Anomaly Coefficient of $SU(N)$ [Srednicki]

Question

In chapter 70 of his book "Quantum Field Theory" (PDF-Link to a free draft) Srednicki defines the anomaly coefficient $A(R)$ for a representation $R$ of the group $SU(N)$ as $$ A(R)\ d^{abc} \equiv \frac{1}{2} \text{Tr}\left( t_R^a \{t_R^b,t_R^c\} \right),\tag{70.33} $$ where $d^{abc}$ are the totally symmetric structure constants of $SU(N)$ and $t_R^a$ is the $a$-th generator of $SU(N)$ in the representation $R$. He writes

We normalize the anomaly coefficient so that it equals one for the smallest complex representation. In particular, for $SU(N)$ with $N \ge 3$, the smallest complex representation is the fundamental, and $A(N) = 1$.

This leads to my question:

1. Why can we even normalize anything here, since the symmetric structure constants are defined via $\{t^a,t^b\}=\frac{1}{N}1\!\!1 \delta^{ab}+d^{abc}t^c$? Edit: Especially when the generator's normalization has already been fixed with the Dynkin index: $\text{Tr}(t^a_Rt^b_R)=T(R)\delta^{ab}$, $T(F)=\frac{1}{2}$.

Additionally, he claims that for the fundamental representation of $SU(3)$, $A(F)=1$, whereas when I do explicit calculations using Mathematica and $t=\frac{1}{2}\lambda$ (Gell-Mann matrices), I get $A(F)=\frac{1}{4}$.

Answer 1

I think I found a solution. Since $$ \begin{align} A(R) d_R^{abc} &= \tfrac{1}{2}\text{Tr} \left( t^a_R \{ t^b_R,t^c_R \} \right)\\ &= \tfrac{1}{2}\text{Tr} \left( t^a_R \frac{1}{N}\delta^{bc}1\!\!\!\,1 \right) + \tfrac{1}{2}\text{Tr} \left( t^a_R d_R^{bcd}t_R^d \right)\\ &= \tfrac{1}{2N}\underbrace{\text{Tr} \left( t^a_R \right)}_{0}\delta^{bc} + \tfrac{1}{2}d_R^{bcd}\text{Tr} \left( t^a_R t_R^d \right)\\ &= \tfrac{1}{2}d_R^{bcd}\ T(R)\delta^{ad}\\ &= \tfrac{1}{2}d_R^{bca}\ T(R)\\ &= \tfrac{1}{2}d_R^{abc}\ T(R)\\ &\quad\leftrightarrow A(R) = \frac{T(R)}{2} \quad\text{ if }d^{abc}\neq 0. \end{align} $$ I suppose Srednicki used the Gell-Mann matrices $\lambda$ instead of $t=\frac{1}{2}\lambda$ as generators, for which the usual normalization is $T(F)_\text{Gell-Mann}=2$ and therefore $A(F)_\text{Gell-Mann}=1$.

[Which is strange, since he wrote $T(F)=\tfrac{1}{2}$ a few pages earlier...]