# Wick rotation and spinors

I am quite familiar with use of Wick rotations in QFT, but one thing annoys me: let's say we perform it for treating more conveniently (ie. making converge) a functional integral containing spinors; when we perform this Wick rotation, in a way we change the metric to $(-,+,+,+)$ to $(+,+,+,+)$, so the invariant group is no more $SO(3,1)$ but $SO(4)$ and ($SO(4)$ being compact and the spinor representation non unitary) spinors don't carry finite dimensional representation of this group. So I feel like we shouldn't be talking anymore about this objects, but only about vectors of $SO(4)$.

Is my fear justified? or where am I wrong in my reasoning?

• You may find these papers interesting: arxiv.org/abs/hep-th/9608174 , arxiv.org/abs/hep-th/9611043 – Qmechanic Feb 21 '12 at 16:48
• Could you specify more precisely where is the problem? Probably, illustrate it with some functional integral. – Misha Feb 21 '12 at 17:06
• Why do you say that there is no finite dimensional spinor representation of SO(4)? What about, for example this discussion? – Henry Deith Aug 22 '17 at 15:30

But perhaps a more practical answer to your concern is that usually when you are doing a loop integral in quantum field theory, the object that you are integrating is a scalar quantity---it's the square of a matrix element. So any spinors inside the expression have contracted with other spinors (with some objects like momenta dotted into $\gamma$/Pauli matrices sandwiched inside).
When I studied at first course and investigated the special theory of relativity the lecturer said about old interpretation of relativity. In this approach instead pseudo-euclidean metric and four-vectors $(t,\bf x)$ people use euclidean metric and four-vectors $(it,\bf{x})$. But it does not mean that we use SO(4) group! We use also SO(3,1) group but we do some change of variables.
• This “change of variables” is imaginary, and this means we can no longer be using $\mathrm{SO}(3,1)$, which is a real group. We can either use the complex version, the rather unfamilliar ${}^{\mathbf C}\mathrm{SO}(3,1) = \mathrm{SO}(4,\mathbf C)$, or choose an appropriate real version of it, which is exactly $\mathrm{SO}(4)$. (You can’t always get away with being cavalier about complexifications.) – Alex Shpilkin Apr 17 '16 at 13:21