# How could we show that these are perpendicular?

I am looking at the following exercise:

Suppose that the first fundamental form of a surface patch $\sigma (u, v)$ is of the form $E(du^2 + dv^2)$.

Prove that $\sigma_{uu} + \sigma_{vv}$ is perpendicular to $\sigma_u$ and $\sigma_v$.



From the first fundamental form we have that $G=E$ and $F=0$.

We also have that $\sigma$ is conformal.



Could you give me a hint how we could show that $\sigma_{uu} + \sigma_{vv}$ is perpendicular to $\sigma_u$ and $\sigma_v$ ?

You know $$\sigma_u\sigma_v = 0$$ which implies $$\sigma_{uu}\sigma_{v}+\sigma_{u}\sigma_{uv}=0\\ \sigma_{vu}\sigma_{v}+\sigma_{u}\sigma_{vv}=0$$

You also know $$\sigma_u^2=\sigma_v^2$$ which implies $$2\sigma_{uv}\sigma_u=2\sigma_{vv}\sigma_{v}\\ 2\sigma_{uu}\sigma_u=2\sigma_{uv}\sigma_{v}$$

Insert the first two implications into the second to arrive at

$$\sigma_u(\sigma_{uu}+\sigma_{vv})=0\\ \sigma_v(\sigma_{uu}+\sigma_{vv})=0$$

• I understand!! Thanks for your help!! Jan 7, 2016 at 19:17
• After that I have to deduce that the mean curvature $H = 0$ everywhere if and only if the Laplacian $\sigma_{uu} + \sigma_{vv} = 0$. I have done the following:  We have that $$H=\frac{LG-2MF+NE}{2(EG-F^2)}=\frac{LN+NE}{2E^2}=\frac{L+N}{2E}=\frac{\sigma_{uu}\cdot \textbf{N}+\sigma_{vv}\cdot \textbf{N}}{2E}=(\frac{(\sigma_{uu}+\sigma_{vv})\cdot \textbf{N}}{2E}$$ We have $H=0$ if and only if $(\sigma_{uu}+\sigma_{vv})\cdot \textbf{N}=0$. Since $\sigma_{uu}+\sigma_{vv}$ is parallel to $\textbf{N}$, we have $(\sigma_{uu}+\sigma_{vv})\cdot \textbf{N}=0$ is and only if $\sigma_{uu}+\sigma_{vv}$. Jan 7, 2016 at 19:18
• Is everything correct? In that way I hav shown both direction of the "if and only if" statement, right? Jan 7, 2016 at 19:18
• @MaryStar after the second $=$ sign it should be $LG+NE$ in the numerator. And yes, that covers both directions. In general you should open a new question instead of disscussing additional question in the comments. Jan 7, 2016 at 19:42
• Oh sorry, I accidentaly wrote $N$ instead of $E$.  Thank you very much!! :-) Jan 7, 2016 at 19:49