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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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up vote 1 down vote accepted

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

$$\left|\frac{f_{zz}}{f_{z\bar{z}}}\right|^2=\frac{ (\tr \Hess(f))^2+(f_{xy})^2}{(\tr \Hess(f))^2}\geq 1. $$

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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
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