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  Why does closed string theory have only one dilaton field instead of $22$?

+ 6 like - 0 dislike

Looking at $5D$ Kaluza-Klein theory, the Kaluza-Klein metric is given by

$$ g_{mn} = \left( \begin{array}{cc} g_{\mu\nu} & g_{\mu 5} \\ g_{5\nu} & g_{55} \\ \end{array} \right) $$

where $g_{\mu\nu}$ corresponds to the ordinary four dimensional metric and $g_{\mu 5}$ is the ordinary four dimensional Maxwell gauge field, $g_{55}$ is the dilaton field.

As there is one dilaton for one extra dimension, I naively would expect that the zero mass states of closed string theory, which can be written as

$$ \sum\limits_{I.J} R_{I.J} a_1^{I\dagger} \bar{a}_1^{I\dagger} ¦p^{+},\vec{p}_T \rangle $$

and the square matrix $R_{I.J}$ can be separated into a symmetric traceless part corresponding to the graviton field, an antisymmetric part corresponding to a generalized Maxwell gauge field, and the trace which corresponds to the dilaton field.

Why is there only one dilaton field given by the trace of $R_{I.J}$, instead of $22$ dilaton fields corresponding to $22$ extra dimensions of closed string theory which has critical dimension $D = 26$? For example, why are there not $22$ dilaton fields needed to parameterize the shape of the 22 extra dimensions if they are compactified?

This post imported from StackExchange MathOverflow at 2014-06-17 09:17 (UCT), posted by SE-user Dilaton
asked Oct 23, 2013 in Theoretical Physics by Dilaton (6,240 points) [ no revision ]
reshown Jun 17, 2014 by Dilaton

Ok, as I am getting unexplained downvotes here too, it is probably appropriate to hide this ...

But the answer is nice, so I am not sure ... ?

@Dilaton You got exactly 1 silent downvote and 6 upvotes. Why would you hide it?

Ok, that was me, sorry, removed it. I was annoyed, because the question was asking about compactification scalars, not string dilatons, and the answer seems clear--- there's compactification scalars and there's a dilaton! They're two different things. But the IIA superstring dilaton is a compactification scalar in disguise, so there can be legitimate confusion. Sorry again.

2 Answers

+ 4 like - 0 dislike

You would actually not expect 22 dilatons. Let me try to explain.

As you have pointed out, a putative field theory limit of the closed bosonic string would consist of a metric, a 2-form (which is the potential for a 3-form) and a dilaton.

Let us assume that such a theory exists and let us dimensionally reduce to four dimensions à la Kaluza-Klein.

The 26-dimensional dilaton $\Phi$ gives a scalar field in four dimensions. The 26-dimensional metric $g_{MN}$ gives the following four-dimesional fields: a metric $g_{\mu\nu}$, 22 gauge fields $g_{\mu n}$ and $\binom{23}2 = 253$ scalars $g_{mn}$. Finally, the 26-dimensional 2-form $B_{MN}$ gives the following: $\binom{22}{2} = 231$ scalars $B_{mn}$, $22$ gauge fields $B_{\mu n}$ and $\binom{4}{2} = 6$ scalars dual to the 2-forms $B_{\mu\nu}$. (In four dimensions, 2-forms can be dualised to scalars: via $\star dB = d\varphi$.)

So from a 4-dimensional perspective the role of the “dilaton” is now the $22 \times 22$ symmetric matrix $g_{mn}$.

This does not imply that you should expect any number of scalars in the massless sector of the closed string. The massless sector of the string theory corresponds to a field theory in 26 dimensions and in order to interpret it in 4-dimensional terms, requires dimensional reduction.

This post imported from StackExchange MathOverflow at 2014-06-17 09:17 (UCT), posted by SE-user José Figueroa-O'Farrill
answered Jun 16, 2014 by José Figueroa-O'Farrill (2,315 points) [ no revision ]
reshown Jun 17, 2014 by Dilaton
+ 2 like - 0 dislike

Because the bosonic string theory is uncompactified. When you compactify you get new scalars.

answered Jun 17, 2014 by Ron Maimon (7,720 points) [ no revision ]

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