I'd like to do the maths for the moduli stabilization of 6D Einstein-Maxwell Gravity
$$
S= \int d^6X \sqrt{-G_6}(M_6^4R_6[G_6]-M_6^2|F_2|^2),
$$

where the 6D metric is specified by
$$
ds^2 = g_{\mu\nu}(x)dx^{\mu}dx^{\nu}+R^2(x)\hat{g}_{mn}(y)dy^mdy^n,
$$
and $\hat{g}_{mn}$ is the metric of a compact 2D manifold with unit volume.

The setup of this model can be found in Denef, Douglas, Kachru, starting on page 10.

Now I want to perform the dimensional reduction of the theory and thereby obtain the correct effective potential $V(R) = \frac{\chi}{R^4}-\frac{n^2}{R^6}$.

I could manage to derive the first part of the potential by rewriting
$$
ds^2= R^2(x)\{\frac{g_{\mu\nu}(x)}{R^2(x)}dx^{\mu}dx^{\nu}+\hat{g}_{mn}(y)dy^mdy^n\}
$$
which allows me to rescale the 6D action by $G_{MN}=R^2\tilde{G}_{MN}$. Then we are left with a product space, i.e. we can write $R_6[\tilde{G}_6]=R_4[\tilde{g}]+R_2[\hat{g}]$, where $\tilde{g}_{\mu\nu}=\frac{1}{R^2}g_{\mu\nu}$. Further rescaling brings a factor $R^{-4}$ for the term $\chi$, while a factor $R^2$ has been absorbed in order to write $M_4^2=M_6^4R^2$.

However, I'm not sure how to obtain the term $\frac{n^2}{R^6}$. I'd appreciate any help.

My idea is as follows:
$$
\int d^6X \sqrt{-G_6}M_6^2 |F_2|^2 = \int d^6X R^6 \sqrt{-\tilde{G}_6}M_6^2 |F_2|^2 = \\
= \int d^4x \sqrt{-\tilde{g}}\int d^2y\sqrt{\hat{g}} R^6M_6^2 |F_2|^2 = \int d^4x \sqrt{-g}\int d^2y\sqrt{\hat{g}} R^2M_6^2 |F_2|^2 \sim \\
\sim M_6^2 \int d^4x \sqrt{-g}R^2 (\frac{1}{R^2})^2 \cdot (\frac{n}{R^2})^2 = M_6^2 \int d^4x \sqrt{-g} \frac{n^2}{R^6}
$$

In the last line, we obtain a factor $R^{-4}$ because of $\gamma^{m\hat{m}}\gamma^{n\hat{n}}F_{mn}F_{\hat{m}\hat{n}}$ and the second factor comes from $F_2 \sim \frac{n}{R^2}$, since
$$
\int_{M_2} F_2 =n.
$$

However, I'm puzzled because the factor of $M_6^2$ remains. If I want to rewrite the action s.t. we have a factor $M_4^2$ in front, then the term I obtain reads $\frac{1}{M_4R} \frac{n^2}{R^6}$ (reminder: $M_4^2 =M_6^4R^2$).

So, where is my mistake?

Furthermore, it is claimed that O3 planes give rise to an additional term $\frac{m}{R^4}$ (after Weyl rescaling), but I've no idea how it arises. I would start with a CS term $\int C_4$ for the action...

It would be great if anyone could help me with these questions. Thanks in advance!

This post imported from StackExchange Physics at 2015-05-22 21:10 (UTC), posted by SE-user psm