# Index Concordia

Posted in AdS/CFT, Relativity, String Theory by Index Guy on July 13, 2008

I have become more comfortable with the “T-dual” transformation and now I am confident on what to consider.

Exact adS can be described by the metric

$ds^2 = \displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right).$

After the T-transformation the metric returns to the same form but now $r = R^2 / \rho$,

$ds^2 = \displaystyle\frac{R^2}{\rho^2} \left(\eta_{mn}dx^m dx^n + d\rho^2\right).$

Both of these spaces can be described by embedding a hyper-surface in $R_{2,d}$. They both have scale invariance.

There are two extra-dimensional spaces that one can obtain by embedding a hyper-surface in $R_{3,d}$ and reduce to adS in some limit.

Case I: $C^2 = 0$

For this case anti-de Sitter space is obtained by setting the extra coordinate equal to $R$. The metric in the original PoincarĂ© coordinates looks like:

$ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 + 2\frac{z}{r}dzdr.$

After the T-transformation (leaving the $z$ coordinate intact) one obtains

$ds^2 = \displaystyle\frac{R^2}{\rho^2} \eta_{mn}dx^m dx^n + \displaystyle\frac{z^2}{\rho^2}d\rho^2 + dz^2 - 2\frac{z}{\rho}dz d\rho.$

Case II: $C^2 = -R^2$

Now adS is reached when the extra coordinate vanishes. In PoincarĂ© coordinates,

$\displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right) + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 + 2\frac{z}{r}dzdr.$

While in the T-coordinates we obtain

$\displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dx^m dx^n + dr^2\right) + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 - 2\frac{z}{r}dzdr.$

Note that both of these cases have scale-invariance in the coordinates $(r, y)$ and $(\rho, x)$. These two cases are atractive because they can be constructed straight from a higher-dimensional reduction. On the other hand one can just modify the adS metric and see what happens. Here are some of the modifications.

• Soft wall: $ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{R^2}{r^2}\left( 1 + A^2(r)\right)dr^2$
• Hard wall: $ds^2 = \displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right)\theta(r - r_{0})$
• Cutoff: $ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{R^2}{r^2 - a^2}dr^2$

I might not want to keep the scale invariance so obvious for the $y$ coordinates.

### 2 Responses

1. Index Guy said, on July 15, 2008 at 1:19 pm

Why insist so much in leaving the extra-coordinate $z$ unchanged? The metric actually diagonalized when one uses $\xi = Rz/r$.

For case I one obtains

$\displaystyle\frac{R^2}{r^2}\eta_{mn}dy^m dy^n + \frac{r^2}{R^2}d\xi^2.$

For case II we have

$\displaystyle \displaystyle\frac{R^2}{r^2}(\eta_{mn}dy^m dy^n + dr^2) + \frac{r^2}{R^2}d\xi^2.$

Notice that for case I the $r$ coordinate will not contain derivatives in the action…

2. Index Guy said, on July 15, 2008 at 1:28 pm

The change of variables $\xi = Rz/r$ should be done in the T-dual space.