In terms of the first two fundamental forms the mean curvature H, can be expressed as
\begin{align*}
H = \frac{LG – 2MF + NE}{2(EG – F^2)}
\end{align*}
Where the first fundamental form is on the form $Edu^2 + 2Fdudv + Gdv^2$
with $E = ||\sigma_{u}||^2$, $F = \sigma_{u} \cdot \sigma_{v}$, $G = ||\sigma_{v}||^2$ for some surface patch $\sigma(u,v)$
Now if $\sigma$ is conformal the mean curvature can be written as
\begin{align*}
H = \frac{L+ N}{2E}
\end{align*}
I understand that a conformal mapping is angle preserving, but I am not sure why the mean curvature $H$ can be expressed like that when $\sigma$ is conformal
Best Answer
The angle between two vectors $x,y$ is $\arccos{( (x \cdot y ) / \sqrt{(x \cdot x) (y \cdot y)} )}$. In particular, $x \cdot y = 0 $ if and only if $x$ and $y$ are perpendicular.
Conformal means that the angle between $a e_1+b e_2$ and $c e_1 + d e_2$ is the same as that between $ a \sigma_u + b \sigma_v $ and $ c \sigma_u + d \sigma_v $, for any $a,b,c,d$. Also, $$ (a \sigma_u + b \sigma_v) \cdot (c \sigma_u + d \sigma_v) = ac E + (ad+bc) F + bd G . $$ We can look at two special cases of this, where the original vectors are perpendicular, so that this should be $0$ and we don't need the denominator in the cosine.
Suppose first that we have $e_1$ and $e_2$, so $a=1,b=0,c=0,d=1$. Then $e_1 \cdot e_2 = 0$, so the original vectors are perpendicular. The transformed vectors are $\sigma_u$ and $\sigma_v$, so the transformed product is $\sigma_u \cdot \sigma_v = F$. Therefore we must have $F=0$.
Now take $e_1+e_2$ and $e_1-e_2$. Then $(e_1+e_2) \cdot (e_1-e_2) = 1 + 0 + 0 - 1 = 0$, so these are also perpendicular. But $$ ( \sigma_u + \sigma_v ) \cdot ( \sigma_u - \sigma_v ) = E + F - F - G = E - G , $$ so for these to be perpendicular, we must also have $E=G$. (Therefore a conformal first fundamental form must have the form $ \Omega^2 (du^2+dv^2) $ for some positive function $\Omega$.)
Putting $F=0$ and $G=E$ in the formula for the mean curvature gives the result.