# Relation between the Gaussian curvature

Let $$f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$$ defined by $$f(v)=\lambda v$$ where $$\lambda >0$$ is constant. Consider a surface $$S \subseteq \mathbb{R}^3$$ and $$S'=f(S)$$. Find a relation between $$K$$ of $$S$$ and $$K'$$ of $$S'$$ where $$K$$ is the Gaussian curvature.

$$\textbf{My attempt : }$$ Let a parametrization of $$S$$, $$\varphi(u,v) : U_0 \subseteq \mathbb{R}^2 \rightarrow S$$. We have a parametrization of $$S'$$ defined by : $$\phi = f \circ \varphi : U_0 \rightarrow S'$$ (because $$f$$ is a diffeomorphism).

So :

$$\phi(u,v) = (f\circ \varphi) (u,v) = \lambda \varphi (u,v)$$

$$\implies \phi_u = \lambda \varphi_u \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_v=\lambda \varphi_v$$ $$\implies \phi_{uu} = \lambda \varphi_{uu} \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_{vv}=\lambda \varphi_{vv} \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_{uv}=\lambda \varphi_{uv}$$

Finally :

$$E' = \langle \phi_u , \phi_u \rangle = \lambda^2 \langle \varphi_u , \varphi_u \rangle = \lambda^2 E$$ and $$L' = \langle \phi_{uu} , n \rangle = \lambda \langle \varphi_{uu} , n \rangle = \lambda L$$

Analogous :

$$F'=\lambda^2 F, G'= \lambda^2 G,M'=\lambda M,N' =\lambda N$$

$$\implies K' = \dfrac{L'N'-(M')^2}{E'G' - (F')^2} = \dfrac{\lambda^2(LN-M^2)}{\lambda^4(EG-F^2)}=\dfrac{1}{\lambda^2}K$$

It´s right? My question is what would happen if I took an arbitrary parameterization of $$S'$$?

No need to use ugly coordinate formulas. If $$N$$ is a normal Gauss field for $$S$$, then a normal Gauss field $$\widetilde{N}$$ for $$\lambda S$$ is given by $$\widetilde{N}(\lambda p) = N(p)$$. The chain rule then gives that $${\rm d}\widetilde{N}_{\lambda p} = (1/\lambda){\rm d}N_p$$. It follows that $$\widetilde{K}(\lambda p)=K(p)/\lambda^2$$ and $$\widetilde{H}(\lambda p)=H(p)/\lambda$$.

• Oh thanks! , the problem is that even in class we don't see that. But your solution is very short, so I'll read what you've put there.
– user411479
Dec 4, 2019 at 4:16
• How come you don't see that? What's the definition of $K$ and $H$ that you have seen in class then? Dec 4, 2019 at 4:17
• Let $K_1,K_2$ the principal curvatures then $K=K_1 K_2$ and $H=\dfrac{K_1 + K_2}{2}$
– user411479
Dec 4, 2019 at 4:22
• Right, so $K(p)=\det(-{\rm d}N_p)$ and $H(p)={\rm tr}(-{\rm d}N_p)/2$. Then you apply $\det$ and ${\rm tr}$ on both sides of ${\rm d}\widetilde{N}_{\lambda p}=(1/\lambda){\rm d}N_p$ to get the correct formulas. You don't even need to waste time finding $k_1$ and $k_2$. Dec 4, 2019 at 4:25
• Right, thanks for your help!
– user411479
Dec 4, 2019 at 4:28