# Why isn't there an $i$ in the Riemannian path integral but there is one in a Pseudo-Riemannian path integral?

In David Skinner's lecture note on AQFT he gives the definition of the path integral as

$$\int_{\mathcal{C}} [\mathcal{D}\phi] \exp\left(-\frac{1}{\hbar}S[\phi]\right)$$

and later states that the when the manifold is Riemannian we'd use $$S[\phi]/\hbar$$ but if it was pseudo-Riemannian then we would use an additional factor of $$i$$. I've seen the Feynman path integral derivation and I don't see why for Riemannian manifolds you would remove the factor of $$i$$. Is there some reason we can remove the factor of $$i$$ while still being a path integral?

• you can treat it as the definition of the QFT on (pseudo)riemannian manifold. Usually though people will write something like $t \to it$ (wick rotation) Commented Sep 7, 2021 at 22:04

I do not know if there are any specific implications for algebraic quantum field theory. But the usual way of explaining that is by integrating over imaginary time $$dt=id\tau$$ which is called Wick rotation. This is an analytic continuation technique (see also the quote from Zee's book in the Wikipedia article), pretty similar to the analytic continuation you can use to compute certain Fourier integrals. Wick rotation is also mentioned in the footnotes on page 5 of Skinner's lecture notes.
By replacing $$dt\to id\tau$$ you turn the Minkowski metric of special relativity into a Euclidean metric. At the same occasion, you turn the path integral formulation of the time evolution of the system ($$exp(iS)$$) into a statistical formulation of the thermal equilibrium of the system ($$exp(-S)$$).