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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)?

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Hint: how would you get $\langle v_1,v_2\rangle/\|v_1\|\|v_2\| = \langle d\phi(v_1),d\phi(v_2)\rangle/\|d\phi(v_1)\|\|d\phi(v_2)\|$ into the form $\langle d\phi(v_2),d\phi(v_2)\rangle = (\mbox{something})^2 \langle v_1,v_2\rangle$? What is "something"? – Neal May 27 '12 at 19:02
@Neal: "something" would be the square root of $\|d\phi(v_1)\|\|d\phi(v_2)\| / \|v_1\|\|v_2\|$. In fact, it seems that $\lambda = \|v\|/\|d\phi(v)\|$ is constant for all $v$, but this is where I got stuck originally... I don't know how to show this. – koletenbert May 27 '12 at 19:23
up vote 2 down vote accepted

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

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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 $a=0$.

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Thanks for your answer. The book from which the exercise was taken is about differential geometry and does not use any complex algebra, so I wonder if this can be shown without complex numbers (preferably just using the properties of the inner product and the differential)? – koletenbert May 27 '12 at 18:35

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