Explicit formula of flattening a boundary of a circle

Question

Let $B(0,1)$ be a unit ball in $\mathbb{R}^{2}$. Then, we know that $\partial B(0,1)$ is smooth. For simplicity, we choose $(0,-1)$ and "flatten" the boundary locally around $(0,-1)$. By definition of smooth boundary, we can find smooth mapping to locally map points around $(0,-1)$ into the upper half of a square $Q:=\{(x',y')\in\mathbb{R}^2\,|\,|x'|\leq 1 \wedge |y'|\leq 1 \}$. Moreover, a portion of $\partial B(0,1)$ (including $(0,-1)$) will be mapped to $\{(x',y')\in\mathbb{R}^2\,|\,|x'|\leq 1 \wedge y'=0 \}$ (boundary of $\partial\mathbb{R}^{2}_{+}$).

My questions is how to find the explicit formula from $B(0,1)$ around $(0,-1)$ to $Q$? Any hint is pretty much appreciated. Thank you!

Answer 1

I don't know where your square comes from, but I suggest you write down precisely what you are visualizing, flattening the arc of the circle into a segment of the line $y=-1$.

Just consider $$f(x,y)=\left(x,y+\sqrt{1-x^2}\right)$$ in a neighborhood of your point $(0,-1)$. The boundary of an open subset of the ball (say with $\|(x,y)-(0,-1)\|<1$) maps to points of the form $(x,0)$ and points inside the ball map above.

Answer 2

$\overline{B(0, 1)}$ is a smooth surface with boundary $\partial B(0, 1) = S^1$. The definition of the boundary of a surface is such that any chart for the surface itself restricts to a chart on the boundary. In this case, take the chart $\phi \colon (0, 1] \times (0, 2\pi) \to \overline{B(0, 1)} \setminus \overline{0(1, 0)}$, $\phi(r, \theta) = r(\cos(\theta), \sin(\theta))$. Then use appropriate diffeomorphism $f \colon (0, 1] \times (0, 2\pi) \to U \subset \mathbb{R}^2_+$ to create a chart $\psi = f \circ \phi^{-1}$ to get the range of $\psi$ to be an open subset of the upper halfspace.