# Minimal surfaces under conformal parametrizations - proof verification

Hi I am studying minimal surfaces from Kuhnel's Curves - Surfaces Manifolds

The theorem states that: If $$f$$ is a conformal parametrization, $$f$$ is a minimal surface if and only if the functions $$\phi_{1},\phi_{2}, \phi_{3}$$ are holomorphic.

So he defines a surface element $$f: U \rightarrow \mathbb{R}^3$$ with components $$f = (f_{1},f_{2},f_{3})$$ and defines the map $$\phi(u+iv) = \frac{\partial f}{\partial u}(u,v) - i\frac{\partial f}{\partial v}(u,v)$$, which is in components for $$j = 1,2,3$$ \begin{align*} \phi_{j} (u +iv) = \frac{\partial f_{j}}{\partial u}(u,v) - i\frac{\partial f_{j}}{\partial v}(u,v) \ (*) \end{align*}

For showing the theorem I assume the result: $$f$$ defines a minimal surface $$\iff$$ the three component functions $$f_{1}, f_{2}, f_{3}$$ of $$f$$ have the relation

\begin{align} \frac{\partial^2 f_{i}}{\partial u_{1}^2} + \frac{\partial^2 f_{i}}{\partial u_{2}^2} = 0 \ (**) \end{align}

So all we have to do is take the second derivatives and show that their sum equals zero Now his proof is:

\begin{align*} \frac{\partial^2 f_{k}}{\partial u^2} &= \frac{\partial}{\partial u}(Re \ \phi_{k}), \ \frac{\partial^2 f_{k}}{\partial v^2} = - \frac{\partial}{\partial v}(Im \ \phi_{k}), \\ \frac{\partial^2 f_{k}}{\partial u \partial v} &= \frac{\partial}{\partial v}(Re \ \phi_{k}) = - \frac{\partial}{\partial u} (Im \ \phi_{k}) \end{align*} And from that he concludes that (**) is satisfied

Now firstly I am confused about if he is differentiating (*)? w.r.t. u and v But in that case I can't really make sense of the calculations, any help would be appreciated

• Please do not vandalise posts, even if they are your own. Sep 30 '19 at 18:06
• no reply, then what can I do Sep 30 '19 at 21:06
• $1$) Wait a little longer than $3$ days. If you were asking about elementary calculus then you could expect a near instantaneous response as there as a lot of people who want to answer those questions. Other questions might need some time for the right person to spot them -- not everyone is on the site all the time. $2)$ If you're not happy with the level of response you can add a bounty to the question to draw more attention. $3)$ you can try asking in some of the chat-rooms (there are general maths one) to draw more attention. Vandalising it guarantees you won't get help though. Oct 1 '19 at 6:20

Starting with the definition: $$\phi_k(u,v) = \frac{\partial f}{\partial u}(u,v) - i \frac{\partial f}{\partial v}(u,v)\qquad \mbox{for }k=1,2,3$$ we see that $$\phi_k$$ is being explicitly written as a sum of real and imaginary parts. So, $$\mbox{Re}(\phi_k) = \frac{\partial f}{\partial u}$$ and $$\mbox{Im}(\phi_k) = -\frac{\partial f}{\partial v}$$ So if we differentiate both sides with respect to $$u$$ for the real part and $$v$$ for the imaginary part, we get $$\frac{\partial f}{\partial u}\mbox{Re}(\phi_k) = \frac{\partial^2 f}{\partial u^2}$$ and $$\frac{\partial f}{\partial v}\mbox{Im}(\phi_k) = -\frac{\partial^2 f}{\partial v^2}$$ and for the mixed derivatives (which are equal thanks to us being $$C^1$$) $$\frac{\partial f}{\partial u}\mbox{Im}(\phi_k) = -\frac{\partial^2 f}{\partial v \partial u} = -\frac{\partial^2 f}{\partial u \partial v} = -\frac{\partial f}{\partial v}\mbox{Re}(\phi_k)$$
Now write out the second derivative and use these relations to substitute in -- you'll find that everything cancels and leaves you with $$0$$. All without ever knowing what $$f$$ and $$\phi$$ explicitly look like!