Milnor surprisingly found an immersion of a circle in the plane which bounds two "incompatible" immersed disks (see [1] for example, picture included). I think I can glue these two disks together to define an immersion of a sphere $f:S^2\to\mathbb{R}^3$ where Milnor's immersed circle (the equator) sits in the $xy$-plane and contracts to a point (in the $\pm z$ directions) by running along either immersed disk.

Does $f(S^2)$ bound an immersed ball, i.e. does $f$ extend to an immersion of a 3-ball?

This question was sparked by glancing at Gromov's book [1] and a related paper of Eliashberg-Mishachev which argues that the projection $S^2\to\mathbb{R}^2$ (composing $f$ above with projection onto $xy$-plane) is not homotopic through folded maps to the "standard" folded map $(x,y,z)\mapsto (x,y)$ on the unit 2-sphere (the fold being the equator).


1 Answer 1


If one assumes a couple of simple conditions on $f$, namely that the restriction of $z$ on $f(S^2)$ is a Morse function with two critical points and that projections of two halves of $f(S^2)$ ($z>0$, $z<0$) onto the plane $z=0$ realize two original different immersions then $f$ can not be extended to an immersion of the ball.

Indeed, assume by contradiction that such an immersion $F:B^3\to \mathbb R^3$ exists. Let $m<0$ be the minimum of $z$ on $f(S^2)$ and $M>0$ be the max. In this case the preimages of planes $z=c$ under $F$ will be $2$-disks in $B^3$ for all $c\in (m,M)$. And by continuity, (changing $c$ from $M$ to 0, or from $m$ to $0$) we will get that the map $F$ applied to the disk $F^{-1}(z=0)$ realizes both immersions, which is absurd.

One can relax the two conditions by requiring that there is a regular function $z'$ on $\mathbb R^3$ such that $z'=0$ is the plane $z=0$ and $z'$ restricts to $f(S^2)$ as a Morse function with two critical points. If I understand correctly this condition is implicit in your construction of $f$. (One should express what it means that the two halves of $S^2$ realize two different immersions of a disk.)

PS. To clarify the above reasoning I'll give a proof in the most restrictive case. This case already contains the main idea. So the setting will be the following

Setting. Let $S^2$ be the unit $2$-sphere, $S^1\subset S^2$ be the equator, $O_+$ be the north pole and $O_-$ the south. Let $S^2_+$ be the open upper half-sphere and $S^2_-$ the lower one. Let us assume that the immersion $f: S^2\mathbb\to \mathbb R^3$ has the following properties.

1) The restriction of $f$ to the equator $S^1\subset S^2$ realizes the Milnor circle.

2) Function $z$ restricts to $f(S^2)$ as a Morse function with two critical points so that $\max_z f(S^2)=1$, $\min_z f(S^2)=-1$ and moreover the poles are sent to critical points $z(f(O_+))=1$, $z(f(O_-))=-1$.

3) Let $\pi: (x,y,z)\to (x,y)$ be the projection. Then the maps $\pi\circ f: S^2_+\to \mathbb R^2$ and $\pi\circ f: S^2_-\to \mathbb R^2$ realize the two Milnor immersions.

So, suppose all these conditions hold. Then each set $F^{-1}(z=c)$ is a disk $D_c\subset B^3$ for $c\in (-1,1)$. Consider the family of immersed disks $\pi(F(D_c))\subset \mathbb R^2$ for $c$ varying from $1$ to $0$. I claim that this family of immersed disks is the same as the following second family of immersed disks:

Second family. Let $D_c^+\subset S_+^2$ be the disk composed of points with $z\ge c$. Then we can consider the family of immersed disks $\pi(F(D_c^+))$.

So the claim is that the two families $\pi(F(D_c))$ and $\pi(F(D_c^+))$ coincide.

But then, the immersion $\pi(F(D_0))$ is the same as $\pi(F(D_0^+))=\pi(F(S_+^2))$.

Just in the same way we prove that $\pi(F(D_0))$ is the same as $\pi(F(S_-^2))$. This contradicts 3).


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