Normalization relation of the generators of the Lie-algebra of the Yang-Mills gauge group At the introduction to Yang-Mills-theory and its gauge group typically a $SU(N)$-group, the generators $t_A$ of the corresponding Lie-algebra are supposed to fulfill the following normalisation:
$$Tr(t_A t_B) = -\delta_{AB}$$
where $Tr$ is the trace of a matrix and $\delta_{AB}$ the Kronecker symbol. Is this trace of 2 generators  actually what in mathematics is called the Killing-form? And why is this normalisation only mentioned in the context of gauge theories while in the discussion of the rotation group or Lorentz group it seems to be of no importance?
 A: $\mathrm{Tr}(t_at_b)$ is indeed the Killing form. The standard definition for the Killing form is $\kappa(x, y) = -\mathrm{Tr}(\mathrm{ad}(x)\circ\mathrm{ad}(y))$, and putting in $x,y\rightarrow T_a,T_b$, the generators of the Lie algebra, you can show that $\kappa_{ab}\equiv\kappa(T_a,T_b)=-\mathrm{Tr}(t_at_b)$, where $t_a$ is the matrix representation of the adjoint action on $T_a$. Through redefinition of the Lie algebra basis, we can diagonalise $\kappa_{ab}$ for compact semisimple Lie algebras (in the case of Yang-Mills theories, the gauge group is $\mathrm{SU}(n)$ for $n>2$, which is compact semisimple). More generally, however, you can show via Schur's Lemma that the trace over these redefined generators in arbitrary representations will be proportional to the identity matrix.
This trace definition primarily comes in handy for perturbative calculations in which trace technology helps to simplify the computation of Feynman diagrams. Since gauge theories with gauge group $\mathrm{SO}(n)$ are not as useful as those with $\mathrm{SU}(n)$, this is normally not stated explicitly for $\mathrm{SO}(n)$.
However, note that during the construction of the quadratic Casimir element in the universal enveloping algebra of any semisimple Lie algebra $\mathfrak{g}$ (as well as the Casimir invariants for the representations of $\mathfrak{g}$), the Killing form is implicitly used since $C_2=\kappa^{ab}T_aT_b$. For example, the quadratic Casimir for $\mathfrak{so}(3)$ is given by $T_a^2+T_b^2+T_c^2=L_x^2+L_y^2+L_z^2$ since the Killing form is $\kappa_{ab}=\delta_{ab}$.
