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The hyperbolic space $H^n$, is defined as follows. Consider in $\mathbb{R}^{n+1}$ a hyperboloid $H$ given by the equation $$(x^{n+1})^2-(x')^2=1,$$ where $x'=(x^1,\ldots,x^n)\in\mathbb{R}^n$ and $x^{n+1}>0$. Show that $H^n$ is a differentiable manifold.

My approach: Obviosuly, we can prove this is a submanifold of $\mathbb{R}^{n+1}$. Let $\mathbb{R}^{n+1}$ a smooth manifold and $f:\mathbb{R}^n\to\mathbb{R}$ a smooth funciont on $\mathbb{R}^{n+1}$ define as $f(x)=(x^{n+1})^2-(x')^2-1$, then if we consider the null set of this function $H^n=\{x\in\mathbb{R}^{n+1}: f(x)=0\}$ we can see that $df\neq0$ on $H^n$, so $H^n$ is a submanifold of dimension $n$.

But how can I show that, this space $H^n$ is a differentiable manifold, using charts?

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    $\begingroup$ It's a graph. Any graph is a submanifold. $\endgroup$
    – Deane
    Apr 12, 2021 at 1:34
  • $\begingroup$ Thanks for your answer! But, how can I show that this is a differentiable manifold using local charts? $\endgroup$
    – user873697
    Apr 12, 2021 at 1:41
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    $\begingroup$ You only need one chart for a graph. $\endgroup$
    – Deane
    Apr 12, 2021 at 1:55
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    $\begingroup$ The chart for the graph of $y=f(x)$ is $x \mapsto (x,f(x))$. $\endgroup$
    – Lee Mosher
    Apr 12, 2021 at 2:48
  • $\begingroup$ Thank you very much for the answer, but do you know another way to define a chart directly on this space? Something similar to defining the stereographic projection on the sphere... $\endgroup$
    – user873697
    Apr 12, 2021 at 2:52

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The chart is $x\to x^\prime.$ You don't need stereographic projection.

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