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Kahler geometry and potentials

  1. Dec 29, 2006 #1
    I'm attempting to reproduce the results in http://arxiv.org/PS_cache/hep-th/pdf/0611/0611332.pdf [Broken], notably the V_F(T) potential involved with the Kahler potential

    [tex]K = -2 \ln \left( \hat{\mathcal{V}} + \frac{\hat{\xi}}{2} \right)[/tex]
    [tex]W = W_{0} + Ae^{-aT}[/tex]
    [tex]\hat{\mathcal{V}} = \gamma (T+\bar{T})^{\frac{3}{2}}[/tex]
    [tex]\hat{\xi} = \xi (S+\bar{S})^{\frac{3}{2}}[/tex]

    Firsly I compute K_ab' using

    [tex]K_{a\bar{b}} = \frac{\partial^{2}K}{\partial \phi^{a} \partial \bar{\phi}^{b}}[/tex]

    This gives me

    [tex]K_{a\bar{b}} = \left(\frac{1}{\hat{\mathcal{V}} + \frac{\hat{\xi}}{2}}\right)^{2} \left( \begin{array}{cc} -\frac{3}{4} \gamma^{\frac{4}{3}} \hat{\mathcal{V}}^{-\frac{1}{3}}(\hat{\xi}-4\hat{\mathcal{V}}) & \frac{9}{4}(\xi \gamma)^{\frac{2}{3}} (\hat{\mathcal{V}}\hat{\xi})^{\frac{1}{3}} \\ \frac{9}{4}(\xi \gamma)^{\frac{2}{3}} (\hat{\mathcal{V}}\hat{\xi})^{\frac{1}{3}} & \frac{3}{4}\xi^{\frac{4}{3}}\hat{\xi}^{-\frac{1}{3}}(\hat{\xi}-\hat{\mathcal{V}}) \end{array} \right)[/tex]

    (god that's a pain to type!)

    The problem is, when computing the inverse of this metric [tex]K^{a\bar{b}}[/tex], I don't get the quoted form of the paper (which is also used in several other papers by Quevado), which gives each entry of the metric being proportional to


    This gives an important result for Westphal, since the potential V_F turns out to have this same singularity and it gives a vacuum uplifting. If only I could get that far :uhh: I'm guessing it's the det of K_ab' which gives that factor, but that's not what I get when I compute |K_ab'| to get K^ab'.

    Where am I going wrong? Am I neglecting a term I shouldn't in section 2 (equations 2.5 through to 2.7) or is it just I've slipped up on the algebra. I've done it about 10 times and still get the same. Given the symmetry between [tex]\hat{\xi}[/tex] and [tex]\hat{\mathcal{V}}[/tex] the Kahler metric has the right form, taking account of the factor of 2 here and there due to the 1/2 in the Kahler potential for xi-hat, so I'm at a lose to see anything obvious I've done wrong unless there's some fundamental result in the theory which I've missed or I've misunderstood the paper itself (both pretty likely knowing me!).

    Thanks for any help :)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 8, 2007 #2
    I realised my error. Westphal doesn't explain that while there's a correct to one of the terms in (2.3) to the form of (2.5), the other terms in (2.3) remain, so the full Kahler potential is not that given in (2.5) but actually the ammendment (2.5) to (2.3) giving

    [tex]K = -2 \ln \left ( \hat{\mathcal{V}} + \alpha '^{3}\frac{\hat{\xi}}{2}\right) - \frac{2}{3}\ln \hat{\xi} - \ln \left( -i \int_{CY_{3}}\bar{\Omega} \wedge \Omega \right)[/tex]

    Dropping the 3rd term in that potential and computing [tex]V_{F}[/tex] gives the results given by Westphal and Quevado. I'm not exactly sure why the 3rd term no longer plays a part. If independent of T and S, it doesn't contribute to the Kahler metric, but it does to the potential, because the potential involves [tex]e^{K}[/tex]. It doesn't alter the value of T which gives [tex]V_{F}=0[/tex] but it does alter [tex]V_{F}[/tex] otherwise, given it multiplies the entire potential.

    Can someone explain to me why I can essentially ignore the [tex]\ln \left( -i \int_{CY_{3}}\bar{\Omega} \wedge \Omega \right)[/tex] term? Thanks.
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