degenerating immersion - MathOverflow

degenerating immersion

2011-11-14T18:44:15Z

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 by Anton Petrunin for degenerating immersion

2011-11-15T02:39:27Z

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 by Nikita Kalinin for degenerating immersion

2011-11-15T19:53:01Z

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 by Sergey Melikhov for degenerating immersion

2011-11-15T20:00:03Z

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 by Sergey Melikhov for degenerating immersion

2011-11-16T13:22:42Z

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).