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Let $\varphi : \mathbb R^2 \longrightarrow \mathbb R^3$ the map defined by $\varphi (x_1,x_2)=(x_1,2x_1,x_1^2+x_2)$. We considerer the surface of $\mathbb R^3$ $S=\varphi(\mathbb R^2)=\{(x_1,2x_1,x_1^2+x_2)\in \mathbb R^3\mid x_1,x_2\in \mathbb R\}$. Then $(S, \{(\mathbb R^2, \varphi\})$ is a differentiable manifold with $\varphi$ a global parametrization.

We considerer now the riemannian metric $\langle~,~\rangle\mid_S$ induced by the scalar product of $\mathbb R^3$.

I have to show if $(S,\langle~,~\rangle\mid_S)$ is geodesically complete.

My attempt

I have calculate the matrix of $\langle~,~\rangle\mid_S$ respect the basis $\mathcal B_{\varphi (x_1,x_2)}=\left\{\left(\dfrac{\partial \varphi}{\partial x_1}\right)_{\varphi(x_1,x_2)},\left(\dfrac{\partial \varphi}{\partial x_2}\right)_{\varphi(x_1,x_2)}\right\}$ of $T_{\varphi(x_1,x_2)}S$: $$\hat G_{\varphi(x_1,x_2)}=\begin{pmatrix}5+4x_1^2 & 2x_1\\ 2x_1 & 1\end{pmatrix}$$

I have thought that it could be proved by the Hopf-Rinow Theorem, proving that $(S, d_S)$ si a complet metric space ($d$ is the distance induced by $\langle~,~\rangle\mid_S$). I have seen this proof in other question. I would like to do the same thing.

  • The fact that $d_S(\varphi(x_1,x_2),\varphi (y_1,y_2))\geq d_{\mathbb R^3}(\varphi(x_1,x_2),\varphi (y_1,y_2))$ I think it would be proof by the same way.
  • The fact that $p\in S$ I think is by the same, because $S$ is closed in $\mathbb R^3$.
  • The fact that $p_n\rightarrow p$ in $(S, d_S)$ I don't know how to prove it.

Can you, please, guide me?

Didier
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GoRza
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1 Answers1

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The third point would follow from the fact that $S$ is a topological subspace of $\mathbb{R}^3$ and the fact that the metric topology induced by $d_S$ coincides with the original topology of $S$.

Let $U\subset S$ be an open set in $S$. We want to show that for large $n$, $p_n\in U$. Since $S$ is a subspace of $\mathbb{R}^3$, $U=V\cap S$ for some open set $V\subset \mathbb{R}^3$. Since $p_n\to p$ in $\mathbb{R}^3$ there is $N\in \mathbb{N}$ such that $n\ge N$ implies $p_n\in V$ and since $\{p_n\}\subset S$ we have $p_n\in V\cap S=U$ for $n\ge N$.

Adam T
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  • But the original topology in $S$ is not the topology hereditary of $\mathbb R^3$, is $T={A\subset S\mid \varphi^{-1}(A) \text{is open in }\mathbb R^2}$ – GoRza Jul 08 '21 at 08:28
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    The map $\varphi$ is a topological embedding so the subspace topology on $S$ coincides with the topology you have listed. – Adam T Jul 08 '21 at 10:57
  • Why you know $\varphi $is a embending? Every global parametrization is an embending? – GoRza Jul 08 '21 at 12:26
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    A topological embedding is a map which is a homeomorphism onto its image. In this case $\varphi$ is continuous as a map into $\mathbb{R}^3$ since its components are polynomials and restricting its codomain to its image, $S$, does not change this. It also has an inverse, $\varphi^{-1}(x,y,z)=(x,z-x^2)$, which is continuous as the restriction to $S$ of a continuous map. – Adam T Jul 08 '21 at 13:59
  • Thanks! I had problem thinking in continuity as with the topology of the surface. But I can seeit now – GoRza Jul 08 '21 at 22:49