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I have been told that the world-sheet described by a closed string is a world-sheet without boundaries. On the contrary, the world-sheet described by an open string has boundaries. I do see why the open case has boundaries but I don't understand the closed case.

I get that the endpoints of the string now are closed so these don't give boundaries, but yet, the surface described by a closed string moving from a given time from another time has boundaries right? Imagine for example a string who happens to be a circle at one time and propagates defining a cylinder. This cylinder does have boundaries. So, what am I not seeing?

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2 Answers 2

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Since the worldsheet theory is conformal, you are allowed to "shrink the boundaries to a point". So the usual viewpoint is that the worldsheet are boundary-less with certain points on them corresponding to the former boundaries.

The cylinder, for instance, becomes a twice-punctured sphere - the punctures are the places where one inserts the vertex operators which correspond to certain string states "on the boundary" by the CFT state-operator correspondence.

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  • $\begingroup$ so, for example, if I have an integral on the worldsheet, and using Stokes' theorem I write it as an integral over the boundary, is it zero? if so, do you know a rigorous proof? $\endgroup$
    – Yossarian
    Commented Sep 25, 2015 at 10:35
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    $\begingroup$ @Scardenalli it is not zero: boundaries have operator insertions which are analogical to poles from the complex variable function theory. $\endgroup$ Commented Nov 24, 2015 at 1:51
  • $\begingroup$ @ACuriousMind It seems that you are overlooking a very important detail here. The conformal map that ''explode'' the punctures to ''boundaries'', also push these ''boundaries'' to infinity, so they are not quite boundaries of the worldsheet (the worldsheet becomes not compact). Conversely, a boundary that is located at finite distance will be always a D-brane or an end point of a open string. $\endgroup$
    – Nogueira
    Commented Aug 21, 2018 at 3:29
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In perturbative closed string theory, the amplitudes are computed by summing over closed Riemann surfaces without boundaries. This is why we don't compute off-shell correlation functions like we do in QFT.

If you write down a world-sheet with a closed string in a generic state that start from some location at the worldsheet and not at the infinity, the Diff plus Weyl gauge symmetries of the Polyakov action will be violated getting something that is gauge dependent. It means that this must be non physical since gauge symmetries are required to act trivially in physical states.

The Diff plus Weyl gauge symmetries can be traced back to the fact that a relativistic string is a very constrained object. If you gauge fix completely all the gauge redundancies, say by light-cone gauge, we end up with a constrained object.

In order to compute something gauge invariant (i.e. compatible with the constraints of a relativistic string) we can do two things:

  1. The first one is to push a boundary to infinity, where the Diff plus Weyl gauge invariance is restored. There will be a conformal transformation that maps this "boundaries at infinity" (locations where the world-sheet becomes noncompact) to punctures at the world-sheet. This will define an asymptotic state.

The state-operator correspondence gives a relation between these asymptotic states and vertex operators. The idea is to "fill" these punctures by local operator insertions, obtaining a compact world-sheet with local insertions instead. This objects are the ingredients to compute the S-matrix for closed strings in flat background for example. You can learn more about that in the chapter 3, section 3.5 of Polchinki.

  1. The second one is to maintain the boundary and try to tune the boundary conditions (i.e. boundary state) such that the Diff plus Weyl gauge symmetry is restored. For the $X^{\mu}$ CFT you can consider boundary conditions such as Dirichlet boundary condition and the Neumann boundary condition:

$$ T(z)=\tilde{T}(\bar{z})\implies (\partial X^{\mu}-\bar\partial X^{\mu})(\partial X^{\mu}+\bar\partial X^{\mu})=0 $$

These boundary conditions define a state of closed strings $|B\rangle$ and you can look at this paper or this one to learn more about this.

The interpretation of the state $|B\rangle$ is a closed strings coming from a object on the target space called D-brane. This is so because each Dirichlet boundary condition fix a value for one component of $X^{\mu}(\sigma)$ along the boundary, i.e.

$$ X^{\mu}(\sigma)= f(\sigma)|\,\sigma \in B $$

where $B$ is the boundary.

This defines a hypersurface on the target space with same dimension as the number of Neumann boundary conditions. Note that if all the boundaries have Neumann conditions, this implies that this hypersurface, the D-brane, is filling all the target space.

An interesting computation in supertrings is to check if the amplitude related to the disk with a puncture in the middle and Neumann condition to the boundary vanishes. This amplitude is related to the tadpole diagrams of QFT for the closed string state described by the puncture.

  1. You can also combine both things. Allow boundaries with particular boundary conditions compatible with gauge symmetries and uncompact regions of worldsheet (places of the wordlsheet where it extends to the infinity).

One trivial example of a combination of these two is an infinity strip attached to the world-sheet. Again, there will be a conformal transformation that map the piece at infinity to a puncture, but now the puncture will be located at the boundary and not at the bulk.

This puncture will describe a open string coming from infinity. The state-operator correspondence will give a vertex operator for this open string, and it will provide the ingredient for computing the amplitudes for closed and open strings.

I would like to finish by saying that as we push the worldsheet to the infinity to become uncompact, in order to the map worldsheet$\rightarrow$target space agree, there must be an uncompact region of the target space associated to it. In other words, the worldsheet$\rightarrow$target space must map uncompact regions of the worldsheet to uncompact regions of the target space.

For euclidean $d$-dimensional flat background we can think about the target space as an $S^{d}$ with the north pole removed, the north pole being the infinity of $\mathbb{R}^{d}$. This means that all the strings associated to vertex operators insertions are coming from the infinity of $\mathbb{R}^{d}$ and the quantum numbers they carry must specify the momentum transverse to the point at infinity. So the states must carry quantum numbers $k^{i}$, with $i=1,\dots, d$.

If the space is a product of $\mathbb{R}^{d}\times \mathcal{M}$, where $\mathcal{M}$ is a compact manifold, then we must have quantum numbers of the form $k^{i}$, $i=1,\dots,d$ and also discrete numbers $n$ describing in which stationary state the $\mathcal{M}$-CFT is at the worldsheet infinity. Note that must be a stationary state!

Finally we can look at the famous background $AdS_5\times S^{5}$. The non-compact regions of this space is at the boundary of $AdS_5$ and have the topology of $S^{4}$ (I am talking about euclidean AdS). This means that we must specify from what point of $S^{4}$ the strings are coming from. This is different from the flat background where there was a single point at the target space infinity.

There are also a single transverse direction that points towards the AdS-bulk. This means that there should be a momentum associated to it. In the AdS/CFT duality this momentum is dual to the dimension of CFT operators. The $S^{5}$ will give the discrete quantum numbers associated to the stationary states of the $S^{5}$ worldsheet CFT.

The quantum numbers of a vertex operator in $AdS_5\times S^{5}$ will be: momentum transverse to the AdS-boundary, points at the boundary, angular momentum along the $S^{5}$, which is the same data of a local operator in $\mathcal{N}=4$ $d=4$ Super-Yang-Mills, with the dictionary:

  • Transverse momentum $\leftrightarrow$ conformal dimension.
  • Angular momentum $\leftrightarrow$ R-charge
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