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 $f^* h = s g$ 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: $$ \cos A = {g(s,t) \over \sqrt{g(s,s)} \sqrt{g(t,t)} $$ 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 $$ {g(u,v) \over \sqrt{g(u,u)} \sqrt{g(v,v)}} = {h(f^* u, f^* v) \over \sqrt{h(f^* u, f^* u)} \sqrt{h(f^* v, f^* v)} $$ 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 $$ {g(u,v) \over \sqrt{g(u,u)} \sqrt{g(v,v)}} = {(f^* h)(u, v) \over \sqrt{(f^* h)(u, u)} \sqrt{(f^* h)(v, v)} $$ 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, $f^* h = s g$ for some scalar field $s$ .