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Let $X$ be a compact (metric) space and $T:X\rightarrow X$ be a continuous map. Let $U_T:C(X)\rightarrow C(X)$ be the linear operator $U_T(f) = f\circ T$.

Then Wikipedia (see http://en.wikipedia.org/wiki/Weak_mixing#Topological_mixing) (vaguely) says that $T$ is weak topological mixing if, whenever $U_t(f)=\lambda f$ for some $\lambda\in\mathbb C$ and $f\in C(X)$, then $f$ is a constant function.

But Terry Tao (see Definition 3 of http://terrytao.wordpress.com/2008/01/28/254a-lecture-7-structural-theory-of-topological-dynamical-systems/ ) says that $T$ is topologically weakly mixing if $T\times T$ is topologically transitive, that is, if $U,V\subseteq X\times X$ are open then there is $n\in\mathbb Z$ with $(T\times T)^n(U)\cap V\not=\emptyset$. I guess this is equivalent to saying that for $A,B,C,D\subseteq X$ open we can find one $n\in\mathbb Z$ with both $T^n(A)\cap B$ and $T^n(C)\cap D$ non-empty.

Unfortunately, if I look in e.g. Brin and Stuck's book, then topologically transitive is defined to mean that for some $(x,y)\in X\times X$, the forward orbit $\{ (T^n(x),T^n(y)) : n\geq 1 \}$ is dense in $X\times X$.

(Edit: Thinking about this, B&S, Prop 2.2.1 shows that Tao's definition implies the B&S definition; and, at least if $T$ is a homeomorphism, the converse holds. But it doesn't seem to if $T$ is not surjective).

Is the definition of mixing involving $U_T$ equivalent to the usual one? If so, can anyone supply a reference or a sketch proof? I am beginning to think that Wikipedia has confused topological and measure-theoretic mixing.

If they are equivalent, do I really need $T$ to be a homeomorphism? (One could either replace $\mathbb Z$ by $\mathbb N$ in Tao's definition, or let $T^{-1}$ be the inverse image, and iterate). Do I need $X$ to be metric?

Edit: The reference Willie found gives the following: Suppose $T$ is a homeomorphism and $T\times T$ is minimal in the B&S sense. Then if $f\in C(X)$ with $f\circ T = \lambda f$, necessarily $\lambda\in\mathbb T$ (as $T$ is a homeomorphism). Consider $g(s,t) = f(s)\overline{f(t)}$, which defines a continuous function on $X\times X$. Then $g(T^ns,T^nt) = f(T^ns)\overline{f(T^nt)} = \lambda^n f(s) \overline{\lambda}^n \overline{f(t)} = f(s)\overline{f(t)} = g(s,t)$. There is $(x,y)$ such that $\{ (T^nx,T^ny):n\geq 1\}$ is dense in $X\times X$, from which it follows that $g$ must be constant (as $g$ is continuous). So in particular $f(x)\overline{f(y)} = |f(x)|^2$ for all $x,y$, so $f$ is constant.

So that's one implication...

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Haven't thought too much yet, but doesn't Exercise 10 in the lecture notes (stated right between Definitions 2 and 3) you linked to say precisely that the two definition of the topological transitivity agree? –  Willie Wong Nov 11 '11 at 13:21
    
@Willie-- right you are-- but Tao's defines $T$ to be a homeomorphism, and I'd quite like to not assume that. –  Matthew Daws Nov 11 '11 at 13:26
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For the second part, see Theorem 2.2 and Theorem 2.3 of ams.org/journals/tran/1969-139-00/S0002-9947-1969-0237748-5/… which seems to say that topologically weakly mixing systems in Terry's sense (which also is the sense defined by the paper) implies weakly topologically mixing in Wikipedia's sense. –  Willie Wong Nov 11 '11 at 13:35

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