Proof that angle-preserving map is conformal

Question

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)?

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

Answer 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 $a=0$.