Dual Coxeter number for the normalized invariant form

Fix one finite basis $x_1,\dots ,x_N$ and its $B$-dual basis $x^1,\dots ,x^N$, so $B(x_i,x^j)={\delta }_i^j$. If $T{x}_i={\sum }_j{t}_{ji}{x}_j$, then $B(T{x}_i,x^i)=t_{ii}$ by duality. Therefore [F3] gives $\mathrm{tr}(T)={\sum }_{i}B(T{x}_i,x^i)$. All choices here are finite and $B$ is nondegenerate by [F1].

For arbitrary $x,y\in \mathfrak{g}$, [F2], symmetry and invariance give $B(Cx,y)={\sum }_{i}B([x_i,[x^i,x]],y)={\sum }_{i}B([x^i,x],[y,x_i])={\sum }_{i}B([x,[y,x_i]],x^i)\mathrm{.}$ Indeed the first conversion uses $B([a,b],c)=B(b,[c,a])$ with $a=x_i$, and the second uses the same identity with $a=x$, followed by symmetry. Step 1.1 identifies the final sum with $\mathrm{tr}(\mathrm{ad}(x)\mathrm{ad}(y))=K_{\mathfrak{g}}(x,y)$. Thus $B(Cx,y)=K_{\mathfrak{g}}(x,y)$, with no compact real form or spectral assertion.

By [F1], $K_{\mathfrak{g}}(x,y)=c^{-1}B(x,y)$. Step 1.2 implies $B(Cx-c^{-1}x,y)=0$ for every $y$, so nondegeneracy yields $Cx=c^{-1}x$ for every $x$. Hence $C={\gamma }_{B}I$ with ${\gamma }_B=c^{-1}>0$. The scalar is unique because $\mathfrak{g}\mathrm{\ne }0$. Its half $h^{\vee }={\gamma }_B\mathrm{/}2$ is therefore well defined and positive without a compact-form existence theorem or a choice axiom.

If a compact real form $\mathfrak{k}$ is supplied, [F1] and [F4] make $Q=-B{\mathrm{\mid }}_{\mathfrak{k}}$ positive definite. Take a finite $Q$-orthonormal real basis $(X_i)$ by [F4]; it is also a complex basis of $\mathfrak{g}$ because $\mathfrak{k}$ is a real form. Its $B$-dual basis is $(-X_i)$, since $B(X_i,-X_j)={\delta }_{ij}$. Thus [F2] gives $C=-{\sum }_i\mathrm{ad}(X_i{)}^2$, and step 2.1 identifies this with $2h^{\vee }I$. This proves the conditional compact-basis formula without constructing a compact real form or inferring its existence from its definition.