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$$F = M \Big|_{A(T^2) \to 0}$$

The above equation is the duality equation between F-theory and M-Theory on a vanishing 2-torus. What's the explanation for this equation?

Is there anything similar to this equation with M-theory and type IIB theory, and how can ones explain it?

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M-theory compactified on a 2-torus is the same as M-theory compactified on a circle and then compactified on another circle because $T^2=S^1\times S^1$.

M-theory compactified on a circle is type IIA string theory with $g_s$ being an increasing power of the radius of the compactified dimension. And if type IIA is compactified on a circle of a small radius, we get type IIB string theory via T-duality. When we connect the M-theory/IIA duality and the IIA/IIB T-duality, we get the $F=MA$ relationship between M-theory and type IIB you mentioned.

One may avoid the type IIA intermediate step, too. M-theory on a two-torus is naively a 9-dimensional theory (the supergravity approximation would lead us to this belief). However, M-theory contains M2-branes, two-dimensional objects, and both of their spatial dimensions may be wrapped on the 2-torus. This produces point-like objects in the remaining 9 large dimensions. These objects are light when $A\to 0$ and they also have bound states of $N$ objects. So one obtains a continuum of new states in the $A\to 0$ limit and they may be reinterpreted as the momentum modes with respect to a new, "emergent", 10th spacetime dimension of the resulting type IIB string theory.

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    $\begingroup$ So, F-Theory compactifyed on a vanishing (axion-dilaton ?) torus gives IIB theory , while M-Theory compactifyed on a vanishing torus gives IIA theory compactifyed on a small radius, that is IIB theory compactifyed on a large radius (by T-Duality). Is it correct ? $\endgroup$ – Trimok Jul 26 '13 at 12:52
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    $\begingroup$ I think so. The process can be described as follow: $M \Big|_{A(S^1) \rightarrow 0} = IIA \Big|_{g_s \rightarrow 0}$ $\Rightarrow M \Big|_{A(T^2) \rightarrow 0} = IIA \Big|_{A(S^1) \rightarrow 0} \rightarrow {T-dualize} \rightarrow IIB \Big|_{A(S^1) \rightarrow \infty} = IIB$ And since $IIB = F$ via elliptical fibration (yes, the torus is described by the axidilaton), therefore $F=M\Big|_{A(T^2) \rightarrow 0}$ $\endgroup$ – Trung Phan Jul 26 '13 at 16:40
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    $\begingroup$ Trimok: heuristically yes. However, one must be careful about the type of compactification you need for F-theory - the signature of the 2-torus isn't really well-defined and there isn't any decompactified 12-dimensional F-theory to start with. $\endgroup$ – Luboš Motl Jul 27 '13 at 14:59

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