Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?

Solution 1:

Yes.

We can consider a sequence of disks $D_1,D_2,\ldots$ around $(\sqrt 2,\pi)$ such that the $n$th disk has precisely $n$ lattice points in its interior (and none on its boundary). Let $z_n$ be the lattice point in $D_n\setminus D_{n-1}$.

Construct accordingly a sequence $C_1, C_2, \ldots$ of simply connected compact sets with smooth boundary such that $C_{n+1}\setminus C_n$ is an annulus: We let $C_1$ a a small closed disk around $f(z_1)$. Given $C_n$, we can connect $f(z_{n+1})$ with $C_n$ via a straight line segment $\ell_n$ that does not pass through any other lattice point (and ends at its first intersection with $\partial C_n$. Then $C_n\cup \ell_n$ is simply connected, hence the complement is (via some map $\phi_n$) biholomorphic to $\{\,z\in\mathbb C:|z|>1\,\}$. Since almost all lattice points are "close" to $f(\infty)=\infty$, the distance $r_n$ between $f(\mathbb Z^2\setminus C_n)$ and $S^1$ is strictly positive. Let $C_{n+1}=C_n\cup \phi_n^{-1}(\{\,z\in\mathbb C:|z|\le 1+\tfrac {r_n}2\,\})$.

Now to construct the desired homeomorphism $h$:

• Since $C_1$ and $D_1$ are disks, we can readily define $h\colon D_1\stackrel \approx \to C_1$
• Assume we have defined $h\colon D_n\stackrel\approx \to C_n$. To extend this, we need only find a homeomorphism between the closed annuli $D_{n+1}\setminus D^\circ_n$ and $\phi_n(C_{n+1}\setminus C^\circ_n)=\{\,z\in\mathbb C:1\le |z|\le 1+\frac{r_n}2\,\}$ that agrees with what we already have as homeomorphis between the inner boundaries. Viewing both sets as $[0,1]\times S^1$, we use the identity on the first factor and the given homeomorphism between the inner boundaries on the second factor.

The resulting map on $\mathbb R^2=\bigcup D_n$ is the desired homeomorphism if we can ensure that $\mathbb R^2=\bigcup C_n$. Right now I am not sure if the construction above warrants this auttomatically. I suppose that one should be more "greedy", that is not only add $\ell n$ to $C_n$ (and then dilate it) but add something more while still maintaining simple connectivity at that step ...

Based on Henning Makholm's and Jim Belk's excellent comments, here's a complete rewrite of the above:

Lemma. Let $Q\subset \mathbb R^2$ be countable, closed, and discrete. Let $q_1,q_2,\ldots$ be an enumeration of $Q$. Then there exists a homeomorphism $h\colon\mathbb R^2\to \mathbb R^2$ with $f(q_n)=(n,0)$ for all $n\in\mathbb N$.

Proof. Pick a point $a$ that is on none of the countably many lines through two points of $Q$. Then for each $n$, the line segment $\ell_n$ from $a$ to $q_n$ is a compact set disjoint from $Q\setminus\{q_n\}$.

For $n\in\mathbb N$, $r>0$ let $C(n,r)$ be the convex hull of $\overline{B(a;r)}\cup\overline{ B(q_n;r)}$; this is the union of two disks and a rectangle, its boundary consists of two arcs and two line segments, it is star shaped around $a$, and every ray originating at $a$ intersects its boundary in exactly one point.

For $m\in\mathbb N$ let $$ r(n,m) = \min\{\,|q_k-y|:k\ge m,k\ne n, y\in \ell_n\,\},$$ which exists and is positive because $\ell_n$ is compact and $Q$ (minus a finte set) is closed. For $0<r<r(n,m)$ we have $C(n,r)\cap Q\subseteq\{q_1,\ldots,q_{m-1}\}\cup\{q_n\}$. Clearly $m'>m$ implies $r(n,m')\ge r(n,m)$. Also $r(n,m)\to\infty$ as $m\to\infty$ because $r(n,m)>r$ for all $m$ implies that $C(n,r)$contans infinitely many elements of $Q$, contradicting discreteness.

This allows us to pick for each $n$ a sequence $\{\rho_{n,m}\}_{m=n}^\infty$ that is strictly increasing and diverges $\to\infty$ as $m\to\infty$ and such that $\rho_{n,m}<r(n,m)$ holds for all $m\ge n$. Now let $$ C_n=\bigcup_{k=1}^n C(k,\rho_{k,n}).$$ Then $C_n$ is starshaped around $a$, compact, has a useer-friendly boundary consisting of finitely many arcs and line segments, and each ray originating in $a$ intersects the boundary in exactly one point. Moreover, $C_n\subset C_{n+1}^\circ$ and $\bigcup_{n=1}^\infty C_n=\mathbb R^2$. We also have $C_n\cap Q=\{q_1,\ldots, q_n\}$ and $\partial C_n\cap Q=\emptyset$. By linear interpolation between the boundaries we obtain a homeomorphism $h_0\colon \mathbb R^2\to\mathbb R^2$ that maps $a\mapsto 0$ and $\partial C_n\to nS^1$. Note that $h_0(q_n)$ is between $(n-1)S^1$ and $nS^1$. It is a simple task to deform each such annulus (and the central disk) in such a way that the boundary remains untouched and $f(q_n)$ moves to $(n-\tfrac12,0)$. With a final translation by $\frac12$ to the right, we obtain our desired homoeomorphism. $_\square$

Now the original problem is solved by applying the lemma to an enumeration $q_1,q_2,\ldots $ of $\mathbb Z\times Z$ and also to the enumeration $f(q_1),f(q_2),\ldots$.

Solution 2:

Here's an idea on how to construct a solution for the bijection you describe. It appears feasible by a diffeomorphism of the plane.

First we solve the following problem : extend the bijection \begin{array}{RCL} \phi:\Bbb Z\times\lbrace-1,0\rbrace & \longrightarrow & \Bbb Z\times\lbrace-1,0\rbrace\\ (n,\epsilon) & \longmapsto & \begin{cases} (n,\epsilon) & \text{if }\epsilon=0\\ (-n,\epsilon) & \text{if }\epsilon=-1\end{cases} \end{array} to a diffeomorphism of the plane using a vector field $X$ defined as follows: it is supported in the open set $$\bigcup_{n\geq 1}~H_n+D_\frac13$$ where $H_n$ is the bottom half circle with diameter the segment $[(-n,-1),(+n,-1)]$, and $D_\frac13$ is the open disk centered at the origin with radius $1/3$. For every $n\geq 1$, the vector field inside $H_n+D_\frac13$ is chosen so that it switches $(-n,-1)$ with $(+n,-1)$ and vice versa after flowing for one second, and, importantly, is chosen so that it is zero on a neighborhood of every point $(n,k)$ with $k\leq -2$. Then $\Phi=\mathrm{Fl}^X_1$ extends $\phi$ to the whole plane.

Now consider the map $\ell:\Bbb R \to\Bbb R^2,(x,y)\mapsto(x,y+1)$. Then the infinite composition

$$\cdots\circ(\ell^{-n} \circ\Phi\circ\ell^n)\circ\cdots\circ(\ell^{-1}\circ\Phi\circ\ell)\circ\Phi$$

EDIT. To make this work one should instead consider the infinite composite in the opposite direction $$\Phi\circ(\ell^{-1}\circ\Phi\circ\ell)\circ\cdots\circ(\ell^{-n} \circ\Phi\circ\ell^n)\circ\cdots$$ This will work and provide a (well-defined this time?!) continuous bijection of $\Bbb R^2$ onto itself, hence a homeomorphism by invariance of domain.

makes sense, and defines a smooth diffeomorphism of the plane that realizes the bijection you describe. The reason this makes sense is that for every $n\geq 1$ the region $\lbrace(x,y)\in\Bbb R^2\mid y\geq-\frac12\rbrace\cup D((0,-1),n+\frac12)$ it stable, and there the infinite composition is actually a finite composite of diffeomorphisms, and is smooth.

I think something similar is possible for any bijection of $\Bbb Z^2$, to get a diffeomorphism that extends it defined as an infinite composition of time one flows of vector fields whose supports tend to infinity, i.e. lie outside of bigger and bigger balls, and hence, after a while, will not do anything on a fixed ball, I might come back to this tomorrow.

I think you could do it like so : first, write $V_\epsilon(A)$ for the $\epsilon$ neighborhood of a set $A\subset\Bbb R^2$. Now enumerate the elements of $\Bbb Z^2$ in a spiral fashion by a bijection $\varphi:\Bbb N\to\Bbb Z\times\Bbb Z$. Set $m_n=\min\lbrace|\phi(n)|,|f(\phi(n))|\rbrace$ and $M_n=\max\lbrace|\phi(n)|,|f(\phi(n))|\rbrace$. For every $n$, construct a compactly supported vector field $X_n$ so that

1. $X_n$ is compactly supported with $$\mathrm{supp}(X_n)\subset\lbrace x\in\Bbb R^2\mid m_n-1\leq|x|\leq M_n+1\rbrace\setminus V_\frac14\left(\Bbb Z^2\setminus\lbrace\phi(n),f(\phi(n))\rbrace\right)$$
2. The time one flow $\Phi_n=\mathrm{Fl}^{X_n}_1$ of $X_n$ sends $\phi(n)$ to $f(\phi(n))$

Then the infinite composition $$\cdots\circ\Phi_n\circ\cdots\circ\Phi_{1}\circ\Phi_{0}$$ makes sense and is smooth because on every compact disk it actually is a finite composite.