# Obtain Lorentzian manifolds from Riemannian ones by Wick rotation

In some cases, Wick rotation of a metric, formally consisting in substituting a coordinate with i times the coordinate itself, allows one to construct a Riemannian manifold starting from a Lorentzian one. The most known example, is Minkowski spacetime, which becomes Euclidean 4-space. Other example are Schwarzschild and Taub-NUT.

I would be interested in knowing which conditions are needed to ensure that by performing a Wick rotation on a Lorentzian manifold one obtains a Riemannian manifold and what problems can arise in the process - i.e. why one cannot always obtain a Riemannian manifold by Wick rotating a Lorentzian one. I am also interested in the reverse process, i.e. under which conditions one can obtain a Lorentzian manifold starting from a Riemannian one.

Are there any properties of the manifold which are preserved by the process?

EDIT: Thanks for the comments and answers so far. Indeed, I do not have a definition of what is meant by a Wick rotation for general manifolds - perhaps I should have asked how and if can such a notion be defined in a general context.

As far as I know twistors theory makes use of complexified manifolds, is anyone aware of whether such a theory has anything to say about the problem?

• Your definition of Wick rotation seems to depend on coordinates. I suppose that one definition of Wick rotation could be that we embed a Lorentzian manifold into a complex manifold with holomorphic metric extending the given metric, and then embed a Riemannian manifold into the same complex manifold, also extending the given metric. Clearly that won't usually be possible, because both metrics would have to be analytic in harmonic coordinates. Although it might have a chance for Einstein metrics. – Ben McKay May 6 '14 at 11:08
• I'm pretty sure any smooth Lorentzian manifold admits a nowhere vanishing tangent vector field, and that is not necessarily true for Riemannian manifolds. For your special cases, Wick rotation is natural because both manifolds in a correspondence can be seen to arise from a single global construction. – S. Carnahan May 6 '14 at 12:04
• What precisely do you mean by Wick rotation? Every smooth manifold admitting a non-vanishing one dimension tangent distribution admits a Lorentzian metric. The usual proof of this fact starts from the fact that any smooth manifold (implicitly paracompact) admit a Riemannian metric. The one dimensional distribution allows us to "rotate" one direction to make it time-like. This, however, does not seem to be quite what you want. Note that an essential feature used in the cases you cited is that the metric has no "shift" in some coordinate system. Those terms will generate terms with imaginary – Willie Wong May 6 '14 at 13:05
• ...coefficients otherwise after the Wick rotation. – Willie Wong May 6 '14 at 13:06
• @GFR: Carnahan is referring to a topological obstruction to the existence of a Lorentz metric. Every manifold has a Riemannian metric. If it also has a Lorentz metric, then we can take both metrics, and the symmetries of the null cone under the linear isometries of a tangent space preserving both metrics form SO(n-1), acting trivially in a time-like direction. Up to replacing the manifold by a 2-1 cover, there is a unique Riemannian-unit vector in the distinguished time-like direction. So it has a nowhere-zero vector field, so Euler characteristic 0. The converse works too. – Ben McKay May 6 '14 at 14:24

## 3 Answers

This is a somewhat different take on Igor's answer, and I offer it just in case you are interested.

First, one doesn't need to have any continuous symmetries in order to have this kind of 'Wick rotation' exist. For example, if $(M,g)$ is a real-analytic Riemannian manifold that admits a nontrivial isometric involution $\iota:M\to M$ that fixes a hypersurface $H\subset M$, then, near $H$, one can write the metric in the form $g = dt^2 + h(t^2)$ where $t$ is the distance from the hypersurface and $h(a)$ is the induced metric on the level sets $t^2 = a$. Then the Wick rotation is just $g' = -d\tau^2 + h(-\tau^2)$ in the sense that this is the Lorentian metric induced on the slice $t = i\tau$ in the complexification $(M^\mathbb{C},g^\mathbb{C})$.

Second, one could generalize things a bit and say that two real-analytic (pseudo-)Riemannian metrics are 'Wick-related' if they are '$\mathbb{R}$-slices' of a common connected holomorphic Riemannian complex $n$-manifold $(M^\mathbb{C},g^\mathbb{C})$. By an '$\mathbb{R}$-slice', I mean a real $n$-manifold $N\subset M^\mathbb{C}$ such that the pullback of $g^\mathbb{C}$ to $N$ is real-valued and nondegenerate. In this terminology, I think that two $\mathbb{R}$-slices $N_1,N_2\subset M^\mathbb{C}$ should be said to be related by a 'Wick rotation' if $N_1\cap N_2$ is a submanifold of dimension $n{-}1$. This is certainly what happens in the case above generated by an isometric involution fixing a hypersurface. (Added note: In fact, I don't know an example of a Wick-rotation in this sense that isn't generated by such an isometric involution. It might be interesting to try to prove that this does give them all or else find a counterexample. I note that, to second order, it is true: The hypersurface $N_1\cap N_2$ is always totally geodesic in each of $N_1$ and $N_2$ (with their induced metrics), so reflection in the hypersurface is an isometry at least up to second order.)

Then the problem becomes how to tell when a given connected holomorphic Riemannian complex $n$-manifold $(M^\mathbb{C},g^\mathbb{C})$ has an $\mathbb{R}$-slice (and, of course, to determine them when they exist). Obviously, the complexification of a real-analytic pseudo-Riemannian $n$-manifold has at least one $\mathbb{R}$-slice, but generically, when $n>1$, this is the only one, so 'most' real-analytic pseudo-Riemannian manifolds are not Wick-related to any other, let alone possess a Wick-rotation.

The reason is that $\mathbb{R}$-slices are the integral manifolds of a very restrictive system of PDE for real submanifolds of $(M^\mathbb{C},g^\mathbb{C})$: If one lets $R\subset \mathrm{Gr}^\mathbb{R}_n(TM^\mathbb{C})$ denote the set of real $n$-planes $E\subset T_pM^\mathbb{C}$ to which $g^\mathbb{C}$ restricts to be real-valued (and, of course, nondegenerate), then $R$ is a smooth manifold of dimension $2n+\tfrac12n(n{-}1)$ with $n{+}1$ components (one for each possible index of $g$ when restricted to the $n$-plane $E$), and the basepoint projection $\pi:R\to M^\mathbb{C}$ given by $\pi(E) = p$ is a smooth submersion. It is easy to show that, for any given $E\in R$, there is at most one $\mathbb{R}$-slice that has $E$ as a tangent space. This is because there is a canonical $n$-plane field $H$ on $R$ with the property that the set of tangent spaces of any $\mathbb{R}$-slice is an $n$-manifold in $R$ that is tangent to $H$ everywhere.

The only time $H$ is Frobenius is when $(M^\mathbb{C},g^\mathbb{C})$ is the complexification of a (real) space form, i.e., a Riemannian manifold of constant sectional curvature. Generically, $H$ has no integral manifolds at all, and, generically, when it does have one, it has only one connected component.

For example, in the case $n=2$, if the Gauss curvature $K$ of $(M^\mathbb{C},g^\mathbb{C})$ is not constant, then it has at most a $1$-parameter family of positive definite $\mathbb{R}$-slices, and this happens only when the $\mathbb{R}$-slices are all isometric and have a symmetry vector field (so they are locally surfaces of revolution).

Now, in fact, each $\mathbb{R}$-slice (of whatever index) lies in the (real) hypersurface in $M^\mathbb{C}$ on which $K$ takes values in $\mathbb{R}$. Set $E = g^\mathbb{C}(\nabla K,\nabla K)$ (everything computed in the holomorphic category). For most metrics $g^\mathbb{C}$, the holomorphic functions $K$ and $E$ will not be functionally dependent, i.e., $\mathrm{d}K\wedge\mathrm{d}E$ will vanish only on a (possibly empty) complex-analytic subvariety $C\subset M^\mathbb{C}$ of complex dimension $1$. Go ahead and define $F = g^\mathbb{C}(\nabla K,\nabla E)$ and $G = g^\mathbb{C}(\nabla E,\nabla E)$.

Obviously, any $\mathbb{R}$-slice $N\subset M^\mathbb{C}$ must lie inside the locus on which the holomorphic functions $K$, $E$, $F$, and $G$ assume real values. Outside of $C$, the set $L\subset M^\mathbb{C}\setminus C$ on which $K$ and $E$ take real values is a (possibly empty) real submanifold of dimension $2$ and its components are the only possible $\mathbb{R}$-slices that lie outside of $C$ (of course, any $\mathbb{R}$-slice can only intersect $C$ in a (real) $1$-dimensional curve at most). A component of $L$ actually is an $\mathbb{R}$-slice if and only if $F$ and $G$ are real-valued on it. Two components are related by a Wick-rotation in the above sense if they intersect (in a real $1$-dimensional curve that necessarily lies inside $C$).

A similar, but slightly more involved analysis can be done for the case in which $K$ and $E$ are dependent everywhere on $M^\mathbb{C}$. Moreover, a similar, but more complicated analysis can be used in higher dimensions, with, say, the symmetric functions of the eigenvalues of the holomorphic Ricci tensor used in the place of $K$.

I'm going to guess that your interest in Wick rotation comes from the role it plays in the formulation of quantum field theories (QFTs) on Miknowski spacetime and some other curved spacetimes, like the examples you mentioned. Of course, it is natural to ask, like you probably have, whether Wick rotation plays a similarly important role in the formulation of QFTs on more general Lorentzian manifolds. And the answer to that question seems to be No. At least, the current state of the art methods in dealing with QFTs in curved spacetimes do not make use of Wick rotation.

Now, on to the question you actually asked. As far as I know, there is no general theory about the conditions under which a Wick rotation can be performed, except the property of being static, where the formal substitution trick $t\mapsto i\tau$ is already sufficient. However, the only way to even attempt at trying to find such general conditions, whatever they are is to make the notion of Wick rotation mathematically precise and coordinate independent. So I'll finish by doing that below.

You need to start with an analytic manifold $M$ with an analytic Lorentzian metric $g$. The manifold $M$ will have a complexification $M_\mathbb{C}$, where it embeds as a real submanifold $M \subset M_\mathbb{C}$. The same can be said about the cotangent bundle $T^*M \to T^*M_\mathbb{C}$ and any tensor powers of it. So, the metric $g$ naturally defines an analytic section of $T^*M_\mathbb{C}$ over the real submanifold $M$, and thus extends uniquely to a section $g_\mathbb{C}$ on all of $M_\mathbb{C}$ by analytic continuation. Of course, there is no reason for $g_\mathbb{C}$ to remain of definite signature or even real off $M$. But let us say that the Lorentzian manifold $(M,g)$ has a Wick rotation if the embedding $M\subset M_\mathbb{C}$ can be smoothly deformed into another embedding $M'\subset M_\mathbb{C}$ such that $g_\mathbb{C}$ restricted to $M'$ is equal to a real Riemannian metric $g'$.

In the case of a static spacetime, the metric can be written as $g = -a\, dt^2 + h$, where $t$ is a special time coordinate so that $a$ and $h$ do not depend on $t$ or $dt$. The deformation $(t,x) \mapsto (e^{i\theta} t,x)$ is a deformation of $M\subset M_\mathbb{C}$ that terminates at a Wick rotation for $\theta=\pi/2$.

• Thanks Igor. So if I understand what you are saying, for a Lorentzian manifold having a timelike, hypersurface orthogonal Killing vector field is a suffcient condition to admit a Wick rotation in the sense you have defined. I believe then the construction an be reversed: a Riemannian manifold having a hypersurface orthogonal vector field can be "Wick rotated" to a Lorentzian manifold, is that correct? – GFR May 6 '14 at 14:22
• @GFR: you forgot Killing in your final sentence. And you probably also want nowhere vanishing. – Willie Wong May 6 '14 at 14:37
• Willie Wong You are right, thanks! For some reasons I cannot edit my comment... – GFR May 6 '14 at 14:38
• +1 for the first paragraph. There are standard ways to make a Lorentzian manifold out of a Riemannian one and vice versa, subject to some small topological obstruction. The emphasis should not be on "whether one can get a Riemannian manifold from a Lorentzian one" but on "whether there exists a process that makes this conversion and, at the same time, preserve such and such property" with the properties being those that you want to be true for your application. – Willie Wong May 6 '14 at 14:40
• Note, though, that, in the case of a static spacetime, as you have defined it, you don't need to assume analyticity at all in order to define a 'Wick rotation' in your sense. Your 'Wick rotation' of $g = -a\ \mathrm{d}t^2 + h$ is just $g_+ = a\ \mathrm{d}t^2 + h$, whether $a$ and $h$ are real-analytic in any coordinates or not. – Robert Bryant May 6 '14 at 14:45

See also the recent paper

A Wick-rotatable metric is purely electric by C. Helleland and S. Hervik, arXiv:1504.01244