Skip to main content
just notation
Source Link
Daniele Zuddas
  • 2.3k
  • 13
  • 19

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$$M = \partial (T \times [-1, 1])$. Let $f : S \to S$$f : M \to M$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$. Let $f : S \to S$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $M = \partial (T \times [-1, 1])$. Let $f : M \to M$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

Improved exposition
Source Link
Daniele Zuddas
  • 2.3k
  • 13
  • 19

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction. Let

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$. Let $f : S \to S$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f^6 = \mathrm{id}$ and $f$ is fixed points freesatisfies the conditions in the question, with $n = 6$.

Here a very explicit construction. Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$. Let $f : S \to S$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f^6 = \mathrm{id}$ and $f$ is fixed points free.

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$. Let $f : S \to S$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

Source Link
Daniele Zuddas
  • 2.3k
  • 13
  • 19

Here a very explicit construction. Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $S = \partial (T \times [-1, 1])$. Let $f : S \to S$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f^6 = \mathrm{id}$ and $f$ is fixed points free.