Lorentz algebra and its generators I'm reading Maggiore's book A Modern Introduction to Quantum Field Theory and I'm getting a bit confused when he writes about Lorentz algebra:
$$K^i = J^{i0},$$ 
$$J^{i}=\frac{1}{2}\epsilon^{ijk}J^{jk},$$
$$[J^{i}, J^{j}] = i\epsilon^{ijk} J^{k},$$
$$[J^{i}, K^j] =i\epsilon^{ijk}  K^k. $$
Then he states that $K^i$ is a spatial vector due to the last commutation relation. Is that the way a spatial vector transform under the $SO(3)$ algebra? If yes why?
 A: From Claude Cohen-Tannoudji, Volume 2, X.D.1:

(...) an observable $\textbf{V}$ is a vector if its three components $V_x, V_y$ and $V_z$ in an orthonormal frame $Oxyz$ satisfy the following commutation relations:
  $$ \tag{4-a} [J_x,V_x] = 0$$
  $$ \tag{4-b} [J_x,V_y] = i \hbar V_z$$
  $$ \tag{4-c} [J_x, V_z] = -i \hbar V_y$$
  as well as those obtained by cyclic permutation of the indices $x,y$ and $z$.

In your notation, these relations can be more compactly written as
$$ \tag{1} [J_i,V_j] = i \epsilon_{ijk} V_k$$
or (in a more formal, less rigorous way)
$$ \textbf{J} \times \textbf{V} = i \hbar \textbf{V}.$$
In other words, (1) are the defining relations of a vector operator $\textbf{V}$.

Other information about vector operators can be found on this wikipedia article and this physics.se answer.
A: You can think of the $J^i$, $i=1,2,3$ as rotation by $\pi/2$, or to be more precise $[J^i, K^j]$ as a rotation of $K^j$ around the $i$th axis by $\pi/2$. Then, e.g., $[J^1,K^2] = iK^3$, which corresponds to the fact that, if you rotate around the $x$ axis, you are rotating vectors on the $y-z$ plane. Hence $K^2$ is rotated onto $K^3$ (similarly, $K^3$ is rotated onto $-K^2$. This is the behaviour of spatial vectors under rotation.
