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  Why are there no branes in heterotic string theory?

+ 4 like - 0 dislike

Why does the heterotic string (or heterotic supergravity) have no brane solutions?

According to David Tong's notes:

the heterotic string doesn’t have (finite energy) D-branes. This is due to an inconsistency in any attempt to reflect left-moving modes into right- moving modes.

I know that the heterotic string does not have the same structure in the left- moving and right- moving sectors (whether one uses the Green Schwarz or the RNS formulation). But why does that imply no branes?

BPS-brane solutions are obtained by starting from the supersymmetry transformations of the fields and solving them in a zero-fermion background. This leads to stable solutions if they exist. Why would such an approach not work with the heterotic supergravity action?

EDIT: It seems that there is a simple qualitative explanation (source). Specifically, a heterotic string must necessarily be closed, and unlike other string theories which have closed strings and open strings (which end on D-branes) there are no heterotic open strings of type $E_8 \times E_8$ (cf. this paper, which I haven't yet read) and so no D-branes.

This post imported from StackExchange Physics at 2015-06-27 21:56 (UTC), posted by SE-user leastaction

asked Jun 22, 2015 in Theoretical Physics by leastaction (425 points) [ revision history ]
edited Jun 27, 2015 by Dilaton

Are you saying the worldsheet theory for E strings has no boundary conditions at all?

1 Answer

+ 4 like - 0 dislike

The problem comes from the boundary conditions to impose on the fermions on the heterotic string worldsheet. At a Dirichlet boundary condition in a two dimensional theory, the scalars have a given imposed value but one also needs to say something about the fermions. The reflection along the boundary reversing the orientation, the only meaningful thing one can do is to relate left and right fermions. It is what happens in ordinary D-brane boundary conditions in type II string. But on the heterotic worldsheet there are only fermions of one chirality so there is nothing to relate them to.

The preceding argument shows that it is likely there is no D-brane in heterotic string theory, i.e. extended objects which are natural boundary conditions for open strings. Nevertheless, as the title of the question contains the word "brane" without other precision, I should mention that there exists some branes, in the general sense of extended objects different from the fundamental string, in heterotic string theory: the NS5-brane, magnetically charged under the B-field, which is half-BPS and so stable and a 6-brane, which is magnetically charged under the gauge field, which is not preserved by supersymmetry and so is presumably unstable. 

answered Jun 28, 2015 by 40227 (5,140 points) [ revision history ]
edited Jun 28, 2015 by 40227

Thanks for the answer. Do you have a reference for the "6-brane"? (I thought it was a fivebrane.)

Is there something like the heterotic string is the boundary of a "Dirac tube" that allows it to be chiral?

@leastaction First it is true that the most well-known heterotic brane is a (supersymmetric, BPS, stable) fivebrane that I called NS5 (probably incorrectly because there is no NS sector in the heterotic string but anyway I mean the thing magnetically charged under the 2-form gauge field and so which looks like the NS5 brane of type II). For the 6-brane the only reference I know is the original paper of Horowitz and Strominger describing black branes from the supergravity point of view: http://ccdb5fs.kek.jp/cgi-bin/img/allpdf?200034882 I guess it has not been studied seriously because it is a nonsupersymmetric, so likely unstable, object and so very difficult to understand with our current tools.

Ryan Thorngren I think I know what is a Dirac tube but I don't see the relation with chirality. Could you explain why you expect such relation?

Then the heterotic string's worldsheet CFT would be like the boundary of a Chern-Simons TQFT with chiral central charge. That it must be realized on the boundary of the disc would explain why it has no boundary conditions of its own.

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