# Computing parametrizations for a differentiable $2$-manifold with boundary

Consider the following subset of $\mathbb{R}^{3}$ $$C=\{(x,y,z)\in\mathbb{R}^{3}\:|\:0\leq x\leq 1,\:0\leq y\leq 1,\:z=x^{2}+y^{2} \}.$$ Intuitively, this looks like a differentiable $2$-manifold with boundary, the boundary being the points at $z=1$.

I am looking to parametrize $C$ using the least possible charts. I think that I'll need two.

I have tried over and over to find such charts, but since I have to get the boundary, the problem got a little hard. I could come up with something like (for most of the manifold, missing a point) $$f_{1}\colon[0,1)\times[0,1)\to\mathbb{R}^{3}\qquad\qquad f_{1}(u,v)=(1-u,\:1-v,\:1-u^{2}-v^{2})$$

But this one has the issue that $[0,1)\times[0,1)$ is not an open set of $\mathbb{H}^{2}=\{(x,y)\in\mathbb{R}^{2}\:y\geq 0\}.$

Any tips? Thank you all.

EDIT

I feel like my failure could be in assuming that I could do that with only two charts. Since the boundary is kind of circular, it doesn't really look like $\mathbb{H}^{2}$.

• I do not think your boundary is correct. There should be three parts and the one you already have is incorrect. – Jean-François Gagnon Nov 12 '15 at 2:03
• I am very doubtful that this is a manifold with boundary. It looks like a manifold with corners to me. – user99914 Nov 12 '15 at 2:21
• @Jean-FrançoisGagnon Yes, as I wrote in the post, $f_{1}$ is not a chart since the set is not open. What should the boundary be? – Shoutre Nov 12 '15 at 2:27
• @JohnMa What is a manifold with corners? Why you do not think it has boundary? It looks like neighborhoods of points at height $z=1$ locally resemble $\mathbb{H}^{2}$. – Shoutre Nov 12 '15 at 2:28
• It is a bit unclear what you are referring to. At $z = 1$, some $(x, y) \in C$ such that $x^2 + y^2 =1$ are lying in the interior: for example $(1/\sqrt 2, 1/\sqrt 2)$. – user99914 Nov 12 '15 at 2:31

One chart is enough. Take $I=[0,1]$ and $$f:I^2\rightarrow\mathbb{R}^3:(u,v)\mapsto (u,v,u^2+v^2)$$.
• This set $I^{2}$ is not open neither in $\mathbb{R}^{2}$ nor in $\mathbb{H}^{2}$. – Shoutre Nov 12 '15 at 2:24
• I believe that the standard definitions ask for the domains of the charts to be open in $\mathbb{H}^{n}$ or $\mathbb{R}^{n}$, where $\mathbb{H}^{n}$ is given the subspace topology. The set $[0,1)$, for instance, is open in $\mathbb{H}^{1}$, because $[0,1)=(-1,1)$$\cap$$\mathbb{H}^{1}$. And you can tell intuitively that the left side of the set $[0,1)$ looks like the "end" of a "curve with boundary". So open sets of $\mathbb{H}^{n}$ ,even not always they look open in $\mathbb{R}^{n}$ . Boundary is a bad word because it has also a topological meaning which is not the same as the one I mean here. – Shoutre Nov 12 '15 at 2:43