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}.



2 Responses

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  1. Index Guy said, on July 2, 2008 at 9:55 pm

    Perhaps another coordinate transformation may bring this result into diagonal form?

  2. Index Guy said, on July 10, 2008 at 12:46 pm

    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 .

    It is basically a change in sign…

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