Suppose that a discrete group $\Gamma$ acts on a compact Hausdorff space $X$ via homeomorphisms. This action induces an action on $C(X)$, the space of all continuous functions from $X$ to $\mathbb{C}$, by $s.f(x)=f(s^{-1}x)$. The action is said to be 'prime' if $C(X)$ doesn't admit any invariant unital $C^*$-subalgebra other than $\mathbb{C}$ and $C(X)$.

There is an equivalent condition for the same which is : $C(X)$ doesn't admit any invariant $C^*$ -sub-algebra other than $\mathbb{C}$ and $C(X)$ iff the only $\Gamma$-equivariant surjective continuous map from $X$ to a compact Hausdorff $\Gamma$-space $Y$, when $Y$ is not a singleton, is one-to-one.

I wonder if there are any examples of such an action for $X$ which doesn't have discrete topology. I couldn't find any example of such an action. It would be great if there any research papers/books which mention of such an action.

Thanks for the help!!

  • $\begingroup$ What about $\Gamma$ the symmetric group $S_n$ and $X=\{1,\ldots, n\}$. Here I assume subalgebra means unital. $\endgroup$ Mar 22, 2018 at 1:11
  • $\begingroup$ Subalgebra is unital. I forgot to mention that I don't want $X$ to be finite. More precisely, I don't want $X$ to have discrete topology. $\endgroup$ Mar 22, 2018 at 1:13
  • $\begingroup$ I think that Thompson's group $V$ acting on the Cantor set $\{0,1\}^{\mathbb N}$ as the topological full group of the one-sided shift gives an example. I am not sure if I have time to check this carefully and write it up today. $\endgroup$ Mar 22, 2018 at 14:10
  • $\begingroup$ Thank you. I will try to check the details and come back again. $\endgroup$ Mar 22, 2018 at 14:13

1 Answer 1


Thompson's group $V$, which is a finitely presented infinite simple group, consists of all homeomorphisms of the Cantor set $\{0,1\}^\mathbb N$ that can be described the following way. A prefix code is a collection of finite words none of which is a prefix of another. A finite prefix code $C$ is maximal if it is not contained in another prefix code. Equivalently, it is maximal if removing the vertices of the binary tree corresponding to $C$ disconnects it.

An element of Thompson's group is given by specifying two maximal prefix codes $C_1=\{u_1,\ldots, u_n\}$ and $C_2=\{v_1,\ldots, v_n\}$ of the same size and a permutation $\sigma\in S_n$. The action of the corresponding element $f_{C_1,C_2,\sigma}$ on $x\in \{0,1\}^n$ is as follows. There is a unique factorization $x=u_iz$ with $u_i\in C_1$. Then $f_{C_1,C_2,\sigma}(x) = v_{\sigma(i)}z$. Note that the prefix codes $C_1,C_2$ are not uniquely determined by the element of $V$.

Let $X=\{0,1\}^\mathbb N$ and let $R$ be an equivalence relation such that $Y=X/R$ is compact Hausdorff and the projection $X\to Y$ is $V$-equivariant. Suppose that $R$ is not the equality relation and has more than one equivalence class. Note that $R$ must be closed as a subset of $X\times X$ by Hausdorffness. Since the complement of $R$ in $X\times X$ is non-empty and open, it follows that there are finite words $u,v$ such that $ux$ and $vy$ are never equivalent for any infinite words $x,y$. By extending the shorter of the two words, we may assume $|u|=|v|=m$ (where $|\cdot|$ is the length).

Since $R$ is not the equality relation we can find two distinct infinite words $x_1,x_2$ which are equivalent under $R$. I claim we can find an element $g\in V$ such that $g(x_1)\in u\{0,1\}^\mathbb N$ and $g(x_2)\in v\{0,1\}^\mathbb N$. This will contradict that $X\to Y$ is $V$-equivariant.

Let $n$ be the length of the longest common prefix of $x_1,x_2$. Let $k>\max\{n,m\}$ and let $C$ be the maximal prefix code consisting of all words of length $k$. Then $x_1,x_2$ have different prefixes $p,q$ of length $k$. Also, since $k>m$, we can find $p',q'\in C$ such that $u$ is a prefix of $p'$ and $v$ is a prefix of $q'$. There is an element $g=f_{C,C,\sigma}\in V$ such that $g$ takes all infinite words $px$ to $p'x$ and $qx$ to $q'x$. It follows that $g(x_1)$ and $g(x_2)$ are not equivalent under $R$, a contradiction.

  • $\begingroup$ So if $\{u_1,u_2,\ldots,u_n$ is a prefix code, then it means that $u_i$ cann't be written as $u_ju_i'$ for all $1 \le j \le n$ and $1 \le i \le n $? $\endgroup$ Mar 22, 2018 at 15:28
  • $\begingroup$ Unless $i=j$ and the suffix is empty. $\endgroup$ Mar 22, 2018 at 18:12
  • $\begingroup$ There will be always an $x$ for which the words from $C_1$ cannot be a prefix since $C_1$ is finite. The action just fixes these $x$?? $\endgroup$ Mar 22, 2018 at 18:27
  • $\begingroup$ I am trying to understand the proof here: Let $\psi: X \to Z$ be a $\Gamma$-equivariant onto map. We want to show that $\psi$ is one-to-one. So we define the relation $R$ by $x \sim y \iff \psi(x)=\psi(y)$. Clearly this is an equivalence relation. Let $Y=(X/\sim)$. We want to show that every equivalence class contains exactly one element. If $R$ is not the equality relation, then I guess you mean to say that equivalence class has more than one element? $\endgroup$ Mar 22, 2018 at 18:56
  • 1
    $\begingroup$ Each infinite word has exactly one prefix in a maximal finite prefix code because removing the code words disconnects the tree and infinite words correspond to infinite paths from the root of the tree. Exactly one is because it is a prefix code. Now my relation R is as you defined. My proof shows that if R identifies at least two distinct elements but not all elements, then one gets that X/R is not Hausdorff. Since $X,Z$ are compact Hausdorff, $\psi$ is closed and so gives $Z$ the quotient topology and so $Z\cong X/R$. This gives the contradiction. $\endgroup$ Mar 22, 2018 at 19:17

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