degenerating immersion

Initial question: I would like to know if there exists a sequence of $C^2$ immersions $f_k : S^2 \rightarrow \mathbb{R}^3$ which converge (in the $C^2$ topology) to $z^2$ except on a finite set of points, i.e. $f_k \rightarrow z^2$ in $C^2_{\text{loc}}(S^2\setminus \{ a_1, \dots , a_n \})$.

Above, $S^2$ is identified with $\hat{\mathbb{C}}$, the Riemann sphere. Hence, the function $z^2: \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}} \sim S^2 \subset \mathbb{R}^3$ makes sense. In fact, my question is about any rational function $P/Q$ where $P$ and $Q$ are two elements of $\mathbb{C}[z]$; but we can start with $z^2$ in order to make it clearer.

The problem 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 am unable to show more.

Further questions: Assuming there exists a sequence of immersions $f_k : S^2 \rightarrow \mathbb{R}^3$ satisfying the conditions in the first question, I would then like to know the following:

1. Can the immersions $f_k$ all extend to immersions from the closed ball to $\mathbb{R}^3$?

2. Can the sequence of immersions $f_k$ be chosen to have curvature bounded above?

3. How to produce such a sequence in the case of a general rational function? This problem reminds me of the sphere eversion: there appears to be no topological obstruction but it is hard to construct an explicit map.

I hope this is clear. Thanks in advance for your contribution.

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I agree with jc's question. Perhaps your maps are immersions into ${\mathbb R}^4$ and you are projecting. In that case the projection of $z\mapsto z^2$ has two branch points --- one at 0 and the other at infinity. There is certainly a sequence of immersions that will do this. You can start from a figure 8 in 3-space and change the crossings for example. But you are interested in curvature properties, so I don't really understand the question. – Scott Carter Nov 15 '11 at 0:45
Sorry, i mean $z^2 : \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}} \sim S^2 \subset \mathbb{R}^3$. – Paul Nov 15 '11 at 9:07
I guess it's very interesting, but not quite clear as it is. What "looks very hard topologically" ? What is "possible" in question 1? A counterexample to what, in question 2? And, what's the regularity of your maps? – Pietro Majer Nov 15 '11 at 14:11
I have edited my post in order to make it more clear. – Paul Nov 15 '11 at 14:22
thank you ! – Pietro Majer Nov 15 '11 at 15:01

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.)

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Could you precise your answer, i don't understand what your last argument? is it work with any polynomial $P/Q$ of $\hat{\mathbb{C}}$. – Paul Nov 15 '11 at 9:05
I add one sentance, see also the answer of Sergey Melikhov. – Anton Petrunin Nov 15 '11 at 20:54
To clarify: my current answer is only a little elaboration on Anton's. I first thought that there's a serious gap in Anton's argument, and tried to fill it by a nontrivial argument which turned out to be wrong. Now I see that there was no real gap after all. – Sergey Melikhov Nov 15 '11 at 22:00

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.

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Thank you, it makes the argument more clear. In fact it looks specific to $z^2$, if i have have well understood it won't works for $z^3$ for instance, because in fact i was looking for an answer for any $P/Q$ where $P$ and $Q$ are two element of $\mathbb{C}[z]$. I will edit my post in this sense. – Paul Nov 16 '11 at 10:01
The same argument works for any branched cover $f$ between surfaces that has at least one branch point $f(z)$ of even index. That is, the composition $M\xrightarrow{f}N\subset\Bbb R^3$ is not $C^1$-close to an immersion. To see this, take a small closed curve $S$ in $N$ going around $f(z)$, and then apply the above argument with $S$ in place of the equator (with precision still smaller than the distance from $S$ to $f(z)$). If all branch points of $f$ have odd indexes, I believe $f$ is $C^1$-close (and hence also $C^\infty$-close) to an immersion. I'll consider $f=z^3$ in a separate answer. – Sergey Melikhov Nov 16 '11 at 13:00

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).

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Ok you have a disc whose boundary is $z^3$ and hence you can be $C^\infty$ closed to $z^3$ on $S^2\setminus \{ S,N\})$ which answer to 2) but can you extend your immersion of $S^2$ to an immersion of $B^3$ OR is your sequence of approximation of $z^3$ get it Gaussian curvature bounded from above, i.e. the blow-up are given by necks and there is no pinching, this will answer to 1). – Paul Nov 16 '11 at 14:10
Paul, you're right, on $S^2\setminus\{S,N\}$. I don't think I fully understand what exactly 1) and 2) ask for. – Sergey Melikhov Nov 16 '11 at 14:43
Sergey, 1) and 2) are my initial question in the first post, i can rephrase them as follow:Thanks to your last answer, we know that there exist a sequence of immersion $f_k :S^2 \rightarrow \R^3$ which converge in $C_{loc}^2(S^2\setminus\{S,N\}$ to $z^3$, my question is: is it sill true if we assume one of following additional properties: i)$f_k$ is the restriction of an immersion of $B^3$. ii)the Gaussian curvature of $f_k(S^2)$ is bounded from above. Of course $z^3$ is example but i look for an answer for any branched covering of the sphere of the form $P/Q$. – Paul Nov 16 '11 at 14:53
OK, this makes it clear enough. I have no idea about (ii), and as to (i) it seems not so easy in general (should be doable for one specific map such as $z^3$). Note that every immersion $S^2\to\Bbb R^3$ is regular homotopic to an embedding, and so bounds an immersed $3$-ball in $\Bbb R^3\times [0,\infty)$. There is some theory on which immersed curves in the plane bound immersed surfaces in that plane, see for instance ams.org/journals/tran/1974-187-00/S0002-9947-1974-0341505-0, projecteuclid.org/euclid.ijm/1256049897, projecteuclid.org/euclid.hmj/1150922487. – Sergey Melikhov Nov 16 '11 at 15:30

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.

"Large circles in selfintersection can't disappear in limit." This is of course not true. For instance consider a generic immersion $f$ approximating the composition $\phi:S^1\times S^1\xrightarrow{2\times 1}S^1\times S^1\subset\Bbb R^3$. Such an $f$ ought to have large self-intersection circles (even though $\phi$ doesn't). – Sergey Melikhov Nov 15 '11 at 20:04
(I guess it depends on your linguistic conventions whether $\phi$ in the above comment is said to have "large self-intersection circles", because its self-intersection is a $2$-manifold; what I wanted to say is that whatever you call it, it's just like for the map in question, $S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$.) – Sergey Melikhov Nov 15 '11 at 20:10
Sorry, I don't understand. The composition $S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$ has infinitely many intersection points, in fact every point except for north and south poles has the same image as some other point. – Sergey Melikhov Nov 16 '11 at 0:48