1
$\begingroup$

In string theory, the Kähler potential of Kähler moduli (e.g. - the volume of a Calabi-Yau manifold) is given by (see, for instance, Becker, Becker, Schwarz: "String Theory and M Theory" p. 498)

$$K = -3\log[-\mathrm{i}(\rho-\bar{\rho})] $$

There are addition irrelevant terms (not coupled to $\rho$) which I am neglecting.

The Kähler metric is Hermitian and is given by,

$$G_{a\bar{b}}= \partial_a\bar{\partial_b}K$$

By the definition of a Hermitian metric $G_{\rho \rho}$ should be 0. But for the given potential it isn't. Why?

$\endgroup$
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/177153/2451 $\endgroup$ – Qmechanic Apr 20 '15 at 12:59
  • 1
    $\begingroup$ Looking at a Hermitian matrix I see no reason why $G_{\rho\rho}$ should be zero, rather, it should be real. Why do you claim that it should be zero by definition? $\endgroup$ – ACuriousMind Apr 20 '15 at 14:58
  • 1
    $\begingroup$ A hermitian matrix by definition has $G_{aa}$ as 0. See the proof in Nakahara or any complex manifold book. $\endgroup$ – sol0invictus Apr 22 '15 at 14:09
1
$\begingroup$

Using the Dolbeault bigrading, the (2,0) and (0,2) components of the Kähler metric $g_{zz}=0$ and $g_{\bar{z}\bar{z}}=0$ do indeed vanish, respectively. In particular, the formula $$g_{z\bar{z}}=\partial_{z}\partial_{\bar{z}}K$$ for the mixed (1,1) components does not generalize to the (2,0) and (0,2) components.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.