I have a problem with proving a certain commutation relation. For my Bachelor's thesis I give a more mathematically rigurous 'treatment' of a select set of chapters of a paper by L.D. Faddeev. Noting that the Hilbert space on which the observables (special class of self-adjoint linear operators in quantum mechanics) act is the tensor product of the N local spaces, which are $C^2$, the full space is $H_N \otimes V_1 \otimes V_2 = h_1 \otimes h_2 \otimes \dots \otimes h_N \otimes V_1 \otimes V_2$ where $V_1$ and $V_2$ are auxiliary spaces and are also $C^2$ spaces - and $H_N$ is the full quantum (Hilbert) space.
$R_{a_1,a_2}(\lambda - \mu) \ T_{a_1}(\lambda) \ T_{a_2}(\mu) = T_{a_2}(\mu) \ T_{a_1}(\lambda) \ R_{a_1,a_2}(\lambda - \mu)$ is a proven relation, "the Fundamental Commutation Relation", where $T$ and $R$ are operators with a complex parameter $\lambda$, the indices denote the auxiliary spaces on which the operator acts non-trivially (note that $T$ also acts non-trivially on $H_N$ while $R$ does not). $T= L_{N,a}(\lambda) L_{N-1,a}(\lambda) \dots L_{1,a}(\lambda)$, $L_{n,a}(\lambda)= (\lambda - i/2)I + iP_{n,a}$ where $P$ denotes the permutation or flip operator, and less important for my question, $R= \lambda I + iP_{a_1,a_2}$.
Now to get to my actual question: taking the limit $\ \lim \mu \to \infty$ should, as Faddeev states at equation 98 of his paper, result in the commutation relation $\left[ T_a(\lambda), \tfrac{1}{2} \sigma^{\alpha} + S^{\alpha} \right] = 0$, using the expansion of $T$: $T_{a}(\lambda)=\lambda^N + i\lambda^{N-1} \sum_{\alpha} S^{\alpha} \otimes \sigma_a^{\alpha} + \mathcal{O}(\lambda^{N-2})$, where $\sigma_{a_1}^{\alpha}$ is one of the three pauli matrices acting on $V_1$. The sum over $\alpha$ in this expression hints to my problem: I can (so far) only prove that the sum $\sum_{\alpha=1}^3 \left[ T_{a_1}(\lambda), \tfrac{1}{2}\sigma_{a_1}^{\alpha} + S^{\alpha} \right] = 0$ is zero, not that the individual commutators are zero.
I have tried using the fact $\tfrac{1}{2}\sigma_{a_1}^{\alpha} + S^{\alpha}$ is hermitian and that they are mutually orthogonal when using the inner product over hermitian matrices given by Tr$(A \hat{B})$ (where the hat in this case denotes the hermitian conjugate). This didn't work (for example $T$ is not hermitian), and I couldn't think of any other way yet. It is however with noting that $T$ is bijective and only differs a constant from a unitary operator. Do any of you have an idea for proving $\left[ T_a(\lambda), \tfrac{1}{2} \sigma^{\alpha} + S^{\alpha} \right] = 0$, reducing the commutator sum over $\alpha$ to a single commutator being zero?
In case people are curious, this is the link to the paper http://arxiv.org/abs/hep-th/9605187