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I am reading Jost's Compact Riemann Surfaces for the derivation of geodesic lines (page 29).

The books does it as follows. Let $\Sigma$ be a compact Riemann surface with a conformal Riemannian metric $\lambda^2(z)dzd\bar z$ (given in local coordinates, and $\lambda$ is smooth). Define the energy of a smooth curve $\gamma:[0,1]\to\Sigma$ to be $$E(\gamma)=\frac{1}{2}\int_0^1\lambda^2(\gamma(t))|\dot\gamma(t)|^2dt$$ Then naturally, one would want to define the geodesic lines as the minimizers of $E$. A necessary condition is for any smooth variation $\gamma+s\eta$ (in local coordinates, with $s\in\mathbb{R}$ and varing in a neighborhood of $0$), $$0=\left.\frac{d}{ds}E(\gamma+s\eta)\right|_{s=0}$$

The computation of the RHS is where I got stuck.

In the book: $$0=\left.\frac{d}{ds}E(\gamma+s\eta)\right|_{s=0}\\ =\frac{1}{2}\int_0^1(\lambda^2(\gamma)(\dot\gamma\dot{\bar\eta}+\dot{\bar\gamma}\dot\eta)+2\lambda(\lambda_\gamma\eta+\lambda_{\bar\gamma}\bar\eta)\dot\gamma\dot{\bar\gamma})dt\\ ={\rm Re}\int_0^1(\lambda^2(\gamma)\dot\gamma\dot{\bar\eta}+2\lambda\lambda_\gamma\dot\gamma\dot{\bar\gamma}\bar\eta)dt$$ (here $\lambda_\gamma=\frac{\partial\lambda}{\partial\gamma})$

My questions:

(1) What does $\lambda_\gamma=\frac{\partial\lambda}{\partial\gamma}$ mean?

(2) How to get to the second line in the equations above?

(3) Why $\lambda_\gamma\eta+\lambda_{\bar\gamma}\bar\eta=2{\rm Re}\lambda_\gamma\bar\eta$? Is it a typo?

My attempts:

(1) I am guessing the directional derivative of $\lambda$ at $\gamma(t)$ along $\dot\gamma(t)$. But I am not sure.

(2) I have got this far: $$\frac{d}{ds}E(\gamma+s\eta)=\frac{1}{2}\int_0^1\frac{d}{ds}\lambda^2(\gamma+s\eta)|\dot\gamma+s\dot\eta|^2dt\\ =\frac{1}{2}\int_0^1\lambda^2(\gamma+s\eta)(\dot\eta(\dot{\bar\gamma}+s\dot{\bar\eta})+\dot{\bar\eta}(\dot\gamma+s\dot\eta))+2\lambda(\gamma+s\eta)\frac{d}{ds}\lambda(\gamma+s\eta)|\dot\gamma+s\dot\eta|^2dt$$ Let $s=0$ and it becomes $$\frac{1}{2}\int_0^1\lambda^2(\gamma)(\dot\eta\dot{\bar\gamma}+\dot{\bar\eta}\dot\gamma)+2\lambda(\gamma)\left.\frac{d}{ds}\lambda(\gamma+s\eta)\right|_{s=0}|\dot\gamma|^2dt$$

Compare this with that in the book, it remains to show $$\left.\frac{d}{ds}\lambda(\gamma+s\eta)\right|_{s=0}=(\lambda_\gamma\eta+\lambda_{\bar\gamma}\bar\eta)$$ This is where I have no idea how to proceed.

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  • $\begingroup$ I agree that notation is ridiculous, but I still need to understand it. $\endgroup$ – trisct Jul 29 at 12:09
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After some struggling I finally figured it out. I am writing down the answer here in case anyone is still interested and also for my future reference.

(1) The notation is confusing. $\gamma$ here is not the curve anymore but the variable taken by $\lambda$. It would be a lot less confusing to write $\lambda_z(z)=\frac{\partial\lambda}{\partial z}$.

(2) Now that it is clear what $\lambda_\gamma$ (I shall replace it by $\lambda_z$ from now on) means, it follows easily that $$d\lambda(z)=\lambda_zdz+\lambda_{\bar z}d\bar z$$ and $$z=\gamma+s\eta\implies dz=\eta\ ds,\ d\bar z=\bar\eta\ ds$$ Hence $$\frac{d}{ds}\lambda(\gamma+s\eta)=\lambda_z(\gamma+s\eta)\eta+\lambda_{\bar z}(\gamma+s\eta)\bar\eta$$

(3) It is a typo. Should be $2{\rm Re}\lambda_{\bar z}\bar\eta$ instead.

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