Question about conformal maps. By definition, a diffeomorphism $\sigma:(M,g)\to (N,h)$ is called conformal if $\sigma^*h=ug$. Another definition I've seen in other contexts is that conformal maps are ones that preserve angles. Now I've considered maps $\sigma$ from $M$ to $N$ that preserve angles (the angle between vectors being defined by $\frac{g(x,y)}{\sqrt{g(x,x)g(y,y)}}$), and I've found that if $\sigma$ preserves angles, then for $\sigma^*h$ to be equal to $ug$ it must be the case that $\frac{\sigma^*h(x_p,x_p)}{g(x_p,x_p)}$ is independent of the vectors. This isn't obvious to me if it is true, and if it isn't true then I'm wondering why conformal maps are defined as they are, as it would seem like an unnecessarily strong definition.
 A: This is really just a linear algebra/geometry problem: once you fix orthonormal bases for $(T_p M, g_p)$ and $(T_{\sigma p} N, h_{\sigma p})$, you're just talking about the conditions for a matrix $d_p \sigma$ to preserve angles. In particular if you choose your bases so that the matrix is diagonal (i.e. SVD) then the question becomes "which diagonal matrices preserve angles?"
The only such matrices are multiples of the identity. (Otherwise, choose two basis vectors $e_i,e_j$ stretched by different amounts and consider the angle between $e_i$ and $e_i + e_j$.)
A: Here's a physics-style (or 18th-century math style) explanation. It can be brought up to modern standards of mathematical rigor, but I think it's clearer without that. Imagine an infinitesimal triangle. Being infinitesimal, it has essentially straight edges.  It gets mapped to another infinitesimal triangle with the same angles. But triangles with the same angles are similar, so the edges are enlarged or shrunk by the same factor.
