Rotations of higher spin and the adjoint representation I am trying to understand the following expression for rotations of higher-spin objects, potentially in greater than 3 dimensions (though 3D is probably enough, in which case you can replace $L_{\mu\nu}\rightarrow S_x, S_y,$ or $S_z$)
$$[L_{\mu\nu}, \mathcal{O}^a(0)] = (S_{\mu\nu})_b^a \mathcal{O}^b(0)$$
where $L_{\mu\nu}$ generates rotations within the $x^\mu,x^\nu$ plane, and therefore satisfies the commutation relations for $SO(d)$, and (this is the part that confuses me) $S_{\mu\nu}$ also satisfy the $SO(d)$ commutation relations. I know how things work out nicely for qubits in 3D; a natural basis is $\mathcal{O}^a \rightarrow \{\sigma^x,\sigma^y,\sigma^z\}$, and $S_{\mu\nu}$ satisfies the commutations for $SO(d)$. More generally, if each $\mathcal{O}^a$ is proportional to an $L_{\mu\nu}$, then $S_{\mu\nu}$ is by definition the adjoint representation of $SO(d)$.
However I don't understand how this generalizes nicely to higher spin or a larger basis of operators. It seems intuitive that $S_{\mu\nu}$ would transform like a spatial rotation is some way, but I can't see this from the math. So my question is the following:
Choose an orthonormal basis of operators $\mathcal{O}^a$, and use the above expression to define $(S_{\mu})_b^a$ (replacing $\mu\nu$ with $\mu$ for conciseness). How can we show that $[S_{\alpha}, S_{\beta}]=if_{\alpha\beta}^\gamma\mathcal{S}_\gamma$, where $f_{\alpha\beta}^\gamma$ are the structure constants of $SO(D)$ (and therefore $[L_\alpha,L_\beta]=if_{\alpha\beta}^\gamma L_\gamma$)?
Or if this expression is not true in general, then what is the correct way to express that $S_\mu$ forms a representation of the same algebra as $L_\mu$?
I came across this confusion when reading about field theory (this paper, Eq. 1.16), but it applies to regular quantum mechanics as well.
 A: You might be faked out by the "rotations" involved in the arbitrary-dimensional Lorentz group, and you had a good instinct to illustrate the expressions in the plain 3D rotation subgroup. In that case, an infinitesimal rotation of an arbitrary operator
$\mathcal{O}^a(0)$ amounts to
$$[L^{j}, \mathcal{O}^a(0)] = (S^j) _{~~b}^a \mathcal{O}^b(0),\tag{1}$$
where, as stated, the three spin matrices $S^j$ obey the exact same commutation relations of so(3) summarized by the above abstract generators,
$$
 [S^j,S^k]=i\epsilon^{jkl} S^l.\tag{2}
$$
The implied row/column indices of these matrices are suppressed here. These matrices may, in general, be $(2s+1)\times (2s+1)$ ones, s being the spin, 1/2,1, 3/2,2,...  and the indices b of your operators take 2s+1 values which these matrices mix among themselves.
For vector operators $\mathcal{O}^a(0)$, transforming like 3-vectors, the indices are three, and the
matrices are the familiar 3$\times$3 generators $\vec L$ of basic QM or classical mechanics. Note the indices of the $\mathcal{O}^a(0)$, undeclared and supressed here, could be anything the kets carry. So, they could be spinorial, as your "qubit" example proffers.
A ready mnemonic is that when kets rotate through R, operators such as $\mathcal{O}^a(0)$ must then rotate adjointly by $R\mathcal{O}^a(0)R^{-1}= {\mathcal D}(R)^a_{~~b} \mathcal{O}^a(0)$, which converts rotations of operator components to a linear combination ( $\leadsto$matrix multiplication) of these components. It is all predicated on Campbell's (1897) fundamental Lemma (1.17), which your  reference oddly dubs "Hausdorff's lemma"-- but then again, many people, including myself, would rather refer to it as "Hadamard's lemma", by dint of educational background. At the Lie algebra level, it maps isomorphically adjoint operation to matrix multiplication, as above, and "integrates"  to the relation of this paragraph.

Edit addressing the comment request for showing (1) $\leadsto$ (2) more explicitly.
The r.h.s. of (1) is a linearly reordered $\mathcal{O}^a(0)$, so we reapply it yet another time, and utilize the Jacobi identity,
$$[L^k,[L^{j}, \mathcal{O}^a(0)]] -[L^j,[L^{k}, \mathcal{O}^a(0)]]= (S^j) _{~~b}^a [L^k,\mathcal{O}^b(0)]- (S^k) _{~~b}^a [L^j,\mathcal{O}^b(0)]\\  
\leadsto [[L^k,L^{j}], \mathcal{O}^a(0)]=\bigl ((S^j) _{~~b}^a (S^k) _{~~c}^b -(S^k) _{~~b}^a (S^j) _{~~c}^b \bigr) {O}^c(0) \leadsto \\
i\epsilon^{kjl} [ L^l, \mathcal{O}^a(0) ]= ([S^j,S^k])_{~~c}^a {O}^c(0)\leadsto \\
\Bigl ( i\epsilon^{kjl} S^l -[S^j,S^k]\Bigr )_{~~c}^a ~{O}^c(0) =0,
$$
leading to (2) through dotting by another $i\epsilon_{kjm}$, up to a rep normalization: the Ss constitute a matrix representation of the abstract Lie algebra elements L, as seen above already. Many good QM texts review these arguments, as does Hamermesh, etc...
