Lorentz transformation of the Spinor Field I'm reading chapter 3 of Peskin and Schroeder and am stuck on page 43 of P&S. They have defined the Lorentz generators in the spinor representation as:
\begin{equation}
S^{\mu \nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu]
\end{equation}
such that a finite transformation is given by:
\begin{equation}
\Lambda_{1/2}=e^{-\frac{i}{2} \omega_{\mu \nu} S^{\mu \nu}}
\end{equation}
where $\gamma^\mu$ are the gamma matrices and $\omega_{\mu \nu}$ are the elements of a real and antisymmetric matrix. According to P&S on page 43 (between equation (3.32) and (3.33), they say that the adjoint of a Dirac spinor transforms as follows:
\begin{equation}
\psi^\dagger \rightarrow \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu}(S^{\mu \nu})^\dagger \right)
\end{equation}
However, I would expect the transformation to be:
\begin{equation}
\begin{aligned}
\psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\&
= \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\&
= \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right)
\end{aligned}
\end{equation}
where in the last line I made use of the fact that $\omega$ is a real and antisymmetric matrix:
\begin{equation}
(\omega_{\mu \nu})^\dagger = (\omega_{\mu \nu})^T = \omega_{\nu \mu} = - \omega_{\mu \nu}
\end{equation}
This implies that according to my calculations, equation (3.33) of P&S should actually be:
\begin{equation}
\overline{\psi} \rightarrow \overline{\psi} \Lambda_{1/2}
\end{equation}
This equation must be wrong because it means that $\overline{\psi} \psi$ does not transform as a scalar and therefore the Dirac Lagrangian is not correct. However, I do not know where my mistake is and was hoping someone could help me out?
 A: The last step you performed is incorrect.
$$\begin{equation}
\begin{aligned}
\psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\&
= \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\&
= \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right)
\end{aligned}
\end{equation}$$
$\omega $ is real, simply means that 
$$\begin{equation}
(\omega_{\mu \nu})^\dagger  = \omega_{\mu \nu}
\end{equation}$$
So we have 
$$\begin{equation}
\begin{aligned}
\psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\&
= \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\&
= \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right)
\end{aligned}
\end{equation}$$
To find the Lorentz invariance of $\bar\psi\psi$ you are missing $\bar\psi = \psi^\dagger\gamma^0$. When this $\gamma^0$ passes through $S^{\mu\nu}$ it fixes the problem. Solve it and share with us again.
A: The mistake you are making is in "daggering" the object $\omega_{\mu\nu}$.  For each $\mu, \nu = 0,\dots 3$, the symbol $\omega_{\mu\nu}$ is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; $(\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}$.
When we say that $\omega_{\mu\nu}$ is an antisymmetric real matrix, we really mean that the matrix with these numbers as components is such a matrix, not that $\omega_{\mu\nu}$ is a matrix for each $\mu$ and $\nu$.
