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I was considering the S-dual of the Type I' String theory (the solitonic Type I string theory).

That is the same as the S-dual of the T-Dual of Type I String theory. Then, that means both length scales and coupling constant are inverted. So, since inverting the length scale of the theory before inverting the coupling constant is the same as inverting the coupling constant before the length scale, I think the S-dual of the T-dual of the Type I String theory is the same as the T-dual of the S-dual of the Type I String theory. The S-dual of the Type I string theory is the Type HO String theory. The T-dual of the Type HO string theory is the Type HE String theory.

Therefore, the S-dual of the Type I' String theory is the Type HE String theory. But the Type HE String theory is S-dual to M-theory compactified on a line segment.

So does this mean that the Type I' String theory is M-theory compactified on a line segment?

Thanks!

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Type I' string theory is equivalent to M-theory compactified on a line segment times a circle, i.e. M-theory on a cylinder.

M-theory on a line segment only is the Hořava-Witten M-theory, a dual description of the $E_8\times E_8$ heterotic string, because every 9+1-dimensional boundary in M-theory has to carry the $E_8$ gauge supermultiplet. The extra compactified circle is needed to break the $E_8\times E_8$ gauge group to a smaller one; and to get the right number of large spacetime dimensions, among other things.

Type I' string theory has D8-branes that come from the end-of-the-world branes in M-theory on spaces with boundaries; it also possesses orientifold O8-planes. Interestingly enough, the relative position of O8-planes and D8-branes in type I' string theory may be adjusted. This freedom goes away in the M-theory limit; the D8-branes have to be stuck at the orientifold planes, those that become the end-of-the-world domain walls of M-theory, and this obligation is explained by the observation that an O8-plane with a wrong number of D8-branes on it is a source of the dilaton that runs. In the M-theory limit, the running of the dilaton becomes arbitrarily fast which sends the maximum tolerable distance between the O8-plane and D8-branes to zero.

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Thanks a lot for the answer but I have 1 more question: M-theory compactified on a cylinder definitely isn't equivalent to M-theory compactified on a line segment. So, there must be some fallacy with my reasoning. Do the T- and S- Dualities not commute? Or is the S-dual of Type HE String theory not M-theory compactified on a line segment (but instead a cylinder)? Thanks! –  Dimensio1n0 May 25 '13 at 16:25
    
There are various fallacies - you use the dualities in a bizarre way. The equivalence of the heterotic strings to M-theory with boundary isn't really a normal S-duality, it's a strong coupling limit and a general equivalence. Even more importantly, the mistake is in the first S-duality between type I' and HE. Type I' is a 9-dimensional theory (counting large dimensions only) so it can't be equivalent to a 10-dimensional one. It can't be hard to trace the number of large dimensions of spacetime and avoid simple mistakes of the sort, can it? –  Luboš Motl May 26 '13 at 4:18
    
Thanks. So does that mean that The Type I' String theory is only T-dual to Type I string theory compactified on a circle, rather than Type I string theory itself? If so what is the T-dual of the actual 10 dimensional string theory called? –  Dimensio1n0 May 26 '13 at 6:42
    
Dear @dimension10, T-duality always requires some dimensions to be compactified on a circle or for type I', on line segment. For 10D string theories, T-duality relates two theories with a circular dimension (of inverse radii) and 8+1 large dimensions. It is nonsense to ask what is the T-dual of a 10-dimensional vacuum. At most, you may understand it as the infinite $R$ limit of some vacua; the T-dual is formally a singular $R=0$ compactification. Let me also mention that the infinite $R$ limit of type I' = type IA looks like type IIA string theory everywhere away from the orientifold planes. –  Luboš Motl May 26 '13 at 7:43
    
I know that the 10D string theories have to be compactified in order for T-duality work. For example, Type IIA string theory compactified on a circle and Type IIB string theory compactified on a circle. What I am asking is, whether 2 dimensions of Type I string theory has to be compactified while only 1 dimension for the Type I' string theory? –  Dimensio1n0 May 26 '13 at 8:24
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