Proof that angle-preserving map is conformal

Solution 1:

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 $v_1, v_2 \in T_{p}S$.

Solution 2:

Since $d\phi$ is a linear map between 2d spaces, the problem boils down to linear algebra. Given a linear map $T\colon \mathbb R^2\to\mathbb R^2$ that preserves angles between vectors, we would like to show that $T$ is a composition of a unitary with dilation.

The most efficient approach may depend on your linear algebra background. My weapon of choice is complex numbers: any linear map $T\colon \mathbb R^2\to\mathbb R^2$ is of the form $z\mapsto az +b\bar z$. The angle-preserving assumption means that $\arg (az+b\bar z)=\phi_0\pm \arg z$ where $\phi_0$ is fixed and the sign in $\pm$ is the same for all $z$. So, either $\arg \frac{az+b\bar z}{z}$ or $\arg \frac{az+b\bar z}{\bar z}$ is constant. In the first case we have $b=0$, in the second $a=0$.