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?")

  • 2
    $\begingroup$ 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? $\endgroup$ – Ben Crowell Sep 1 '17 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!

I hope this helped!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.