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  Can a 4-dimensional equivalent of M-theory be obtained through T-duality?

+ 1 like - 0 dislike

I've learned that M-theory compactified on \(T^2\) is equivalent to Type IIB compactified on \(S^1\), because of T-duality, for example see the "PhysicsOverflow" post "A duality between M-theory and F-theory?". 

I wonder if this could be extended so that one has a 4-dimensional theory equivalent to M-theory? For instance, M-theory uncompactified (\(R\to\infty\)) is equivalent to a 10-dimensional theory, that is the toroidal compactification of the T-dual of M-theory at \(R\to0\) compactification radius, let's call this \(M'\)-theory. Then this \(M'\)-theory uncompactified (\(R'\to\infty \)) is equivalent to a 9-dimensional theory, the toroidal compactification of its T-dual at \(R'\to0\) compactification radius, let's call this the  \(M''\) theory, and so on.

Then wouldn't the 4-dimensional theory \(M'''''''\)-theory (7 "prime"s) be equivalent to the 11-dimensional M-theory?

So my question is, "Is it possible to obtain a 4-dimensional theory from M-theory, through T-duality?".

Thank you in advance.

asked Feb 5, 2015 in Theoretical Physics by (-1,-1,1,1,1,1,1,1,1,1,1,1,1,1) [ revision history ]
edited Feb 5, 2015 by dimension10

2 Answers

+ 4 like - 0 dislike
There is no theory T-dual of M-theory. M-theory is the only known consistent theory of gravity with 11 spacetime dimensions so if M-theory were T-dual of some theory, it should be of itself but M-theory uncompactified looks at low-energy as 11 dimensional supergravity whereas M-theory compactified on a small circle is type IIA superstring theory at weak coupling: these two theories at clearly different. The fact that M-theory does not exhibit T-duality is not astonishing because T-duality really uses strings wrapping the circle and there is no string in M-theory. There are branes but this leads to a different kind of relation: M-theory on a small circle is type IIA superstring (the M2 branes wrapping the circle give the type IIA string). In the derivation of the fact that M theory on a torus is equivalent to type IIB superstring on a circle, what is used is the T-duality between type IIA and type IIB and not a hypothetic T-duality of M-theory.
answered Feb 5, 2015 by 40227 (5,140 points) [ revision history ]
+ 0 like - 0 dislike

11-dimensional M-theory has \(\mathcal{N}=8\) supersymmetry, and \(\mathcal{N}=8\) supersymmetry is observable only at very high energies; Usually, \(\mathcal{N}=1 \) supersymmetry, or "minimal supersymmetry" is considered the most "natural" amount of supersymmetry in string theory.

The formulation you propose, would be fact exactly equivalent to M-theory, and would also have \(\mathcal{N}=8\) supersymmmetry. In general, compactifying on a class of manifolds known as \(G(2)\) manifolds (the 7-dimensional counterparts of Calabi-Yau manifolds) reduces the supersymmetry of M-theory to \(\mathcal{N}=1 \).

So, well, yes, you could derive a 4-dimensional theory related to M-theory by a series of T-duality relations (this is absolutely wrong, as 40227 points out above) but it's going to have \(\mathcal{N}=8\) supersymmetry, which is not exactly an ideal solution.

answered Feb 5, 2015 by dimension10 (1,985 points) [ revision history ]
edited Apr 11, 2015 by dimension10

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