Question: suppose $M\subset \mathbb{R}^3$ is a minimal surface (mean curvature $H \equiv 0$), show that after conformal change $$\tilde{g}=-Kg$$
the Gaussian curvature $\tilde{K}\equiv 1$.
Since $H \equiv 0$ implies $k_1=-k_2$, where $k_1,k_2$ are principal curvature and we assume $k_1\geq k_2$, we get $K=-k_1^2<0$, so $\tilde{g}=-Kg$ is a conformal change indeed.
I try to verify above statement by direct computation via moving frame:
Let $\{\omega^i\},\{\tilde{\omega}^i\}$ be orthogonal coframes, and suppose $g = (\omega^1)^2 + (\omega^2)^2$, and $\tilde{g} = (\tilde{\omega}^1)^2 + (\tilde{\omega}^2)^2$, then $$\tilde{\omega}^i=\sqrt{-K}\omega^i$$ Also, we suppose connection 1-form $\omega_1^2 = p\omega^1+q\omega^2$.
\begin{align*}
d \tilde{\omega}^1 &= d(\sqrt{-K}\omega^1)=-(\sqrt{-K})_2 \omega^1 \wedge \omega^2+\sqrt{-K}d\omega^1\\
&=-(\sqrt{-K})_2 \omega^1 \wedge \omega^2+\sqrt{-K}\omega^2\wedge\omega_2^1\\
&=-(\sqrt{-K})_2 \omega^1 \wedge \omega^2+p\sqrt{-K} \omega^1\wedge \omega^2
\end{align*}
\begin{align*}
d \tilde{\omega}^2 &= d(\sqrt{-K}\omega^2)=(\sqrt{-K})_1 \omega^1 \wedge \omega^2+\sqrt{-K}d\omega^2\\
&=(\sqrt{-K})_1 \omega^1 \wedge \omega^2+\sqrt{-K}\omega^1\wedge\omega_1^2\\
&=(\sqrt{-K})_1 \omega^1 \wedge \omega^2+q\sqrt{-K}\omega^1\wedge\omega^2
\end{align*}
where $d(\sqrt{-K})=(\sqrt{-K})_1 \omega^1 + (\sqrt{-K})_2 \omega^2 $.
Then according to structure equation: $d\omega^i=\omega^j\wedge\omega_j^i$ and $d\tilde{\omega}^i=\tilde{\omega}^j\wedge\tilde{\omega}_j^i$, we get
$$
\tilde{\omega}_2^1 = \omega_2^1 + \frac{(\sqrt{-K})_2}{\sqrt{-K}}\omega^1 – \frac{(\sqrt{-K})_1}{\sqrt{-K}}\omega^2
$$
Take exterior differential
\begin{align*}
d\tilde{\omega}_2^1 &= d \omega_2^1-\left[\left(\frac{(\sqrt{-K})_1}{\sqrt{-K}}\right)_1+ \left(\frac{(\sqrt{-K})_2}{\sqrt{-K}}\right)_2 \right]\omega^1\wedge \omega^2 \quad \quad \quad \quad (1)
\end{align*}
By Cartan structure equation:
$$\Omega_2^1 = d \omega_2^1 = R_{1212}\omega^1 \wedge \omega^2 = K\omega^1 \wedge \omega^2 \quad \quad \quad \quad (2) $$
Similarly, $$\tilde{\Omega}_2^1 = d \tilde{\omega_2^1 }= \tilde{K}\tilde{\omega}^1 \wedge \tilde{\omega}^2 = -K \tilde{K}\omega^1 \wedge \omega^2 \quad \quad \quad \quad (3) $$
Then combine $(1),(2),(3)$:
$$\tilde{K}=-1 + \frac{1}{K}\left[\left(\frac{(\sqrt{-K})_1}{\sqrt{-K}}\right)_1+ \left(\frac{(\sqrt{-K})_2}{\sqrt{-K}}\right)_2 \right]$$
So the problem converts to verify the following PDE of Gaussian curvature $K$ is true:
\begin{equation}
\frac{1}{K}\left[\left(\frac{(\sqrt{-K})_1}{\sqrt{-K}}\right)_1+ \left(\frac{(\sqrt{-K})_2}{\sqrt{-K}}\right)_2 \right]=2 \quad \quad \quad (4)
\end{equation}
Since we haven't really use the fact that $M$ is a minimal surface (except for $K<0$), I guess the above equation follows from $M$ is a minimal surface. But, frankly, I have no idea how to preceed.
So I try to compute a concrete example of minimal surface to get some hints, for example, the helicoid
$$
x(u,v)=(a \sinh(v) \cos(u),a \sinh(v) \sin(u),au)
$$
then my computation shows that $K=-\frac{1}{\cosh^2(v)}$, plug into $\frac{1}{K}\left[\left(\frac{(\sqrt{-K})_1}{\sqrt{-K}}\right)_1+ \left(\frac{(\sqrt{-K})_2}{\sqrt{-K}}\right)_2 \right] = 1$ instead of $2$, so I got puzzled. Is my computation for $\tilde{K}$ having something wrong or the question itself having something wrong. Could you please help me with that? Thank you in advance!
Best Answer
$\require{AMScd}$
Yes, you're correct that we need to make use of the fact that we're starting with a minimal surface. So we will work with the full moving frames arsenal. Let $e_1,e_2$ be the principal directions, and let the principal curvatures be $\pm k$ with $k>0$. Then we have $$\omega_3^1 = k\omega^1 \quad\text{and}\quad \omega_3^2 = -k\omega^2.$$ When when you make your conformal metric change you have $$\tilde\omega^1 = k\omega^1 = \omega_3^1 \quad\text{and}\quad \tilde\omega^2 = k\omega^2 = -\omega_3^2.$$ Now the rest just follows from the structure equations. We have \begin{CD}\begin{align*} d\tilde\omega^1 &= \tilde\omega^2\wedge\tilde\omega_2^1 \\ @| & \\ d\omega_3^1 &= \omega_3^2\wedge\omega_2^1 = -\tilde\omega^2\wedge\omega_2^1, \end{align*}\end{CD} and similarly for $\tilde\omega^2$. It follows immediately that $\tilde\omega_2^1 = -\omega_2^1$. Thus, $$\tilde K\tilde\omega^1\wedge\tilde\omega^2 = d\tilde\omega_2^1 = -d\omega_2^1 = -K\omega^1\wedge\omega^2,$$ and since $\tilde\omega^1\wedge\tilde\omega^2 = k^2\omega^1\wedge\omega^2 = -K\omega^1\wedge\omega^2$, we conclude that $\tilde K = 1$, as desired.