8
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Yes, seems strange. Usually, it seems that there is no $M_6^2$ term in front of the $|F_2|^2$ term, and there is also a cosmological constant $\Lambda$ (see for instance page $18$ and $19$, of this presentation, with $S^2$ as the internal space). Here the ansatz is a little bit different - with a $\psi$ field - so your potential could be understood as the limit of $V(\psi)$ when $\psi \to 0$, with $\Lambda = 0$. However, we see also clearly the flux term $\sim n^2$. $\endgroup$ – Trimok Dec 26 '13 at 21:32
  • $\begingroup$ Thank you, Trimok! Unfortunately I think I made a little mistake when deriving the potential for $F_2=0$. Now, I come up with $S=\int d^4x \sqrt{-h}M_6^4(R_4[h]-\frac{\chi}{R^4})$. The strategy is the same as sketched in the question, except that I need to do another Weyl rescaling in the end. But the factor $M_6^4$ contains $R^2$. However, Denef, Douglas and Kachru pulled a factor of $R^2$ out of the integral, even though $R=R(x)$ - see eq (3) of the paper in my question. $\endgroup$ – psm Dec 27 '13 at 9:18
  • $\begingroup$ A problem with all this stuff is the mass dimensionality of all the used terms and lagrangians. For instance, $V(R) = \dfrac{\chi}{R^4}-\dfrac{n^2}{R^6}$ seems curious, if $n$ is dimensionless, because $R$ has the dimension of a length. So maybe one needs some redefinition of $F$ ? $\endgroup$ – Trimok Dec 27 '13 at 10:26
  • $\begingroup$ True, $V(R)$ like that seems very strange. However, I believe this should represent just a schematic dependence. So, prefactors should be included (and that's what Zwiebach does in his introductory book, but unfortunately w/o derivation). A redefinition of $F$ with factors of $R$ might further spoil the result I've derived in my question (which makes sense to me and looks good up to $M_6^4$). Maybe I should ignore any fundamental masses and include them afterwards? $\endgroup$ – psm Dec 27 '13 at 11:07
  • $\begingroup$ Sounds a good idea, because it is headsick, to check mass dimensionality... $\endgroup$ – Trimok Dec 27 '13 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.