= Solution
Choose the following <Pauli matrix> realization, with its spinor <basis> fixed for the later <Dirac adjoint>:
$$
\boxed{r(K_1)=\frac{\sigma_3}{2}=\frac12\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad
r(K_2)=-\frac{\sigma_2}{2}=\frac12\begin{pmatrix}0&i\\-i&0\end{pmatrix},\quad
r(J)=-\frac{i\sigma_1}{2}=\frac12\begin{pmatrix}0&-i\\-i&0\end{pmatrix}.}
$$
The <Pauli matrix> multiplication rule verifies
$$
[r(J),r(K_1)]=r(K_2),\qquad
[r(J),r(K_2)]=-r(K_1),\qquad
[r(K_1),r(K_2)]=-r(J).
$$
Extending linearly therefore gives a two-dimensional <Lie algebra representation>. For an <irreducible representation>, it suffices to rule out an invariant line. Any such line would have to be an <eigenline> of $r(K_1)$, whose two <eigenvalues> are distinct. Those two coordinate lines are interchanged by $r(K_2)$, so neither is invariant. There is no proper nonzero <invariant subspace>.
This is a spinorial <representation> rather than a single-valued two-dimensional <representation> of the identity component of the specified orthogonal <group>:
$$
e^{2\pi J}=I_3,\qquad e^{2\pi r(J)}=-I_2.
$$
The mismatch explicitly exhibits the need for the spin double cover. It causes no obstruction to the requested infinitesimal construction.
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