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 $\mathbb{R}^3$.

-
Do you have a question? Your 2nd sentence does not quite make sense. – Ryan Budney May 15 '12 at 22:24
This result is false for dimension 2, but the question is about dimension 3. – checkmath May 15 '12 at 22:29
Sorry but show that something is an immersion is more reasonable, but show that it is not is a totally different approach. – checkmath May 20 '12 at 14:49

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!)

-
I don't see how this contradicts the cone being immersed (it does rule out it being embedded). – user31373 May 15 '12 at 22:56
Mariano, I don't understand this argument. Doesn't the cone $z^2 = x^2 + y^2$ have dimension $2$ at every point except the origin? – treble May 16 '12 at 0:32
I took three curves $t(0,1,1)$; $t(1,0,1)$ and $t(1,1,\sqrt{2})$; they are continuous into the $R^3$ but I could not show about the cone with a topology that is not the induced. – checkmath May 16 '12 at 0:36

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.

-
This only shows that the map $(x,y)\mapsto (x,y,\sqrt{x^2+y^2})$ is not an immersion. What if there is another map of plane onto the cone, which is an immersion? – user31373 May 16 '12 at 0:56
That is not the way I interpreted the OP's question. It is true that I did not show that there does not exist an immersion with image equal to the cone, the construction does however indicate what the problem is. – treble May 16 '12 at 0:58
In fact such an immersion would have to be a local embedding (every immersion is locally one-to-one by the inverse/implicit function theorem) near the origin. One can then show that the cone doesn't have a well-defined $2$ dimensional tangent plane at the origin to get a contradiction, which is perhaps what Mariano was going for. – treble May 16 '12 at 1:04
I'm not convinced. The infinity sign $\infty$ is an immersed smooth manifold (namely, an immersed circle). Self-intersection is not a problem for immersions. – user31373 May 16 '12 at 1:11
That's true; however self-intersections do not play a role here (since the cone has no self-intersections). The problem is that, since the derivative of an immersion has full rank (or is injective, however you like to say it), if we pick a point $p$ in the parameter space and consider a small neighborhood $U$ containing $p$, then the image $f(U)$ under the immersion must have a (in this case) two-dimensional tangent plane at $f(p)$. This is another way of saying that $Df$ has full rank. The vertex of the cone however does not have a $2$ dimensional tangent plane. – treble May 16 '12 at 1:18