1
$\begingroup$

I've spent some time studying some definition in smooth manifolds theory in order to give a proper definition of a worldsheet in classical string theory at least. My attempt is the following:

Definition. Let $\Sigma$ be a $2$-dimensional smooth manifold with or without boundary and $X: \Sigma \to \mathbb{R}^D$ an embedding. Then if $\Sigma$ is diffeomorphic to $\mathbb{R} \times [0,1]$, $\Sigma$ is called an open string worldsheet. If $\Sigma$ is diffeomorphic to $\mathbb{R} \times S^1$, then $\Sigma$ is called a closed string worldsheet. In both cases $X$ is called a string embedding.

My definition seems to be fine because it matches the elements used in Polchinski's book, because the usual coordinate domain he uses are $-\infty<\tau<\infty$ and $0\le \sigma <\ell$. However there exists this article on nLab about Polyakov action he defines $\Sigma$ to be a compact smooth manifold. Well, the manifolds I have defined above are not compact. Is anything possibly wrong with my definition, or the conflict between mine and nLab's definition are just a matter of taste?

EDIT: So far I came up to the conclusion that my definitions are actually acceptable, although I don't know why nLab's definition require it to be compact. However Now I want to know two things:

  1. Can be the worldsheet definition be extended to the definition of a Riemann Surface? Because I've read, don't remember when, that the terminology "Riemann Surfaces" in String Theory is somewhat used impricesely, in the sense that we just impose the natural complex structure on the manifold.

  2. If is possible to extended the definition to a Riemann Surface, what ingredients should I add to mine?

$\endgroup$
3
  • $\begingroup$ Technically I believe one wants an immersion rather than embedding, but this is a little pedantic. $\endgroup$ May 19, 2023 at 10:02
  • $\begingroup$ @SvenForkbeard Why do you think that? $\endgroup$ May 20, 2023 at 11:29
  • $\begingroup$ Basically because you are allowed intersections. For example, for open strings a Klein bottle is needed (even in D = 3+1). Similarly the path integral over a particle's worldline includes a figure of 8. These situations are atypical in the path integral however. $\endgroup$ May 22, 2023 at 16:24

1 Answer 1

3
$\begingroup$

That's not how anyone uses the term "worldsheet", even though you are correct that $\mathbb{R}\times I$ and $\mathbb{R}\times S^1$ are the "worldsheets" one would assign to a classical string moving from the infinite past to the infinite future.

The nLab article where a worldsheet is just a compact manifold is closer to how people use the term in practice. Ultimately, no one cares about the theory of classical strings, what string theorists actually care about is the "sum over worldsheets" in the definition of the string scattering amplitude. That sum is over all compact manifolds, and to understand why people call that a "sum over worldsheets" we need to look back at QFT:

In QFT, the perturbative scattering amplitude can be written as a sum over Feynman diagrams, and a graph - like a Feynman diagram is - can be interpreted to be a bunch of 1d manifolds (lines) stitched together at the vertices of the graph. We then notice that ordinary QFT has particles as its fundamental objects, and so these lines might be viewed as the "worldlines" of particles. So the sum over Feynman diagrams becomes a "sum over worldlines", even though strictly speaking a Feynman diagram with a vertex isn't really the worldline of any particle (but we might imagine this as some reaction happening at the vertex and then following the worldlines of the products of that reaction, please don't take this literally, it's just a picture).

So, in analogy, we define the "sum over worldsheets" e.g. in closed string theory to be a sum over cylinders (the worldsheets of closed strings) stitched together. In 2 dimensions, stitching stuff together is less singular than in one dimensions, and you can get "worldsheets" like the pair of pants which you might imagine as representing a closed string splitting into two strings. Now, the "open ends" of a sheet like the pairs of pants are associated with incoming and outgoing scattering states, like the external legs of a Feynman diagram, and these states are created in the CFT that lives on the world sheet, and the conformal state-operator correspondence means that these states hence correspond to operators "inserted" at these ends.

Finally, there are conformal maps that map these "$n$-holed worldsheets" to Riemann surfaces with $n$ marked points: The prototypical example is mapping the cylinder to the sphere: $$\mathbb{R}\times S^1 \to S^2 \cong \mathbb{C}\cup \{\infty\}, (\sigma,\tau)\mapsto \mathrm{e}^{\tau - \mathrm{i}\sigma}, $$ which sends the infinite past and future on the actual worldsheet to the two poles of the sphere at $z=0$ and $z=\infty$. You can show that similar maps exist for all the worldsheet topologies gained from stitching cylinders together - if the worldsheet has a hole (i.e. a string splits in two and then later merges again along it), then the target of the map is not the sphere but the torus, and generally the Riemann surface of genus $g$ for a sheet with $g$ holes.

So, ultimately, what we sum over in the scattering amplitude are the compact Riemann surfaces with marked points/vertex operators inserted, and hence it is often these Riemann surfaces that get called the worldsheets.

$\endgroup$
3
  • $\begingroup$ There is one more thing that maybe a source of confusion that I forgot to write in the question (or maybe write another one). If I've defined the classical world sheets to be compact, then they would admit a lorentzian metric iff their genus was 1 (Euler characteristic =0). Is this important at the end of the day since this hole process of mapping the manifolds to Riemann surfaces occurs under the wick rotation (or complexification in general)? $\endgroup$ May 18, 2023 at 16:40
  • 1
    $\begingroup$ @Geni Note that the image of the map is a punctured Riemann surface, not the full Riemann surface (the $z=\infty$ and $z =0$ are not in the image of the map to the sphere I wrote down), so the Lorentzian metric from the actual worldsheet gets mapped to a Lorentzian metric on the punctured surface (which is non-compact and so the theorem about Lorentzian metrics on Riemann surfaces is not relevant). $\endgroup$
    – ACuriousMind
    May 18, 2023 at 16:56
  • $\begingroup$ Thanks. It's pretty funny that I learned with you more in 30 minutes than in a week trying to answer this in other places $\endgroup$ May 18, 2023 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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