 3.1, step 4.1, A1] ∎