# 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 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. Nov 15, 2011 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, 2011 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? Nov 15, 2011 at 14:11
• I have edited my post in order to make it more clear.
– Paul
Nov 15, 2011 at 14:22

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.)

• 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, 2011 at 9:05
• 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. Nov 15, 2011 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.

• 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, 2011 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. Nov 16, 2011 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).

• 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, 2011 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. Nov 16, 2011 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, 2011 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. Nov 16, 2011 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). Nov 15, 2011 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, $S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$.) Nov 15, 2011 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. Nov 16, 2011 at 0:48