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


1 Answer 1


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.

  • $\begingroup$ Don't you mean G = E = 1? $\endgroup$
    – Chengdu
    Sep 16, 2019 at 9:43
  • $\begingroup$ Rescaling all vectors by the same amount doesn't change the angles between them. ($x \mapsto 2x$ is conformal.) $\endgroup$
    – Chappers
    Sep 16, 2019 at 10:04
  • $\begingroup$ sorry my mistake thank you for the detailed answer I appreciate it $\endgroup$
    – Chengdu
    Sep 16, 2019 at 10:25
  • $\begingroup$ Hello I hope you don't mind me asking you again about this fact. I have been studying geometry now for 3-4 weeks and this fact keeps being useful again and again, can you think of any literature which proves this statement? $\endgroup$
    – Chengdu
    Sep 26, 2019 at 12:57
  • 1
    $\begingroup$ I've had a look through half-a-dozen differential geometry texts, and they all define a conformal metric to be one of the form $\Omega^2 (du^2+dv^2)$. For example, do Carmo's Differential Geometry of Curves and Surfaces (p. 299). See also p. 204 for the definition and calculation of mean curvature in "isothermal" coordinates, which are the same thing. $\endgroup$
    – Chappers
    Sep 26, 2019 at 14:17

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