Prove a subset is not a submanifold using dimension of the Tangent space

174 Views Asked by At

Let's consider the set : $$\mathcal{A}=\left\{(x,y,z)\in \mathbb{R^3},x^2+y^2-z^2=0\right\}$$ I'm struggling to demonstrate it's not a submanifold of $\mathbb{R^3}$ for the following reasons :

A common way to demonstrate a subset is not a subvariety of global space would be to demonstrate it's tangent space has not the same dimension as $\mathcal{A}$.

Defining the function $f:(x,y,z)\rightarrow x^2+y^2-z^2$ , i would like to say the dimension of the manifold , would be 2, but f does not define a submersion of $\mathbb{R}^3$ in $\mathbb{R}$ as it's differential is equal to the null function in $(0,0,0)$.

But even if i knew the potential dimension of the manifold would be $2$, i don't really understand what property i should use to determine it . To what i could understand , i should be able to show the Tangent space in $0$ of $\mathcal{A}$ is a cone , but i have no idea how to demonstrate it properly.

The reader can see another question on the subject has been asked

Why cone of revolution is not a submanifold?

However , it seems it doesn't use the same criterion i'd like to use for the contradiction.

1

There are 1 best solutions below

2
On BEST ANSWER

Lets supposed for the moment that $\mathcal{A}$ is a submanifold of $\mathbb{R}^3$. Lets consider the tangent space at the point $(1,0,1)\in \mathcal{A}$. At this point, the curves $\alpha(t)=(1+t,0,1+t)$ and $\beta(t)=(\cos t,\sin t,1)$ for $t\in (-\epsilon,\epsilon)$ are tangent to $\mathcal{A}$. The tangent vector associated to $\alpha$ and $\beta$ at the point $p:=(1,0,1)$ is $$ \dot{\alpha}(0)=\frac{\partial}{\partial x}|_p+\frac{\partial}{\partial z}|_p $$ and $$ \dot{\beta}(0)=\frac{\partial}{\partial y}|_p $$ So if $\mathcal{A}$ is a submanifold, its tangent space must be of dimension at least 2. Since $\dim \mathcal{A}\ge 2$, locally $\mathcal{A}$ looks like $\mathbb{R}^k$ for some $k\ge 2$. Hence, if we remove a point from $\mathcal{A}$, $\mathcal{A}$ must remain connected since $\mathbb{R}^k-\{pt\}$ is connected for $k\ge 2$. However, the set $\mathcal{A}-\{(0,0,0)\}$ consists of two components (upper and lower cones). Therefore, $\mathcal{A}$ can't be a submanifold.