Lobachevskii Journal of Mathematics Vol. 6, 2000, 39–46 \begin{matrix}c \\ ◯ \end{matrix} S. A. Grigorian, R. N. Gumerov, A. V. Kazantsev

S. A. Grigorian, R. N. Gumerov, A. V. Kazantsev

GROUP STRUCTURE IN FINITE COVERINGS OF COMPACT SOLENOIDAL GROUPS

(submitted by M.A.Malakhaltsev)

Abstract. Let p : X → G be an n - fold covering of a compact solenoidal group G by a connected topological space X. We prove that there exists a group structure in X turning p into a homomorphism between compact abelian groups

Introduction

It is well known that the technique of covering maps is widely used in the theory of algebraic equations with functional coefficients (see, e. g., [3] and [4]). The questions considered in this note arose in the study of equations with coefficients in Banach algebras of generalized analytic functions. In this case one deals with solenoidal groups (see [1], [5]) and their coverings. The problem on the existence of a group structure in a covering space is also motivated by the well-known result on covering groups in classical topological group theory [8, §51]. But in the present note we assume neither arcwise connectedness nor local connectedness of spaces.

Throughout the note we shall denote by G a compact solenoidal group. Recall that this means the existence of a continuous homomorphism τ from the additive group of the real numbers ℝ into G such that an one - parameter subgroup τ(ℝ) = {g_{t} ∈ G/t ∈ ℝ} is dense in G.

For a positive integer n let p : X → G be an n - fold covering of G by a connected topological space X. That is, every point g of G has a neighborhood W such that the inverse image p^{−1}(W) can be written as the union of nonempty disjoint open sets V_{1}, V_{2}, …, V_{n} in X and for each V_{k} the restriction of p to V_{k} is a homeomorphism of V_{k} onto W. Briefly, W is said to be evenly covered by p.

The purpose of this note is the construction of a multiplication in X turning it into a compact abelian group and p : X → G into homomorphism of groups. To do it, we shall use a lifting of a curve {gg_{t}/t ∈ ℝ} passing through a point g in G to a curve in X passing through a point of p^{−1}(g). The note consists of two sections. In the first section preliminary lemmas are given. The second section contains the main result.

Preliminary lemmas

Lemma 1. Let p : X → G be an n - fold covering of a compact group G by a topological space X. Then X can be imbedded in G × ℂ^{m} for some positive integer m.

Proof. Consider a finite open covering {W_{1}, W_{2}, …, W_{m}} of the group G by evenly covered sets. Let {ϕ_{1}, ϕ_{2}, …, ϕ_{m}} be a partition of unity on G dominated by this covering.

Define a family of continuous functions f̂_{k} : X → ℂ for k = 1, 2, …, m by the equation f̂_{k}(x) = ϕ̂_{k}(x)f_{k}(x), where ϕ̂_{k}(x) = ϕ_{k}(p(x)) for x ∈ X, and

f_{k}(x) = \{ \begin{matrix}0, & if x ∉ p^{−1}(W_{k}) = ⨆_{l=1}^{n} V_{l}^{k} \\ exp[i\frac{2π}{n}(l − 1)], & if x ∈ V_{l}^{k}, l = 1, 2, …, n. \end{matrix}

Note that it is easy to check that the functions ϕ̂_{k} : X → [0, 1] for k = 1, 2, …, m form a partition of unity on X dominated by the family {p^{−1}(W_{1}), p^{−1}(W_{2}), …, p^{−1}(W_{m})}.

Let us define a continuous mapping Ψ : X → G × ℂ^{m} by the rule Ψ(x) = (p(x), f̂_{1}(x), f̂_{2}(x), …, f̂_{m}(x)) for x ∈ X.

We assert that Ψ is an imbedding. To prove this we need only to show that Ψ is injective (because X is compact). Suppose that Ψ(x_{1}) = Ψ(x_{2}) for x_{1}, x_{2} ∈ X. Then f̂_{k}(x_{1}) = f̂_{k}(x_{2}) for each k = 1, 2, …, m, p(x_{1}) = p(x_{2}) and x_{1}, x_{2} ∈ p^{−1}(g) for some g ∈ G. Since the family {ϕ_{1}, ϕ_{2}, …, ϕ_{m}} is a partition of unity on G, ϕ_{j}(g) > 0 for some j. It follows that g ∈ W_{j} and x_{1}, x_{2} ∈ p^{−1}(W_{j}). Thus, if x_{1} and x_{2} were distinct elements of X, then they would belong to different slices of p^{−1}(W_{j}) and f̂_{j}(x_{1}) wouldn’t equal f̂_{j}(x_{2}). This completes the proof of the lemma.

We now define the function F : X → ℂ^{m} by the rule F (x) = (f̂_{1}(x), f̂_{2}(x), …, f̂_{m}(x)) for x ∈ X, where the f̂_{k}’s are constructed in the proof of Lemma 1.

For an arbitrary element g ∈ G with p^{−1}(g) = {x_{1}(g), x_{2}(g), …, x_{n}(g)} we set

δ(g) = min{∥F (x_{i}(g)) − F (x_{j}(g))∥ : i < j, i, j = 1, 2, …, n},

where

∥F (x) − F (y)∥ = ∑_{k=1}^{m} |f̂_{k}(x) − f̂_{k}(y)|

for x, y ∈ X. Since the mapping Ψ is injective , we have the positive function δ : G → (0, +∞) : g ↦ δ(g). Obviously, it is continuous, so that δ_{0} = min{δ(g)/g ∈ G} > 0.

Let us fix an element e ∈ G. It is easily seen that in the proof of Lemma 1 a covering {W_{1}, W_{2}, …, W_{m}} of G can be taken such that

e ∈ W_{1} \ ⋃_{k=2}^{m} W_{k} (1)

This condition allows us to choose a point y_{0} ∈ p^{−1}(e) satisfying f̂_{1}(y_{0}) = 1, and also guarantees the correctness of the assertion of the following lemma.

Lemma 2. Let x be a point in X such that the inequality

|f̂_{1}(x) − 1| < ε_{0} = \frac{sin \frac{π}{n}}{1 + sin \frac{π}{n}}

holds for n > 1. Then x ∈ V_{1}^{1}.

Proof. Since |f̂_{1}(x)| > 1 − ε_{0} > 0, it follows that x ∈ p^{−1}(W_{1}). We also have the inequality

|ϕ̂_{1}(x) − 1| = ||f̂_{1}(x)| − 1| ≤ ε_{0}, (2)

which implies

ϕ̂_{1}(x) > 1 − ε_{0} (3)

To obtain a contradiction, suppose that x ∉ V_{1}^{1}. Then f_{1}(x) = exp(i\frac{2π}{n}k) for some k such that 1 ≤ k ≤ n − 1. Using (2) and (3) , we have

|f̂_{1}(x) − 1| ≥ |f̂_{1}(x) − ϕ̂_{1}(x)| − |ϕ̂_{1}(x) − 1| =

= ϕ̂_{1}(x)|exp(i\frac{2π}{n} k) − 1| − |ϕ̂_{1}(x) − 1| >

> (1 − ε_{0})2 sin \frac{π}{n} − ε_{0} = ε_{0}.

This contradiction completes the proof.

We now define the function ρ : G × G → ℝ by the formula

ρ(g_{1}, g_{2}) = ∏_{σ}{ ∑_{x∈p^{−1}(g)} ∥F (x) − F (σ(x))∥}

for g_{1}, g_{2} ∈ G, where σ runs through all bijections between sets p^{−1}(g_{1}) and p^{−1}(g_{2}). Let us index elements of this sets:

p^{−1}(g_{j}) = {x_{1}(g_{j}), x_{2}(g_{j}), …, x_{n}(g_{j})}, j = 1, 2.

Then we can rewrite ρ(g_{1}, g_{2}) as follows:

ρ(g_{1}, g_{2}) = ∏_{σ∈S_{n}}{∑_{l=1}^{n} ∥F (x_{l}(g_{1})) − F (x_{σ(l)}(g_{2}))∥}, (4)

where the same letter σ is used to denote an element of symmetric group S_{n} of degree n. Note that ρ(g_{1}, g_{2}) = ρ(g_{2}, g_{1}), ρ(g, g) = 0 for all g_{1}, g_{2} ∈ G and ρ is continuous on G × G. Using the representation (4) and standard arguments, it is easily seen that for any ε > 0 and any g ∈ G there exists a neighborhood W of g such that ρ(g_{1}h, g_{2}h) < ε for all g_{1}, g_{2} ∈ W and all h ∈ G.

Now we establish one inequality for the function ρ, which will be used later. For let α_{0} = \frac{1}{2}min\{δ_{0}, ε_{0}\}. It is clear that α_{0} > 0. We claim that if

∑_{l=1}^{n} ∥F (x_{l}(g)) − F (x_{l}(h))∥ = \frac{α_{0}}{2} for some g, h ∈ G,

then

ρ(g, h) ≥ (\frac{α_{0}}{2} )^{n!} (5)

In fact, if σ is not the identity permutation Id, then we have

∑_{l=1}^{n} ∥F (x_{l}(g)) − F (x_{σ(l)}(h))∥ ≥ ∑_{l=1}^{n} ∥F (x_{l}(g)) − F (x_{σ(l)}(g))∥ −

− ∑_{l=1}^{n} ∥F (x_{σ(l)}(g)) − F (x_{σ(l)}(h))∥ ≥

≥ α_{0} − \frac{α_{0}}{2} = \frac{α_{0}}{2} .

This estimate implies the desired inequality.

Recall that for each element g ∈ G there exists a dense curve {gg_{t}/t ∈ ℝ} in G. Using the lifting path lemma ( see, e. g., [7, Lemma 15.1] ), for each x ∈ X, we obtain a curve {T_{t}(x)/t ∈ ℝ} in X such that pT_{t}(x) = p(x)g_{t} and T_{0}(x) = x. Also, for each t ∈ ℝ we have a map T_{t} : X → X : x ↦ T_{t}(x).

Lemma 3. Let e be an element of G. Let {W_{1}, W_{2}, …, W_{m}} be a covering of G by evenly covered sets such that the condition (1) holds. Let p^{−1}(e) ∩ V_{1}^{1} = {y_{0}}. Then there exists a neighborhood V_{0} ⊂ V_{1}^{1} of point y_{0} satisfying the following property:

if T_{t_{0}}(V_{0}) ∩ V_{0} ≠ ∅ for some t_{0} ∈ ℝ, then T_{t_{0}}(V_{0}) ⊂ V_{1}^{1}.

Proof. A partition p^{−1}(W_{1}) = ⨆_{l=1}^{n} V_{l}^{1} determines the enumeration of elements

of a set p^{−1}(g) for each g ∈ W_{1}, namely, x_{l}(g) = (p|_{V_{l}^{1}})^{−1}(g), l = 1, 2, …, n, so that, y_{0} = x_{1}(e).

As we have pointed out earlier, there exists a neighborhood W_{0} ⊂ W_{1} of e such that

ρ(g_{1}h, g_{2}h) < (\frac{α_{0}}{2} )^{n!} (6)

for all g_{1}, g_{2} ∈ W_{0} and h ∈ G.

We set V_{0} = V_{1}^{1} ∩ p^{−1}(W_{α_{0}}), where

W_{α_{0}} = \{g ∈ W_{0} : ∑_{l=1}^{n} ∥F (x_{l}(e)) − F (x_{l}(g))∥ < \frac{α_{0}}{4} \}

is obviously an open set. Thus, V_{0} is a neighborhood of y_{0} which is contained in V_{1}^{1}.

We now fix arbitrary elements g, h ∈ W_{α_{0}}. Define the continuous function r : ℝ → ℝ by the formula

r(t) = ∑_{l=1}^{n} ∥F (T_{t}(x_{l}(g))) − F (T_{t}(x_{l}(h)))∥.

We shall show that the inequality r(t) < α_{0}/2 holds for each t ∈ ℝ. (Note, it follows from the definition of W_{α_{0}} that r(0) < α_{0}/2 ). To obtain a contradiction, suppose that r(t_{0}) = α_{0}/2 for some t_{0} ∈ ℝ. In view of the inequality (5) and the equations ρ(T_{t}(x_{l}(g))) = gg_{t}, ρ(T_{t}(x_{l}(h))) = hg_{t}, we obtain

ρ(gg_{t}, hg_{t}) ≥ (\frac{α_{0}}{2} )^{n!}

a contradiction of the inequality (6). Thus, r(t) < α_{0}/2 for all t ∈ ℝ.

Finally, let x ∈ T_{t_{0}}(V_{0}) ∩ V_{0} for some t_{0} ∈ ℝ. Take an arbitrary y ∈ V_{0}. We shall show that

|f̂_{1}(T_{t_{0}}(y)) − 1| < α_{0} (7)

In view of Lemma 2, this inequality will imply that T_{t_{0}}(y) ∈ V_{1}^{1}, so that, Lemma 3 will be proved.

Obviously, y = x_{1}(h) for some h ∈ W_{α_{0}}, so that the left-hand side of (7) can be rewritten and we have

|f̂_{1}(T_{t_{0}}(x_{1}(h))) − f̂_{1}(x_{1}(e))| ≤ |f̂_{1}(T_{t_{0}}(x_{1}(h))) − f̂_{1}(T_{t_{0}}(x_{1}(g)))| +

+|f̂_{1}(x_{1}(g′)) − f̂_{1}(x_{1}(e))| (8)

This estimate holds because f̂_{1}(y_{0}) = 1, y_{0} = x_{1}(e) and x = x_{1}(g′) = T_{t_{0}}(x_{1}(g)) for some g, g′ ∈ W_{α}. Further, the first term on the right-hand side of (8) is at most r(t_{0}). From the definition of the neighborhood W_{α_{0}} follows that the second term on the right-hand side of (8) is less than α_{0}/4. Since r(t_{0}) < α_{0}/2, the left-hand side of (8) is less than α_{0}. Thus, the estimate (7) is valid. This completes the proof of Lemma 3.

Corollary. For each t ∈ ℝ the mapping T_{t} : X → X : x → T_{t}(x) is a homeomorphism.

Proof. One can readily verify that for each t ∈ ℝ the composition of T_{t} and T_{−t} is the identity mapping on X. So that, it sufficies to show the continuity of T_{t}.

Let us fix points t ∈ ℝ and x ∈ X. Of course, the constructions from the proof of the inequality r(t) < α_{0}/2 are applied to an element p(x) = e ( with α instead of α_{0} ). It follows that for each α ∈ (0, α_{0}) there exists a neighborhood W_{α} of e such that the inequality

∑_{l=1}^{n} ∥F (T_{t}(x_{l}(g))) − F (T_{t}(x_{l}(e)))∥ < \frac{α}{2}

holds for every g ∈ W_{α}.

Since x = x_{k}(e) for some k = 1, 2, …, n, we have

∥F (T_{t}(y)) − F (T_{t}(x))∥ < \frac{α}{2}

for all y ∈ V_{k}^{1}∩p^{−1}(W_{α}). This implies that the composition F ∘T_{t} is continuous on X. Thus the diagonal mapping ψ ∘ T_{t} = (p ∘ T_{t})△(F ∘ T_{t}) is continuous as well. From this the desired conclusion follows.

Further, we set ℝ_{x} = {T_{t}(x) : t ∈ ℝ}. Since X is a compact topological space, it follows that p is a closed map [2, Corollary 3.1.11], so that p(ℝ̄_{x}) = G.

Lemma 4. A set ℝ_{x} is dense in a connected space X for each x ∈ X.

Proof. We shall show that ∂ℝ̄_{x} = ∅.

By the definition of ℝ_{x}, we have T_{t}(ℝ_{x}) = ℝ_{x} for each t ∈ ℝ, hence, T_{t}(∂ℝ̄_{x}) = ∂ℝ̄_{x}.

To obtain a contradiction, suppose that there exists a point y ∈ ∂ℝ̄_{x}. This implies the following inclusion for interiors: Intℝ̄_{y} ⊂ Int∂ℝ̄_{x} = ∅. But, by the Baire category theorem, we have that Intℝ̄_{y} is non-empty. This contradiction completes the proof of the lemma.

Remark. Certainly, the assertion of Lemma 4 is equivalent to the following. For each neighborhood V in X and each point x ∈ X there exists a point t ∈ ℝ such that x ∈ T_{t}(V).

Main result

Theorem. Let p : X → G be an n - fold covering of a compact solenoidal group G by a connected topological space X. Then there exists a group structure in X turning p : X → G into a homomorphism between compact abelian groups.

Proof. In what follows, let e be an identity of G.

Obviously, there exists a symmetric neighborhood W of e, such that W^{2} ⊂ W_{α_{0}}. We set U_{0} = p_{1}^{−1}(W) ⊂ V_{1}^{1}, where p_{1} = p|_{V_{1}^{1}}. Note that U_{0} ⊂ V_{0} = p^{−1}(W_{α_{0}}) ∩ V_{1}^{1}.

We now define a multiplication on the set U_{0} which takes values in V_{0} by the formula

xy = p_{1}^{−1}(p(x)p(y)). (9)

Further, we shall extend this operation on X. For any elements x and y in X, by Lemma 4, there exist t_{1}, s_{1} ∈ ℝ and x_{1}, y_{1} ∈ U_{0} such that x = T_{t_{1}}(x_{1}) and y = T_{s_{1}}(y_{1}). In this case we set

x · y = T_{t_{1}+s_{1}}(x_{1}y_{1}). (10)

First of all we must show that this algebraic operation is well-defined. For let x = T_{t_{2}}(x_{2}), y = T_{s_{2}}(y_{2}); x_{2}, y_{2} ∈ U_{0}, t_{2}, s_{2} ∈ ℝ. We shall show that

T_{t_{1}+s_{1}}(x_{1}y_{1}) = T_{t_{2}+s_{2}}(x_{2}y_{2}).

Clearly, x_{1} = T_{t_{2}−t_{1}}(x_{2}) and y_{1} = T_{s_{2}−s_{1}}(y_{2}). Then we have

T_{t_{1}+s_{1}}(x_{1}y_{1}) = T_{t_{1}+s_{1}}(T_{t_{2}−t_{1}}(x_{2}y_{1})) =
= T_{s_{1}+t_{2}}(x_{2}y_{1}) = T_{s_{1}+t_{2}}(x_{2} T_{s_{2}−s_{1}}(y_{2})) =
= T_{s_{1}+t_{2}}(T_{s_{2}−s_{1}}(x_{2}y_{2})) = T_{t_{2}+s_{2}}(x_{2}y_{2}).

Here, we have used Lemma 5, which, in order to avoid interrupting the proof of the required equation, is stated and proved below.

Lemma 5. Let x, y, z ∈ U_{0} and y = T_{t_{0}}(x) for some t_{0} ∈ ℝ. Then

y z = T_{t_{0}}(x z).

Proof. Using (9) and the definition of T_{t_{0}}, it is easily checked that

p(y z) = p(T_{t_{0}}(x z)).

Therefore, it is sufficient to show that T_{t_{0}}(x z) ∈ V_{1}^{1}. But this follows immediately from Lemma 3, because y = T_{t_{0}}(x) ∈ T_{t_{0}}(V_{0}) ∩ V_{0}. Thus, Lemma 5 is proved.

Continuation of the proof of the theorem. Note, that if x and y belong to U_{0}, then x · y = T_{0}(x y) = x y. Hence, it is unnecessary to use the notation x · y.

Obviously, y_{0}x = xy_{0} = x for each x ∈ X. In other words, y_{0} is the identity element with respect to the multiplication introduced in X.

For x ∈ U_{0} we set x^{−1} = p_{1}^{−1}(p(x)^{−1}). Then we have

x x^{−1} = x^{−1} x = y_{0} (11)

Since W is symmetric, x^{−1} ∈ U_{0} for each x ∈ U_{0}. It is also clear that there exists a unique element x^{−1} in U_{0} satisfying (11).

Further, we shall prove that the multiplication in X is an associative operation. For let x, y, z ∈ X. By Lemma 4, y = T_{t}(y_{1}) for some y_{1} ∈ U_{0} and t ∈ ℝ. Note that the set y_{1}^{−1}U_{0} is open. Again, by Lemma 4, we have x = T_{r}(x_{1}) and z = T_{s}(z_{1}) for some r, s ∈ ℝ and x_{1}, z_{1} ∈ y_{1}^{−1}U_{0} ∩ U_{0}, so that x_{1}y_{1}, y_{1}z_{1} ∈ U_{0}. Then, in view of the associativity of multiplication in G, we have

(xy)z = (T_{r}(x_{1})T_{t}(y_{1}))T_{s}(z_{1}) = T_{r+t+s}((x_{1}y_{1})z_{1}) =
= T_{r+t+s}(x_{1}(y_{1}z_{1})) = x(yz).

We now claim that each element x in X has an inverse element relative to y_{0}. Indeed, by Lemma 4, x = T_{t}(y) for some y ∈ U_{0} and t ∈ ℝ. It is easily checked that the element x^{−1} defined by

x^{−1} = T_{−t}(y^{−1})

satisfies (11).

Thus, we have the set X with the associative multiplication ( see (9) and (10)) and each element x ∈ X has at least one inverse element relative to the identity element y_{0}. It is well known that in this case we have a group structure in X.

Finally, it is straightforward to check that X is an abelian topological group and p : X → G is a homomorphism. This completes the proof of the theorem.

References

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46 S. A. GRIGORIAN, R. N. GUMEROV, A. V. KAZANTSEV

[5] S. A. Grigorian, Generalized analytic functions, (in Russian), Uspekhi Matem. Nauk, V. 49, No. 2 (1996), 3–42.

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Kazan State University, Kremlevskaya, 18, Kazan, 420008: Russia E-mail address: Renat.Gumerov@ksu.ru

Received May, 25, 2000