# Is the Dirac adjoint in the representation dual to Dirac spinor?

As seen in this Wikipedia page, the Lorentz group is not compact and the Dirac spinor (spin $$\frac{1}{2}$$) representation is NOT unitary.

Therefore, the complex conjugate representation does NOT coincide with the dual representation in general, cf. this Wiki page.

Now, I am looking at the definition of the Dirac adjoint. Using the notations in this link, I think that $$\psi^{\dagger}$$ is the complex conjugate representation of the Dirac spinor $$\psi$$.

Then, the Dirac adjoint $$\overline{\psi}:= \psi^\dagger \gamma^0$$ transforms as follows: $$$$\overline{\psi} \to \overline{\psi} \lambda^{-1}$$$$ for any Lorentz transformation $$\lambda$$ in the spin $$\frac{1}{2}$$ representation.

This transformation law seems to be in the form of the dual representation to $$\lambda$$. That is,

The Dirac adjoint $$\overline{\psi}:= \psi^\dagger \gamma^0$$ is in the representation dual to the Dirac spinor $$\psi$$,

Is this correct? After all, $$\overline{\psi} \psi$$ is well-known to be a Lorentz scalar. So, I guess these all looks consistent. Still, I would like to check for sure.

In order for $$\overline{\psi} \psi$$ to be a Lorentz scalar, we should have $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$ or equivalently (assuming $$(\gamma^0)^2 = 1$$) $$\lambda^{-1} = \gamma^0\lambda^{\dagger}\gamma^0$$ what else do you need to know?
The non-unitarity is caused by $$\gamma^0$$ in $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$. Without $$\gamma^0$$, $$\lambda$$ as in $$\lambda^{\dagger}\lambda = 1$$ would be unitary.