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D-brane book-keeping and non-abelianity

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In Becker's book String Theory and M-Theory in the chapter about T-duality and D-brane (Chapter 6) the following comment is made

The Chan–Paton factors associate $N$ degrees of freedom with each of the end points of the string. For the case of oriented open strings, which is the case we have discussed so far, the two ends of the string are distinguished, and so it makes sense to associate the fundamental representation $N$ with the $\sigma = 0$ end and the antifundamental representation $N$ with the $\sigma =\pi$ end, as indicated in Fig. 6.3. In this way one describes the gauge group $U(N)$.

Chan-Paton

  • How do you know it is $U(N)$? Ok you have $N$ possibilities for the Chan-Paton of each end, but why not the fundamental of $O(N)$ for example that also acts on $N$-dim vectors?

  • I'm also confused about what the representation acts on: these are vectors with $N$ entries Do I have to picture an end as represented by a vector with a non-zero entry that 'labels' the D-Brane where it is connected? And that a $U(N)$ matrix gives the result of 'some interaction' where the end shifts to another D-Brane on the coincident stack.

  • How can you label consistently $N$ D-branes lying at the same location? Does this actually makes sense? I mean these D-branes fluctuate due to the massless scalar excitations. How can you disentangle them?


This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Anne O'Nyme

asked Oct 10, 2014 in Theoretical Physics by Anne O'Nyme (170 points) [ revision history ]
recategorized Oct 11, 2014 by dimension10

1 Answer

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  • Analyzing the spectrum of the strings, one finds that it contains $N^2$ massless vector states, which is precisely the number of gauge fields corresponding to a $U(N)$ group. Note that this is only true for massless oriented open strings; the unoriented case yields $SO(N)$ or $Sp(N)$.
  • As is described in the same chapter of the book, open string states can be described in terms of a basis $|\phi,k,ij\rangle$, where $i$ and $j$ are integers going from 1 to N (denoting the branes on which the string ends), $\phi$ represents the Fock space of the string (the information about which modes are excited) and $k$ is the momentum. A specific combination for $i$ or $j$ represents a state the corresponding endpoint can be in. In terms of this basis, any string state can be built as a linear combination given by $$|\phi,k,\lambda\rangle=|\phi,k,ij\rangle\lambda_{ij},$$ where the so-called Chan-Paton matrices $\lambda_{ij}$ form a representation of $U(N)$. This means that an actual state is a superposition of basis states corresponding to possible values for $i$ and $j$.
  • The fact that it seems hard to distinguish between the three branes is not so much of a coincidence: after all, a gauge symmetry is a redundancy in the description. Performing a symmetry transformation (for example a change of the Chan-Paton indices $i$ and $j$) does not change anything about the physical position of the string and its spectrum.
This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Frederic Brünner
answered Oct 10, 2014 by Frederic Brünner (1,060 points) [ no revision ]

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