Commutator with generator of Lorentz group shows transformation property The generators of the Lorentz group in spinorial representation are
\begin{align}
M^{\mu\nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu]
\end{align}
where $[\gamma^\mu,\gamma^\nu]$ is the commutator of two dirac matrices. In a exercise sheet of my course on quantum field theory I should show, that
\begin{align}
[M^{\mu\nu},\gamma^\rho] = i(g^{\nu\rho}\gamma^\mu-g^{\mu\rho}\gamma^\nu)
\end{align}
which is not too hard. My question here is the following. Its says in a note on the sheet, that this means, that the gamma matrices $\gamma^\rho$ transform as a vector. I can not see why this commutation relation gives information about the transformation property of $\gamma^\rho$.
 A: Let $G$ be some group. A representation of $G$ is a group homomorphism $D:G\to{\rm GL}(V)$ where $V$ is a vector space. This means that $D$ is a map that assings a matrix $D(g)$ for each $g\in G$ acting as a linear operator on the vector space $V$. Moreover it has to obey $$D(gg')=D(g)D(g')\tag{1}$$
so that matrix multiplication reproduces the abstract composition pattern of the group. This means we are realizing the group elements as matrices.
Now, when $G$ is a Lie group and $g\in G$ is close to the identity we can write it in terms of an element of the Lie algebra $$g=1+i\theta^a X_a+O(\theta^2)\tag{2}$$
where $X_a\in \mathfrak{g}$ form a basis of the Lie algebra and $\theta^a$ are the parameters of that particular group element. At the level of the representation this is realized as $$D(g)=1+i\theta^a T_a+O(\theta^2)\tag{3}$$
where $T_a$ are now matrices acting on $V$ called generators of the representation. These matrices obey the exact same Lie algebra that the abstract $X_a$ does:
$$[X_a,X_b]=if_{ab}^{\phantom {ab} c}X_c\Longrightarrow[T_a,T_b]=if_{ab}^{\phantom {ab} c}T_c\tag{4}$$
where $[X_a,X_b]$ is the $\mathfrak{g}$ Lie bracket and $[T_a,T_b]$ the standard matrix commutator. A more precise statement is that $D:G\to{\rm GL}(V)$ induces one Lie algebra representation $\pi:\mathfrak{g}\to\mathfrak{gl}(V)$ called the derived representation.
Next let us consider the Lorentz group and consider the vector representation, which is its defining representation. In that case $D(\Lambda)=\Lambda$ is the Lorentz transformation matrix itself acting on $\mathbb{R}^{1,3}$ column vectors. It is not hard to show that if $$\Lambda^\mu_{\phantom\mu \nu}= \delta^\mu_{\phantom\mu \nu}+\omega^\mu_{\phantom \mu\nu}\tag{5},$$
with infinitesimal $\omega^\mu_{\phantom \mu\nu}$ then $\omega_{\mu\nu}=\eta_{\mu\sigma}\omega^\sigma_{\phantom\sigma\nu}$ must be skew-symmetric. As a result you may show that (5) admits to be written as $$\Lambda^\mu_{\phantom\mu \nu}= \delta^\mu_{\phantom \mu\nu}+\dfrac{i}{2}\omega_{\alpha\beta}(\mathcal{J}^{\alpha\beta})^{\mu}_{\phantom \mu\nu},\quad (\mathcal{J}^{\alpha\beta})^{\mu}_{\phantom \mu\nu}=i(\eta^{\mu\beta}\delta^{\alpha}_{\phantom\alpha\nu}-\eta^{\mu\alpha}\delta^\beta_{\phantom\beta \nu}),\tag{6}$$
and therefore the matrices $\mathcal{J}_{\alpha\beta}$ work as the generators $T_a$ of the vector representation. Observe that the generators act on vectors as
$$(\mathcal{J}^{\alpha\beta})^{\mu}_{\phantom \mu\nu} v^\nu = i(\eta^{\mu\beta}v^\alpha-\eta^{\mu\alpha}v^\beta)\tag{7}.$$
Now we can answer your question. Observes that $[M^{\mu\nu},\gamma^{\rho}]$ reproduces (7), i.e., it we can write
$$[M^{\mu\nu},\gamma^\rho]= ({\cal J}^{\mu\nu})^\rho_{\phantom\rho \sigma}\gamma^\sigma\tag{8}.$$
This means that the adjoint action of $M^{\mu\nu}$ on the gamma matrices is exactly the same as transforming the gamma matrices as a vector would under the defining representation of the Lorentz group. This is why we can say that grouping $\gamma^\mu$ in a column array gives you something that behaves as a Lorentz vector.
