- #1
Mak182
- 1
- 0
Hi!
I have a question about Kähler manifolds. Of course there are many books (I prefer Nakahara) and lecture notes on this topic, but as a physicist I need a very hands-on way of dealing with metrics, etc.
Given a metric, what is the simplest way to find the Kähler form and to prove the Kähler property? I mean, e.g. Nakahara describes Kähler, but is there something simpler one can do (perhaps just in some cases)?
An example I have encountered is
\begin{equation}
ds^2 = (r-u) \left(\frac{\text d r^2}{F(r)} -\frac{\text d u^2}{G(u)} \right) + \frac{1}{r-u} \left(F(r) (\text d t +u \text d z)^2 - G(u) (\text d t+ r \text d z)^2 \right) ~,
\end{equation}
where $F$ and $G$ may be any functions of one variable. This metric is claimed to be Kähler with Kähler form
\begin{equation}
\Omega = \text d (r+ u) \wedge \text d t + \text d (r u) \wedge \text d z ~,
\end{equation}
which is obviously closed.
Following Nakahara, one needs to change coordinates to make the metric Hermitian, but can the Kähler form be read off more directly?
Cheers!
I have a question about Kähler manifolds. Of course there are many books (I prefer Nakahara) and lecture notes on this topic, but as a physicist I need a very hands-on way of dealing with metrics, etc.
Given a metric, what is the simplest way to find the Kähler form and to prove the Kähler property? I mean, e.g. Nakahara describes Kähler, but is there something simpler one can do (perhaps just in some cases)?
An example I have encountered is
\begin{equation}
ds^2 = (r-u) \left(\frac{\text d r^2}{F(r)} -\frac{\text d u^2}{G(u)} \right) + \frac{1}{r-u} \left(F(r) (\text d t +u \text d z)^2 - G(u) (\text d t+ r \text d z)^2 \right) ~,
\end{equation}
where $F$ and $G$ may be any functions of one variable. This metric is claimed to be Kähler with Kähler form
\begin{equation}
\Omega = \text d (r+ u) \wedge \text d t + \text d (r u) \wedge \text d z ~,
\end{equation}
which is obviously closed.
Following Nakahara, one needs to change coordinates to make the metric Hermitian, but can the Kähler form be read off more directly?
Cheers!