Lorentz transformation of four vector field For a 4-vector field $V^\mu (x)$, the Lorentz transformed 4-vector field $V'^\mu(x')$ can be written as
$$V'^\mu(x')={\Lambda^\mu} _\nu V^\nu(\Lambda^{-1}x')={\Lambda^\mu}_\nu V^\nu(x).$$
This can be seen as two transformations combined: one transformation on the 4-vector $V^\mu$ and another transformation on the coordinates $x$.
In the Physics from Symmetry book, pg. 87-88, it was said that we can  use the infinite-dimensional representation of the Lorentz group
$$M_{\mu\nu}=i(x^\mu\partial^\nu-x^\nu\partial^\mu)$$ to find the transformation on the coordinates, i.e. $$V^\nu(\Lambda^{-1}x')=e^{-i\frac{w_{\mu\nu}}{2}M_{\mu\nu}}V(x),$$
where $w_{\mu\nu}$ are constants.
However, from the first equation, we can see that $V^\nu(\Lambda^{-1}x')=V^\nu(x)$. So the exponential factor seems redundant. What is wrong with my reasoning here?
 A: Well they are not exactly saying that.  Notice how they have masterfully used the arrow $\rightarrow$ instead of the equal sign to basically tell you "you figure it out".
The last equation, 3.251, reads:
$$ \Phi_a(x) \rightarrow \left ( \mathrm{e}^{-\mathrm{i}\frac{\omega^{\mu\nu}}{2}M^{\mu\nu}} \right ) ^b_a \Phi_b(x).$$
Now, remember that you can write the Lorentz transformation matrix $\boldsymbol{\Lambda}$ in term of its generators $L_i$ (rotations) and $K_i$ (boosts) as:
$$ \boldsymbol{\Lambda} = \mathrm{e}^{\xi_i K_i + \theta_i L_i} = \mathrm{e}^{-\mathrm{i}\frac{\omega^{\mu\nu}}{2}M^{\mu\nu}}, $$
so the bit in the brackets is just a Lorentz transformation matrix.
So the first expression becomes:
$$ \Phi_a(x) \rightarrow \Lambda^b_a \,\Phi_b(x).$$
What does $\Phi_a(x)$ go to? Well it goes to $\Phi_a'(x')$:
$$ \Phi_a(x) \rightarrow  \color{red}{\Phi'_a(x') =} \Lambda^b_a \,\Phi_b(x).$$
This is the same expression that you start with in your question, with the substitution $\Phi_a \rightarrow V^\mu$.
