Is a Wick rotation a change of coordinates?

Question

My understanding is that a Wick rotation is a change of coordinates from $(t,x) \rightarrow (\tau , x)$ where $\tau = i t$. In the $(t,x)$ coordinate system, the Minkowski metric has components $ \eta_{\mu \nu} = \mathrm{diag}(1,-1,-1,-1)$. Using the formula for the transformation of the components under a coordinate change:

$$ \eta_{\alpha \beta} = \frac{\partial x^\mu}{\partial x'^\alpha}\frac{\partial x^\nu}{\partial x'^\beta}\eta_{\mu \nu} $$

we find in the $(\tau,x)$ coordinate system, the metric has components $\eta_{\alpha \beta} = \mathrm{diag}(-1,-1,-1,-1)$.

In QFT for the gifted amateur by Lancaster and Blundell equation 25.4, it is stated that under a Wick rotation, the magnitude of a vector is given by

$$ x^2 = - x_E^2 $$

where $x$ is the Minkowski vector and $x_E$ is the corresponding Euclidean vector. Now I am confused by this statement, because the objects $x$ and $x_E$ are coordinate representations of a vector, say $X$, which is a geometric object independent of the coordinate system we choose, so we should expect

$$ |X|^2 = \eta_{\mu \nu} x^\mu x^\nu = \eta_{\alpha \beta} x^\alpha_E x^\beta_E$$

in other words, the magnitude of the vector $X$ should not depend on which coordinate system we use. So under a simple Wick rotation, how could the magnitude of a vector change?

I was thinking, maybe a Wick rotation is an active rotation into the complex plane but the book states that the metric tranforms too so we can use the Euclidean metric. If we transform both the vector and the metric then that suggests a change of coordinates, but if only the vector changes then it suggests some sort of active transformation.

My Question

Is a Wick rotation simply a change of coordinates or is it an active rotation of the vector into the complex plane?

Answer 1

[The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.]

One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a particular metric) can be viewed as hypersurfaces lying in an underlying four-dimensional complex manifold. For example, in the complex manifold $\mathbb{C}^4$ with the obvious metric, you can find hypersurfaces with four (real) dimensions that are diffeomorphic to Euclidean four-space, and hypersurfaces with four (real) dimensions that are diffeomorphic to Minkowski space. The reason that Wick rotations are often successful in flat spacetime is because the functions we're looking at are generally holomorphic, and so they can be analytically continued from one "cross-section" to another.

In this picture, a vector lying in a Euclidean cross-section of $\mathbb{C}^4$ must be actively "rotated" into the Lorentzian cross-section. Simply changing the coordinates on your cross-section will not magically "pull in" a vector that doesn't already lie in that cross-section.

This picture, by the way, does not necessarily carry over to the analysis in curved spacetimes. We might think that if the Lorentzian metric is of the form $$ ds^2 = - f(x^i,t) dt^2 + g^{ij} dx_i dx_j $$ in some set of coordinates, then we could define a Euclidean analog $$ ds_E^2 = f(x^i,t) dt^2 + g^{ij} dx_i dx_j $$ and do the analysis there. However, there is no guarantee that there exists a complex manifold having these two cross-sections, and so we cannot rely on the Euclidean results to tell us anything about the Lorentzian physics.