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Pertaining to the answer within link.

Why is it the case, that for Lorentz invariant Lagrangian $\mathcal{L}$, after Wick rotation, the $O(4)$ invariance is established, thus manifesting itself as having Euclidean metric? Is that a consequence of requiring the four vector fields to transform as $A_0^E = iA_0$ and $A_j^E=E_j$ or a result of it? So which premise comes first?

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As long as the Minkowski action is constructed from Lorentz-covariant tensors, then under Wick rotation [where the contravariant and covariant $0$-components of the tensors are Wick-rotated in opposite ways], the corresponding Euclidean action becomes constructed from the corresponding $O(4)$-covariant tensors, cf. e.g. this Phys.SE post.

Note in particular that the Minkowski metric tensor [with the signature convention $(-,+,+,+)$] is Wick rotated to the Euclidean metric tensor.

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  • $\begingroup$ How does one postulate that? I get that after the wick rotation one gets $dx_1^2+ dx_2^2 _\dots$ so it certainly looks like Euclidean metric. But how can one infer from this information alone that the Lagrangian itself must be also one that adapts euclidean symmetry ? The fields themselves are not uniquely defined, but is the Lagrangian by analytic continuation uniquely determined, so as to leave $O(4)$ invariance as the only possible outcome ? $\endgroup$
    – 2000mg Haigo
    Commented Jul 26, 2022 at 19:04

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