The cone is not immersed in $\mathbb{R}^3$

The problem is to show that the cone $z^2= x^2+y^2$ is not an immersed smooth manifold in $\mathbb{R}^3$.

Solutions Collecting From Web of "The cone is not immersed in $\mathbb{R}^3$"

Consider the vector space generated by the velocity vectors at $0$ of all curves passing through the origin: it is of dimension $3$. If the cone were an inmersed submanifold, its dimension should therefore be $3$ and then it would have to have a non-empty interior.

(This is the idea of a proof: you should actually prove every statement I made!)

Recall that a function $f:\mathbb R^2 \to \mathbb R^3$ is called an immersion if $f$ is differentiable and the derivative $Df$ is injective at every point of $\mathbb R^2$. There are then a few problems here, the first being that $z$ is not a function of $x$ and $y$ (it has two values, $-\sqrt{(x^2 + y^2)}$ and $\sqrt{(x^2 + y^2)}$). Setting this aside, you may take the “upper half” of the cone $$z = \sqrt{(x^2 + y^2)}$$

so that now $z$ is a function of $x$ and $y$, however this is still not an immersion since $Dz$ is not defined at the point $(0,0)$. One way to see this is by writing down the definition (using a limit) of $Dz$ and showing that the limit doesn’t exist. However it is easier to see this geometrically. You can make a paper model: cut out a disk with some paper, then cut out a wedge (“pizza slice”) from the disk. Then take the disk with the sliced removed and glue along the edges where you cut out the wedge. You have constructed the image of the function $z$, and you can see that it has a singular nonsmooth point, being the vertex of the cone.