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We know that in Lorentzian signature, fermions are representations of \begin{equation} Spin(3,1)\cong SL(2,\mathbb{C})\cong SU(2)\times SU(2)^* \end{equation} where crucially left/right handed fermions are exchanged under complex conjugation. This is what one means by the $SU(2)^*$ above. From an operational perspective, we write undotted and dotted indices, $\psi^\alpha$ and $\bar{\chi}_{\dot{\alpha}}$ to keep track of the left/right handed Weyl spinors. The statement that left/right handed spinors are related by conjugation is then captured by $(\psi_\alpha)^\dagger=\bar{\psi}_{\dot{\alpha}}$. An important consequence here is that one can’t write a $U(1)$ invariant mass term such as $\psi \bar{\psi}$.

Now if we move to Euclidean signature, fermions transform under \begin{equation} Spin(4)\cong SU(2)\times SU(2) \end{equation} and this time we have two independent copies of $SU(2)$. This means that left/right handed spinors now do not transform into each other under conjugation. This leads me to the following two questions:

  1. Why can I not write a $U(1)$ invariant mass term $\psi \psi^\dagger$ in Euclidean signature? This to me seems fine as it is just the contraction of a fundamental/anti-fundamental $SU(2)$ index?

  2. How does this conjugation difference between the two signatures get resolved under Wick rotation? Are there any issues with degrees of freedom on the two sides?

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