Chevalley basis and real structure constants

Let $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ and $H\in \mathfrak{h}$. By [L1] $[x,y]\in {\mathfrak{g}}_0=\mathfrak{h}$, and $[y,H]=-[H,y]=\alpha (H)y$ because $y\in {\mathfrak{g}}_{-\alpha }$; invariance [L2] therefore gives $B([x,y],H)=B(x,[y,H])=\alpha (H)B(x,y)=B(B(x,y)H_{\alpha },H)$ by [L3]. As $H\in \mathfrak{h}$ was arbitrary, $[x,y]-B(x,y)H_{\alpha }$ is $B$-orthogonal to $\mathfrak{h}$, hence zero by the nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$; that is, $[x,y]=B(x,y)H_{\alpha }$.

(Cartan integers) Let $\alpha ,\beta \in \mathrm{\Phi }$ be nonproportional and put $n:=\beta (h_{\alpha })=2(\alpha ,\beta )\mathrm{/}(\alpha ,\alpha )$ and $n^{\mathrm{\prime }}:=\alpha (h_{\beta })=2(\alpha ,\beta )\mathrm{/}(\beta ,\beta )$; both are integers by [L4] and [L5]. Then $n{n}^{\mathrm{\prime }}=4(\alpha ,\beta {)}^2\mathrm{/}((\alpha ,\alpha )(\beta ,\beta ))$ is an integer, while Cauchy-Schwarz [L6] and nonproportionality give $0\le (\alpha ,\beta {)}^2<(\alpha ,\alpha )(\beta ,\beta )$, so $n{n}^{\mathrm{\prime }}\in \{0,1,2,3\}$. Consequently $n=0$ if and only if $n^{\mathrm{\prime }}=0$; $n^{\mathrm{\prime }}\mathrm{/}n=(\alpha ,\alpha )\mathrm{/}(\beta ,\beta )$ whenever $n\mathrm{\ne }0$; and $\mathrm{\mid }n\mathrm{\mid }\le 3$.

By step 1.1 and [L3], $[x,y]=h_{\alpha }$ holds for $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ exactly when $B(x,y)=2\mathrm{/}(\alpha ,\alpha )$, where $(\alpha ,\alpha )=B(H_{\alpha },H_{\alpha })=\alpha (H_{\alpha })\mathrm{\ne }0$; moreover $B(x,y)\mathrm{\ne }0$ for all nonzero $x,y$, because the pairing of [L2] is nondegenerate and both ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ are one-dimensional by [L1], so that the nonzero functional $B(x,\cdot )$ on the line ${\mathfrak{g}}_{-\alpha }$ vanishes only at $0$.

(Strings and their bounds) Let $\alpha ,\beta \in \mathrm{\Phi }$ with $\alpha +\beta \in \mathrm{\Phi }$, let $\{-p,\dots ,q\}$ be the $\alpha$-string through $\beta$ (so $q\ge 1$ because the index $1$ lies in the string) and let $\{-p^{\mathrm{\prime }},\dots ,q^{\mathrm{\prime }}\}$ be the $\beta$-string through $\alpha$ (so $q^{\mathrm{\prime }}\ge 1$), as supplied by [L4]; then $n=p-q$ and $n^{\mathrm{\prime }}=p^{\mathrm{\prime }}-q^{\mathrm{\prime }}$ by [L4]. The pair $(\alpha ,\beta +q\alpha )$ is nonproportional, since $\beta +q\alpha \in \mathbb{R}\alpha$ would force $\beta \in \mathbb{R}\alpha$ and hence $\beta =\pm \alpha$ by the reducedness recorded in [L5], contradicting $\alpha +\beta \in \mathrm{\Phi }$ and $\alpha +\beta \mathrm{\ne }0$; its Cartan integer is $(\beta +q\alpha )(h_{\alpha })=\beta (h_{\alpha })+2q=(p-q)+2q=p+q$, so step 1.2 gives $p+q\le 3$. The same argument with $\alpha$ and $\beta$ interchanged gives $p^{\mathrm{\prime }}+q^{\mathrm{\prime }}\le 3$.

(First normalization) Choose one representative in each pair $\{\alpha ,-\alpha \}\subseteq \mathrm{\Phi }$ and put $c_{\alpha }:=1$ for the chosen representatives while $c_{-\alpha }:=2/((\alpha ,\alpha )B(e_{\alpha }^0,e_{-\alpha }^0))$ for their negatives, a nonzero number by step 2.1; then $e_{\alpha }:=c_{\alpha }{e}_{\alpha }^0$ satisfies $[e_{\alpha },e_{-\alpha }]=c_{\alpha }{c}_{-\alpha }B(e_{\alpha }^0,e_{-\alpha }^0)H_{\alpha }=h_{\alpha }$ for every root $\alpha$ by step 1.1, so (i) of the statement holds for a rescaling of the given basis.

(String-length identity) With the notation of step 2.2, $q(\alpha +\beta ,\alpha +\beta )=(p+1)(\beta ,\beta )$. Indeed $q\ge 1$, $p\ge 0$ and $p+q\le 3$ by step 2.2, so $(p,q)$ is one of $(0,1)$, $(1,1)$, $(2,1)$, $(0,2)$, $(1,2)$, $(0,3)$; writing $A=(\alpha ,\alpha )$, $P=(\alpha ,\beta )$ and $C=(\beta ,\beta )$ and using $P=nA\mathrm{/}2$ with $n=p-q$, the six cases are as follows. If $(p,q)$ is $(0,1)$ or $(1,2)$, then $n=-1$, so $A+2P=0$ and both sides of the identity equal $qC=(p+1)C$. If $(p,q)=(1,1)$, then $n=0$, so $P=0$ and the identity becomes $A+C=2C$, that is $A=C$, which holds because the Cartan integer $2(\alpha ,\alpha +\beta )\mathrm{/}(\alpha +\beta ,\alpha +\beta )=2A\mathrm{/}(A+C)$ lies in $\mathbb{Z}$ by [L5] and in the open interval $(0,2)$, hence equals $1$. If $(p,q)=(2,1)$, then $n=1$, so $P=A\mathrm{/}2$ and the identity becomes $A+2P=2C$, that is $A=C$; here $n^{\mathrm{\prime }}=2P\mathrm{/}C>0$ satisfies $n^{\mathrm{\prime }}\in \{1,2,3\}$ by step 1.2 and also $n^{\mathrm{\prime }}=p^{\mathrm{\prime }}-q^{\mathrm{\prime }}\le p^{\mathrm{\prime }}+q^{\mathrm{\prime }}-2\le 1$ by step 2.2, so $n^{\mathrm{\prime }}=1$ and $C=2P=A$. If $(p,q)=(0,2)$, then $n=-2$ while $n^{\mathrm{\prime }}\mathrm{/}n=A\mathrm{/}C>0$ gives $n^{\mathrm{\prime }}<0$, so $n{n}^{\mathrm{\prime }}\in \{0,1,2,3\}$ of step 1.2 forces $n^{\mathrm{\prime }}=-1$, hence $C=nA\mathrm{/}{n}^{\mathrm{\prime }}=2A$ and $P=nA\mathrm{/}2=-A$, and the identity becomes $2(A+2P+C)=2(A-2A+2A)=2A=C$. If $(p,q)=(0,3)$, then $n=-3$ and again $n^{\mathrm{\prime }}<0$ with $n{n}^{\mathrm{\prime }}\in \{0,1,2,3\}$, so $n^{\mathrm{\prime }}=-1$, hence $C=3A$ and $P=-3A\mathrm{/}2$, and the identity becomes $3(A+2P+C)=3(A-3A+3A)=3A=C$.

(The string module is irreducible) Keep the notation of step 2.2 and put $V:={⨁}_{i=-p}^q{\mathfrak{g}}_{\beta +i\alpha }$. By [L1] the brackets with $e_{\alpha }$ and $e_{-\alpha }$ shift the index $i$ by $\pm 1$ and vanish past the ends of the string, while $h_{\alpha }$ preserves each weight space ${\mathfrak{g}}_{\beta +i\alpha }$, on which it acts by the scalar $\beta (h_{\alpha })+2i$; hence $V$ is a finite-dimensional module for the triple $(e_{\alpha },e_{-\alpha },h_{\alpha })$ of [L7]. Its weights are $\beta (h_{\alpha })+2i=p-q+2i$ for $i=-p,\dots ,q$, that is the arithmetic progression $-(p+q),-(p+q)+2,\dots ,p+q$, each occurring on a one-dimensional space by [L1]. A decomposition of $V$ into irreducibles [L7] has distinct highest weights, because every weight space is one-dimensional, and the top weight $m:=p+q$ occurs exactly once; the irreducible summand of highest weight $m$ has all the weights $m,m-2,\dots ,-m$ with multiplicity one, so it exhausts the weight multiset of $V$, and $V$ itself is irreducible of highest weight $m$.

Any further rescaling $e_{\alpha }\mapsto t_{\alpha }{e}_{\alpha }$ with nonzero $t_{\alpha }$ satisfies $[t_{\alpha }{e}_{\alpha },t_{-\alpha }{e}_{-\alpha }]=t_{\alpha }{t}_{-\alpha }{h}_{\alpha }$ by step 3.1, so it preserves the relations (i) exactly when $t_{\alpha }{t}_{-\alpha }=1$ for all $\alpha$.

(Base and generators) Let $\mathrm{\Delta }=\{{\alpha }_1,\dots ,{\alpha }_r\}$ be the base of a positive system of $(E,\mathrm{\Phi })$; such a base exists because regular vectors exist (Positive systems and simple roots) and it is a basis of $E$ by [L5]. Put $e_i:=e_{{\alpha }_i}$ and $f_i:=e_{-{\alpha }_i}$ for the vectors of step 3.1. Then $[e_i,f_i]=h_{{\alpha }_i}=h_i$, while $[h_i,e_i]={\alpha }_i(h_i)e_i=2e_i$ and $[h_i,f_i]=-2f_i$ by [L1] and ${\alpha }_i(h_i)=2$ of [L3]; thus $(e_i,f_i,h_i)$ is a root ${\mathfrak{s}\mathfrak{l}}_2$ triple with $h_i$ the coroot of ${\alpha }_i$.

(Coefficients along the string) Let $v:=e_{\beta +q\alpha }\mathrm{\ne }0$. By step 3.3 the vector $v$ is a highest weight vector: $[e_{\alpha },v]\in {\mathfrak{g}}_{\beta +(q+1)\alpha }=0$ by [L1] and the maximality of $q$. Since the weight spaces of $V$ are one-dimensional, for each $i\in \{-p,\dots ,q\}$ there are nonzero $c_i\in \mathbb{C}$ with $e_{\beta +i\alpha }=c_i{f}^{q-i}v$, where $f:=e_{-\alpha }$, and $c_q=1$. Induction on $k\ge 1$, using $[e,f]=h_{\alpha }$, $[h_{\alpha },f]=-2f$ and $[h_{\alpha },v]=mv$ with $e:=e_{\alpha }$, gives $[e,f^{k}v]=k(m-k+1)f^{k-1}v$; hence for $-p\le i\le q-1$ we get $[e,e_{\beta +i\alpha }]=c_i(q-i)(p+i+1)f^{q-i-1}v$, that is $N_{\alpha ,\beta +i\alpha }=(c_i\mathrm{/}{c}_{i+1})(q-i)(p+i+1)$.

(Chevalley involution) The assignment $e_i\mapsto -f_i$, $f_i\mapsto -e_i$, $h_i\mapsto -h_i$ preserves the relations of the Serre algebra of [L8]: it fixes the relations $[h_i,h_j]=0$ up to sign, interchanges the relations $[h_i,e_j]=a_{ij}{e}_j$ and $[h_i,f_j]=-a_{ij}{f}_j$, preserves $[e_i,f_j]={\delta }_{ij}{h}_i$ in the form $[f_i,e_j]=-{\delta }_{ij}{h}_i$, and sends the Serre relations $(\mathrm{ad}{e}_i{)}^{1-a_{ij}}{e}_j=0$ and $(\mathrm{ad}{f}_i{)}^{1-a_{ij}}{f}_j=0$ into each other, because $(\mathrm{ad}(-f_i){)}^{1-a_{ij}}(-f_j)=\pm (\mathrm{ad}{f}_i{)}^{1-a_{ij}}{f}_j$ and $(\mathrm{ad}(-e_i){)}^{1-a_{ij}}(-e_j)=\pm (\mathrm{ad}{e}_i{)}^{1-a_{ij}}{e}_j$ both vanish. By the universal property of the presented algebra $\mathfrak{g}(A)$ (Lie algebra presented by generators and relations), this relation-preserving assignment extends to a Lie algebra homomorphism ${\omega }_A$; it is bijective because ${\omega }_A^2$ fixes the generators $e_i,f_i,h_i$ of $\mathfrak{g}(A)$ and hence equals $\mathrm{id}$, so ${\omega }_A$ is an automorphism. Transported through the isomorphism $\mathfrak{g}(A)\to \mathfrak{g}$ of [L8], it defines an automorphism $\omega$ of $\mathfrak{g}$ with ${\omega }^2={\mathrm{id}}_{\mathfrak{g}}$, $\omega (e_i)=-f_i$, $\omega (f_i)=-e_i$ and $\omega (h_i)=-h_i$.

(The opposite string) Applying steps 3.3 and 4.3 to the pair $(-\beta ,\alpha )$ in place of $(\beta ,\alpha )$ — the $\alpha$-string through $-\beta$ is $\{-\beta -j\alpha :-p\le j\le q\}$, because $-\beta -j\alpha =-(\beta +j\alpha )$ — produces a highest weight vector $w:=e_{-\beta +p\alpha }$ and nonzero $c_j^{\mathrm{\prime }}$ with $e_{-\beta -j\alpha }=c_j^{\mathrm{\prime }}{f}^{p+j}w$ and $c_{-p}^{\mathrm{\prime }}=1$; here the bracket with $e_{-\alpha }=f$ lowers the $f$-power without an extra coefficient, so $N_{-\alpha ,-\beta -j\alpha }=c_j^{\mathrm{\prime }}\mathrm{/}{c}_{j+1}^{\mathrm{\prime }}$ for $-p\le j\le q-1$.

Since $\omega (h_i)=-h_i$ for all $i$ and the $h_i=h_{{\alpha }_i}$ form a basis of $\mathfrak{h}$ by [L5], $\omega$ acts as $-\mathrm{id}$ on $\mathfrak{h}$. If $x\in {\mathfrak{g}}_{\alpha }$ and $H\in \mathfrak{h}$, then $[H,\omega x]=[\omega (-H),\omega x]=\omega ([-H,x])=-\alpha (H)\omega x$, so $\omega x\in {\mathfrak{g}}_{-\alpha }$; as $\omega$ is bijective, $\omega ({\mathfrak{g}}_{\alpha })={\mathfrak{g}}_{-\alpha }$ for every root $\alpha$. In particular each $\omega (e_{\alpha })$ is a nonzero element of the line ${\mathfrak{g}}_{-\alpha }$, say $\omega (e_{\alpha })=c_{\alpha }{e}_{-\alpha }$ with $c_{\alpha }\in {\mathbb{C}}^{\times }$, and applying $\omega$ twice gives $c_{\alpha }{c}_{-\alpha }=1$.

(Pairing of the two strings) By the invariance of $B$, $B(fx,y)=-B(x,fy)$ for all $x,y\in \mathfrak{g}$ [L2]. Applying this $q-i$ times and using $B({\mathfrak{g}}_{\gamma },{\mathfrak{g}}_{\delta })=0$ unless $\gamma +\delta =0$ [L2] together with the weights of $v$ and $w$ gives $B(e_{\beta +i\alpha },e_{-\beta -i\alpha })=c_i{c}_i^{\mathrm{\prime }}(-1{)}^{q-i}B(v,f^{m}w)$, and $f^{m}w=(c_q^{\mathrm{\prime }}{)}^{-1}{e}_{-\beta -q\alpha }$ because $e_{-\beta -j\alpha }=c_j^{\mathrm{\prime }}{f}^{p+j}w$ at $j=q$. Also $B(e_{\gamma },e_{-\gamma })=2\mathrm{/}(\gamma ,\gamma )$ for every root $\gamma$: by step 1.1 the bracket $[e_{\gamma },e_{-\gamma }]$ equals both $B(e_{\gamma },e_{-\gamma })H_{\gamma }$ and, by (i), the coroot $h_{\gamma }=2H_{\gamma }\mathrm{/}(\gamma ,\gamma )$ of [L3]. Comparing the two expressions for $B(e_{\beta +i\alpha },e_{-\beta -i\alpha })$ therefore gives $c_i{c}_i^{\mathrm{\prime }}(-1{)}^{q-i}(c_q^{\mathrm{\prime }}{)}^{-1}\mathrm{/}(\beta +q\alpha ,\beta +q\alpha )=1\mathrm{/}(\beta +i\alpha ,\beta +i\alpha )$, that is $c_i{c}_i^{\mathrm{\prime }}=(-1{)}^{q-i}{c}_q^{\mathrm{\prime }}(\beta +q\alpha ,\beta +q\alpha )\mathrm{/}(\beta +i\alpha ,\beta +i\alpha )$.

(Second normalization) By step 4.1 the rescalings with $t_{\alpha }{t}_{-\alpha }=1$ are exactly those preserving (i). Choose $t_{\alpha }\in {\mathbb{C}}^{\times }$ with $t_{\alpha }^2=-1\mathrm{/}{c}_{\alpha }$ for the numbers $c_{\alpha }\mathrm{\ne }0$ of step 6.1 and put $t_{-\alpha }:=1\mathrm{/}{t}_{\alpha }$, so that $t_{\alpha }{t}_{-\alpha }=1$. For $e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}:=t_{\alpha }{e}_{\alpha }$ we then have $[e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}]=h_{\alpha }$ by step 4.1, and $\omega (e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }})=t_{\alpha }{c}_{\alpha }{e}_{-\alpha }=t_{\alpha }{c}_{\alpha }{t}_{\alpha }{e}_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}=c_{\alpha }{t}_{\alpha }^2{e}_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}=-e_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}$, because $t_{\alpha }{t}_{-\alpha }=1$ gives $e_{-\alpha }=t_{\alpha }{e}_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ and because $c_{\alpha }{t}_{\alpha }^2=-1$.

(Chevalley's identity) Dividing the relation of step 6.2 for the indices $i$ and $i+1$, the unknown $c_q^{\mathrm{\prime }}$ cancels and the sign flips, so $c_i{c}_i^{\mathrm{\prime }}\mathrm{/}(c_{i+1}{c}_{i+1}^{\mathrm{\prime }})=-(\beta +(i+1)\alpha ,\beta +(i+1)\alpha )\mathrm{/}(\beta +i\alpha ,\beta +i\alpha )$; multiplying by the coefficient formulas of steps 4.3 and 5.2 gives $N_{\alpha ,\beta +i\alpha }{N}_{-\alpha ,-\beta -i\alpha }=-(q-i)(p+i+1)\frac{(\beta +(i+1)\alpha ,\beta +(i+1)\alpha )}{(\beta +i\alpha ,\beta +i\alpha )}=-(p+i+1{)}^2$, the last equality by the string-length identity 3.2 applied to the pair $(\alpha ,\beta +i\alpha )$, whose string is $\beta +i\alpha +n\alpha$ for $-p-i\le n\le q-i$; in particular, at $i=0$, $N_{\alpha \beta }{N}_{-\alpha ,-\beta }=-(p+1{)}^2$.

(Sign relation) Let $\alpha ,\beta \in \mathrm{\Phi }$ with $\alpha +\beta \in \mathrm{\Phi }$. Applying $\omega$ to $[e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{\beta }^{\mathrm{\prime }\mathrm{\prime }}]=N_{\alpha \beta }{e}_{\alpha +\beta }^{\mathrm{\prime }\mathrm{\prime }}$ and using $\omega (e_{\gamma }^{\mathrm{\prime }\mathrm{\prime }})=-e_{-\gamma }^{\mathrm{\prime }\mathrm{\prime }}$ for all $\gamma$ gives $[e_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{-\beta }^{\mathrm{\prime }\mathrm{\prime }}]=N_{\alpha \beta }(-e_{-\alpha -\beta }^{\mathrm{\prime }\mathrm{\prime }})$, that is $N_{-\alpha ,-\beta }=-N_{\alpha \beta }$; when $\alpha +\beta \mathrm{\notin }\mathrm{\Phi }\cup \{0\}$ both brackets $[e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{\beta }^{\mathrm{\prime }\mathrm{\prime }}]$ and $[e_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{-\beta }^{\mathrm{\prime }\mathrm{\prime }}]$ vanish by [L1], so $N_{\alpha \beta }=N_{-\alpha ,-\beta }=0$. Hence $N_{\alpha \beta }=-N_{-\alpha ,-\beta }$ for all $\alpha ,\beta \in \mathrm{\Phi }$ with $\alpha +\beta \mathrm{\ne }0$, which is (ii) of the statement, the excluded case $\beta =-\alpha$ being the one in which the constants are not defined.

(Integrality) The product $N_{\alpha \beta }{N}_{-\alpha ,-\beta }$ is unchanged by the rescaling of step 7.1: that rescaling replaces $N_{\alpha \beta }$ by $(t_{\alpha }{t}_{\beta }\mathrm{/}{t}_{\alpha +\beta })N_{\alpha \beta }$ and $N_{-\alpha ,-\beta }$ by $(t_{-\alpha }{t}_{-\beta }\mathrm{/}{t}_{-\alpha -\beta })N_{-\alpha ,-\beta }=(t_{\alpha +\beta }\mathrm{/}(t_{\alpha }{t}_{\beta }))N_{-\alpha ,-\beta }$, because $t_{\gamma }{t}_{-\gamma }=1$, so the product is multiplied by $1$. Hence step 7.2 gives $N_{\alpha \beta }{N}_{-\alpha ,-\beta }=-(p+1{)}^2$ also for the vectors $e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ of step 7.1, and combining this with $N_{-\alpha ,-\beta }=-N_{\alpha \beta }$ of step 8.1 gives $N_{\alpha \beta }^2=(p+1{)}^2$, so $N_{\alpha \beta }=\pm (p+1)$ for all $\alpha ,\beta \in \mathrm{\Phi }$ with $\alpha +\beta \in \mathrm{\Phi }$; this is (iii) of the statement, and together with the convention $N_{\alpha \beta }=0$ otherwise it shows that all constants $N_{\alpha \beta }$ are integers. Moreover each $e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}=t_{\alpha }{c}_{\alpha }{e}_{\alpha }^0$ is a nonzero complex multiple of the given vector $e_{\alpha }^0$, so the family produced is obtained from the given root-vector basis by rescaling, as required.

(Real form and integral structure) Let ${\mathfrak{g}}_0:={\mathrm{span}}_{\mathbb{R}}\{h_{\alpha },e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}:\alpha \in \mathrm{\Phi }\}$. It is closed under brackets: $[{\mathfrak{h}}_{\mathbb{R}},{\mathfrak{h}}_{\mathbb{R}}]=0$; $[h_{\alpha },e_{\beta }^{\mathrm{\prime }\mathrm{\prime }}]=\beta (h_{\alpha })e_{\beta }^{\mathrm{\prime }\mathrm{\prime }}$ with $\beta (h_{\alpha })\in \mathbb{Z}$ by [L4]; $[e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{\beta }^{\mathrm{\prime }\mathrm{\prime }}]=N_{\alpha \beta }{e}_{\alpha +\beta }^{\mathrm{\prime }\mathrm{\prime }}$ with $N_{\alpha \beta }\in \mathbb{Z}$ by step 9.1; $[e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }},e_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}]=h_{\alpha }$; and all brackets with $\alpha +\beta \mathrm{\notin }\mathrm{\Phi }\cup \{0\}$ vanish by [L1]. Since $\mathfrak{h}={\mathfrak{h}}_{\mathbb{R}}\oplus i{\mathfrak{h}}_{\mathbb{R}}$ by [L5] and the vectors $e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ form a basis of ${⨁}_{\alpha \in \mathrm{\Phi }}{\mathfrak{g}}_{\alpha }$, we get $\mathfrak{g}={\mathfrak{g}}_0\oplus i{\mathfrak{g}}_0$, so that ${\mathfrak{g}}_0$ is a real form of $\mathfrak{g}$. With respect to the basis $\{h_{{\alpha }_1},\dots ,h_{{\alpha }_r}\}\cup \{e_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}\}$ the structure constants are integers, because $h_{\alpha }\in {\sum }_i\mathbb{Z}{h}_{{\alpha }_i}$ by the coroot-lattice statement of [L5]; and for $H\in {\mathfrak{h}}_{\mathbb{R}}$ the operators ${\mathrm{ad}}_H$ are diagonalizable over $\mathbb{R}$ with eigenvalues $\alpha (H)\in \mathbb{R}$ by [L1] and [L5]. Hence ${\mathfrak{g}}_0$ is a split real form of $\mathfrak{g}$.

(Dual normalization) Put ${\lambda }_{\alpha }=((\alpha ,\alpha )\mathrm{/}2{)}^{1\mathrm{/}2}>0$, a positive real number, and $X_{\alpha }:={\lambda }_{\alpha }{e}_{\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ for all $\alpha$; then also $X_{-\alpha }={\lambda }_{\alpha }{e}_{-\alpha }^{\mathrm{\prime }\mathrm{\prime }}$, because ${\lambda }_{-\alpha }={\lambda }_{\alpha }$. By step 3.1 and (i), $[X_{\alpha },X_{-\alpha }]={\lambda }_{\alpha }^2{h}_{\alpha }=\frac{(\alpha ,\alpha )}{2}\cdot \frac{2H_{\alpha }}{(\alpha ,\alpha )}=H_{\alpha }$ and $B(X_{\alpha },X_{-\alpha })={\lambda }_{\alpha }^2\cdot 2\mathrm{/}(\alpha ,\alpha )=1$. Hence for $\alpha +\beta \in \mathrm{\Phi }$ the constants defined by $[X_{\alpha },X_{\beta }]=C_{\alpha \beta }{X}_{\alpha +\beta }$ are $C_{\alpha \beta }=({\lambda }_{\alpha }{\lambda }_{\beta }\mathrm{/}{\lambda }_{\alpha +\beta })N_{\alpha \beta }$, real numbers because ${\lambda }_{\gamma }$ and $N_{\alpha \beta }=\pm (p+1)$ are real; and $C_{\alpha \beta }=-C_{-\alpha ,-\beta }$ because ${\lambda }_{-\gamma }={\lambda }_{\gamma }$ and $N_{\alpha \beta }=-N_{-\alpha ,-\beta }$.