Following Onishchik's "Topology of Transitive Transformation Groups" section 3.10 the Dynkin index, $j_{\rho}$, of a Lie group homomorphism $\rho: H \rightarrow G$ of connected simple compact Lie groups is given as follows:
Choose an invariant non-dengenerate bilinear form $(\cdot,\cdot)_{\mathfrak{g}} $ on $\mathfrak{g} = \text{Lie}(G)$ such that $(\alpha,\alpha) = 2$ for a root of maximal length (where the form is defined on roots via the isomorphism given by $(\cdot,\cdot)$ between the Cartan subalgebra and it's dual) and similarly define $(\cdot,\cdot)_{\mathfrak{h}}$ on $\mathfrak{h} = \text{Lie}(H)$.
Then for any $X,Y \in \mathfrak{h}$, the Dynkin index of the map $\rho$ is given as follows:
$(\rho_\ast(X),\rho_\ast(Y))_{\mathfrak{g}} = j_\rho \cdot(X,Y)_{\mathfrak{h}}$.
I am particularly interested in the case where $\mathfrak{g}$ is a simple Lie algebra, $\alpha$ is a root of $\mathfrak{g}$ and $\{H_{\alpha},X_{\alpha},Y_{\alpha} \} \cong \mathfrak{sl}(2) \subset \mathfrak{g}$ is a subalgebra of $\mathfrak{g}$ isomorphic to $\mathfrak{sl}(2)$ (so $\rho$ is just the inclusion map), the following post says that for a simple root $\alpha$ the Dynkin index is given by $\frac{(\alpha_\text{max},\alpha_\text{max})}{(\alpha,\alpha)}$ where $\alpha_{\max}$ is a long root of $\mathfrak{g}$.
https://mathoverflow.net/questions/90124/3rd-homotopy-group-of-a-compact-simple-lie-group
I tried to argue this as follows:
Let $\kappa = \kappa_{\mathfrak{g}}$ be the Killing form on $\mathfrak{g}$, and let $\alpha_{max}$ be a root of maximal length in $\mathfrak{g}$, then define the invariant non- degen bilinear form $(\cdot,\cdot)_{\mathfrak{g}}$ on $\mathfrak{g}$ by
$$(\cdot,\cdot)_{\mathfrak{g}} = 2\frac{\kappa(\cdot,\cdot)}{\kappa(\alpha_{\text{max}},\alpha_{\text{max}})}$$
Now since $\{H_\alpha,X_\alpha,Y_\alpha\}$ is a subalgebra of $\mathfrak{g}$ with maximal length root $\alpha$ we can define the required invariant non-degen bilinear form on this subalgebra using the same Killing form $\kappa$ (which I will denote $(\cdot,\cdot)_{\mathfrak{sl}(2)})$ by
$$(\cdot,\cdot)_{\mathfrak{sl}(2)} = 2\frac{\kappa(\cdot,\cdot)}{\kappa(\alpha,\alpha)}.$$
Now working through the definition of the Dynkin index, $j$, we get
\begin{align} (H_\alpha,H_\alpha)_{\mathfrak{g}} &= j \cdot (H_\alpha,H_\alpha)_{\mathfrak{sl}(2)} \\ \iff 2\frac{\kappa(H_\alpha,H_\alpha)}{\kappa(\alpha_{\text{max}},\alpha_{\text{max}})} &= 2j\frac{\kappa(H_\alpha,H_\alpha)}{\kappa(\alpha,\alpha)} \\ \iff j &= \frac{(\alpha,\alpha)}{(\alpha_\text{max},\alpha_\text{max})}\end{align}
So somehow I am getting the reciprocal of what I want, and I cannot work out what mistake I am making. However, as the length of the longest root of $\mathfrak{g}$ will be at least as long as the longest root of the subalgebra I am thinking that under the above construction the Dynkin index should be defined as
$$ j_\rho \cdot (\rho_\ast(X),\rho_\ast(Y))_{\mathfrak{g}} = (X,Y)_{\mathfrak{h}}.$$
This fix would solve my issue however I want to make sure I am not fixing one mistake by making another. Finally, I should note that I am perfectly happy with the fact that the Dynkin index should be $\geq 1$ so I know something has gone wrong.
Any help would be much appreciated as I am just going in circles trying to work this out, thanks!