quantum mechanics - Commutation of abstract $O(3)$ generators and vectors

I've been given the following problem, and I'm quite lost with it.

Let $L_1$, $L_2$, and $L_3$ denote the abstract $o(3)$ algebras. You are given that $\vec{A} = (A_1, A_2, A_3)$ and $\vec{B} = (B_1, B_2, B_3)$ transform as vector operators of $o(3)$.

Show that $[L_j, \vec{A} \cdot \vec{B}] = 0$

I know that $L_j = \varepsilon_{jlm} q_l p_m$, and I can obviously determine the dot product, but I'm not sure where to go from there.

I do, however, know that $\vec{A} = \frac{1}{Ze^{2}\mu}(\vec{L} \times \vec{p}) + (\frac{1}{r})\vec{r}$, but I'm not sure how to integrate that into this problem.

Answer

A collection $\{V_1, V_2, V_3\}$ of operators on a vector space $V$ is called an $\mathfrak{o}(3)$ vector operator with respect to a representation $\rho$ of $\mathfrak{o}(3)$ acting on $V$ provided

$$\begin{align} [V_i, L_j] = i\epsilon_{ijk}V_k \end{align}$$ where $L_1, L_2, L_3$ are the representatives of the standard basis on $\mathfrak{o}(3)$. This means, in your case, that merely knowing that $A_i$ and $B_i$ are the components of vector operators tells us that $$\begin{align} [A_i, L_j]=i\epsilon_{ijk}A_k, \qquad [B_i, L_j]=i\epsilon_{ijk}B_k \end{align}$$ These commutation relations alone are sufficient to demonstrate the desired result using commutator identities.

As a tangential note, you will probably find the Wikipedia page on tensor operators to be generally, conceptually helpful for understanding this stuff. Also, a while back I asked the following question on physics.SE dealing with how to generalize and formalize the notion of tensor operators in a less basis-dependent way than how I defined them at the beginning of this answer. In case you're interested in math, here is that question:

Tensor Operators