Regarding the Weyl spinor and its transformation properties I am trying to prove the Lorentz invariance of the (left-handed) Weyl Lagrangian:
$$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi$$
A Lorentz transformation is realized as
$\psi\to M\psi$, where $M=\exp[-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}]$, and $S$ are the generators of the representation, i.e., $S^{0i}=-\frac{i}{2}\sigma^i$ and $S^{ij}=\frac{1}{2}\epsilon^{ijk}\sigma^k$. Therefore, the Lagrangian transforms as
$$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi\to i\psi^\dagger M^\dagger\bar\sigma^\mu M\partial_\mu \psi$$
so that if it actually is invariant, we must have $M^\dagger\bar\sigma^\mu M=\bar\sigma^\mu$ (and, in particular, $M^\dagger M=\mathbb I$). I think this is a nice relation, which expresses the invariance of the symbol $\bar\sigma^\mu$.
Now the thing is, I think this relation is not true. Take, for example, a boost in the $z$-direction with rapidity $\omega$. This means
$$M=\exp[-i\omega S^{03}]=\exp\left[\frac{\omega}{2}\begin{pmatrix}-1& 0\\0&1\end{pmatrix}\right]=\begin{pmatrix}e^{-\omega/2}&0\\0&e^{\omega/2}\end{pmatrix}$$
In this case, it is easly seen that $M^\dagger M\neq\mathbb I$. My question is: what went wrong in my argument?
 A: Lorentz invariance refers to the action $S=\int\mathcal{L}(x)\,\mathrm{d}x$, not to the Lagrangian. To determine the condition on the Lagrangian which we must have, we make the coordinate change $x\to \Lambda x=:x'$ (a Lorentz transformation) and use the general fact that the Jacobian of a Lorentz transformation is unity, so
$$\int\mathcal{L}(x)\,\mathrm{d}x\to\int\mathcal{L}(x')\,\mathrm{d}x'=\int\mathcal{L}(\Lambda x)\,\mathrm{d}x$$
We must thus show $\mathcal{L}(x)\to\mathcal{L}(\Lambda x)$, where $\mathcal{L}(x)=\mathrm{i}\xi^\dagger(x)\bar\sigma^\mu\partial_\mu\xi(x)$.
The Weyl spinors transform as $\xi(x)\to D(\Lambda^{-1})\xi(\Lambda x)$, so the first step is$$\mathrm{i}\xi^\dagger(x)\bar\sigma^\mu\partial_\mu\xi(x)\to\mathrm{i}\xi^\dagger(\Lambda x)D^\dagger(\Lambda^{-1})\bar\sigma^\mu\partial_\mu D(\Lambda^{-1})\xi(\Lambda x)$$
The Pauli 4-vector $\bar\sigma^\mu =(1,\vec\sigma)$ transforms as $\bar\sigma^\mu\to D^\dagger(\Lambda)\bar\sigma^\mu D(\Lambda)=\Lambda^\mu{}_\nu\bar\sigma^\nu$, so the transformed Lagrangian is
$$\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu(\Lambda^{-1})^\mu{}_\nu\partial_\mu \xi(\Lambda x)$$
Finally, the partial derivative transformation rule is $\partial_\mu=\Lambda^\nu{}_\mu\partial_\nu'$, and the transformed Lagrangian is
$$\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu(\Lambda^{-1})^\mu{}_\nu\Lambda^\nu{}_\mu\partial_\nu' \xi(\Lambda x)=\mathrm{i}\xi^\dagger(\Lambda x)\bar\sigma^\nu\partial_\nu' \xi(\Lambda x)=\mathcal{L}(\Lambda x)$$
This shows the Lagrangian $\mathrm{i}\xi^\dagger\bar\sigma^\mu\partial_\mu\xi$ leads to a Lorentz-invariant action, as desired.
