Let $\phi: S \to \bar{S}$ be a diffeomorphism between two surfaces in $\mathbb{R^3}$. Such a map is called conformal if for all $p \in S$, and $v_1, v_2 \in T_p(S)$ (the tangent plane) we have
$$\langle d\phi_p(v_1), d\phi_p(v_2) \rangle = \lambda^2 \langle v_1, v_2 \rangle_p$$
for some nowhere-zero function $\lambda$.
$\phi$ is said to be angle-preserving, if
$$\cos(v_1, v_2) = \cos(d\phi_p(v_1), d\phi_p(v_2)),$$
which I take to mean
$$\frac{\langle v_1, v_2\rangle}{\lVert v_1 \rVert \lVert v_2 \rVert} =
\frac{\langle d\phi(v_1), d\phi(v_2)\rangle}{\lVert d\phi(v_1) \rVert \lVert d\phi(v_2) \rVert}
$$
From do Carmo, "Differential Geometry of Curves and Surfaces", 4.2/14:
Prove that $\phi$ is locally conformal if and only if it preserves angles.
The "only if" part is obvious, but how can the "if" portion be proved (i.e. how does preserving angles imply conformality)?
Best Answer
Let $e_1$, $e_2$ be an orthonormal basis of $T_{p}S$. Let:
\begin{align*} \langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \\ \langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \mu \\ \langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_2 \end{align*}
Now take:
\begin{align*} v_1 &= e_1 \\ v_2 &= \cos\theta\ e_1 + \sin\theta\ e_2 \end{align*}
The equation in your question implies that:
$$ \cos\theta = \frac{\lambda_1 \cos\theta + \mu \sin\theta}{\sqrt{\lambda_1\left(\lambda_1\cos^2\theta + 2\mu\sin\theta\cos\theta + \lambda_2\sin^2\theta\right)}} $$
Take $\theta = \frac{\pi}{2}$ to get $\mu = 0$. This implies that:
$$ \lambda_1 = \lambda_1 \cos^2\theta + \lambda_2\sin^2\theta $$
Or $\lambda_1 = \lambda_2$. Hence:
\begin{align*} \langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \langle e_1, e_1 \rangle_{p} \\ \langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_2, e_2 \rangle_{p} \\ \langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_1, e_2 \rangle_{p} \qquad (= 0) \end{align*}
Since both $\langle, \rangle_{p}$ and $\langle d\phi_{p}(), d\phi_{p}() \rangle$ are bilinear forms, the above is true for all $v_1, v_2 \in T_{p}S$.