# Index Concordia

## In search for anti-de Sitter space (III)

Posted in Relativity by Index Guy on July 2, 2008

Let us start by considering the space consisting of the product of a 3-d Minkowski manifold and $d$-dimensional Minkowski manifold. The metric reads as:

$ds^2 = -dZ_{0}^2 + dZ^{2}_{1} + dZ^{2}_{2} - dY_{0}^2 + dY_{1}^2 + ... + dY_{d - 1}^2.$

After imposing $X^2 = -R^2$ one can use coordinates:

$\sqrt{2} Z_{+} = Z_{0} + Z_{1} = \displaystyle\frac{R^2}{r}$   ,   $\sqrt{2} Z_{-} = Z_{0} - Z_{1} = r + \displaystyle\frac{r z^2}{R^2} + \frac{y^2}{r}$   ,   $Z_{2} = z$   and   $Y^{m} = \displaystyle\frac{R}{r}y^{m}$.

Once we plug-in these coordinates, the metric takes a surprisingly simple form:

$ds^2 = \left(\displaystyle\frac{R^2 + z^2}{r^2}\right)dr^2 + \displaystyle\frac{R^2}{r^2}dy^2 + \displaystyle\frac{2 z}{r}dz dr + dz^2.$

Now we concentrate in the case where $d = 4$. Then the space described by the above metric is six-dimensional. We can obtain a five-dimensional space by fixing one of the coordinates. It turns out that by fixing $z = R_{0}$ one gets some interresting results.

For $R_{0} \neq 0$ we have a funky version of 5-d anti-de Sitter space:

$ds^2 = \displaystyle \frac{R^2}{r^2}dy^2 + \frac{\rho^{ 2}}{r^2} dr^2$   with   $\rho^{2} = R^2 + R_{0}^2.$

This could be useful for something. Exact 5-d adS is obtained when $R_{0} = 0.$

The metric is certainly more attractive than my previous blunder. But I still do not know how to understand the off-diagonal terms.

In any case, this metric still needs some work. In the work of Alday and Maldacena, they do a “T-dual” transformation on the $y$ coordinates. This transformation is defined by:

$\partial_{a} x^{m} = i \displaystyle\frac{R^2}{r^2}\epsilon_{ab}\partial_{b}y^{m}.$

(Alday & Maldacena use $x$ for the initial coordinates and $y$ for the T-dually-transformed coordinates. In my work I use the opposite convention.) Furthermore, after redefining $\rho = R^{2}/r$ we can write the metric in AdS form again, with a slight change in the second term:

$ds^2 = \displaystyle\frac{R^2}{\rho^2}\left(dx^2 + d\rho^2\right) + \left(dz - \displaystyle\frac{\rho z}{R^2}d\rho\right)^{2}.$

In this form it certainly looks interesting. Alas, I do not understand this T-duality transformation. A quote from the Alday-Maldacena paper:

In the regime under consideration the T-dual coordinates are real and the worldsheet is Euclidean. In addition, the boundary condition for the original coordinate $y^{m}$, which is that the carry momentum $k^{m}$, translates into the condition that $y^{m}$ has “winding” $\Delta y^{m} = 2 \pi k^{m}.$

Indeed.

### 2 Responses

Once again I have made a stupid mistake. Looks like after the change $r = R^2 / \rho$ one obtains:
$ds^2 = \displaystyle\frac{R^2}{\rho^2}(dx^2 + d\rho^2) + \left(dz - \displaystyle\frac{z}{\rho}d\rho\right)^2 .$