Reduction of the Low Energy String Effective Action

The method discussed in the previous section can be generalised to include more dimensions and more complicated fields. We can use it to reduce our ten-dimensional string effective action (3.7) to four dimensions. But let's consider the general case of an effective action in $D$-dimensions that we want to reduce to ($D-d$) dimensions. So in the simplest case we assume that space-time is of the form $M\times K$, where $M$ has $D-d$ dimensions and $K$ has $d$ dimensions and all fields are independent of the coordinates $y^m$ of $K$.

This the simplest way to compactify the extra dimensions and of course there are more interesting ways to do this compactification which have more realistic features, but they are also far more difficult. For the moment we will stick to the case in which the fields are taken to be independent of the extra $K$ coordinates.

We will use the following notation: Local coordinates of $M$ are $x^\mu$ ( $\mu=0,1,\ldots,D-d-1$), internal coordinates of $K$ are $y^m$, ($m=1,\ldots,d$). The total space $M\times K$ has signature ($+-\ldots -$) and fields in this $D$-dimensional space are denoted with a hat, as well as their coordinates ($\hat\Phi$, $\hat g_{\hat\mu\hat\nu}$, etc.), ( $\hat\mu=0,1,\ldots ,D-1$). Quantities without a hat are then the ($D-d$)-dimensional ones.

We begin with the first terms of the low energy string effective action (3.6), the Einstein term coupled to the Dilaton:

$$S_E = \int_M d^{D-d}x \int_K d^dy \sqrt{\hat g} e^{-\hat\Phi... ...tial _{\hat\mu} \hat\Phi \partial ^{\hat\mu}\hat\Phi \right\}.$$ (3.18)

If we assume that the internal space $K$ is compact, we can perform the integration over the internal coordinates: $\int_K d^dy =1$.

It is very useful to do this reduction with the aid of so called vielbeins (vierbeins or tetrads in four dimensions, vielbeins in any other dimension). With the use of these vielbeins we make the connection between curved coordinate systems and local Lorentz (flat) coordinates. See appendix A for a short introduction of these vielbeins.

On the same transformation grounds that we saw in the previous section we can determine the reduction of these vielbeins. Let again the greek indices denote the curved indices, ( $\hat\mu=\{\mu,m\}$) where $m$ labels the internal coordinates. And the hatted roman indices will denote the Lorentz indices in $D$-dimensions, ( $\hat a=\{a,n\}$). We assume that the internal space is flat, so there is no difference between the internal 'curved' indices and the internal Lorentz indices. The vielbeins reduce in the following way [17]:

$$\hat e^{\hat a}{}_{\hat\mu} = \left( \begin{array}{cc} e^a{}... ...-e_a{}^\rho A_\rho{}^m \\ &\\ 0 & E^n{}_m \end{array} \right).$$ (3.19)

The internal metric is $G_{mn} = E^k{}_m \delta_{kl} E^l{}_n$ and the space-time metric is $g_{\mu\nu} = e^a{}_\mu \eta_{ab} e^b{}_\nu$. We can express the $D$-dimensional metric in these quantities as we did in previous section [20]:

$$\hat g_{\hat\mu\hat\nu} = \left( \begin{array}{cc} g_{\mu\nu... ...u{}^m G_{mn} \\ &\\ A_nu{}^n G_{mn} & G_mn \end{array} \right)$$ (3.20)

and its inverse

$$\hat g_{\hat\mu\hat\nu} = \left( \begin{array}{cc} g^{\mu\nu... ...{\nu m} & G^{mn} + A^{\rho m} A_\rho{}^n \end{array} \right).$$ (3.21)

So when we reduce $d$ dimensions, the $D$-dimensional metric will reduce to a ($D-d$)-dimensional metric, $d$ abelian vector fields $A_\mu{}^m$, ($m=1,\ldots,d$) and a scalar matrix $G_{mn}$. Because a metric is a symmetric tensor, this scalar matrix $G_{mn}$ will consist of ${\textstyle{1\over 2}}d (d+1)$ independent scalar fields.

We can put these expressions into the Einstein part of our effective action (3.18) and after a tedious calculation one finds

$$S = \int_M d^{D-d}x \sqrt{\vert g\vert} e^{-\phi} \left\{ -R + g^{\mu\nu} \partial _\mu \phi \partial _nu \phi + \right.$$

$$\left. + {\textstyle{1\over 4}}g^{\mu\nu} \partial _\mu G_{mn} \part... ...{mn} + {\textstyle{1\over 4}}G_{mn} F_{\mu\nu}{}^m(A) F^{\mu\nu n}(A) \right\},$$ (3.22)

where we have made a shift in the Dilaton field

$$\phi = \hat\Phi - {\textstyle{1\over 2}}\log \det G_{mn}$$

and of course $F_{\mu\nu}{}^m = \partial _\mu A_\nu{}^m - \partial _\nu A_\nu{}^m$.

We also have to reduce the other part of our string effective action involving the anti-symmetric field tensor $B_{\mu\nu}$

$$S_{\hat B} = -{\textstyle{1\over 12}} \int_M d^{D-d}x \int_K... ...t H_{\hat\mu\hat\nu\hat\rho} \hat H^{\hat\mu\hat\nu\hat\rho} ,$$ (3.23)

where $\hat H_{\hat\mu\hat\nu\hat\rho} = \partial {[\hat\mu} \hat B_{\hat\nu\hat\rho]}$.

We can derive that the following reduction of the $D$-dimensional anti-symmetric field tensor gives us the correct transformation properties in $(D-d)$ dimensions

$$\hat B_{\hat\mu\hat\nu} = \left( \begin{array}{cc} B_{\mu\n... ... -B_{\nu m} + B_{mn} A_{\nu}{}^n & B_{mn} \end{array} \right).$$ (3.24)

So the $D$-dimensional anti-symmetric field tensor gives us in $D-d$ dimensions again a anti-symmetric field tensor, $d$ abelian vector fields $B_{\mu m}$ and a $d\times d$ scalar matrix $B_{mn}$, which because of anti-symmetry has ${\textstyle{1\over 2}}d(d-1)$ independent components.

Note that $\hat B_{\hat\mu\hat\nu}$ is a gauge tensor field: $\delta\hat B_{\hat\mu\hat\nu} = \partial _{[\mu} \Lambda_{\nu]}$. This means that the ($D-d$)-dimensional fields will have the following transformation properties:

$$\delta B_{\mu\nu} = \partial _{[\mu} \Lambda_{\nu]}$$

$$\delta B_{\mu m} = \partial _{\mu} \Lambda_m$$

We can use the vielbeins to convert the curved indices of $\hat H$ to Lorentz indices $\hat H_{\hat a \hat b \hat c}$ and then use the reduced vielbeins ($e^a{}_\mu$ and $E^n{}_m$) to convert them back to $(D-d)$-dimensional curved indices:

$$\hat {\textstyle{1\over 12}} H^2 = {\textstyle{1\over 12}}\hat H_{\hat a \hat b\hat c} \hat H^{\hat a \hat b\hat c} =$$

$$= {\textstyle{1\over 12}} H_{mnk} H^{mnk} + {\textstyle{1\over 4}}H... ...mu\nu m} H^{\mu\nu m} + {\textstyle{1\over 12}} H_{\mu\nu\rho} H^{\mu\nu\rho} .$$ (3.25)

Here $H_{mnk} =0$ since $B_{mn}$ is independent of the $K$ coordinates. We can also calculate the other terms:

$$H_{\mu mn} = e^r{}_\mu \hat e_r{}^{\hat \mu} \hat H_{\mu mn} = \partial _\mu B_{mn},$$ (3.26)

$$H_{\mu\nu m} = e^r{}_\mu e^s{}_\nu \hat e_r{}^{\hat\mu} \hat e_s{}^{\hat\nu} \hat H_{\hat\mu\hat\nu m}$$

$$= F_{\mu\nu m}(B) - B_{mn} F_{\mu\nu}{}^n(A),$$ (3.27)

$$H_{\mu\nu\rho} = e^r{}_\mu e^s{}_\nu e^t{}_\rho \hat e_r{}^{\hat\mu} \hat e_s{}^{\hat\nu} \hat e_{t}{}^{\hat \rho} \hat H_{\hat\mu\hat\nu\hat\rho}$$

$$= \partial _\mu B_{\nu\rho} - {\textstyle{1\over 2}}( A_\mu{}^r F_{\nu\rho r}(B) + B_{\mu r} F_{\nu\rho}{}^r(A) ) + \mbox{cycl.},$$ (3.28)

where $F_{\mu\nu m}(B) = \partial _\mu B_{\nu m} - \partial _{\nu} B_{\mu m}$.

Due to the dimensional reduction there arise extra terms in the definition in the field strength tensor $H_{\mu\nu\rho}$. These are the so called Abelian Chern-Simons terms.

We can put all these terms back into the action. The total reduced effective action, consisting of the Einstein part as well as the part with the anti-symmetric field tensor, can be written in a very symmetric form [17]. The claim is that there is a $O(d,d)$ global symmetry that keeps this action invariant. (see Appendix B on the symmetry group $O(d,d)$.)

First we have to introduce the $2d\times 2d$ scalar matrix $M$,

$$M=\left( \begin{array}{cc} G^{-1} & -G^{-1}B \\ BG^{-1}&G-BG^{-1}B \end{array} \right),$$ (3.29)

where $G$ is the $d\times d$ metric scalar matrix $G_{mn}$ and $B$ is the scalar matrix $B_{mn}$. Furthermore we have to introduce the matrix $L$:

$$L = \left( \begin{array}{cc} 0&I_d\\ I_d&0 \end{array} \right),$$ (3.30)

which is the identity matrix of the group $O(d,d)$ in a basis rotated from the one with a diagonal identity (B). If we also put the field strength tensors of the vector fields in a $d+d$ doublet,

$${\cal F}^i_{\mu\nu} = \left( \begin{array}{c} F_{\mu\nu}{}^m... ...partial _\mu {\cal A}_\nu{}^i - \partial _\nu {\cal A}_mu{}^i,$$ (3.31)

where $i=1,\ldots,2d$, the total action can be written as [19][17]

$$S = \int_M d^{D-d}x \sqrt{\vert g\vert} e^{-\phi} {\cal L},$$ (3.32)

with ${\cal L}={\cal L}_1+{\cal L}_2+{\cal L}_3+{\cal L}_4$, where

$${\cal L}_1 = -R + (\partial \phi)^2,$$

$${\cal L}_2 = -{\textstyle{1\over 8}} g^{\mu\nu} \mbox{tr} (\partial _\mu ML \partial _\nu ML) ,$$

$${\cal L}_3 = {\textstyle{1\over 4}}g^{\mu\mu'}g^{\nu\nu'} {\cal F}_{\mu\nu}^i (LML)_{ij} {\cal F}_{\mu'\nu'}^j,$$

$${\cal L}_4 = -{\textstyle{1\over 12}} g^{\mu\mu'}g^{\nu\nu'}g^{\rho\rho'} H_{\mu\nu\rho} H_{\mu'\nu'\rho'},$$

where $H_{\mu\nu\rho} = \partial _{[\mu} B_{\nu\rho]} + 2{\cal A}_{[\mu}{}^i L_{ij} {\cal F}_{\nu\rho]}^j$. This action is obviously invariant under the $O(d,d)$ transformation:

$$M \to \Omega M \Omega^T, \quad {\cal A}_\mu{}^i \to \Omega_{ij} {\cal A}_{\mu}{}^j,$$ (3.33)

keeping the other fields invariant and where $\Omega \in O(d,d)$ satisfying

$$\Omega^T L \Omega = L$$

So the reduction of a Low energy String Effective Action in $D$ dimensions to $D-d$ dimensions results allways in a action (3.32) which has an $O(d,d)$ symmetry. This dual symmetry is called T-Duality, where the 'T' stands for Target-space. In the case of the string effective action of the bosonic sector of the Heterotic string (3.7) we have to reduce from $D=10$ to four dimensions. This means $d=6$ and the reduced action will be invariant under $O(6,6)$ transformations. This also means there will be 12 vector fields and $M$ will be a $12\times 12$ matrix with 36 independent scalars.

As we mentioned before, we expect that compactification on more complicated manifolds $K$ will give more realistic features in four dimensions. In fact there are very many ways to perform this compactification and it is not clear yet which of them should be the correct way.