Equivalent Idempotents in the Ellis Semigroup

Let $(X,T)$ be a dynamical system where $T$ is a (at least countably infinite) group acting on a compact Hausdorf space $X$, and let $E(X)$ be the Ellis semigroup of this system (if we abuse notation and take $T\subset X^X$, where the latter is with the topology of pointwise convergence, then $E(X):=\overline{T}$, and can be shown to be a sub-semigroup of $X^X$ with respect to composition of functions). We have that minimal (edit: right) ideals in $E(X)$ exist, and have idempotents (elements such that $u\circ u= u$). We define two idempotents $u,v$ to be equivalent iff $u\circ v=v, \ v\circ u=u$ (i.e. $u\circ v(x)=v(x)\ \forall x\in X$, etc).

Suppose $E(X)$ has three minimal ideals, we have that there are 3 distinct equivalent idempotents belonging to each of the ideals (call them $u,v,w$). Do we have that there is $x\in X$ such that $ux\neq vx\neq wx\neq ux$ (again abusing notation a bit)? I think the answer should be 'no', but any thoughts/references would be appreciated.

• An Ellis semigroup has a unique minimal ideal. Do you mean it has 3 minimal left ideals and u,v,w each belong to one of these ideals and they are equivalent to each other? – Benjamin Steinberg Sep 26 '17 at 18:42
• Maybe you want a unique minimal right ideal? – Benjamin Steinberg Sep 26 '17 at 19:14
• An Ellis semigroup may have many different, disjoint, minimal right ideals. In particular, it may have uncountably many, as is the case of the full shift on the binary biwords (i.e. the system $(\{0,1\}^\mathbb{Z},\sigma)$ and its respective Ellis semigroup which is 'the same as' $\beta\mathbb{Z}$). I mostly care about the case when there are finitely many such ideals. – Simon_Peterson Sep 28 '17 at 17:23
• You originally had said minimal ideal without right. That is why I asked. In semigroup theory ideal usually means two sided ideal by default. – Benjamin Steinberg Sep 28 '17 at 17:32
• Thanks for the note, have amended the question. When considering the Ellis semigroup, ideal usually means left or right, depending on which way you take composition of functions (whether $f\circ g$ means apply $f$ first or apply $g$ first), and since specifying sometimes causes confusion, I originally omitted the side. – Simon_Peterson Sep 28 '17 at 18:37