While studying the following paper An $\mathcal{N}=1$ Lagrangian for the rank 1 $E_6$ superconformal theory, on page 10 a stumbled upon the definition of embedding index
Consider an embedding of $H$ into $G$ and choose a representation $R_G$ of $G$ that decomposes to $\sum R_{H_i}$ representations of H under the embedding. Then the embedding index is defined as $$I_{I \hookrightarrow G} = \frac{\sum_i T_{R_{H_i}}}{T_{R_G}}$$ where $T_R$ stands for the Dynkin index of the representation $R$.
This definition is borrowed from another paper S-duality in $\mathcal{N}=2$ supersymmetric gauge theories where the authors do even some examples in appendix C.
The definition is clear to me, but what does not follow is the result in equation $(18)$ of the first paper where they evaluate $I_{U(1)\hookrightarrow SU(6)}$ from the embedding $\mathfrak{su}(5)\otimes\mathfrak{u}(1) \subset \mathfrak{su}(6)$ decomposing the fundamental of $SU(6)$ using, I imagine, the branching rule $6\rightarrow (1,-5)\oplus(5,1)$ where the notation stands for $(R_{SU(5)},R_{U(1)})$. This branching rule was found using the LieART package for mathematica.
What the authors find is the following
$$ I_{U(1)\hookrightarrow SU(6)} = \frac{5\times (2/3)^2+(-10/3)^2}{1/2}$$
which I really do not understand where it came from! In particular the squares.
I worked out a possible answer if someone will need it in the future. If we think about the definition of the Dynkin index of a representation of a group $G$ with generators $t^a$ where $a=1,\ldots,\dim G$ we have that $$\operatorname{Tr}t^a_Rt^b_R = T_R\delta^{ab}$$
Therefore in the case of the abelian $U(1)$, where we have only one generator $t$, we have that
$$\operatorname{Tr}t^2 = t^2 = T$$ where $t^2$ is just the value of the assigned hypercharge (sorry for the physics, non rigorous, term). This explains the squares in the formula of the embedding index. The values differ from mine simply for the assignment of the hypercharges, where the author used the branching rule $6 \rightarrow (5,2/3)\oplus(1,-10/3)$.