3
$\begingroup$

I am looking for some example (other than identity) of homeomorphism in the Torus $\mathbb{T}^2$ so that it has a set of fixed points with non-empty interior. It's possible?

I appreciate any reference.

$\endgroup$
1
  • 1
    $\begingroup$ Intuitively there is a continuum of such homeomorphisms. $\endgroup$
    – lanskey
    May 30, 2017 at 4:22

4 Answers 4

1
$\begingroup$

Pick the compact set $F \subset \mathbb{T}$ of points that you want fixed (making sure it has open interior, or any other property you require). Perturb the identity so that any point in the complement of $F$ moves by a vector of magnitude proportional to its distance to $F$.

Depending on your choice for $F$, and vector directions, it should not be difficult to ensure your perturbed map is a homeomorphism. For instance:

  • $F$ is an annulus, so the complement is an annulus too, and the vector can point in the direction of the core loop.
  • $F$ is the union of a meridian and a parallel of $\mathbb{T}$. Then the complement is a topological disk, and the vector can point in the direction of a specific boundary point.
$\endgroup$
3
  • $\begingroup$ The key point being how to «Perturb the identity so that…», no? $\endgroup$ May 30, 2017 at 5:10
  • $\begingroup$ I gave two examples... $\endgroup$ May 30, 2017 at 5:13
  • $\begingroup$ A systematic way of doing this is to pick a continuous function which vanishes on the set $F$, multiply it by a nowhere vanishing vector field and consider the function I constructed. $\endgroup$ May 30, 2017 at 5:25
0
$\begingroup$

Think first about the map $f:\mathbb{C}\rightarrow\mathbb{C}$ given by $$f(re^{i\theta})=(\min\{r, r^2\})e^{i\theta}$$ (for $\theta$ real and $r$ positive and real). This map is the identity outside the unit disc, and "stretches" the inside of the unit disc; and is an autohomeomorphism of $\mathbb{C}$ with the usual topology.

We can apply this same picture to the torus. Specifically, let $S$ be the set of complex numbers whose real and imaginary part each lie in $[-2, 2]$; this is a square of side length $4$ centered at the origin. The torus can be thought of as the quotient gotten by identifying opposite edges in $S$. Then the map $f$ described above transfers to an autohomeomorphism of the torus which is the identity on a set with nonempty interior, and is non-identity on a set with nonempty interior.

By changing the exponent, we get lots of similar examples; and of course this isn't the only way to do this sort of thing, by a long shot.

$\endgroup$
0
$\begingroup$

Pick any vector field $X$ on the torus $T$ which vanishes on an open set, and consider the "time 1" map for the flow it generates, that is, find the function $\Phi:\mathbb R\times T\to T$ such that $\Phi(0,x)=x$ for all $x\in T$ and $\frac{\mathrm d}{\mathrm dt}\Phi(t,x)=X_{\Phi(t,x)}$ and consider the function $f:x\in T\mapsto \Phi(1,x)\in T$, which is a diffeomorphism. Its fixed point set contains the vanishing locus of the field $X$.

$\endgroup$
0
$\begingroup$

The torus contains a homeomorphic copy of a closed disc together with an injection map $i: D\rightarrow\mathbb{T}^2$. Now take any non-trivial homeomorphism of a closed disc $\varphi: D\rightarrow D$ that fixes $\partial D$. For example $z\mapsto e^{i(1-|z|)\theta}z$ for some $\theta\neq 2k\pi, k\in\mathbb{Z}$. We can now construct our homeomorphism of the torus taking $i\circ\varphi\circ i^{-1}$ on $i(D)$ and identity on $\mathbb{T}^2\setminus i(D)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.