# Topological problems with Lorentzian metric on worldsheet

In string theory we study maps $X: \Sigma \to M$, where $\Sigma$ is the two dimensional worldsheet of the string and $M$ is the target manifold. When studying non-linear sigma models, for instance when we are looking at the Polyakov action for string theory, $\Sigma$ is often endowed with the two dimensional Minkowski metric.

From a topological perspective, however, it is well-known that a (compact) manifold admits a Lorentzian metric iff the Euler characteristic vanishes. This should mean that we can only take the metric on $\Sigma$ to be the Minkowski metric for genus 1 worldsheets. What do we do for other genus worldsheets?

(If the answer is something along the lines of "Wick rotate so you have a Riemannian signature", then my question is "Why is this a sensible thing to do?")

• a (compact) manifold admits a Lorentzian metric iff the Euler characteristic vanishes Why would $\Sigma$ be compact? I'm ignorant of string theory, but wouldn't a typical topology be $S^1\times\mathbb{R}$, which is noncompact?
– user4552
Sep 1, 2017 at 3:35

In string theory, the world-sheet is typically taken to be that swept out by a set of $m$ incoming strings that have propogated "from infinity" and $n$ outgoing strings that will propogate "to infinity." The resultant world sheet $\Sigma$, at each level in perturbation theory, will be a homeomorphic to a genus $g$ surface with $m+n$ punctures, which is non-compact. The simplest example is a free string world sheet, which is homeomorphic to a 2-sphere with two punctures (i.e. a cylinder). Since this surface is noncompact, your theorem doesn't apply and you're safe!