In Calabi-Yau 3-fold, the Kähler metric is given in terms of the Kähler potential $\kappa$ :

$ g_{i\bar{j}} = \partial_i \partial_{\bar{j}} \kappa$,

where $i, \bar{j}$ = 1,2,3 $ ( the Holomorphic coordinates) . While the complex structure metric is given by:

$ G_{a\bar{b}} = \partial_a \partial_{\bar{b}} \kappa$,

where $ a, \bar{b} = 1, …, h_{2,1} $, where $ h_{2,1} $ is the number of complex structure moduli.

See for instance:


The Question:

  • first if I in 5d, N=2 supergravity, for instance , ( after dimensional reduction from 11d SUGRA, can I find an explicit form for the Kähler potential ?

  • Can i after that calculate $G_{a\bar{b}}$ and $ g_{i\bar{j}} $ ? and what about the degrees of freedom?

I mean for example, N=1 , D= 4, SUGRA , see for instance:


the Kähler potential is given by

$ \kappa = \phi_i \phi^{i *} $,

and the Kähler metric is given by:

$ g_{i j^*} = \frac{\partial^2 \kappa}{\partial \phi_i \partial \phi_{j^*}} $ where $\phi_i$ are scalar fields.

Any help appreciated!

  • 1
    $\begingroup$ In the linked article for $N=1, D=4$ SUGRA, I don't think it says that the Kähler potential is given by $\kappa = \phi_i\phi^{i\ast}$, only that this is a particularly simple example, that is worked out as a toy model. $\endgroup$ – doetoe Sep 9 '19 at 16:53
  • $\begingroup$ Look at the article carefully pls. What do you mean by I don’t think! ! It’s will known in 4d supergravity models , the Kähler potential is given in terms of scalars!!! $\endgroup$ – Dr. phy Sep 9 '19 at 17:09
  • 1
    $\begingroup$ Yes, always, not only in 4D. You claim however that it is given by $\kappa = \phi_i\phi^{i\ast}$. The authors only say "A simple form of K [...] which will be used in the next subsection as a toy model, is given by $K = \phi_i\phi^{i\ast}$. $\endgroup$ – doetoe Sep 9 '19 at 17:23
  • $\begingroup$ @doetoe, you don’t understand the question as a whole!! $\endgroup$ – Dr. phy Sep 9 '19 at 21:22

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