# Ngô Quốc Anh

## July 15, 2014

### Why should we call ” f=φg ” conformal change?

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:07

I asked this question to Professor Alice Chang when I met her during a conference in the University of Notre Dame this June. Loosely speaking, given a Riemannian manifold $(M,g)$, why frequently we follow the rule $f = \varphi g$ to change our metric in conformal geometry?

Professor Chang told me that it is because under the new metric, angles are preserved. The aim of this note is to make her answer clearer.

The best way to see this is to make use of the vector formulation of the law of cosines. Indeed, let us take two vectors $X$ and $Y$ sitting in the same tangent space, say $T_pM$ for some $p \in M$. Then the angel between these two vectors can be estimated as follows

$\displaystyle \cos_g(X,Y)=\frac{g(X,Y)}{\|X\|_g\|Y\|_g}$

where $\|\cdot\|_g$ is the norm evaluated with respect to the metric $g$. Under the new metric $f$ given by $\varphi g$, we first obtain

$\displaystyle \cos_f(X,Y)=\frac{f(X,Y)}{\|X\|_f\|Y\|_f}.$

Clearly, $f(X,Y) = \varphi g(X,Y)$. Moreover, if we write $X$ in local coordinates as $X=X^i \frac{\partial}{\partial x^i}$, we then have

$\displaystyle {\left\| X \right\|_f} = \sqrt {{{\left\langle {{X^i}\frac{\partial }{{\partial {x^i}}},{X^j}\frac{\partial }{{\partial {x^j}}}} \right\rangle }_f}} = \sqrt {{X^i}{X^j}} \sqrt {{f_{ij}}} = \sqrt \varphi \sqrt {{X^i}{X^j}} \sqrt {{g_{ij}}} .$

Therefore,

$\displaystyle {\left\| X \right\|_f}{\left\| Y \right\|_f} = \varphi {\left\| X \right\|_g}{\left\| Y \right\|_g},$

which immediately shows that $\cos_f(X,Y)=\cos_g(X,Y)$. In other words, the angle between the two vectors $X$ and $Y$ is preserved under the change of metrics.