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

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?

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I fail to see why this is a math question. As you explained in the middle paragraph, the trace is the dilaton field. And the trace is one dimensional. Should this not be in fact a physics question at the physics stackexchange? –  Willie Wong Oct 24 '13 at 7:48