1. ## Metric transformation

I'm trying to understand the paper by Hitchin called ''Polygons and gravitons", Polygons and gravitons - INSPIRE-HEP. I'm stuck at page 471. At this point, he does some computations and obtains a metric:
$\displaystyle $$\gamma dz d\bar{z}+\gamma^{-1}\left(\dfrac{2dy}{y}+\bar{\delta}dz \right)\left(\dfrac{2d\bar{y}}{\bar{y}}+\delta d\bar{z} \right)$$$
where
$\displaystyle \gamma=\sum \dfrac{1}{\Delta_i}=\sum \dfrac{1}{\sqrt{(b-b_i)^2+|\bar{z}+a_i|^2}}$
$\displaystyle \delta=\sum \dfrac{(b-b_i)-\Delta_i}{\Delta_i(\bar{z}+a_i)}$
and $\displaystyle b$is defined implicitly by
$\displaystyle $$\prod ((b-b_i)+\Delta_i)=y\bar{y}$$$
We have that $\displaystyle$z,y$$are complex coordinates. He proceeds by saying: If we return to the form of the metric (4.4), and the description of the space of real quadratic polynomials as Euclidean 3-space with the metric given by the discriminant, the we obtain the metric which describes the gravitationals multi-instantons of Gibbons and Hawking: \displaystyle$$\gamma d\vec{x} d\vec{x}+\gamma^{-1}(d\tau+ \vec{\omega}d\vec{x})$$where \displaystyle \gamma=\sum \frac{1}{|x-x_i|} and \displaystyle curl \omega=grad \gamma$$.
The form of the metric (4.4) mentioned is
$\displaystyle $$\gamma^2(b'^2+a'\bar{a}')+\left(Im\left(\frac{2A '}{A}-\delta a' \right) \right)^2$$$
The problem I have is that I don't know how to go from the first metric to the metric of Gibbons and Hawking. Is a change of variables? Which one?
PS: Maybe this helps to understand (4.4),
$\displaystyle $$A\bar{A}=\prod((b-b_i)+\Delta_i)$$$

2. ## Re: Metric transformation

Hey spink.

If there is indeed a change of co-ordinate system (basically a substitution that retains the information involved) then they usually give these things.

I guess if you give this information and try and do the change of basis/space we can comment on that.

3. ## Re: Metric transformation

Hi chiro,

I did tried to do the change, but I'm stuck.

I think that the first logic step is this:
$\displaystyle x_1=-Re(\bar{z}) \hspace{1cm} x_2=-Im(\bar{z}) \hspace{1cm} x_3= b$
$\displaystyle x_{1i}=Re(a_i) \hspace{1cm} x_{2i}=Im(a_i) \hspace{1cm} x_{3i}= b_i$
Doing this, we have that then that
$\displaystyle \gamma=\dfrac{1}{\sqrt{(x_1-x_{1i})^2+(x_2-x_{2i})^2+(x_3-x_{3i})}}$
Like we have in Hawking ansatz. We can obtain also this relation $\displaystyle dx_3=db=\gamma^{-1}Re\left(\frac{2d\bar{y}}{\bar{y}}+\delta d\bar{z}\right)$.

It will remain to know what is $\displaystyle \tau$ in terms of $\displaystyle y,z$.

We could equalize

$\displaystyle d\tau +w_1dx_1+w_2dx_2+w_3dx_3=Im\left( (\frac{2d\bar{y}}{\bar{y}}+\delta d\bar{z})\right)$

But I don't know how to continue. I don't know why we obtain the condition that $\displaystyle curl \omega=grad \gamma$...

4. ## Re: Metric transformation

Unfortunately I haven't had enough training in vector calculus (presently) to answer the question of your last statement.

Hopefully others can come along and provide more help than myself.

5. ## Re: Metric transformation

Thank you for your interest anyway chiro

I hope some will help me...