# Show that hyperbolic space is differentiable manifold

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?

• It's a graph. Any graph is a submanifold. Apr 12, 2021 at 1:34
• Thanks for your answer! But, how can I show that this is a differentiable manifold using local charts?
– user873697
Apr 12, 2021 at 1:41
• You only need one chart for a graph. Apr 12, 2021 at 1:55
• The chart for the graph of $y=f(x)$ is $x \mapsto (x,f(x))$. Apr 12, 2021 at 2:48
• 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...
– user873697
Apr 12, 2021 at 2:52

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