DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X)

LI-SHU CHEN, JYH-SHYANG JEANG, AND NGAI-CHING WONG

Abstract. We study disjointness preserving (quasi-)n-shift operators on C_{0}(X), where X is locally compact and Hausdorff. When C_{0}(X) admits a quasi-n-shift T, there is a countable subset of X_{∞} = X ∪ {∞} equipped with a tree-like structure, called φ-tree, with exactly n joints such that the action of T on C_{0}(X) can be implemented as a shift on the φ-tree. If T is even an n-shift, then the φ-tree is dense in X and thus X is separable. By analyzing the structure of the φ-tree, we show that every (quasi-)n-shift on c_{0} can always be written as a product of n (quasi-)shifts. Although it is not the case for general C_{0}(X) as shown by our counter examples, we may do so after dilation.

1. Introduction

A linear operator S from a Banach space E into itself is called an n-shift if

(a) S is injective and has closed range;

(b) S has corank n;

(c) The intersection ⋂_{m=1}^{∞} S^{m}E of the range spaces of all powers S^{m} of S is zero.

S is called a quasi-n-shift if S satisfies conditions (a) and (b). When n = 1, we will simply call S a shift or a quasi-shift accordingly. Crownover [3] showed that S is a shift on a Banach space if and only if it is similar to the unilateral shift on a sequence space. In fact, every n-shift on a Banach space is similar to an operator on a sequence space shifting the first n coordinates of a vector to the right (Proposition 4.3).

Let X and Y be locally compact Hausdorff spaces. Let C_{0}(X) and C_{0}(Y) be Banach spaces of continuous (real- or complex-valued) functions defined on X and Y vanishing at infinity, respectively. In the papers of Gutek et. al. [6] and Farid and Varadarajan [4], isometric shifts and quasi-shifts on C(X) (= C_{0}(X)) are studied for compact Hausdorff spaces X. When the underlying scalar field is the complex ℂ, Haydon [7] provided examples to demonstrate such shifts do exist in some compact connected Hausdorff space as well as in the Cantor set. This is an interesting complement to the fact found by Holub [9] that the real Banach space C(X, ℝ) of continuous real-valued functions defined on X admits no shift at all if X is compact and connected. More recently, Rajagopalan [14] and Araujo and Font [2] discussed related questions in this direction.

Date: August 12, 2000; submitted to Can. J. Math.

1991 Mathematics Subject Classification. 46J10, 47B30, 47A53.

Key words and phrases. shifts, quasi-shifts, disjointness preserving operators.

Partially supported by Taiwan National Science Council grants: NSC 86-2115-M110-002, 37128F.

2 CHEN, JEANG, AND WONG

Since most of the interesting examples of shift operators in the literature so far are those on function spaces, it is reasonable to study disjointness preserving shifts. Recall that T is disjointness preserving or separating if T f · T g = 0 whenever f · g = 0. Disjointness preserving shifts on Banach lattices are studied in [6], where the authors apply results in [1] and others to obtain the non-existence of such operators on Dedekind complete Banach lattices with at most finitely many atoms. However, they did not discuss disjointness preserving shifts on general C_{0}(X); except for the special case when X is an extremely disconnected compact Hausdorff space. The authors of [6, 4] do not seem to be aware of the recent development of the theory of disjointness preserving linear operators. In particular, similar tools as those provided in [8], a major reference of [6, 4], have been established, especially the one that such operators are exactly weighted composition operators (see e.g., [10, 5, 11]).

In Sections 2, we shall discuss n-shift (resp. quasi-n-shift) operators. In [12], it was proved that every disjointness preserving Fredholm linear operator from C_{0}(X) into C_{0}(Y) with closed range is automatically continuous and can be written as a very special weighted composition operator. Since disjointness preserving quasi-n-shifts on C_{0}(X) are Fredholm, they are automatically continuous. Moreover, tools developed in [12] is used to give a full description of them (Theorem 2.3). In fact, every disjointness preserving quasi-n-shift on C_{0}(X) is implemented by a shift on a countable set with a tree-like structure, called φ-tree, with exactly n joints in the one-point compactification X_{∞} of X (Theorem 2.5). The φ-tree arising from an n-shift is proved to be dense in X. In particular, X is separable whenever any disjointness preserving n-shift on C_{0}(X) exists (Theorems 2.11).

In Section 3, we shall verify that all disjointness preserving (quasi-)n-shifts on c_{0} ≅ C_{0}(ℕ) can be written as a product of n (quasi-)shifts (Theorem 3.4). It is, however, not the case in general. We shall provide a counter example in Section 4 that some disjointness preserving isometric n-shifts cannot be written as products of n disjointness preserving shifts (Example 4.4). There is also a compact connected Hausdorff space X such that C(X) admits a quasi-n-shift but not any quasi-k-shift for n ≥ 2 and k = 1, 2, …, n − 1 (Example 4.5). Nevertheless, we show that a disjointness preserving quasi-n-shift can be dilated to a product of n quasi-shifts and corank one injections, provided for example that X is compact (Theorem 4.13).

We shall apply results in this paper and [12] to the study of isometric (quasi-)n-shifts on C_{0}(X) in [13], which extends [6, 4, 2].

2. Disjointness preserving n-shifts and the related φ-tree structure

Let X be a locally compact Hausdorff space. Let X_{∞} be the one-point compactification of X; namely X_{∞} = X ∪ {∞}. The point ∞ is an isolated point in X_{∞} if and only

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 3

if X is compact. In this case, we write C(X) for C_{0}(X) = {f ∈ C(X_{∞}) : f(∞) = 0} as usual.

Definition 2.1. Let X be a locally compact Hausdorff space and let φ be a continuous map from X_{∞} onto X_{∞} with φ(∞) = ∞. Define an equivalence relation ∼ in X_{∞} by

x ∼ x′ if and only if φ(x) = φ(x′).

(1) We call a point x in X_{∞} a φ-vanishing point if φ(x) = ∞.

(2) We call x a φ-merging point and φ(x) a φ-merged point if the equivalence class [φ(x)] = φ^{−1}{φ(x)} contains at least two points. Denote by M_{φ} the set of all φ-merging points, and thus by φ(M_{φ}) the set of all φ-merged points in X_{∞}.

(3) A φ-branch originated at a point x in X_{∞} is defined to be the set

B_{x} = ⋃{φ^{−n}(x): n = 0, 1, 2, … },

where φ^{0}(x) = {x} and φ^{−n}(x) = {y ∈ X: φ^{n}(y) = x} for n = 1, 2, . . ..

(4) The φ-tree is a directed graph in X_{∞}, whose vertex set is the union ⋃{B_{c} : c ∈ φ(M_{φ})} of all φ-branches originated at φ-merged points, and there is a directed edge from a to b if and only if φ(a) = b.

(5) The crown of the φ-tree is the union ⋃{B_{a} : a ∈ M_{φ}} of all φ-branches originated at φ-merging points.

(6) The number #(M_{φ}) − #(φ(M_{φ})) is called the number of joints of the φ-tree.

(7) A φ-tree is said to be rooted at ∞ if the φ-tree coincides with the φ-branch B_{∞} originated at ∞.

We are interested in the φ-tree associated to a disjointness preserving quasi-n-shift T on C_{0}(X). In fact, every such T gives rise to a unique map φ such that the action of T can be visualized as a shift on the φ-tree in X_{∞}, which has exactly n joints. Let us consider an example first.

Example 2.2. Let T be the disjointness preserving isometric 3-shift on c_{0} (≅ C_{0}(ℕ)) defined by

T ((x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, …)) = (0, x_{1}, x_{1}, x_{2}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, …).

Every null sequence (x_{n}) in c_{0} can be considered as a continuous function f on ℕ_{∞} such that f(∞) = ∞ and f(n) = x_{n} for all n in ℕ. Write T f = f ∘ φ where the action of φ: ℕ_{∞} → ℕ_{∞} can be visualized in the following φ-tree in which a directed edge b ← a indicating φ(a) = b.

[figure: 4; 7; 10; ⋯; 2; ∞; 1; 5; 8; 11; ⋯; 3; 6; 9; 12; ⋯] .

Note that the set of all φ-merging points is M_{φ} = {∞, 1, 2, 3, 4, 5}, and the set of all φ-merged points is φ(M_{φ}) = {∞, 1, 2}. There are exactly #(M_{φ}) − #(φ(M_{φ})) = 3 joints in the φ-tree at ∞, 1 and 2, respectively. Moreover, the φ-tree coincides with its crown and is rooted at ∞. In this case,

φ: (ℕ_{∞}, {∞, 1, 2, 3, 4, 5}) → (ℕ_{∞}, {∞, 1, 2})

is a relative homeomorphism, and the induced map

φ̃: ℕ^{∞}⧸∼ → ℕ_{∞}

is a homeomorphism. In fact, ℕ^{∞}⧸∼ = {[∞], [2], [4], [6], [7], [8], [9], … } and φ̃([∞]) = ∞, φ̃([2]) = 1, φ̃([4]) = 2, φ̃([6]) = 3, φ̃([7]) = 4, φ̃([8]) = 5, φ̃([9]) = 6, . . ..

Denote by δ_{x} the evaluation at a point x in X. The following theorem is a special case of the results in [12].

Theorem 2.3. Every disjointness preserving quasi-n-shift T on C_{0}(X) is continuous. Let

X_{0} = {x ∈ X: δ_{x} ∘ T = 0} and X_{c} = X \ X_{0}.

(1) There exist a continuous map φ from X_{∞} onto X_{∞} and a continuous bounded and away from zero scalar function h on X_{c} such that φ(X_{0} ∪{∞}) = {∞}, φ(X_{c}) = X, and

T f_{|X_{c}} = h · f ∘ φ,
T f_{|X_{0}} ≡ 0.

(2) The set M_{φ} of all φ-merging points in X_{∞} is finite. In fact,

#(M_{φ}) − #(φ(M_{φ})) = n.

(3) The map

φ: (X_{∞}, M_{φ}) → (X_{∞}, φ(M_{φ}))

is a relative homeomorphism, and the induced map

φ̃: X^{∞}⧸∼ → X_{∞}

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 5

is a homeomorphism. Consequently, the finite set X_{0} = φ^{−1}(∞) ∩ X consists of isolated points in X when X is compact, and X_{0} is empty when X is compact and connected.

The following example borrowed from [12] says that the last assertion in Theorem 2.3(3) can be false when X is not compact.

Example 2.4. Let X be the disjoint union in ℝ^{2} of I_{n}^{+} = {(n, t): 0 < t ≤ 1} and I_{n}^{−} = {(n, t) : −1 < t < 0} for n = 1, 2, . . .. Let p be the point (1, 1) and let X_{1} = X \ {p}. Let φ be the homeomorphism from X_{1} onto X by sending the intervals I_{1}^{+} \{p} onto I_{1}^{−}, I_{n+1}^{+} onto I_{n}^{+}, and I_{n}^{−} onto I_{n+1}^{−} in a canonical way for n = 1, 2, . . .. Then the disjointness preserving isometric quasi-shift T f = f ∘ φ on C_{0}(X) has exactly one vanishing point, i.e. p, which is not an isolated point in X. In a similar manner, one can also construct an example in which X is connected (by adjoining each I_{n}^{±} a common base point, for example).

From Theorem 2.3(2), we know that all equivalence classes in X_{∞} induced by the relative homeomorphism φ are finite and at most finitely many of them consist of more than one points. Let all the possibly exceptional classes be

[∞] = {p_{1}, p_{2}, …, p_{k}, ∞},
[a_{l_{1}}^{(1)}] = {a_{1}^{(1)}, a_{2}^{(1)}, …, a_{l_{1}}^{(1)}} ,
⋮
[a_{l_{j}}^{(j)}] = {a_{1}^{(j)}, a_{2}^{(j)}, …, a_{l_{j}}^{(j)}} .

In other words, we have

φ (p_{1}) = φ (p_{2}) = ⋯ = φ (p_{k}) = φ (∞) = ∞,
φ (a_{1}^{(1)}) = φ (a_{2}^{(1)}) = ⋯ = φ (a_{l_{1}−1}^{(1)}) = φ (a_{l_{1}}^{(1)}) = c_{1},
⋮
φ (a_{1}^{(j)}) = φ (a_{2}^{(j)}) = ⋯ = φ (a_{l_{j}−1}^{(j)}) = φ (a_{l_{j}}^{(j)}) = c_{j},

for some distinct c_{1}, c_{2}, …, c_{j} in X. Then

M_{φ} = {∞, p_{1}, p_{2}, …, p_{k}, a_{1}^{(1)}, a_{2}^{(1)}, …, a_{l_{1}}^{(1)}, …, a_{1}^{(j)}, a_{2}^{(j)}, …, a_{l_{j}}^{(j)}} .

In case [∞] = φ^{−1}(∞) = {∞}, we have

M_{φ} = {a_{1}^{(1)}, a_{2}^{(1)}, …, a_{l_{1}}^{(1)}, …, a_{1}^{(j)}, a_{2}^{(j)}, …, a_{l_{j}}^{(j)}}

instead.

The following theorem is again a consequence of the results in [12].

6 CHEN, JEANG, AND WONG

Theorem 2.5. C_{0}(X) admits a disjointness preserving quasi-n-shift if and only if X_{∞} admits a φ-tree with exactly n joints. In this case, let

X_{0} = {x ∈ X: φ(x) = ∞} and X_{c} = X \ X_{0}.

For any bounded and away from zero scalar function h on X_{c}, the disjointness preserving operator T defined by T f_{|X_{c}} = h · f ∘ φ and T f_{|X_{0}} = 0 is a quasi-n-shift on C_{0}(X). In above notations, we have

ran(T) = {g ∈ C_{0}(X): g(p_{1}) = ⋯ = g(p_{k}) = 0 and

\frac{g(a_{1}^{(i)})}{h(a_{1}^{(i)})} = \frac{g(a_{2}^{(i)})}{h(a_{2}^{(i)})} = ⋯ = \frac{g(a_{l_{i}}^{(i)})}{h(a_{l_{i}}^{(i)})}, i = 1, 2, …, j\}.

In the following example, there are a quasi-shift (not shift) and a shift on c_{0} ≅ C_{0}(ℕ) such that they give rise to the same φ-tree. A necessary and sufficient condition on the weight function h to ensure T f = h · f ∘ φ defining an n-shift is given to this particular φ-tree.

Example 2.6. Let T: c_{0} → c_{0} be a disjointness preserving linear operator defined by

T (x_{1}, x_{2}, …) = (2x_{1}, x_{1}, x_{2}, …).

Then h(1) = 2, h(n) = 1, φ(1) = 1, and φ(n) = n − 1 for n ≥ 2. The φ-tree is

[figure: 1; 2; 3; ⋯].

It is clear that the φ-tree is the whole space ℕ, coincides with its crown and has one joint at 1. However, T is just a quasi-shift but not a shift, since (1, \frac{1}{2}, \frac{1}{4}, …, \frac{1}{2^{n}} , …) ∈ ⋂_{i=1}^{∞} ran(T^{i}). On the other hand, the operator sending (x_{1}, x_{2}, …) to (x_{1}, x_{1}, x_{2}, …) is a shift on c_{0} giving rise to the same φ-tree.

In general, let h in C(ℕ) be bounded and away from zero. Then the weighted composition operator S on c_{0} defined by Sf = h · f ∘ φ is a quasi-shift. We shall show that S is a shift on c_{0} if and only if

lim sup_{i→∞} |\frac{h(i + 1) ⋯ h(2)}{h(1)^{i}} | > 0.

Note that

S^{i}f(1) = h(1) · S^{i−1}f(1) = h(1)^{2} · S^{i−2}f(1) = ⋯ = h(1)^{i} · f(1)

and

S^{i}f(i + 1) = h(i + 1) · S^{i−1}f(i) = h(i + 1) · h(i) · S^{i−2}f(i − 1)
= ⋯ = h(i + 1) ⋯ h(2) · f(1).

Hence \frac{S^{i}f(i+1)}{h(i+1)···h(2)} = f(1) = \frac{S^{i}f(1)}{h(1)^{i}} for all i in ℕ. Note also that if g ∈ ran(S^{i}) then g ∈ ran(S^{j}) for all 1 ≤ j ≤ i. It follows that

ran(S^{i}) = \{g ∈ C_{0}(ℕ): g(2) = \frac{h(2)}{h(1)}g(1),

g(3) = \frac{h(3)h(2)}{h(1)^{2}}g(1),

⋮

g(i + 1) = \frac{h(i + 1) ⋯ h(2)}{h(1)^{i}}g(1)\}.

Therefore, g ∈ ⋂_{i=1}^{∞} ran(S^{i}) if and only if

g = (g(1), \frac{h(2)}{h(1)}g(1), \frac{h(3)h(2)}{h(1)^{2}}g(1), …, \frac{h(i + 1) ⋯ h(2)}{h(1)^{i}}g(1), …) .

Consequently,

⋂_{i=1}^{∞} ran(S^{i}) = \{0\} if and only if lim sup_{i→∞} |\frac{h(i + 1) ⋯ h(2)}{h(1)^{i}} | > 0.

We are interested in the question of which φ-trees do provide us with a disjointness preserving n-shift regardless of the choice of the weight functions h. As a supplement to [6, Theorem 2.4], the following result states that every dense φ-tree rooted at ∞ does.

Theorem 2.7. Suppose that a φ-tree is rooted at ∞, dense in X_{∞} and has exactly n joints. Then for any bounded and away from zero continuous scalar function h on X_{c} = X \ {p_{1}, p_{2}, …, p_{k}}, where p_{1}, …, p_{k} are all φ-vanishing points, the operator T, defined by T f_{|X_{c}} = h · f · φ and T f(p_{1}) = T f(p_{2}) = ⋯ = T f(p_{k}) = 0, is a disjointness preserving n-shift on C_{0}(X).

Proof. By Theorem 2.5, T is a quasi-n-shift. We only need to verify ⋂_{m=1}^{∞} ran(T^{m}) = {0}. Suppose g = T^{m}f for some f in C_{0}(X) and m ≥ 1. Then g vanishes at φ^{−r}(p_{i}) for r = 0, 1, 2, …, m − 1 and i = 1, 2, …, k. Consequently, every continuous function in ⋂_{m=1}^{∞} ran(T^{m}) vanishes on the whole φ-tree which is dense in X_{∞}. Hence, ⋂_{m=1}^{∞} ran(T^{m}) = {0} as asserted. Therefore, T is an n-shift.

In the following example, we see that there are some φ-trees which provide us with no n-shift at all.

Example 2.8. Let

X = {(−n, 0), (n, 1), (n, −1): n = 1, 2, 3, … } ∪ {(0, 0)}

8 CHEN, JEANG, AND WONG

in ℝ^{2}. Let φ: (X, {(1, 1), (1, −1)}) → (X, {(0, 0)}) be the relative homeomorphism defined by φ(1, ±1) = (0, 0), φ(n + 1, ±1) = (n, ±1) and φ(−n + 1, 0) = (−n, 0) for n = 1, 2, . . .. The φ-tree has one joint at (0, 0), and is given below:

\xymatrix{ & (1, 1) \ar@{-->}[dl] & (2, 1) \ar[l] & (3, 1) \ar[l] & ⋯ \ar[l] \\ (0, 0) \\ & (1, −1) \ar@{-->}[ul] & (2, −1) \ar[l] & (3, −1) \ar[l] & ⋯ \ar[l]} .

We shall show that there is not any disjointness preserving shift T on C_{0}(X) associated with this φ-tree; no matter how we define T f = h · f ∘ φ for any bounded and away from zero continuous scalar function h on X. To this end, we first note that

ran(T) = \{g ∈ C_{0}(X): \frac{g(1, 1)}{h(1, 1)} = \frac{g(1, −1)}{h(1, −1)}\}

by Theorem 2.5. Similarly,

ran(T^{2}) = \{g ∈ C_{0}(X): \frac{g(1, 1)}{h(1, 1)} = \frac{g(1, −1)}{h(1, −1)} and \frac{g(2, 1)}{h(2, 1)h(1, 1)} = \frac{g(2, −1)}{h(2, −1)h(1, −1)}\} .

In the same manner, we have for each positive integer m that

ran(T^{m}) = \{g ∈ C_{0}(X): \frac{g(1, 1)}{h(1, 1)} = \frac{g(1, −1)}{h(1, −1)},

\frac{g(2, 1)}{h(2, 1)h(1, 1)} = \frac{g(2, −1)}{h(2, −1)h(1, −1)},

⋮

\frac{g(m, 1)}{h(m, 1) ⋯ h(1, 1)} = \frac{g(m, −1)}{h(m, −1) ⋯ h(1, −1)}\}.

It is then easy to see that the nonzero continuous function g_{0} in C_{0}(X), defined by g_{0}(0, 0) = 1 and g_{0} = 0 elsewhere, does belong to all ran(T^{m}) for m = 1, 2, . . .. In fact, ⋂_{m=1}^{∞} ran(T^{m}) has infinite codimension in C_{0}(X).

In dealing with the range space ran(T^{m}) of a power of a disjointness preserving quasin-shift T on C_{0}(X), it is useful to consider the notion of “h-equipotential functions”. Suppose T f = h · f ∘ φ on X_{c}. Denote

h ∘ φ^{k!}(x) = h(x)h(φ(x)) ⋯ h (φ^{k−1}(x)) , ∀x ∈ X_{∞}, ∀k = 1, 2, ….

We set h_{|X_{0}∪{∞}} = 1 for convenience.

Definition 2.9. A function g in C_{0}(X) is said to be h-equipotential on the φ-tree at level k, for k = 1, 2, …, if \frac{g(a)}{h∘φ^{k!}(a)} = \frac{g(b)}{h∘φ^{k!}(b)} whenever a, b are vertices in the φ-tree such that φ^{k}(a) = φ^{k}(b).

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 9

Examples 2.6 and 2.8 are two demonstrations of the following lemma which is a consequence of Theorem 2.5.

Lemma 2.10. Let T be a disjointness preserving quasi-n-shift on C_{0}(X) and m be a positive integer. Then

ran(T^{m}) = {g ∈ C_{0}(X): g is h-equipotential on the φ-tree arising from T up to level m}.

The following theorem says that C_{0}(X) admits no disjointness preserving n-shift if X is inseparable for any n = 1, 2, . . ..

Theorem 2.11. Let T be a disjointness preserving n-shift on C_{0}(X). Then the crown of the φ-tree arising from T is dense in X. In particular, X is separable.

Proof. Recall that the crown of the φ-tree is the union of all φ-branches originated at φ-merging points. Suppose g in C_{0}(X) vanishes on the crown of the φ-tree. Then g is in ran(T^{m}) for m = 1, 2, …, by Lemma 2.10. As a result, ⋂_{m=1}^{∞} ran(T^{m}) contains the subspace {g ∈ C_{0}(X): g vanishes on the crown of the φ-tree}. If T is an n-shift, then ⋂_{m=1}^{∞} ran(T^{m}) = {0}, and thus, the crown of the φ-tree is dense in X. In this case, X is separable since the crown of the φ-tree is a countable set.

3. Writing (quasi-)n-shifts on c_{0} as products of n (quasi)-shifts

This section is devoted to a comprehensive study of disjointness preserving n-shifts on c_{0} (≅ C_{0}(ℕ)). In the following two lemmas, φ: ℕ_{∞} → ℕ_{∞} will be a continuous surjective map such that φ(∞) = ∞.

Lemma 3.1. Suppose a φ-tree in ℕ_{∞} has exactly n joints and its crown contains ℕ. Then

(1) the set {φ^{l}(x): l ∈ ℕ} is finite for all x in ℕ;

(2) ℕ is the disjoint union ⋃ B_{a_{i}} of the branches B_{a_{i}} = {φ^{−n}(a_{i}): n ∈ ℕ} of the φ-tree originated at some merging points a_{i} for which either φ(a_{i}) = ∞ or φ^{m}(a_{i}) = a_{i} for some m in ℕ.

Proof. Suppose the set {φ^{l}(x): l ∈ ℕ} is infinite for some x in ℕ. Since the number of merging points is finite, there exists N in ℕ such that {φ^{l}(x): l ≥ N} ∩ M_{φ} = ∅. As the crown of the φ-tree contains ℕ, we have φ^{N}(x) ∈ B_{a} = ⋃{φ^{−l}(a): l ∈ ℕ}, the branch of the φ-tree originated at some merging point a in M_{φ}. Then φ^{N}(x) ∈ φ^{−m}(a) for some m in ℕ. Hence φ^{N+m}(x) = a ∈ M_{φ}, a contradiction. Thus {φ^{l}(x): l ∈ ℕ} is finite for all x in ℕ. This gives (1), and in particular, {φ^{l}(a): l ∈ ℕ} is finite for all merging points a. It is easy to see if there is no other merging point in {φ^{l}(a): l ∈ ℕ} and φ(a) ≠ ∞, then there exists a positive integer m such that φ^{m}(a) = a. We take a_{1}, a_{2}, …, a_{k} to be

10 CHEN, JEANG, AND WONG

such merging points together with those a ≠ ∞ such that φ(a) = ∞. Finally, we note that if a_{i} and a_{j} are two distinct merging points satisfying a_{i} ∉ B_{a_{j}} and a_{j} ∉ B_{a_{i}} then B_{a_{i}} ∩ B_{a_{j}} = ∅. This gives (2).

Lemma 3.2. Suppose a φ-tree in ℕ_{∞} has n joints. Then the crown of the φ-tree contains ℕ if and only if any (and thus every) weighted composition operator T f = h · f ∘ φ is an n-shift on c_{0}, where h is a unimodular function on ℕ, i.e. |h(x)| ≡ 1 for all x in ℕ.

Proof. By Theorem 2.11, we need to verify the necessity only. Suppose that the crown of the φ-tree contains ℕ. By Theorem 2.5, it is enough to show that ⋂_{l=1}^{∞} ran(T^{l}) = {0}. By Lemma 3.1, ℕ is a disjoint union of the φ-branches originated at some φ-merging points a_{1}, …, a_{j}, where either φ(a_{i}) = ∞ or φ^{m}(a_{i}) = a_{i} for some m > 0.

Let g ∈ ⋂_{i=1}^{∞} ran(T^{i}). If φ(a_{1}) = ∞ then g vanishes on the branch of the φ-tree originated at a_{1} by Lemma 2.10. Suppose otherwise φ^{m}(a_{1}) = a_{1} for some positive integer m. Choose distinct b_{1} = a_{1} and b_{i} ∈ φ^{−i+1}(a_{1}) with φ(b_{i+1}) = b_{i}. A part of the branch of the φ-tree originated at b_{1} = a_{1} looks like:

\xymatrix{ & b_{m−2} \ar@{.>}[dl] & b_{m−1} \ar[l] \\ ⋮ \ar@{.>}[dr] &  &  & b_{m} \ar[ul] & b_{m+1} \ar[l] & b_{m+2} \ar[l] & b_{m+3} \ar[l] & ⋯ \ar[l] \\  & b_{2} \ar[r] & b_{1} \ar[ur]}

Then φ^{i}(b_{i}) = φ^{i}(b_{m+i}) = b_{m} for all i in ℕ. By Lemma 2.10 again, |g(b_{i})| = |g(b_{m+i})| for all i in ℕ. Note that {b_{nm+i}}_{n=1}^{∞} converges to ∞ for i = 1, …, m. This implies g(b_{i}) = 0 for i = 1, 2, 3, . . .. We thus conclude again in this case that g vanishes on the branch of the φ-tree originated at a_{1}. In a similar manner, we assert that g vanishes on the branches of the φ-tree originated at all other merging points a_{2}, …, a_{j}, and thus on the crown of the φ-tree which contains ℕ. This gives g = 0. Consequently, T is an n-shift.

Corollary 3.3. Isometric disjointness preserving shifts T on c_{0} are exactly those in one of the following forms:

T ((x_{1}, x_{2}, …, x_{m}, …)) = (0, λ_{2}x_{1}, λ_{3}x_{2}, …, λ_{m+1}x_{m}, …),

or

T ((x_{1}, x_{2}, …, x_{m}, …)) = (λ_{1}x_{m}, λ_{2}x_{1}, λ_{3}x_{2}, …, λ_{m+1}x_{m}, …), m = 1, 2, 3, …,

after reordering the standard basis of c_{0}, if necessarily, where |λ_{k}| = 1 for k = 1, 2, 3, . . ..

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 11

Proof. It is indeed a direct consequence of Lemmas 3.1 and 3.2. More precisely, if T f = h · f ∘ φ then the φ-tree will be either rooted at ∞ or the one with a loop of m elements. In other words, the φ-tree of T is in either one of the following two forms.

[figure: ∞; a_{1}; a_{2}; a_{3}; ⋯],

or

\xymatrix{ & a_{m−2} \ar@{.>}[dl] & a_{m−1} \ar[l] \\ ⋮ \ar@{.>}[dr] &  &  & a_{m} \ar[ul] & a_{m+1} \ar[l] & a_{m+2} \ar[l] & a_{m+3} \ar[l] & ⋯ \ar[l] \\  & a_{2} \ar[r] & a_{1} \ar[ur]}

After reordering the standard basis of c_{0}, if necessarily, and then setting λ_{k} = h(k) for k = 1, 2, 3, …, we will arrive at the desired conclusion.

Theorem 3.4. Let T be an isometric disjointness preserving n-shift on c_{0}. Then T can be written as a product of n isometric disjointness preserving shifts on c_{0}.

Proof. Let φ: ℕ_{∞} → ℕ_{∞} be a continuous surjective map with φ(∞) = ∞ such that T f_{|X_{c}} = h · f ∘ φ, where X_{c} = {p ∈ ℕ: φ(p) ≠ ∞} and h is continuous on X_{c} with |h(x)| ≡ 1. Since ℕ is discrete, we may extend h continuously to ℕ by setting h_{|ℕ\X_{c}} ≡ 1. By Theorem 2.5 and Lemma 3.2, the φ-tree has exactly n joints and the crown of the φ-tree contains ℕ.

We claim that there exist n continuous maps φ_{1}, …, φ_{n} from ℕ_{∞} onto ℕ_{∞} sending ∞ to ∞ such that every φ_{i}-tree has exactly 1 joint, the crown of the φ_{i}-tree contains ℕ and φ = φ_{n} ∘ ⋯ ∘ φ_{1}.

By Lemma 3.1, there exist n disjoint sequences {a_{m}^{(i)}}_{m=1}^{∞}, i = 1, …, n, such that

a_{m}^{(i)} = φ (a_{m+1}^{(i)}) and ℕ = ⋃_{i=1}^{n} {a_{m}^{(i)} : m ∈ ℕ}. Note that the φ-tree has n joints and we may need to make a cut at each subsequent merged point in those initial φ-branches given in Lemma 3.1. Let φ^{∗} be the continuous map from ℕ_{∞} \ {a_{1}^{(1)}} onto ℕ_{∞} defined by

φ^{∗}(a_{j}^{(i)}) = a_{j}^{(i−1)}, i = 2, …, n and j ∈ ℕ,
φ^{∗}(a_{j+1}^{(1)}) = a_{j}^{(n)}, j ∈ ℕ,

and

φ^{∗}(∞) = ∞.

It is easy to see that φ^{∗} is bijective,

(φ^{∗})^{n}(a) = φ(a) for all a ≠ a_{1}^{(i)}, (1)

and

(φ^{∗})^{(i−1)}(a_{1}^{(i)}) = a_{1}^{(1)} for i = 2, …, n. (2)

Now define φ_{i}: ℕ_{∞} → ℕ_{∞}, i = 1, …, n, by

φ_{i}_{|ℕ_{∞}\{a_{1}^{(1)}}} = φ^{∗}

and

φ_{i}(a_{1}^{(1)}) = \begin{cases}∞, & if φ(a_{1}^{(i)}) = ∞, \\ (φ^{∗})^{−(n−i)}(b_{i}), & if φ(a_{1}^{(i)}) = b_{i} ≠ ∞. \end{cases}

By (1), to see φ = φ_{n} ∘ ⋯ ∘ φ_{2} ∘ φ_{1} we only need to check φ_{n} ∘ ⋯ ∘ φ_{1}(a_{1}^{(i)}) = φ(a_{1}^{(i)}) for all i = 1, …, n. In fact, if φ(a_{1}^{(i)}) = ∞ then by (2) we have

φ_{n} ∘ ⋯ ∘ φ_{1}(a_{1}^{(i)}) = φ_{n} ∘ ⋯ ∘ φ_{i}(a_{1}^{(1)}) = φ_{n} ∘ ⋯ ∘ φ_{i+1}(∞) = ∞.

If φ(a_{1}^{(i)}) = b_{i} ≠ ∞, then

φ_{n} ∘ ⋯ ∘ φ_{1}(a_{1}^{(i)}) = φ_{n} ∘ ⋯ ∘ φ_{i}(a_{1}^{(1)}) = φ_{n} ∘ ⋯ ∘ φ_{i+1}((φ^{∗})^{−(n−i)}(b_{i}))
= (φ^{∗})^{n−i}((φ^{∗})^{−(n−i)}(b_{i})) = b_{i}.

Hence, φ = φ_{n} ∘ ⋯ ∘ φ_{2} ∘ φ_{1}.

It is clear that φ_{i} is continuous from ℕ_{∞} onto ℕ_{∞} satisfying that φ_{i}(∞) = ∞, the φ_{i}-tree has exactly 1 joint at φ_{i}(a_{1}^{(1)}), and the crown of the φ_{i}-tree contains ℕ. Now define T_{i}: c_{0} → c_{0}, i = 1, …, n, by

T_{1}f(x) = \begin{cases}h(x) · f ∘ φ_{1}(x), & if φ_{1}(x) ≠ ∞, \\ 0, & if φ_{1}(x) = ∞, \end{cases}

and

T_{i}f(x) = \begin{cases}f ∘ φ_{i}(x), & if φ_{i}(x) ≠ ∞, \\ 0, & if φ_{i}(x) = ∞, \end{cases} i = 2, 3, …, n.

By Lemma 3.2, T_{1}, …, T_{n} are isometric disjointness preserving shifts. It is plain that T = T_{1} ∘ ⋯ ∘ T_{n}.

The following example demonstrates the idea we employed in the proof above.

Example 3.5. Let T be a 5-shift on c_{0} defined by

T (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}, x_{10}, x_{11}, x_{12}, x_{13}, …) =
(0, x_{13}, x_{1}, x_{2}, x_{2}, x_{3}, x_{3}, x_{4}, x_{5}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}, x_{10}, x_{11}, x_{12}, x_{13}, …).

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 13

Then the φ-tree is

[figure: ∞; 1; 3; 6; 11; 16; 21; ⋯; 7; 12; 17; 22; ⋯; 2; 4; 8; 13; 18; 23; ⋯; 5; 9; 14; 19; 24; ⋯; 10; 15; 20; 25; ⋯] .

First, we relabel the φ-tree as in the following. Note that the vertices a_{1}, a_{2}, a_{3}, a_{4} and a_{5} are the pivots in our machinery.

[figure: ∞; a_{1}; a_{6}; a_{11}; a_{16}; a_{21}; a_{26}; ⋯; a_{2}; a_{7}; a_{12}; a_{17}; ⋯; a_{3}; a_{8}; a_{13}; a_{18}; a_{23}; a_{28}; ⋯; a_{4}; a_{9}; a_{14}; a_{19}; a_{24}; ⋯; a_{5}; a_{10}; a_{15}; a_{20}; ⋯] .

Let φ^{∗} be the continuous map from ℕ_{∞} \ {a_{1}} onto ℕ_{∞} defined by

φ^{∗}(∞) = ∞ and φ^{∗}(a_{n+1}) = a_{n}, ∀n ∈ ℕ.

Observe that φ(a_{1}) = ∞, φ(a_{2}) = a_{6}, φ(a_{3}) = a_{18}, φ(a_{4}) = a_{3} and φ(a_{5}) = a_{4}. Following the proof of Theorem 3.4, we let φ_{1}(a_{1}) = ∞, φ_{2}(a_{1}) = (φ^{∗})^{−3}(a_{6}) = a_{9}, φ_{3}(a_{1}) = (φ^{∗})^{−2}(a_{18}) = a_{20}, φ_{4}(a_{1}) = (φ^{∗})^{−1}(a_{3}) = a_{4} and φ_{5}(a_{1}) = a_{4}. Moreover, we set φ_{i} = φ^{∗} elsewhere for i = 1, 2, 3, 4, 5. The φ_{i}-trees are given below.

φ_{1} :  [figure: ∞; a_{1}; a_{2}; a_{3}; a_{4}; ⋯],
φ_{2} :  \xymatrix{a_{1} \ar[rrrrr] & a_{2} \ar[l] & a_{3} \ar[l] & ⋯ \ar[l] & a_{8} \ar[l] & a_{9} \ar[l] & a_{10} \ar[l] & ⋯ \ar[l]},
φ_{3} :  \xymatrix{a_{1} \ar[rrrrr] & a_{2} \ar[l] & a_{3} \ar[l] & ⋯ \ar[l] & a_{19} \ar[l] & a_{20} \ar[l] & a_{21} \ar[l] & ⋯ \ar[l]},
φ_{4} :  \xymatrix{a_{1} \ar[rrr] & a_{2} \ar[l] & a_{3} \ar[l] & a_{4} \ar[l] & a_{5} \ar[l] & ⋯ \ar[l]},
φ_{5} :  \xymatrix{a_{1} \ar[rrr] & a_{2} \ar[l] & a_{3} \ar[l] & a_{4} \ar[l] & a_{5} \ar[l] & ⋯ \ar[l]}.

It is easy to see that φ = φ_{5} ∘ φ_{4} ∘ φ_{3} ∘ φ_{2} ∘ φ_{1}. In its original notations, we have

φ_{1} :  [figure: ∞; 1; 7; 2; 5; ⋯],
φ_{2} :  \xymatrix{1 \ar[rrrrr] & 7 \ar[l] & 2 \ar[l] & ⋯ \ar[l] & 4 \ar[l] & 9 \ar[l] & 15 \ar[l] & ⋯ \ar[l]},
φ_{3} :  \xymatrix{1 \ar[rrrrr] & 7 \ar[l] & 2 \ar[l] & ⋯ \ar[l] & 19 \ar[l] & 25 \ar[l] & 16 \ar[l] & ⋯ \ar[l]},
φ_{4} :  \xymatrix{1 \ar[rrr] & 7 \ar[l] & 2 \ar[l] & 5 \ar[l] & 10 \ar[l] & ⋯ \ar[l]},
φ_{5} :  \xymatrix{1 \ar[rrr] & 7 \ar[l] & 2 \ar[l] & 5 \ar[l] & 10 \ar[l] & ⋯ \ar[l]}.

Let T_{i}f = f ∘ φ_{i} for i = 1, 2, …, 5. Then we have 5 isometric disjointness preserving shifts on c_{0} such that T = T_{1} ∘ T_{2} ∘ T_{3} ∘ T_{4} ∘ T_{5}.

Corollary 3.6. Let T f = h · f ∘ φ be a disjointness preserving n-shift on c_{0}, or a quasin-shift on c_{0} such that the crown of the φ-tree contains ℕ. Then there exist n isometric disjointness preserving shifts S_{1}, …, S_{n} on c_{0} such that

T = h · S_{1} ∘ ⋯ ∘ S_{n}.

Proof. Apply Lemma 3.2 and Theorem 3.4 to the isometry f ↦ f ∘ φ.

In Example 2.6, there is a disjointness preserving quasi-shift on c_{0}, which is not a shift but the crown of its φ-tree contains ℕ. On the other hand, in Example 4.2 below, we shall have an isometric disjointness preserving quasi-2-shift on c_{0}, which can be written as a product of two isometric disjointness preserving shifts on c_{0} but its φ-tree does not contain the whole of ℕ. In particular, the converse of Corollary 3.6 does not hold.

4. Simplifying (quasi-)n-shifts on C_{0}(X)

By definition, we have

Proposition 4.1. The product of a disjointness preserving quasi-m-shift and a disjointness preserving quasi-n-shift on C_{0}(X) is a disjointness preserving quasi-(m + n)-shift on C_{0}(X).

In contrast to Proposition 4.1, the following example tells us that the product of any n 1-shifts may not be an n-shift.

Example 4.2. Let S_{1} and S_{2} be the isometric disjointness preserving shifts on c_{0} defined by

S_{1}(x_{1}, x_{2}, x_{3}, x_{4}, ⋯) = (x_{2}, x_{1}, x_{2}, x_{3}, x_{4}, ⋯),
S_{2}(x_{1}, x_{2}, x_{3}, x_{4}, ⋯) = (x_{2}, x_{1}, x_{1}, x_{3}, x_{4}, ⋯).

Then

(S_{1} ∘ S_{2})(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, …) = (x_{1}, x_{2}, x_{1}, x_{1}, x_{3}, x_{4}, x_{5}, x_{6}, …)

is a quasi-2-shift, but not a 2-shift, since (0, 1, 0, 0, 0, …) ∈ ⋂_{m=1}^{∞} ran(T^{m}).

Modifying the proof given for the case n = 1 in [3], we have the following result which says every n-shift on a Banach space E is similar to a ‘classical’ n-shift on a sequence space E_{S}.

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 15

Proposition 4.3. Suppose T is an n-shift on the Banach space E. Then there exists a Banach space E_{S} of scalar sequences, isomorphic and isometric to E, such that on E_{S} the n-shift T corresponds to the operator T_{S} defined by

T_{S}(a_{1}, a_{2}, …) = (\underbrace{0, ⋯, 0}_{n}, a_{1}, a_{2}, …).

Proof. Since T has closed range and corank n, there exist n elements x_{1}, x_{2}, …, x_{n} in E linear independent modulo T E such that E is the Banach space direct sum

E = span{x_{1}, x_{2}, …, x_{n}} ⊕ T E.

Let y ∈ E. Then there exist n unique scalars a_{1}(y), a_{2}(y), …, a_{n}(y) and an element y_{1} in E such that

y = a_{1}(y)x_{1} + a_{2}(y)x_{2} + ⋯ + a_{n}(y)x_{n} + T y_{1}.

Since T is injective, the choice of y_{1} is unique. Similarly, there exist another n unique scalars a_{n+1}(y), a_{n+2}(y), …, a_{2n}(y) and a unique element y_{2} in E such that

y_{1} = a_{n+1}(y)x_{1} + a_{n+2}(y)x_{2} + ⋯ + a_{2n}(y)x_{n} + T y_{2}.

Thus,

y =a_{1}(y)x_{1} + a_{2}(y)x_{2} + ⋯ + a_{n}(y)x_{n}
+ a_{n+1}(y)T x_{1} + a_{n+2}(y)T x_{2} + ⋯ + a_{2n}(y)T x_{n} + T^{2}y_{2}.

By induction, there exist a unique sequence of scalars {a_{m}(y)}_{m=1}^{∞} and a unique sequence of vectors {y_{m}}_{m=1}^{∞} in E such that for m = 1, 2, …, we have

y = ∑_{k=0}^{m} (a_{kn+1}(y)T^{k}x_{1} + a_{kn+2}(y)T^{k}x_{2} + ⋯ + a_{kn+n}(y)T^{k}x_{n}) + T^{m+1}y_{m+1}. (3)

Let E_{S} denote the vector space of sequences {a_{m}(y)}_{m=1}^{∞}. The mapping y ↦ {a_{m}(y)}_{m=1}^{∞} is linear and maps E onto E_{S}. Since ⋂_{k=1}^{∞} ran(T^{k}) = {0}, no non-zero vector is maped to the zero sequence. Thus the correspondence is a linear isomorphism.

Let ∥{a_{m}(y)}_{m=0}^{∞}∥ be defined as ∥y∥. Then the two spaces are isometric, and E_{S} is a Banach space. Equation (3) implies that

T y = ∑_{k=0}^{m} (a_{kn+1}(y)T^{k+1}x_{1} + a_{kn+2}(y)T^{k+1}x_{2} + ⋯ + a_{kn+n}(y)T^{k+1}x_{n}) + T^{m+2}y_{m+1}.

Therefore, the corresponding sequence for T y is \{\underbrace{0, 0, …, 0}_{n}, a_{1}(y), a_{2}(y), … \}. Thus T

is similar to the ‘classical’ n-shift T_{S} on E_{S}.

It is plausible that T = T_{1}^{n} where T_{1} is induced by the unilateral shift sending (x_{1}, x_{2}, …) to (0, x_{1}, x_{2}, …). However, it is not necessarily true that (0, x_{1}, x_{2}, …) belongs to E_{S} when (x_{1}, x_{2}, …) does. Even if it is the case, the shift operator T_{1} need

16 CHEN, JEANG, AND WONG

not be disjointness preserving on E when T is. Thus, this idea may not be implementable in some cases. We shall see in the following two examples that such a hope is indeed fruitless.

Example 4.4. This example tells us that there exists a compact Hausdorff space X such that C(X) admits an isometric disjointness preserving 2-shift which cannot be written as a product of two disjointness preserving shifts.

Let X = \{(\frac{1}{n}, i): n ∈ ℕ and i = 0, 1, 2\} ∪ \{(0, 0), (0, 1), (0, 2)\}. Then X is a compact Hausdorff space contained in ℝ^{2}. Note that ∞ is an isolated point in X_{∞} = X ∪ {∞}. Define φ: X_{∞} → X_{∞} by

φ(1, 0) = φ(1, 2) = φ(∞) = ∞,

φ(\frac{1}{n + 1}, 0) = (\frac{1}{n}, 0), φ(\frac{1}{n}, 1) = (\frac{1}{n}, 2) and φ(\frac{1}{n + 1}, 2) = (\frac{1}{n}, 1), ∀n ∈ ℕ,

and

φ(0, 0) = (0, 0), φ(0, 1) = (0, 2), φ(0, 2) = (0, 1). (4)

Define T: C(X) → C(X) by

T f(x) = f(φ(x)), ∀x ≠ (1, 0), (1, 2),
T f(1, 0) = 0 and T f(1, 2) = 0.

By Theorem 2.7, T is a disjointness preserving 2-shift. We shall show that T cannot be written as a product of two disjointness preserving shifts.

We first make some general observations. Let ψ: (X_{∞}, M_{ψ}) → (X_{∞}, ψ(M_{ψ})) be a relative homeomorphism induced by a shift on C(X), where M_{ψ} = {a, b} is the set of all two ψ-merging points with b ≠ ∞. Since ψ maps cluster points to cluster points, we have

{ψ(0, 0), ψ(0, 1), ψ(0, 2)} ⊆ {(0, 0), (0, 1), (0, 2)}. (5)

We are going to show that ψ maps {(0, 0), (0, 1), (0, 2)} onto itself without fixing any point.

We claim

M_{ψ} = {a, b} ⊄ {(0, 0), (0, 1), (0, 2)}. (6)

If it is not the case then X_{∞} \ {a, b} has only one cluster point while X_{∞} \ {ψ(b)} has two cluster points. It is impossible since ψ is a homeomorphism from X_{∞} \ {a, b} onto X_{∞} \ {ψ(b)}. As a consequence, the equality in (5) holds.

Since the ψ-tree contains X \{(0, 0), (0, 1), (0, 2)} (Theorem 2.11), a similar argument as in the proof of Lemma 3.1 will give us that {ψ^{n}(x): n ∈ ℕ} is a finite set for every x

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 17

in X. Hence we can assume ψ^{m}(a) = a for some positive integer m. Note that a can be the isolated point ∞. The ψ-tree is exactly the branch originated at a, i.e.,

\xymatrix{⋮ \ar[rr] & & a \ar[dl] \\ & ψ(a) \ar[ul] & b \ar[l] & ψ^{−1}(b) \ar[l] & ψ^{−2}(b) \ar[l] & ⋯ \ar[l]} . (7)

In this case, b must be an isolated point in X. In fact, if b ∈ {(0, 0), (0, 1), (0, 2)} then ψ^{i}(b) ∈ {(0, 0), (0, 1), (0, 2)} for all i in ℕ by (5). As a = ψ^{m}(a) = ψ^{m}(b), we have {a, b} ⊂ {(0, 0), (0, 1), (0, 2)}, a contradiction to (6). Since ψ: (X_{∞}, {a, b}) → (X_{∞}, {ψ(b)}) is a relative homeomorphism and both b and ∞ are isolated, ψ: X \{b} → X is a homeomorphism. Let Ψ be the inverse of ψ_{|X\{b}}. Then Ψ is a homeomorphism from X onto X \ {b}.

Now, we claim

ψ(x) ≠ x if x = (0, 0), (0, 1) or (0, 2).

Suppose not and assume, for example, that ψ(0, 0) = (0, 0) and thus Ψ(0, 0) = (0, 0). Let

A_{0} = \{(\frac{1}{n}, 0): n ∈ ℕ\} ∪ \{(0, 0)\}.

By the continuity of Ψ, there are at most finitely many points z_{1}, …, z_{k} in the open set A_{0} such that

Ψ(x) ∈ A_{0}, for x ∈ A_{0} \ {z_{1}, …, z_{k}}. (8)

Recall that the ψ-tree {a, ψ(a), …, ψ^{m−1}(a)} ∪ {b, Ψ(b), Ψ^{2}(b), … } displayed in (7) contains X \ {(0, 0), (0, 1), (0, 2)}. Then there exist positive integers N_{0} and N_{1} with N_{1} > N_{0} such that

z_{1}, …, z_{k} ∈ {a, ψ(a), …, ψ^{m−1}(a)} ∪ {b, Ψ(b), …, Ψ^{N_{0}}(b)} (9)

and

Ψ^{N_{1}}(b) ∈ A_{0}.

It follows from (8) and (9) that

Ψ^{N_{1}+k}(b) ∈ A_{0}, ∀k = 1, 2, ….

This implies the whole ψ-tree is contained in A_{0} eventually. Thus it is not dense in X, a contradiction.

At this moment, we arrive at the conclusion that for every relative homeomorphism ψ arising from a disjointness preserving shift on C(X) either one of the following two alternatives holds; namely,

ψ(0, 0) = (0, 1), ψ(0, 1) = (0, 2), ψ(0, 2) = (0, 0) (10) or

ψ(0, 0) = (0, 2), ψ(0, 1) = (0, 0), ψ(0, 2) = (0, 1). (11)

We are now ready to verify that T cannot be written as a product of two disjointness preserving shifts on C(X). Suppose, on the contrary, there were two disjointness preserving shifts S_{1} and S_{2} on C(X) such that T = S_{1} ∘ S_{2}. Let ψ_{i}: X_{∞} → X_{∞} be the relative homeomorphism induced by S_{i} for i = 1, 2. This gives φ(x) = ψ_{2}(ψ_{1}(x)). However, this cannot be true by (4), (10) and (11). Hence T cannot be written as a product of two disjointness preserving shifts.

In Example 4.4, although the 2-shift T cannot be written as a product of two shifts, there are anyway two quasi-shifts (not shifts) T_{1} and T_{2} on C(X) such that T = T_{1} ∘ T_{2}. In fact, let φ_{1} : X \ {(1, 0)} → X and φ_{2} : X \ {(1, 2)} → X be homeomorphisms defined by

φ_{1}_{|A_{0}} = φ_{|A_{0}} and φ_{1}(x) = x, ∀x ∈ X \ A_{0},

and

φ_{2}_{|X\A_{0}} = φ_{|X\A_{0}} and φ_{2}(x) = x, ∀x ∈ A_{0}.

Then the weighted composition operators T_{i}f = f ∘ φ_{i}, i = 1, 2, are quasi-shifts on C(X). It is easy to see that T = T_{1} ∘ T_{2}. Nevertheless, we can have a situation even worse than this.

Example 4.5. This example tells us that there is a compact connected Hausdorff space X such that C(X) admits an isometric disjointness preserving quasi-2-shift but no disjointness preserving quasi-shift at all. As a result, a disjointness preserving quasi-2-shift need not be a product of two disjointness preserving quasi-shifts.

For x, y in ℝ^{2}, let

l(x, y) = {tx + (1 − t)y: 0 ≤ t ≤ 1}

be the line segment joining x and y in ℝ^{2}. Denote by re^{iθ} the point (r cos θ, r sin θ) in ℝ^{2} and by

arc(re^{iθ_{1}}, re^{iθ_{2}}) = {re^{i(tθ_{1}+(1−t)θ_{2})} : 0 ≤ t ≤ 1}

the circular arc joining re^{iθ_{1}} to re^{iθ_{2}} in ℝ^{2}. Let O denote the origin (0, 0) in ℝ^{2}. We are going to construct a compact connected space X contained in the closed united disk in ℝ^{2}. Let

A_{1} = l(O, e^{i \frac{3π}{4}} ) ∪ l(\frac{1}{2}e^{i \frac{3π}{4}} , e^{i \frac{5π}{8}} ) ∪ l(\frac{1}{2}e^{i \frac{3π}{4}} , e^{i \frac{7π}{8}} ),
A_{2} = l(O, \frac{1}{2}e^{i \frac{3π}{8}} ) ∪ l(\frac{1}{4}e^{i \frac{3π}{8}} , \frac{1}{2}e^{i \frac{5π}{16}} ) ∪ l(\frac{1}{4}e^{i \frac{3π}{8}} , \frac{1}{2}e^{i \frac{7π}{16}} ),

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 19

and, in general,

A_{n} = l(O, \frac{1}{2^{n−1}} e^{i \frac{3π}{2^{n+1}}} ) ∪ l(\frac{1}{2^{n}} e^{i \frac{3π}{2^{n+1}}} , \frac{1}{2^{n−1}} e^{i \frac{5π}{2^{n+2}}} ) ∪ l(\frac{1}{2^{n}} e^{i \frac{3π}{2^{n+1}}} , \frac{1}{2^{n−1}} e^{i \frac{7π}{2^{n+2}}} ),

for n = 1, 2, . . .. Similarly, we let

B_{1} = l(O, e^{−i \frac{3π}{4}} ) ∪ l(\frac{1}{2}e^{−i \frac{3π}{4}} , e^{−i \frac{5π}{8}} ) ∪ l(\frac{1}{2}e^{−i \frac{3π}{4}} , e^{−i \frac{7π}{8}} ) ∪ arc(e^{−i \frac{5π}{8}} , e^{−i \frac{7π}{8}} ),
B_{2} = l(O, \frac{1}{2}e^{−i \frac{3π}{8}} ) ∪ l(\frac{1}{4}e^{−i \frac{3π}{8}} , \frac{1}{2}e^{−i \frac{5π}{16}} ) ∪ l(\frac{1}{4}e^{−i \frac{3π}{8}} , \frac{1}{2}e^{−i \frac{7π}{16}} ) ∪ arc(\frac{1}{2}e^{−i \frac{5π}{16}} , \frac{1}{2}e^{−i \frac{7π}{16}} ),

and, in general,

B_{n} = l(O, \frac{1}{2^{n−1}} e^{−i \frac{3π}{2^{n+1}}} ) ∪ l(\frac{1}{2^{n}} e^{−i \frac{3π}{2^{n+1}}} , \frac{1}{2^{n−1}} e^{−i \frac{5π}{2^{n+2}}} ) ∪ l(\frac{1}{2^{n}} e^{−i \frac{3π}{2^{n+1}}} , \frac{1}{2^{n−1}} e^{−i \frac{7π}{2^{n+2}}} )
∪ arc(\frac{1}{2^{n−1}} e^{−i \frac{5π}{2^{n+2}}} , \frac{1}{2^{n−1}} e^{−i \frac{7π}{2^{n+2}}} ),

for n = 1, 2, . . .. The following is the picture of A_{1}, A_{2}, A_{3}, B_{1}, B_{2} and B_{3}.

A_{1} B_{1}  O A_{2} B_{2} A_{3} B_{3} -1 -0.5 0 0.5 1-1  -0.5  0  0.5  1

Set

X = ⋃_{n=1}^{∞} A_{n} ∪ B_{n}.

It is clear that each pair of A_{1}, A_{2}, …, B_{1}, B_{2}, … intersects exactly at the origin O. Let

φ: (X, \{e^{i \frac{5π}{8}} , e^{i \frac{3π}{4}} , e^{i \frac{7π}{8}} \}) → (X, \{e^{−i \frac{3π}{4}} \})

be a relative homeomorphism such that φ is onto X, and one-to-one from X except for

φ(e^{i \frac{5π}{8}} ) = φ(e^{i \frac{3π}{4}} ) = φ(e^{i \frac{7π}{8}} ) = e^{−i \frac{3π}{4}} .

Moreover, we assume that φ(A_{1}) = B_{1}, φ(A_{n+1}) = A_{n}, and φ(B_{n}) = B_{n+1} for n = 1, 2, . . .. Then the φ-tree has exactly two joints (both at e^{−i \frac{3π}{4}} ) and the composition operator T f = f ∘ φ is an isometric disjointness preserving quasi-2-shift on C(X).

On the other hand, there is no disjointness preserving quasi-shift on C(X) at all. In fact, suppose there were one. By Theorem 2.3(3), there would be two points a and b in X such that the quotient space X⧸∼_{a,b} is homeomorphic to X, where the equivalence relation ∼_{a,b} in X is defined by identifying a and b. But this is impossible.

With a trivial modification, one can also obtain examples of compact connected Hausdorff spaces X_{n} such that C(X_{n}) admits isometric disjointness preserving quasin-shifts but not any disjointness preserving quasi-k-shift for k = 1, 2, …, n − 1 and n = 2, 3, . . ..

Remark 4.6. When X does not contain isolated points, it is shown in [13] that every isometric quasi-n-shift on C_{0}(X) is disjointness preserving. Therefore, Example 4.5 gives also an example of an isometric quasi-n-shift which cannot be written as a product of n isometric quasi-shifts.

Question 4.7. How can we study (quasi-)n-shifts in term of (quasi-)shifts?

For a partial answer to Question 4.7, we show below that every “simple” disjointness preserving quasi-n-shifts on C_{0}(X) can be dilated to a product of n quasi-shifts.

Definition 4.8. A φ-tree is said to be simple if all φ-vanishing points in X are isolated points. A disjointness preserving quasi-n-shift T is said to be simple if its associated φ-tree is simple.

We note that all disjointness preserving quasi-n-shifts on a compact Hausdorff space are simple by Theorem 2.3(3).

Lemma 4.9. Let T be a simple disjointness preserving quasi-n-shift on C_{0}(X) with exactly n vanishing points. Let

X̃ = X ∪ ℕ (disjoint union),

and thus C_{0}(X̃) = C_{0}(X) ⊕ c_{0}. Then the simple quasi-n-shift T̃ = T ⊕ I can be written as a product of n simple quasi-shifts on C_{0}(X̃). In case T is an isometry, we can assume that these quasi-shifts are also isometries.

Proof. Let X_{0} = {p ∈ X: δ_{p} ∘ T = 0} = {p_{1}, …, p_{n}} and X_{c} = X \ X_{0}. Write f̃ in C_{0}(X̃) = C_{0}(X) ⊕ c_{0} as f ⊕ (f_{k}); namely,

f̃_{|X} = f and f̃(k) = f_{k} for k in ℕ.

Let s be the unilateral shift on c_{0}, i.e.,

s((x_{1}, x_{2}, …)) = (0, x_{1}, x_{2}, …).

Define S_{1} : C_{0}(X̃) → C_{0}(X̃) by S_{1} = I ⊕ s, i.e.,

S_{1}(f̃) = f ⊕ (f_{k−1}),

where we set f_{0} = 0 for convenience. Define S_{2} : C_{0}(X̃) → C_{0}(X̃) by

(S_{2}f̃)_{|X_{c}} = (T f)_{|X_{c}},
(S_{2}f̃)(p_{1}) = 0,
(S_{2}f̃)(p_{k+1}) = f_{k} for k = 1, 2, …, n − 1,

and

S_{2}f̃(k) = f_{n+k−1} for k = 1, 2, ….

Clearly, both S_{1} and S_{2} are simple disjointness preserving quasi-shifts on C_{0}(X̃). S_{1} is always an isometry, and so is S_{2} whenever T is. Observe that S_{1}^{n−1} = I ⊕ s^{n−1}. It then follows

(S_{2}S_{1}^{n−1}f̃)_{|X_{c}} = (T (S_{1}^{n−1}f̃)_{|X})_{|X_{c}} = (T f)_{|X_{c}},
(S_{2}S_{1}^{n−1}f̃)(p_{1}) = 0,
(S_{2}S_{1}^{n−1}f̃)(p_{k+1}) = S_{1}^{n−1}f̃(k) = 0 for k = 1, 2, …, n − 1,

and

(S_{2}S_{1}^{n−1}f̃)(k) = (S_{1}^{n−1}f̃)(n + k − 1) = f_{k} = f̃(k) for k = 1, 2, ….

Hence

S_{2}S_{1}^{n−1} = T ⊕ I = (\begin{matrix}T & 0 \\ 0 & I \end{matrix})

in C_{0}(X̃) = C_{0}(X) ⊕ c_{0}.

Lemma 4.10. Let T be a simple disjointness preserving quasi-n-shift on C_{0}(X) with m vanishing points p_{1}, …, p_{m}. Let l = n − m. Let X̃ = X ⋃ ℕ be a disjoint union. Then the simple quasi-n-shift T ⊕ I on C_{0}(X̃) = C_{0}(X) ⊕ c_{0} can be written as

T ⊕ I = T_{l}S_{1}^{m},

where S_{1} is a simple isometric quasi-shift on C_{0}(X̃) and T_{l} is a quasi-l-shift on C_{0}(X̃) without vanishing points. In case T is an isometry, we can assume that T_{l} is an isometry as well.

Proof. Let X_{c} = X \ {p_{1}, …, p_{m}}. Define T_{l}: C_{0}(X̃) → C_{0}(X̃) by

(T_{l}f̃)_{|X_{c}} = (T f)_{|X_{c}},
(T_{l}f̃)(p_{k}) = f_{k} for k = 1, …, m,

and

(T_{l}f̃)(k) = f_{m+k} for k = 1, 2, ….

Let S_{1} = I ⊕ s as in Lemma 4.9. Then

(T_{l}S_{1}^{m}f̃)_{|X_{c}} = T (S_{1}^{m}f̃_{|X})_{|X_{c}} = T f_{|X_{1}},
(T_{l}S_{1}^{m}f̃)(p_{k}) = S_{1}^{m}f̃(k) = 0 for k = 1, 2, …, m,

and

(T_{l}S_{1}^{m}f̃)(k) = S_{1}^{m}f̃(m + k) = f̃(k) for k = 1, 2, ….

Hence

T_{l}S_{1}^{m} = T ⊕ I = (\begin{matrix}T & 0 \\ 0 & I \end{matrix}) .

Finally, we note that T_{l} is a quasi-l-shift without vanishing point, and isometric whenever T is.

Remark 4.11. If l = 0 in Lemma 4.10 then

T ⊕ I = T_{0}S_{1}^{n};

that is, every simple disjointness preserving quasi-n-shift on C_{0}(X) with exactly n vanishing points can be dilated to a product of an invertible (composition) operator T_{0} and n copies of the isometric quasi-shift S_{1} = I ⊕ s. We note that S_{2} = T_{0} ∘ S_{1} is the one given in Lemma 4.9.

Recall that a bounded linear operator T between Banach spaces is an injection if and only if it is injective and has closed range.

Lemma 4.12. Let T be a disjointness preserving injection (resp. isometry) from C_{0}(X) into C_{0}(Y) of corank n. Suppose there is no vanishing points of T. Then T can be written as a product of n disjointness preserving injections (resp. isometries) of corank 1.

Proof. By a result in [12], we can suppose T f = h · f ∘ φ for some continuous map φ from Y onto X and continuous bounded and away from zero scalar function h on Y. Moreover, if M_{φ} = {y ∈ Y : #φ^{−1}(φ(y)) ≥ 2} is the set of all merging points of T then #(M_{φ}) − #φ(M_{φ}) = n. Fix two distinct points a and b in M_{φ} with φ(a) = φ(b). Let Y⧸∼_{a,b} be the quotient space of Y by identifying a and b. Define φ̃^{a,b}: Y⧸∼_{a,b} → X by φ̃^{a,b}([y]) = φ(y). Let M_{φ^{a,b}} = {[y] ∈ Y⧸∼_{a,b}: #(φ̃^{a,b})^{−1}(φ̃^{a,b}([y])) ≥ 2}. Then M_{φ^{a,b}} ⊂ [M_{φ}] and #(M_{φ^{a,b}}) − #φ̃^{a,b}(M_{φ^{a,b}}) = n − 1. On the other hand, we define φ_{1} : Y → Y⧸∼_{a,b} by φ_{1}(y) = [y]. Note that M_{φ_{1}} = {a, b} is the set of all φ_{1}-merging points in Y. Clearly,

φ = φ̃^{a,b} ∘ φ_{1}.

DISJOINTNESS PRESERVING n-SHIFTS ON C_{0}(X) 23

Let g be a continuous scalar function on Y satisfying either one of the following conditions:

g(a) = |\frac{h(b)}{h(a)}| , g(b) = 1, and |\frac{h(b)}{h(a)}| ≤ g ≤ 1 when |\frac{h(b)}{h(a)}| ≤ 1;

g(a) = 1, g(b) = |\frac{h(a)}{h(b)}| , and 1 ≥ g ≥ |\frac{h(a)}{h(b)}| when |\frac{h(b)}{h(a)}| ≥ 1.

Define h_{2}(y) = |h(y)|g(y) and h_{1}(y) = \frac{h(y)}{h_{2}(y)} for y in Y. Then

h_{2}(a) = h_{2}(b) and h_{1}(y) · h_{2}(y) = h(y), ∀y ∈ Y.

Define a scalar function h̃_{2}^{a,b} on Y⧸∼_{a,b} by h̃_{2}^{a,b}([y]) = h_{2}(y). Then h_{1} and h̃_{2}^{a,b} are continuous, bounded and away from zero on Y and Y⧸∼_{a,b}, respectively. Moreover,

h_{1}(y) · (h̃_{2}^{a,b} ∘ φ_{1})(y) = h_{1}(y)h̃_{2}^{a,b}([y]) = h(y), ∀y ∈ Y.

Define T̃^{a,b}: C_{0}(X) → C_{0} (Y⧸∼_{a,b}) by

T̃^{a,b}f = h̃_{2}^{a,b} · f ∘ φ̃^{a,b},

and Q̃^{a,b}: C_{0} (Y⧸∼_{a,b}) → C_{0}(Y) by

Q̃^{a,b}f̃^{a,b} = h_{1} · f̃^{a,b} ∘ φ_{1}.

Then, Q̃^{a,b} ∘ T̃^{a,b}: C_{0}(X) → C_{0}(Y) satisfies that

(Q̃^{a,b} ∘ T̃^{a,b})f = h_{1} · (T̃^{a,b}f) ∘ φ_{1} = h_{1} · (h̃_{2}^{a,b} · f ∘ φ̃^{a,b}) ∘ φ_{1}
= h_{1} · (h̃_{2}^{a,b} ∘ φ_{1}) · f ∘ (φ̃^{a,b} ∘ φ_{1})
= h · f ∘ φ = T f, ∀f ∈ C_{0}(X).

Hence T = Q̃^{a,b} ∘ T̃^{a,b}.

Clearly Q̃^{a,b} is a disjointness preserving injection of corank one, and T̃^{a,b} is a disjointness preserving injection of corank n − 1. Both Q̃^{a,b} and T̃^{a,b} will be isometries whenever T is. The above construction can be applied to further decompose T̃^{a,b} into n − 1 disjointness preserving injections (resp. isometries) of corank one.

Theorem 4.13. Let T be a simple disjointness preserving quasi-n-shift on C_{0}(X) with m vanishing points. Let l = n − m and let X̃ = X ∪ ℕ (disjoint union). Then T ⊕ I on C_{0}(X̃) = C_{0}(X) ⊕ c_{0} is a product of m copies of the isometric disjointness preserving quasi-shift S_{1} = I ⊕s and l corank one disjointness preserving injections Q_{1}, Q_{2}, …, Q_{l}, i.e.

(\begin{matrix}T & 0 \\ 0 & I \end{matrix}) = Q_{1}Q_{2} ⋯ Q_{l}S_{1}^{m}.

Here, s is the unilateral shift on c_{0}. In case m = n, the right hand side becomes QS_{1}^{n} for some invertible composition operator Q on C_{0}(X̃). Moreover, all Q_{1}, Q_{2}, …, Q_{l} can be chosen to be isometries whenever T is.

Proof. It follows from Lemmas 4.9, 4.10 and 4.12.

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Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung, Taiwan, 80424, Republic of China.

E-mail address: wong@math.nsysu.edu.tw