Dimensional Reduction to Three Dimensions

We already know how dimensional reduction works, and the general formulation we gave in section 3.4 also applies for the reduction to three dimensions. However, we want to keep track of the different components of the four-dimensional fields, so we will perform the reduction from four to three dimensions explicitly. In that way we can find out which extra symmetries there are for our special class of solutions.

We will reduce the time-direction and not one of the space-like direction. This is because we want to preserve spherical symmetry, that would be broken if we would reduce one of the compact space-directions. At first sight this does not look like a good idea, because the time-direction is not compact. We can not perform the integration over the reduced direction as in the Kaluza-Klein case, but we do not have to do this. We are not looking for a physical theory in three dimensions, we are only looking for extra symmetries. The extra integration in the action we can just leave out.

We will use the following notation: hatted quantities will denote four-dimensional ones, hatted greek indices will be the four-dimensional indices. The normal greek indices thus will be three-dimensional, $\mu=0,1,2$ and $\hat\mu=\{t,\mu\}$.

We can divide the four-dimensional action in three different parts, the Einstein part, scalar part and vector part. In this sense the action is

$$S^{(4)} = \int d^4x \sqrt{-g} \left\{ \hat{\cal L}_E + \hat{\cal L}_S + \hat {\cal L}_V \right\},$$ (7.1)

$$\hat {\cal L}_E = -\hat R,$$ (7.2)

$$\hat {\cal L}_S = ,{\textstyle{1\over 2}}(\partial \hat\phi)^2 + {\textstyle{1\over... ...tial \hat\Psi)^2 + {\textstyle{1\over 4}}\hat\sigma^{-2}(\partial \hat\sigma)^2$$ (7.3)

$$\hat {\cal L}_V = -{\textstyle{1\over 4}}e^{-\hat\phi} \left[ \hat\sigma \hat F(\ha... ...right] - {\textstyle{1\over 2}}\hat\Psi \hat F(\hat A) \tilde{\hat F} (\hat B).$$ (7.4)

Again we will use the tetrad formalism. The tetrads reduce in the following manner:

$$\hat e^{\hat a}{}_{\hat \mu} = \left( \begin{array}{cc} e^... ...ho D_\rho \\ \\ 0 & (\sqrt{\tau})^{-1} \end{array} \right).$$ (7.5)

This gives us the reduction of the four-dimensional metric:

$$\hat g_{\hat\mu\hat\nu} = \left( \begin{array}{cc} g_{\mu\n... ...u \\ \\ -D^\nu & \frac{1}{\tau} + D^2 \end{array} \right).$$ (7.6)

Although we are reducing the time-direction, we reduce the fourth coordinate, this is just more convenient. Note that the reduced coordinate in the metric has positive signature, and thus is time-like as we wanted.

The watchful reader will have noticed that the metric has a redundant term. In the chapter on Black Holes, we postulated a general form of the metric for a static spherically symmetric solution, which is diagonal. If we put that in our reduced metric (7.6), we find that the Kaluza-Klein vector, $D_\mu$, has to be equal to zero. We will see later this is indeed will happen, but for generality we will have to take it in to account.

With this we can reduce the first parts of the action (7.1). First the metric determinant reduces as

$$-\hat g = -\mbox{det}(\hat g_{\hat\mu\hat\nu}) = -\mbox{det}(g_{\mu\nu})\tau=-g\tau,$$

and the Einstein and scalar parts

$${\cal L}_E = -R + {\textstyle{1\over 4}}\tau F(D)^2,$$ (7.7)

$${\cal L}_S = ,{\textstyle{1\over 2}}(\partial \hat\phi)^2 + {\textstyle{1\over... ...tial \hat\Psi)^2 + {\textstyle{1\over 4}}\hat\sigma^{-2}(\partial \hat\sigma)^2$$ (7.8)

But again we want to work in the Einstein frame, so we have to make the conformal transformation

$$g_{\mu\nu} \to \tau^{-1} g_{\mu\nu}$$

to get rid of the scalar factor that originated from the four-dimensional metric determinant. The two terms then become

$${\cal L}_E = -R + \frac{1}{\tau^2}(\partial \tau)^2 - {\textstyle{1\over 4}} \tau^2 F(D)^2,$$ (7.9)

but the scalar part does not change, the contributions from the conformal transformation in the metric determinant and the metric in the scalar terms cancel each other.

We are going to have to reduce vectorfields, so we have to know how a vector reduces. If we are going to reduce the $t$-direction a vector reduces as:

$$\hat A_{\hat \mu} = \left( \begin{array}{c} \varphi_A \\ A_\mu + \varphi_A D_\mu \end{array} \right)$$ (7.10)

And with this we can calculate the vector term in (7.1). The vectors appear in two ways:

$$\hat F^2 = \hat F_{\hat\mu\hat\nu} \hat F^{\hat\mu\hat\nu} = \hat F_{\hat a\hat b} \hat F^{\hat a\hat b}$$

$$= \hat F_{ab}\hat F^{ab} + 2 \hat F_{a3} \hat F^{a3}$$

and

$$\hat F\hat{\tilde F} = \hat F_{ab} \hat{\tilde F}^{ab} + 2 \hat F_{a3} \hat{\tilde F}^{a3}.$$ (7.11)

The components of the field tensors in Lorentz-indices are:

$$\hat F_{ab} = \hat e_a{}^M \hat e_b{}^N \hat F_{MN}$$

$$= \hat e_a{}^\mu \hat e_b{}^\nu \partial _{[\mu}A_{\nu]} +e_a{}^\mu... ...{}^3(\partial _\mu\varphi) - \hat e_a{}^3 \hat e_b{}^\nu (\partial _\nu\varphi)$$

$$= F_{ab}(A) + \varphi F_{ab}(D)$$ (7.12)

$$\hat F_{a3} = \hat e_a{}^M \hat e_3{}^N \hat F_{MN}$$

$$= \hat e_a{}^\mu \hat e_3{}^3 \hat F_{\mu 3}$$

$$= \frac{1}{\sqrt{\sigma}} \partial _a \varphi$$ (7.13)

and the duals

$$\hat{\tilde F}_{ab} = {\textstyle{1\over 2}}\epsilon_{ab}{}^{\hat c \hat d} \hat F_{\hat c\hat d}$$

$$= \epsilon_{ab}^c \frac{1}{\sqrt{\tau}} \partial _c \varphi,$$ (7.14)

$$\hat{\tilde F}_{a3} = {\textstyle{1\over 2}}\epsilon_{a3}{}^{\hat c\hat d} \hat F_{\hat c\hat d}$$

$$= {\textstyle{1\over 2}}\epsilon_{abc} \left( F^{bc} + \varphi F^{bc}(D) \right).$$ (7.15)

The part of the action (7.1) involving the vectors then becomes, after the conformal transformation:

$${\cal L}_V = -{\textstyle{1\over 2}}e^{-\phi}\tau^{-1} \left[ \sigma(\partial \varphi_A)^2 + \sigma^{-1}(\partial \varphi_B)^2 \right]$$

$${\textstyle{1\over 4}}e^{-\phi}\tau \left[ \sigma\left( F(A) + \v... ..._A F(D) \right)^2 + \sigma^{-1}\left( F(B) + \varphi_B F(D) \right)^2 \right] +$$

$${\textstyle{1\over 2}}\frac{1}{\sqrt{\vert g\vert}} \epsilon^{\mu... ...u\nu}(B) + \varphi_B F_{\mu\nu}(A) + \varphi_A \varphi_B F_{\mu\nu}(D) \right].$$ (7.16)

In the last term which includes the axion we have made use of a partial integration. So the action in three dimensions is of the form

$$S^{(3)} = \int d^3x \sqrt{\vert g\vert} \left\{ {\cal L}_E + {\cal L}_S + {\cal L}_V \right\}.$$ (7.17)

The axion in four dimensions originated from the dualisation of a 2-form, the anti-symmetric field tensor. If we would have reduced the dual four-dimensional action, the four-dimensional anti-symmetric field tensor, would have resulted in a vector and again a field tensor in three dimensions. In three dimensions has the anti-symmetric field tensor no degrees of freedom, or equivalently, it is dual to a constant. The field strength tensor of the 2-form, $H_{\mu\nu\rho}$, is therefore equal to zero. But the vector remains and is precisely the dual of the axion.

In three dimensions, vectors and scalars are dual to each other. So we can redualise the axion to a vector $\Psi_\mu$, which will be useful in the next sections. The fact that vectors and scalars are duals will play an important role later.

The parts of the three-dimensional action (7.17) which involve the axion are

$${\cal L}_{Axion} = {\textstyle{1\over 2}}e^{2\phi} (\partial... ...i_B F_{\mu\nu}(A) + \varphi_A \varphi_B F_{\mu\nu}(D) \right].$$ (7.18)

We can dualise the scalar $\Psi$ with the same procedure we used before and after some calculation the term can written in the form:

$${\cal L}_{Axion} = - {\textstyle{1\over 4}}e^{-2\phi} \left(... ..._A F(B) + \varphi_B F(A) + \varphi_A \varphi_B F(D) \right)^2.$$ (7.19)

If we look at the three-dimensional action, it is not easy to find extra symmetries. Luckily we already know some of the symmetries involved in reduction schemes, but we will come to that in the next section.

Important to note is, however, that the electric/magnetic duality, which was only a symmetry of the equations of motion in four dimensions, is now a symmetry of the three-dimensional action. We do not mean by this a duality relation in three dimensions, but the four-dimensional duality expressed in three dimensional components. The three-dimensional action thus is invariant under $SL(2,\hbox{\mybb R})$ transformations.