proof of criterion for conformal mapping of Riemannian spaces

In this attachment, we prove that the a mapping $f$ of Riemannian (or pseudo-Riemannian) spaces $(M,g)$ and $(N,h)$ is conformal if and only if $f^{*}h=sg$ for some scalar field $s$ (on $M$).

The key observation is that the angle $A$ between curves $S$ and $T$ which intersect at a point $P$ is determined by the tangent vectors to these two curves (which we shall term $s$ and $t$) and the metric at that point, like so:

$$\cos A=\frac{g(s,t)}{\sqrt{g(s,s)}\sqrt{g(t,t)}}$$

Moreover, given any tangent vector at a point, there will exist at least one curve to which it is the tangent. Also, the tangent vector to the image of a curve under a map is the pushforward of the tangent to the original curve under the map; for instance, the tangent to $f(S)$ at $f(P)$ is $f^{*}s$. Hence, the mapping $f$ is conformal if and only if

$$\frac{g(u,v)}{\sqrt{g(u,u)}\sqrt{g(v,v)}}=\frac{h(f^{*}u,f^{*}v)}{\sqrt{h(f^{*}u,f^{*}u)}\sqrt{h(f^{*}v,f^{*}v)}}$$

for all tangent vectors $u$ and $v$ to the manifold $M$. By the way pushforwards and pullbacks work, this is equivalent to the condition that

$$\frac{g(u,v)}{\sqrt{g(u,u)}\sqrt{g(v,v)}}=\frac{(f^{*}h)(u,v)}{\sqrt{(f^{*}h)(u,u)}\sqrt{(f^{*}h)(v,v)}}$$

for all tangent vectors $u$ and $v$ to the manifold $N$. Now, by elementary algebra, the above equation is equivalent to the requirement that there exist a scalar $s$ such that, for all $u$ and $v$, it is the case that $g(u,v)=s{h}^{*}(u,v)$ or, in other words, $f^{*}h=sg$ for some scalar field $s$.

Title	proof of criterion for conformal mapping of Riemannian spaces
Canonical name	ProofOfCriterionForConformalMappingOfRiemannianSpaces
Date of creation	2013-03-22 16:22:03
Last modified on	2013-03-22 16:22:03
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	7
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 30E20