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My problem is:

Prove that the Gauss map of a minimal surface $S$ in the Euclidean space $\mathbb{R}^3$ is conformal.

My intuition tells me that this is true but I don't know how to attack the problem. I've looked at the differential of Gauss map but no ideea, also i don't know how to use the fact that $S$ is minimal.

EDIT: By This question I know that the gauss map is anti-holomorphic. Thus my question becomes, is any anti-holomorphic function a conformal function? I know that this is true for holomorphic functions.

Hurjui Ionut
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    Holomorphic maps and antiholomorphic maps between surfaces are weakly conformal. You need to make sure $df$ is never zero. – user10354138 May 16 '19 at 09:41
  • @user10354138 weakly conformal? I don't know the term. For me, conformal map is just angle-preserving transformation. – Hurjui Ionut May 16 '19 at 09:46
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    Conformal at a point $p$ means the derivative $df(p)$ is invertible and preserves orthogonality (equivalently, preserves angles). Weakly conformal means conformal or $df(p)=0$. You need weakly conformal to deal with the cases such as $z\mapsto z^2$ on $\mathbb{C}$ which is clearly holomorphic, but isn't conformal at $0$. – user10354138 May 16 '19 at 10:03
  • @user10354138 Got you! Can you point me to a good ref for this elementary complex analysis? – Hurjui Ionut May 16 '19 at 10:08
  • Holomorphic functions are analytic, the complex conjugate of an anti-holomorphic function is holomorphic. If $f$ is analytic then (weak) conformality of $f(z)$ or $\bar{f(z)}$ is easy to see – reuns May 25 '19 at 21:51

2 Answers2

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When $N: S\rightarrow \mathbb{S}^2$ is unit out normal, then assume that $S$ is a minimal surface whose principal curvatures are $k,\ -k$ at a point $p$. Hence $dN : T_pS\rightarrow T_{N(p)}\mathbb{S}^2$ is a linear map, i.e. $dN $ is a diagonal matrix $(-k,k)$, which is a composition of reflection and scaling.

HK Lee
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Essentially, you want to know that if $\mathbf x$ is a surface patch of your surface, then $\mathbf x$ and $\mathbf n\circ\mathbf x$ have proportional fundamental form. Say the first fundamental form for $\mathbf x$ is $\begin{bmatrix}E&F\\F&G\end{bmatrix}$. That of $\mathbf n\circ\mathbf x$ is $$\begin{bmatrix}\mathbf n_u\cdot\mathbf n_u&\mathbf n_u\cdot\mathbf n_v\\\mathbf n_u\cdot\mathbf n_v&\mathbf n_v\cdot\mathbf n_v\end{bmatrix}$$ If $S$ is the shape operator, then $\mathbf n_u=-S\mathbf x_u$ so $$\mathbf n_u\cdot\mathbf n_u=S\mathbf x_u\cdot S\mathbf x_u=\mathbf x_u\cdot S^2\mathbf x_u$$ Now, $$S=\begin{bmatrix}a&c\\b&d\end{bmatrix}\implies S^2=\begin{bmatrix}a^2+bc&c(a+d)\\b(a+d)&d^2+bc\end{bmatrix}=\lambda I$$ as $a+d=0$. This completes the proof.

Hope this helps. :>

ultralegend5385
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