Find an atlas for $H=\{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 : x_1+x_2^2=x_3^2+x_4=1\}$

Question

Find an atlas for $H=\{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 : x_1+x_2^2=x_3^2+x_4=1\}$

Let $F:\mathbb{R}^4 \rightarrow \mathbb{R}^2$ s.t. $(x_1,x_2,x_3,x_4) \mapsto (x_1+x_2^2-1,x_3^2+x_4-1)$.

Then, $H=F^{-1}(0,0)$.

Since $DF_a:\mathbb{R}^4 \rightarrow \mathbb{R}^2$, $(x_1,x_2,x_3,x_4) \mapsto (x_1+2a_2x_2,2a_3x_3+x_4)$, is surjective, $H$ is a subvariety of $\mathbb{R}^4$ and its dimension is $4-2=2$.

Hence $H$ admits an atlas.

Is there any easy strategy to give an atlas of such varieties (ie, given by fibers)?

Thanks in advance!

Answer 1

$$\phi_1(u, v)=(u, \sqrt{1-u}, v, \sqrt{1-v})$$ $$\phi_2(u, v)=(u, \sqrt{1-u}, v, -\sqrt{1-v})$$ $$\phi_3(u, v)=(u, -\sqrt{1-u}, v, \sqrt{1-v})$$ $$\phi_4(u, v)=(u, -\sqrt{1-u}, v, -\sqrt{1-v})$$