# Definition of coroots in a semisimple Lie algebra

For $$\mathfrak{g}$$ a semisimple Lie algebra, $$\mathfrak{h}$$ a Cartan subalgebra, $$\Phi$$ a root system with respect to $$\mathfrak{h}$$, the coroot $$\check{\alpha}$$ associated to a root $$\alpha \in \Phi$$ is usually defined as $$\check{\alpha} = \frac{2}{(\alpha,\alpha)}\alpha,$$ where $$(\cdot,\cdot)$$ is a bilinear form defined to be the dual of the Killing form $$\kappa$$, which I understand to be defined on $$\mathfrak{h}^*$$ by setting $$(\gamma, \delta) = \kappa (t_\gamma, t_\delta)$$, where $$t_\gamma \in \mathfrak{h}$$ for which $$\gamma(X) = \kappa(t_\gamma, X)$$ for all $$X \in \mathfrak{h}$$.

In my textbook (Goodman-Wallach) the definition of coroot is slightly different: $$\check{\alpha} = \kappa(e_\alpha, f_\alpha) \alpha,$$ where $$\{e_\alpha, f_\alpha, h_\alpha\}$$ is an $$\mathfrak{sl}_2$$-triple in $$\mathfrak{g}$$. However from my working out, I have that $$\kappa(h_\alpha, h_\alpha) = 2 \kappa (e_\alpha, f_\alpha),$$ so that on rearranging the given definition I obtain $$\check{\alpha} = \frac{\kappa(h_\alpha,h_\alpha)}{2} \alpha = \frac{(\alpha,\alpha)}{2}\alpha.$$

So is the definition in my textbook wrong, and should we instead define $$\check{\alpha} = \frac{1}{\kappa(e_\alpha, f_\alpha)} \alpha,$$ or have I misunderstood something in the theory?

Half of my confusion stems from the fact that Goodman-Wallach (unlike e.g. Humphreys) do not explicitly write out what form $$h_\alpha$$ should take: $$h_\alpha = \frac{2}{\kappa(t_\alpha, t_\alpha)}t_\alpha,$$ and finding appropriate $$e_\alpha$$ and $$f_\alpha$$ will still give an $$\mathfrak{sl}_2$$-triple. Using this explicit formula, it is easy to check that my identity $$\kappa(h_\alpha, h_\alpha) = 2 \kappa (e_\alpha, f_\alpha)$$ implies $$\check{\alpha} = \frac{2}{(\alpha, \alpha)}\alpha.$$ The other half of my confusion was that in my second-last line, $$\kappa(h_\alpha, h_\alpha) \neq (\alpha, \alpha)$$. Once there is an explicit form for $$h_\alpha$$, everything else is clarified.