0
$\begingroup$

There exists a connected topological space such that the permutations group $S_3$ acting without fixed points? I try to consider matrice spaces.

1 Answers 1

3

For a trivial example, you can take $S_3$ itself with the indiscrete topology, with the group acting by translation.

Assuming you want a less trivial example (say, something like a nice subset of $\mathbb{R}^n$), here's a hint on how to find one. Notice that $S_3$ acts on $\mathbb{C}^3$ by permuting the coordinates. Unfortunately, this action has some fixed points, so it doesn't quite work. But if you remove all the points of $\mathbb{C}^3$ which are fixed by a non-identity element of $S_3$, can you prove that the resulting space is still connected?

(As an extra exercise, you can try and see why this wouldn't work for $\mathbb{R}^3$: if you remove the fixed points of non-identity elements, you are left with a disconnected space.)

  • 0
    Why $R^3$ is not good? If we want that the action to be fixed point free we must to consider $R^3$ without the line generated by (1,1,1). This space is connected, no? – 2017-02-08
  • 0
    Acctualy my definition for fixed point was wrong. Now I see why $R^3$ is not good. We must to get out an hyperplane so what remain is not conncted. – 2017-02-08