Regular surfaces

[Tsakanikas Nickos]

Consider the sets

• \( \displaystyle S_{1} = \left\{ (x,y,z) \in \mathbb{R}^{3} \Big| x^2 + y^2 \leq 1 , z=0 \right\} \)
• \( \displaystyle S_{2} = \left\{ (x,y,z) \in \mathbb{R}^{3} \Big| x^2 + y^2 = z^2 \right\} \)
• \( \displaystyle S_{3} = \left\{ (x,y,z) \in \mathbb{R}^{3} \Big| -x^2 -y^2 +z^2 = 1 \right\} \)

Which of these are regular surfaces? Justify your answers.

[Papapetros Vaggelis]

For \(\displaystyle{S_3}\) : We define \(\displaystyle{f:\mathbb{R}^3\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x,y,z)=-x^2-y^2+z^2-1}\) .

The function \(\displaystyle{f}\) is smooth on \(\displaystyle{\mathbb{R}^3}\) and we observe that \(\displaystyle{S_{3}=f^{-1}\left(\left\{0\right\}\right)}\) .

Also, \(\displaystyle{f(0,0,1)=0\iff 0\in f(\mathbb{R}^3)}\) . For each \(\displaystyle{\left(x,y,z\right)\in\mathbb{R}^3}\) holds :

\(\displaystyle{\dfrac{\partial{f}}{\partial{x}}(x,y,z)=-2\,x\,\,,\dfrac{\partial{f}}{\partial{y}}(x,y,z)=-2\,y\,\,,\dfrac{\partial{f}}{\partial{z}}(x,y,z)=2\,z}\) with

\(\displaystyle{\dfrac{\partial{f}}{\partial{x}}(x,y,z)=\dfrac{\partial{f}}{\partial{y}}(x,y,z)=\dfrac{\partial{f}}{\partial{z}}(x,y,z)=0\iff \left(x,y,z\right)=\left(0,0,0\right)}\)

but \(\displaystyle{\left(0,0,0\right)\notin f^{-1}\left(\left\{0\right\}\right)}\) since \(\displaystyle{f(0,0,0)=-1\neq 0}\) . So, the number \(\displaystyle{0}\) is

a regular value of \(\displaystyle{f}\) and \(\displaystyle{S_{3}=f^{-1}\left(\left\{0\right\}\right)}\) is a regular surface of \(\displaystyle{\mathbb{R}^3}\) .

For \(\displaystyle{S_{2}}\) : Suppose that \(\displaystyle{S_{2}}\) is a regular surface of \(\displaystyle{\mathbb{R}^3}\) . Then, in a neighborhood of

\(\displaystyle{\left(0,0,0\right)\in S_{2}}\), this surface is a graph-surface and takes one of the following forms :

\(\displaystyle{z=f(x,y)\,,x=g(y,z)\,,y=h(x,z)}\). If \(\displaystyle{x=g(y,z)}\) or \(\displaystyle{y=h(x,z)}\), then restricted to the \(\displaystyle{yz\,,xz}\) planes,

we have \(\displaystyle{z=\pm \left|y\right|}\) or \(\displaystyle{z=\pm \left|x\right|}\), and these functions are not one to one. If \(\displaystyle{z=f(x,y)}\), then

\(\displaystyle{z=\pm \sqrt{x^2+y^2}\,,x\,,y\in\mathbb{R}}\) and this function is not differentiable at \(\displaystyle{\left(x,y\right)=\left(0,0\right)}\).

Therefore, \(\displaystyle{S_{2}}\) is not a regular surface.

For the set \(\displaystyle{S_{1}}\) i am not so sure. I am waiting for your opinion.


I would like to ask a question in the same spirit to this one you proposed here :

Prove that \(\displaystyle{S_{3}}\) is not a connected regular surface.

[Tsakanikas Nickos]

The set \( \displaystyle S_{1} \) is not a regular surface. However, it is not clear to me either that this is the case! Here is one attempt to prove this:

Suppose that \( \displaystyle S_{1} \) is a regular surface and consider a point \( \displaystyle p \in \partial S_{1} \). Then, in a neighborhood of \( \displaystyle p \), \( \displaystyle S_{1} \) is the graph of a differentiable function \( \displaystyle f \) defined on an open subset of \( \displaystyle \mathbb{R}^2 \) and having one of the following forms \( \displaystyle z=f(x,y), y=f(x,z), x=f(y,z) \). Because the projections of \( \displaystyle S_{1} \) over the \( \displaystyle xz \) and \( \displaystyle yz \) planes are clearly not one-to-one, \( \displaystyle f \) is of the form \( \displaystyle z=f(x,y) \). We now observe that \( \displaystyle f \equiv 0 \) and therefore the domain of \( \displaystyle f \) cannot be an open subset of \( \mathbb{R}^2 \). We have thus reached a contradiction.