How we show that $Z := \{(x, y, z) ∈ R^3 | x^2 + y^2 − z^2 = 0\}$ is not a manifold (even it is not a topology manifold)

Question

can anyone help me to prove that $Z := \{(x, y, z) ∈ \mathbb{R}^3 | x^2 + y^2 − z^2 = 0\}$ is not a manifold (even it is not a topology manifold)?

Answer 1

Notice the following: Which are the open neighborhoods of the point (0,0,0)? If $A=\{(x, y, z) \in\mathbb{R}^3 | x^2 + y^2 − z^2 = 0, z>0\}$ $B=\{(x, y, z) \in\mathbb{R}^3 | x^2 + y^2 − z^2 = 0, z<0\}$ Then $Z=A\cup B\cup \{ (0,0,0)\}$

Let $B_Z$ be an open nhood of $Z$ Then $B_Z-\{(0,0,0)\}$ is disconnected. That means that $B_Z$ can be disconnected by removing one single point. So if it isomorphic to some open set of some $\mathbb{R}^n$ is must be some interval of $\mathbb{R}$. But an interval can be disconnected by removing ANY point (at least an open one), while $B_Z$ remains connected if we remove any other point that is not the origin

Answer 2

$Z\setminus\{(0,0,0)\}$ is no longer connected. With a manifold, this can happen only if it is 1-dimensional. But looking at a nighbourhood of $(1,0,1)$, say, $Z$ "should" be two-dimensional.

Answer 3

The idea for this sort of thing (where it's still low-dimensional enough) is to try and draw it, and see where the problem is on the drawing, and what that problem corresponds to. Once you see that, you can do a formal, precise argument by basing what you're looking for on the drawing (although of course the drawing itself isn't enough !)

To draw it, notice that if you fix $z$, then the "$(x,y)$-slice" is a circle of radius $\sqrt {z^2}$, so your "surface" is just a bunch of circles moving along the $z$ axis, with radius moving "linearly", so it should look something like

Now if you stare at this you'll think it looks like a manifold everywhere, except at the $z=x=y=0$ point, so that's what you have to analyse. Now what's wrong with this specific point ?

Well if you take a small ball around it and intersect it with the surface, it still looks like the surface itself (a smaller version). So the point is to see that the surface itself isn't homeomorphic to $\mathbb R^n$ for any $n$. And why is that ?

Well that's where the formal argument comes from, and now you can just pick one of the arguments that work; the one I would use is that if you take a small ball around the point, intersect it with the surface and remove the point, you get two connected components ! The only $n$ such that $\mathbb R^n$ has that property is $n=1$.

But now it's also clear that the surface is connected (if it's not clear to you, try to prove that it's path-connected), so if there was a $1$-dimensional neighbourhood around this $(0,0,0)$ point, and the surface was a manifold, then every point would have a $1$-dimensional neighbourhood. Now you can simply look at any other point and see that it actually has a $2$-dimensional neighbourhood (with the usual tricks, e.g. the differential of $x^2+y^2-z^2$ is a submersion at other points), which is a contradiction.

Or you can try to find another proof for why the neighbourhood of $(0,0,0)$ is not homeomorphic to $\mathbb R$ (there are probably many arguments)