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 $\widetilde{H}(\lambda p)=H(p)/\lambda$.

  • $\begingroup$ 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. $\endgroup$ – Juan Daniel Valdivia Fuentes Dec 4 '19 at 4:16
  • $\begingroup$ How come you don't see that? What's the definition of $K$ and $H$ that you have seen in class then? $\endgroup$ – Ivo Terek Dec 4 '19 at 4:17
  • $\begingroup$ Let $K_1,K_2$ the principal curvatures then $K=K_1 K_2$ and $H=\dfrac{K_1 + K_2}{2}$ $\endgroup$ – Juan Daniel Valdivia Fuentes Dec 4 '19 at 4:22
  • $\begingroup$ 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 $k_2$. $\endgroup$ – Ivo Terek Dec 4 '19 at 4:25
  • 1
    $\begingroup$ Right, thanks for your help! $\endgroup$ – Juan Daniel Valdivia Fuentes Dec 4 '19 at 4:28

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