Question

Hi, I would like to know if it exists a sequence of $C^2$ immersion $f_k: S^2 \rightarrow \mathbb{R}^3$ which converge (in C^2) to $z^2$ except on a finite set of point, i.e $f^k \rightarrow z^2$ in $C^2_{loc}(S^2\setminus { a_1, \dots , a_n })$.

Here $S^2$ is identified to $\hat{\mathbb{C}}$ the Riemann sphere, hence $z^2: \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}} \sim S^2 \subset \mathbb{R}^3$ makes sense. In fact my question is about $P/Q$ where $P$ and $Q$ are two element of $\mathbb{C}[z]$, but we can start with $z^2$ in order to make it more clear.

It looks very hard topologically. For instance if I assume "embedded" instead of "immersed" it is not very difficult to prove that such a sequence doesn't exist. But I can't show more, especially, would like to know if

1) it exists, assuming we have have a sequence of immersion from a ball to $\mathbb{R}^3$ which satisfies the same hypothesis on the boundary or if the curvature if bounded from above?

2) how to produce such a sequence in the general case? In fact looking at a proof it seems to looks like something like the sphere eversion: no topological obstruction but no way to see the effective map.

I hope to be clear, Thanks in advance for your contribution.

Answer 1

There is no such sequence.

For an immersion $f_k\colon \mathbb S^2 \rightarrow \mathbb{R}^3$ (after a small perturbation) the set of self-intersections is formed by some number of closed curves $\gamma_1,,\gamma_2,\dots \gamma_n,$ in $\mathbb R^3$. So any plane which intercets all $\gamma_i$ transversally, has to intersect them at even number of points.

On the other hand the the equator plane say $\Pi$ (or its small perturbation) has to itersect it odd number of times. Indeed, the curves in $f_k^{-1}(\Pi)$ is close to equator $\mathbb S^2$; the turning number of its image in $\Pi$ is $2$; so it has odd number of self-intersections. (This works for if $f_k$ is $C^1$-close to $z^2$ near $\Pi$, which is easy to arrange.)

Answer 2

The answer is no. Two 2-dim smooth immersed in $\mathbb R^3$ objects generically intersect by line, so if intersection is a point then it can be eliminated. But it is clear that near $z^2$ there are no embeddings.

Therefore what do you want it is a immersions with self-intersections as a small circles and these circles collapse to points when $k\to\infty$. But if a selfintersection is a small circle, it can be eliminated too. Large circles in selfintersection can't disappear in limit.

added. Sorry, this answer is about absolutely different problem.

Answer 3

The idea of Anton Petrunin can be made into an accurate proof. One does not need $C^2$ convergence, $C^1$ convergence is enough. That is, I claim that there is no $C^1$ immersion sufficiently $C^1$-close to the composition $\phi:S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$. (By the way, any map $S^2\to\Bbb R^3$ is $C^0$-close to a $C^\infty$ immersion, according to the $C^0$-dense $h$-principle and using that $S^2$ immerses in $\Bbb R^3$.)

Let $f:S^2\to\Bbb R^3$ be a self-transverse map (not necessarily an immersion) that is $C^1$-close to $\phi$. The image of $f$ lies in a tubular neighborhood $S^2\times\Bbb R$ of the image of $\phi$. Consider the composition $\psi:S^2\xrightarrow{f}S^2\times\Bbb R\xrightarrow{\text{projection}}S^2$. It is $C^1$-close to $\phi$, so it is equivalent to $\phi$ by a change of coordinates outside a small neighborhood of the poles (which are the singular points of $\phi$).

So we may assume that, outside of a small neighborhood of the poles, $f$ is a vertical lift of $\phi$ (with respect to the projection $S^2\times\Bbb R\to S^2$). Then, in particular, $f$ sends the equator of $S^2$ into the plane $\Pi$ in $\Bbb R^3$ that contains the equator of $S^2$. This equatorial map is a $C^1$-approximation to the composition $S^1\xrightarrow{\text{double covering}}S^1\subset\Pi$, so it is an immersion and has an odd number of double points. But then the double point set of $f$ cannot be a union of closed curves. So $f$ cannot be an immersion.

Answer 4

New answer to the generalized question. It's shown in previous answers that for $z^2$, and some other branched coverings, there are no immersions that are $C^1$-close except at the branch points. (I believe this should also imply that there are no immersions that are $C^1$-close except on a finite set.)

But $z^3:S^2\to S^2$ is arbitrarily $C^\infty$-close, except at the two branch points, to a $C^\infty$ immersion in $\Bbb R^3$. (Also, any $C^\infty$ map $S^2\to S^2$ that is equivalent to $z^3$ by a $C^0$ change of coordinates is $C^\infty$-close on the entire $S^2$ to an immersion in $\Bbb R^3$). To see this, pick a generic lift $f:S^1\to S^1\times\Bbb R$ of the $3$-fold covering $S^1\to S^1$. It suffices to show that the composition $f':S^1\xrightarrow{f} S^1\times\Bbb R\subset S^2$ bounds an immersion of a $2$-disk in a $3$-ball. Equivalently, we want to find a regular homotopy from $f'$ to an embedding. But it is an exercise that that there are only two regular homotopy classes of immersions $S^1\to S^2$, distinguished by the parity of the number of double points (in the case of self-transverse immersions).