Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 1 like - 0 dislike
893 views

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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...