# 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:

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

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 |