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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?

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    $\begingroup$ You may find these papers interesting: arxiv.org/abs/hep-th/9608174 , arxiv.org/abs/hep-th/9611043 $\endgroup$
    – Qmechanic ♦
    Commented Feb 21, 2012 at 16:48
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    $\begingroup$ Could you specify more precisely where is the problem? Probably, illustrate it with some functional integral. $\endgroup$
    – Misha
    Commented Feb 21, 2012 at 17:06
  • $\begingroup$ Why do you say that there is no finite dimensional spinor representation of SO(4)? What about, for example this discussion? $\endgroup$
    – Henry Deith
    Commented Aug 22, 2017 at 15:30

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I don't think I follow your statement:

spinors don't carry finite dimensional representation of this group.

I follow up in this comment to the original question.

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).

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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.

The Wick rotations is the same thing, it is only change of variables no more.

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    $\begingroup$ 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.) $\endgroup$
    – Alex Shpilkin
    Commented Apr 17, 2016 at 13:21

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