# Special Morse function on a Riemann surface

Let $f: S \to \Bbb R$ be a Morse function on a Riemann surface. Let $x_0$ be a saddle point of $f$. Since $x_0$ is a critical point of $f$, it makes sense to talk about the bilinear forms $f_{z\overline{z}}(x_0)$, $f_{zz}(x_0)$ and $f_{\overline{z}\overline{z}}(x_0)$ as components of the real Hessian of $f$ at $x_0$ (if one only allows only complex coordinates).

Can $f$ be arranged so that the Hermitian form $f_{z\overline{z}}$, which can be defined on all $S$, is arbitrarily small (compared to some fixed Hermitian metric) and $|f_{zz}(x_0)|$(as a number) is arbitrarily big in a fixed coordinate system around $x_0$?

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First, let me point out that only the differential form $\newcommand{\pa}{\partial}$ $\newcommand{\bpa}{\bar{\partial}}$ $\pa\bpa f$ is well defined globally on $S$. Locally

$$\pa\bpa f= f_{z\bar{z}} dz\wedge d\bar{z},$$

but the coordinate $z$ is only locally defined. If you choose different a different local coordinate $u$ you have an equality

$$f_{z\bar{z}} dz\wedge d\bar{z}= \pa\bpa f=f_{u\bar{u}} du\wedge d\bar{u} \tag{A}$$

yet

$$f_{z\bar{z}} \neq f_{u\bar{u}}.$$

The equality (A) implies that

$$f_{u\bar{u}}= f_{z\bar{z}} \cdot \left| \frac{dz}{du} \right|^2.$$

On the other hand,

$$\frac{d}{du} =\frac{dz}{du}\frac{d}{dz}$$

so that

$$\frac{d^2}{du^2}= \frac{d^2 z}{du^2} \frac{d}{dz}+ \left(\frac{dz}{du}\right)^2\frac{d^2}{dz^2}.$$

Hence, if $z= au+bu^2+ O(u^3)$$,$$ a\in \mathbb{C}\setminus 0$$, we deduce$$ f_{u\bar{u}}= |a|^2 f_{z\bar{z}}, f_{uu}= a^2 f_{zz}. $$In particular, at the critical point x_0 we have$$\left| \frac{f_{uu}}{f_{u\bar{u}}}\right|= \left| \frac{f_{zz}}{f_{z\bar{z}}}\right|. $$This proves that you cannot find a holomorphic coordinate u near x_0 that makes f_{u\bar{u}} very small and f_{uu} very large. Update. \newcommand{\ii}{\boldsymbol{i}} If z=x+\ii y, then using the equalities$$\pa_z=\frac{1}{2}(\pa_x-\ii\pa_y),\;\;\pa_{\bar{z}}=\frac{1}{2}(\pa_x+\ii\pa_y), $$we deduce \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\Hess}{Hess}$$ f_{z\bar{z}}=\frac{1}{4} (f_{xx}+f_{yy}) =\frac{1}{4}\tr \Hess(f),  f_{zz}=\frac{1}{4}( f_{xx}+f_{yy}+2\ii f_{xy}). $$As I mentioned in one of my comments, you can choose f so that the partials f_{xx},f_{yy}, f_{xy} have any prescribed values at x_0, but observe that$$\left|\frac{f_{zz}}{f_{z\bar{z}}}\right|^2=\frac{ (\tr \Hess(f))^2+(f_{xy})^2}{(\tr \Hess(f))^2}\geq 1.$$- Thanks for the response, but you got me wrong on this one I think. I was asking if$f$can be arranged, not if the coordinates can be arranged ... if I am not mistaken, you proved here that the elements of the Hessian transform well. – The Common Crane Jul 18 '12 at 21:54 Given a point$x_0\in S$and an arbitrary symmetric bilinear form$H$on the tangent space$T_{x_0}S$there exist a smooth function on$S$whose Hessian at$x_0$is$H$. – Liviu Nicolaescu Jul 19 '12 at 1:15 I forgotvto add:$x_0$is a critical point of the above funtion$f$. – Liviu Nicolaescu Jul 19 '12 at 1:18 You are right again. Thanks for your insistance. But I want is that$f_{z\overline{z}}$is arbitrarily small all over$S$and$f_{zz}$is arbitrarily big at$x_0$only. – The Common Crane Jul 22 '12 at 11:53 Sorry for getting back so late. I think you still don't completely understand me, I want to minimize$f_{z\bar z}$not just at$x_0$but all over$S$. What you wrote was very clear for me, and it shows that you can do this at$x_0$. But to minimze$f_{z\bar z}\$ one has to be more careful. Thanks for your willingness to help, I will accept your answer. – The Common Crane Nov 7 '12 at 6:16