Ehresmann Connection on Foliations Generated by Rn

Ehresmann connection on foliations generated by Rn

Pyotr N. Ivanshin

Abstract. We find sufficient conditions for the existence of the Ehresmann connection on manifold foliated by the locally free action of a commutative Lie group H in case codimension of the foliation equals 1. The connection constructed here is invariant with respect to the modified action of H.

M.S.C. 2000: 37C85, 53C12. Key words: Ehresmann connection, foliation, Lie group, transversality.

§ 1. Introduction. The question of the existence of the Ehresmann connection on the foliated manifold was considered by number of authors . Usually it is assumed that this object already exists [1, 25, 18, 19, 20]. Nevertheless the existence problem was discussed before by number of authors. Note for example extensive work of G. Cairns on totally geodesic foliations [4] and monograph of P. Molino [13] which deals mainly with Rie- mannian foliations. There exists a number of criteria on existence of the Ehresmann connection. All of them naturally assume existence of some other structure on the foliated manifold — symplectic structure [22], vanishing cycles [24], absence of Reeb components (especially on manifolds of dimension 3) [16, 5], absence of so-called limit cycles [3], existence of the special Riemannian metric [12]. An abundant review of the existing results one can find also in [17]. Here we try to find conditions on the foliated manifold that somewhat generalise that explored before (absence of Reeb components or limit cycles). Note that in this paper we consider foliations whose properties agree to the duality (totally geodesic foliations — Riemannian foliations) given in [22]. Let us define first Ehresmann connection on foliated manifold. Consider a manifold M with foliation F . Consider also the distribution E ⊂ TM complement to F . We call a curve σ : [0, 1] → M horizontal one if dσ/dt ∈ E. A curve γ : [0, 1] → M we call vertical if γ([0, 1]) belongs to one leaf. Rectangle is the mapping Π : [0, 1] × [0, 1] → M, such that for any s ∈ [0, 1] Π(·, s) is a horizontal curve, and for any t ∈ [0, 1] Π(t, ·) is a vertical curve. The curves Π(·, 0), Π(0, ·) are called initial curves of the rectangle Π.

Deﬁnition 1. [1]The distribution E is called Ehresmann connection if for any horizontal curve σ and vertical curve τ we can ﬁnd necessarily unique rectangle for which σ and τ are the initial curves.

Differential Geometry - Dynamical Systems, Vol. 9, 2007, pp. 58-81. °c Balkan Society of Geometers, Geometry Balkan Press 2007. Ehresmann connection on foliations 59

Deﬁnition 2. Let us call a transversal PH-complementary to the foliation F if T f {hP }h∈H is the foliation on M complementary to F . Or, equivalently, if hP P 6= implies hP = P .

Deﬁnition 3. We say that the transversal P is homotopic to H-complementary if we can deﬁne new group structure on H so that under it P becomes H-complementary.

Consider first a map R : P ×H → M, R(p, h) = ph. It defines the set B = R−1(P ). Theorem 2. Let (M,F ) be 2-dimensional manifold foliated by the action of ∼ H = R. Let there exist a total connected transversal P such that π0 : B → P (π0(p, h) = p) is a trivial covering. Then P is homotopic to H-complementary to the foliation F . Description of the set B as a locally trivial fibre bundle is close to one of the graph of the foliation [23, 25]. Consider for each x ∈ P the set Hx = {h ∈ H | hx ∈ P }. Note that Hx ×{x} ⊂ B. 0 Now let us define a translation of h ∈ Hx by g ∈ Hx as follows: Let (gx, h ) ∈ B 0 belong to the same connected component as (x, h) ∈ B. Since h belongs to Hgx we 0 0 have gh ∈ Hx. Put Πgh = gh . In terms of functions of whose graphs consists B,

Πgh = fh(gx) + fg(x). (4)

Note that if foliation F on a manifold M does not have limit cycles [3] then translation of the element h ∈ Hx by g ∈ Hx is defined for any pair h, g ∈ Hx. The converse is evident for 2-dimensional manifolds. Theorem 3. If B is a trivial covering and for all x ∈ P and h ∈ Hx the translation n Πg is restriction of the shift in R then P is homotopic to H-complementary. Further on we consider partial case in which total transversal is diffeomorphic to S1. Theorem 4. If a transversal P is compact then π0 : B → P is a covering. It seems clear that in case we have a Reeb component on the foliated manifold M the transversal can not be compact; also if we have two of them then there is no connected total transversal. Statement 4. For any point x ∈ P the set of translations Φx = {Πγ,h}, here Πγ,h are defined by the formula (4) is a transformation group of the set Hx. The group Φx acts transitively on Hx. ∼ Statement 6. If π0 : B → P is locally trivial bundle and H = R then π0 : B → P is globally trivial. Corollary 1. Consider a 1-dimensional foliation F on a 2-dimensional manifold. If there exists a compact total connected transversal P then P is homotopic to H- complementary. Theorem 5. Let a manifold M with foliation F meet the following conditions: 1) codimF = 1. 2) The foliation F is generated by locally free action of Rn. 3) There exists a compact connected transversal on (M,F ). Then there exist a transversal homotopic to H-complementary one and an Ehres- mann connection on (M,F ). Theorem 6. Let a manifold M with foliation F meet following conditions: 1) codimF = 1. 60 Pyotr N. Ivanshin

2) Foliation F is generated by action of Rn. 3) (M,F ) is transversally orientable. Then there exist a transversal homotopic to H-complimentary and an Ehresmann connection on (M,F ).

§ 2. Basic existence theorem.

2.1 General assumptions Consider a manifold M with foliation F generated by the action of locally free action of n-dimensional commutative Lie group H: L : G×M → M, L :(g, p) → Lg(p) = pg. Let ρ : He → H be a universal covering. Action L induces locally free action Le of a universal covering space He:gp ˜ = ρ(˜g)p such that the orbits of actions coinside. Thus further on we assume that H is 1-connected, i.e. H =∼ Rn [2]. Also we assume that 1) codimF = 1. 2) There exists a closed transversal P ⊂ M of the foliation F [13].

2.2 Stable subgroup of the transversal. Since leaves of F are orbits of the action of a Lie group H each transversal P of the foliation F and each h ∈ H gives us a submanifold hP transversal to F . Thus the action of H gives rise to the set of transversals {hP }h∈H to F . Let P be some transversal of the foliation F (it exists by assumption 2) of a paragraph 1.1.) Put for each x ∈ P

Hx = {h ∈ H | hx ∈ P }. Lemma 1. The following statements are equivalent: i) P is H-complementary to F transversal; ii) there exists a subgroup H0 of H such that H0 = Hx for all x ∈ P .

Proof. i) ⇒ ii) Let us show that for any x, y ∈ P we have Hx = Hy. By deﬁnition of H-complementary transversal h ∈ Hx implies hP = P , hence h ∈ Hy. Now put

H0 = Hx0 for some x0 ∈ P . ii) ⇒ i)

ii) implies that if h belongs to Hx0 for some x0 ∈ P then hP ⊂ P . Thus in this case by assumption h belongs to Hx for any x ∈ P . f Let hP ∩ P 6= . hen there exists x0 such that hx0 = y0 ∈ P . Hence hP ⊂ P and h−1P ⊂ P thus hP = P . This completes the proof.

2.3 H-complementary transversal and Ehresmann connection. Let a foliation F possess an H-complementary transversal P . Let us deﬁne a 1- dimensional distribution E on M as follows: First note that since the transversal P intersects all leaves of a foliation F (being orbits of action of H) we have for each x ∈ M an element p ∈ P and h ∈ H such that x = hp. Put

E(x) = dhp(TpP ). Ehresmann connection on foliations 61

Let us show that this definition is correct. Let x = hp = h0p0, here h, h0 ∈ H, p, p0 ∈ P . Then p0 = (h0)−1hp, hence by definition of H-complementary transversal 0 −1 0 −1 we get an equality (h ) hP = P . This implies a relation d((h ) h)p(TpP ) = Tp0 P , 0 thus dhp0 (Tp0 P ) = dhp(TpP ). Let us show that the distribution E is differentiable. Fist note that since action of the group H is locally free each point p0 ∈ P possess a neighbourhood V (p0) in P and a neighbourhood W (0) in H such that the mapping α : V (p0) × W (0) → U(p0), (p, h) → hp, here U(p0) is a neighbourhood of p0 in M is diffeomorphism. It seems clear that α maps differentiable distribution TpV (p0) on V (p0) × W (0) into the restriction of the distribution E on U(p0). Thus a distribution E is differentiable i the neighbourhood of each point p ∈ P . Now note that by construction a distribution E is H-invariant. Thus E is differentiable in each point of M. The distribution E is tangent to submanifolds hP by construction. Thus the set of submanifolds {hP } gives rise to 1-dimensional foliation on M.

Theorem 1. The distribution E deﬁnes an Ehresmann connection on M.

Proof. The distribution E is complementary to the distribution TF by construction. Let us check that each horizontal curve σ and vertical curve τ define a rectangle for which σ and τ become initial curves (cf. definition [1]). Let Π : [0, 1] × [0, 1] → M be a rectangle. Then H-invariance of the distribution E implies that for any h ∈ H the mapping hΠ : [0, 1] × [0, 1] → M also defines a rectangle. Note now that for any point p ∈ M there exists h ∈ H such that h p belongs to P moreover the shift by h maps arbitrary horizontal (vertical) curve beginning in p into horizontal (vertical) curve starting at h p. Thus it suffices to prove that there exists a rectangle Π with initial curves σ and τ for any vertical curve σ and horizontal curve τ such that σ(0) = τ(0) = p0 ∈ P . Since the mapping φ : H → L0, h → hp0 is a universal covering (cp. [6]), we get that there exists a unique curve h : [0, 1] → H such that h(0) = 0 ∈ H and τ(t) = h(t)p0. Put

Π : [0, 1] × [0, 1] → M, Π(s, t) = h(t)σ(s).

It seems clear that Π is rectangle.

2.4 Conditions of the existence of the H-complementary transversal.

Statement 1. The mapping R : P ×H → M, R(h, p) = ph is a local diﬀeomorphism.

Proof. It suffices to show that for any h ∈ H, p ∈ M the differential dR : T(p,h)(P × H) → TphM is isomorphism. Consider following diffeomorphisms:

Rh : M → M, p → ph 0 0 Reh : P × H → P × H, (p, h ) → (p, hh ) 62 Pyotr N. Ivanshin

It seems clear that the followwing diagram is commutative:

P × H −−−−→R M     Rehy yRh

P × H −−−−→R M Hence it suﬃces to prove that

dRp,e : Tp,eP × H → TpM is isomorphism. The deﬁnition of R implies that the following diagram is commutative: dRp,e T(p,e)(P × H) −−−−→ TpM     y y ∼ A TpP ⊕ TeH = TpP ⊕ g(H) −−−−→ TpP ⊕ TpL(p) here L(p) is the leaf of the foliation passing through p and vertical arrows are isomor- phisms. Then the linear operator A has the following properties:

∀V ∈ TpPA(V ) = V, ∀a ∈ g(H) A(a) = σ(a).

Since the action of H on M is locally free the mapping g(H) → TpM, a → σ(a) is monomorphism. Then the dimensional considerations give us that A is isomorphism. Hence dR(p,e) is also an isomorphism. Corollary 1.1. The set B = R−1(P ) is a submanifold of P × H of dimension equal to that of P . Proof. The mapping R : P × H → M is local diﬀeomorphism. Hence its images are transversal to any submanifold of M. Hence the inverse image of the submanifold is a submanifold (cf. Theorem 3.2 of the Chapter 1 from [8]).

2.4.1 Canonical covering.

The set B naturally possess two smooth maps: π0, π1 : B → P , π0(p, h) = p, π1(p, h) = ph.

Statement 2. The map π0 : B → P is a local diﬀeomorphism. Proof. It suﬃces to show that

dπ0 : T(p,h)B → TpP is an isomorphism. The local diﬀeomorphism R maps H ino the leaf transversal to P , hence B is transversal to H it the point (p, h) ∈ B. Thus ∼ T(p,h)B ⊂ T(p,h)(P × H) = TpP ⊕ ThH is transversal ∼ T(p,h)H ⊂ T(p,h)(P × H) = TpP ⊕ ThH.

Hence, ker dπ0 = {0}. The statement now follows from dimensional considerations. Ehresmann connection on foliations 63

2.4.2 Conditions of existence of the H-complementary transversal on 2- dimensional manifold. Our aim is to determine conditions under which the given transversal is homotopic to H-complementary.

Theorem 2. Let (M,F ) be a 2-dimensional manifold foliated by locally free action of a commutative Lie group H =∼ R. Let there exist a total connected closed transversal P on M for which π0 : B → P is globally trivial. Then P is homotopic to H-complementary to the foliation F . ∼ Proof. Fix a point x ∈ P . Let us show that Hx = Z. −1 a) By assumption π0 : B → B is a covering hence the leaf π0 (x) = {(x, h) | h ∈ Hx} is discrete in B. Thus this set is discrete also in {x} × H. So, Hx is discrete in H =∼ R. b) Let us prove that Hx is infinite. Consider x ∈ P and h ∈ Hx. Let γ : [0, 1] → P , γ(0) = x, γ(1) = hx. Let also γe : [0, 1] → B be such that γ(0) = (x, h), 0 0 γ(1) = (hx, h ), here h ∈ Hhx. Since π0 : B → P is globally trivial and B0 = {(x, 0) | x ∈ P } is a connected 0 0 component of B the inequality h 6= 0 holds true. Assume that h > 0, hence h1 > h. 0 00 Consider h1 = h + h , then h1 ∈ Hx. Thus we construct h ∈ Hh1x and h2 = 00 0 00 h1 + h ∈ Hx by point hx and h1. As (x, h), (hx, h ), (h1x, h ) belong to the same 00 connected component of B which does not coincide with B0 h > 0 and h2 > h1. Continuing this process we obtain an increasing sequence of the elements of Hx. Hence, Hx is infinite. c) Let us prove that the set Hx does not have either minimum or maximum. To show this note that we have a natural order on Hx ⊂ R. Assume that there exists a minimal element hmin for x ∈ P . Let hmin = 0. Consider h > 0 from Hx. Then −h ∈ Hhx. Now consider a curve γ : [0, 1] → P such that γ(0) = hx, γ(1) = x. Let γe : [0, 1] → B be the lift of γ such that γe(0) = (hx, −h), γe(1) = (x, h0). Then h0 < 0 since the connected component of 0 B which contains γe does not intersect B0. Since by construction h ∈ Hx we arrive to the contradiction with the minimality of hmin = 0. Now let hmin < 0. As in b) we construct the sequence hk ∈ Hx by point x and element hmin. Then this sequence decreases. This again contradicts the assumption of minimality of hmin. S d) Let us construct a diffeomorphism φ0 : P × R → P × R, φ(B) = P × {i}. i∈Z Paragraphs a)-c) imply that B ⊂ P × R consists of disjoint union of the graphs of functions fi, i ∈ Z. Let us construct a mapping Φ1 : P × R → P × R,(x, y) 7→ (x, φ1(x, y)). The map φ1(x, ·) for any fixed x ∈ P is diffeomorphism of R which meets the following conditions:   0, if y = 0, φ (x, y) = y/f (x), if y ≥ f (x), 1  1 1 y/|f−1(x)|, if y ≤ f−1(x)

Then Φ1((x, f1(x)) = (x, 1), Φ1((x, f−1(x)) = (x, −1). Consider now a diﬀeomor- phism Φ2 : P × R → P × R,(x, y) 7→ (x, φ2(x, y)). Here φ2 is again a diﬀeomorphism 64 Pyotr N. Ivanshin of R for each x ∈ P which meets the following conditions:   y, if − 1 ≤ y ≤ 1, φ (x, y) = 2f (x)y/f (x), if y ≥ f (x)/f (x), 2  1 2 2 1 2|f−1(x)|y/|f−2(x)|, if y ≤ f−2(x)/|f−1(x)|

Note that Φ2(Φ1)|{(x,y)|f−1(x)≤y≤f1(x)} = Φ1|{(x,y)|f−1(x)≤y≤f1(x)}. Let us construct diﬀeomorphisms Φ3, Φ4,... in similar way. It follows from the construction that for each point (x, y) there exists a natural number n such that

Φk(x, y) ··· Φn+1Φn ··· Φ1(x, y) = Φn ··· Φ1(x, y).

The desired diffeomorphism now is φ0 = Φ1 ◦ Φ2 ◦ · · · . −1 e) Let us define now new group action. First put h ? x = π1(φ0 (x, h)) for h ∈ R and x ∈ P . Now consider any y ∈ M. Then there exist x0 ∈ P and h0 ∈ H such that h0x0 = y. Put h ? y = (h + h0) ? x0. Let us prove that this definition is correct, i.e. it does not 0 0 0 0 00 00 00 0 depend on the choice of h and x . Assume that y = h x = h x . Then h −h ∈ Hx00 and x0 = (h00 − h0)x00. Hence

(h + h00)x00 = ((h + h0) + (h00 − h0))x = (h + h0)x0 so the action ? is correctly deﬁned. Finally let us show that the transversal P is H-complementary with respect to the action ?. Let h ? P ∩ P 6= f. Then there exist x ∈ P and h ∈ H such that h ? x ∈ P . Let −1 0 φ0 (x, h) = (x, h ) then

−1 0 h ? x = π1(φ0 (x, h)) = h x ∈ P.

0 0 0 Hence h x ∈ P , so h ∈ Hx. Since (x, h) = φ0(x, h ) we get h ∈ Z. −1 0 0 0 Now for any x ∈ P , φ0 (x, h) = (x, h ), here h ∈ Hx. Then h ? x = h x ∈ P . Thus h ? P = P .

2.4.3 Suﬃcient condition of the existence of the H-complementary transversal in case dim H > 1 Now we assume that dim H > 1 and the bundle B is trivial. Then B ⊂ P ×H consists ∼ n of the set of graphs of the vector-functions P → H = R . Let us ﬁx a point x0 ∈ P . Let us denote by fh(x) a vector-function whose graph is a connection component of

B passing through (x0, h) for any h ∈ Hx0 . Then fh(x0) = h. 0 Now let us deﬁne a translation of h ∈ Hx by g ∈ Hx as follows: Let (gx, h ) ∈ B 0 belong to the same connected component as (x, h) ∈ B. Since h belongs to Hgx we 0 0 have gh ∈ Hx. Put Πgh = gh . In terms of functions of whose graphs consists B,

Πgh = fh(gx) + fg(x).

Theorem 3. If B is a trivial covering and for all x ∈ P and h ∈ Hx the translation n Πg is the restriction of a shift in R then P is homotopic to H-complementary. Ehresmann connection on foliations 65

Proof. a) Fix x ∈ P . Since Πg is a shift for each g ∈ Hx, we get Πg(h) = h + q(g, x). Let us put in the last equation h = 0. Then

q(g, x) = Πg(0) = f0(gx) + fg(x) = fg(x), since f0(x) = 0 for any x. Henceforth

(2.1) fh(gx) + fg(x) = Πg(h) = h + fg(x), thus

fh(gx) = h = fh(x) for any g ∈ Hx. b) Condition (2.1) implies that

(2.2) fg+h(x) = fg(x) + fh(x), since h = fh(x).

Let us choose a base h1, h2, ... , hq of the group Hx0 . Then (2.2) implies that for any x ∈ P fh1 (x), fh2 (x), ... , fhq (x) is a base of Hx. Note that the system {fi(x)} is linearly independent over R for any x. Now assume that fq+1(x), ... , fn(x) is the set of vector-functions such that

fh1 (x), fh2 (x), . . . , fhq (x), fq+1(x), . . . , fn(x)(2.3) is a base of H for any x. q c) Now we construct a diﬀeomorphism φ0 : P × H → P × H, φ0(B) = P × Z using base (2.3) similarly to the paragraph d) of the proof of theorem 2. Put

−1 h ? x = π1(φ0 (x, h)), x ∈ P, h ∈ H

Now take y ∈ M. Then there exist x0 ∈ P and h0 ∈ H such that h0x0 = y. Put h ? y = (h + h0) ? x0. The correctness of the deﬁnition is proved as theorem 3.

Note 1. We may not use the translation to be a shift. We may use only triviality of 2 B and commutativity of the set of translations Πg. If H ' R then we are able to 2 construct a curve γ : R → R passing throughhi, i ∈ Z without selfintersections. The construction of such a curve is inductive. We note first that after the addition of the 2 n-s point to the curve γ the set R \ γn remains 1-connected. Thus we can addT the next point so that the curve γ keeps desired properties. Hence, discreteness of L P and closedness of P imply that the curve γ passes through {∞} in the extended plane R ∪ {∞} ∼ C. = S Jordan theorem then says that this curve divides R2 {∞} into two different 1- connected sets. Hence, Riemann theorem gives us the transformation f : C → C such that these sets map into halfplanes and a curve γ — to the line separating them. Since different translations commute curves corresponding to different hi intersect only by 0. 66 Pyotr N. Ivanshin

2.4.4 Condition of triviality of the covering π0 : B → P

−1 Let π0 : B → P be a covering. Then the curve γ : [0, 1] → P and b ∈ π0 (γ(0)) possess the lift γe : [0, 1] → P , γe(0) = b. Hence we get a parallel translation of the points of the leaf along path in P .

Statement 3. Assume that for any elements x ∈ P , g ∈ Hx and the path γ : [0, 1] → P such that γ(0) = x, γ(1) = gx, the parallel translation along γ is a shift. Then π0 : B → P is a trivial covering.

Proof. If P =∼ R then any covering is trivial. ∼ 1 Assume now that P = S . Since B0 maps into itself under parallel translation the shift is necessarily trivial. Hence holonomy group is trivial as is the bundle.

2.4.5 Examples

Let a covering π0 : B → P be not necessarily trivial. Let us define a translation of the −1 points of the leaf Hx = π0 (x) along paths γ connecting x with gx, here g ∈ Hx. In 0 this case we have a unique path γe in B starting into h ∈ Hx and ending into h ∈ Hgx. 0 Put Πγ,g(h) = h g ∈ Hx. Thus Πγ,g : Hx → Hx. Let a leaf L of the foliation F be diffeomorphic to Rn, and intersect P only in finite set of points. Then for any x ∈ L ∩ P neither of Πγ,g is a shift. Assume on the k contrary that the set {(Πγ,g) }k∈N is infinite then the set L ∩ P is also infinite. If a transversal P is H-complementary and a leaf L of the foliation F is diffeomorphic to Rn then L ∩ P either consists only of one point or infinite. Assume that there exists x ∈ P such that for some h ∈ H hx ∈ P then h belongs to Hx. But Hx = HP by lemma 1. Hence hkx belongs to L ∩ P for any k ∈ N. Since L =∼ Rn the points hkx are different. Example 1. Consider a foliation on M = R2 ×S1 as the natural foliation with the leaf R2. Consider a trajectory of a rational flow on S1 × S1 as transversal P . Since the set of intersection points of the leaf with P is finite, the transversal P does not give rise to the Ehresmann connection invariant under the action of R2. Note 2. If we do not want to deform action of a group then we may consider a set of transversals which may converge (pointwise) to the transversal which defines Ehresmann connection. Assume as in the proof of the previous statement that x ∈ P , h1 ∈ Hx. Put P1 = h1P . Then P2 = (h1 + 1/2h(t))P1, here h(0) = h12(0), h−1(x)x ∈ h P for x ∈ P and h(t ) = e for P (t ) = h x; we assume also that 12 T 1 0 0 1 P −1 f [0, 1]h12 P2 = . And so on. This process converges if and only if the series ai i∈N with general summand ai = hi,i+1(x) converges for any leaf of the foliation F .

1 Example 2. Consider a foliation on S ×R, generated by images of curves lb: y = ax+b, (the constant a 6= 0), b ∈ R under natural mapping R2 → S1 × R,(x, y) 7→ ([x], y), here [x] ∈ S1 is the equivalence class of x ∈ R with respect to the relation x ∼ x + k, k ∈ Z. Consider the transversal P given by the following equation in the coordinate chart U =∼ (−1/2, 1/2)×R, x ∈ R, here y ∈ (−1/2, 1/2), (−1/2, 1/2) is the coordinate Ehresmann connection on foliations 67 chart on S1:   0 x ≤ 0;  x x ∈ (0, 1/4]; y =  −x + 1/2 x ∈ (1/4, 1/2];  0 x > 1/2 This subset of M is not a smooth manifold (though it can be approximated by a smooth manifold). Nevertheless it suﬃces to consider this set for the illustration of the method. This transversal converges to the curve given by y = 0. Example 3. Consider a foliation on R2 \{0} generated by circles with common center 2 2 ih/|x| in 0. Let us deﬁne action of the group R as follows: R : R×R → R : Rhx = e x. Then there is no connection invariant under the action of the group H. Note that at the same time we are able to deform action of R (following theorem 2) to the following: 0 ih Rhx = e x. Let us show that there is no invariant Ehresmann connection. Assume that it exists. Then the tangent vector at the point x ∈ R2 to the transversal under ∼ the action of an element from isotropy group Hx = Z of the leaf Lx, comprising x maps d ih/r 1 ih/r to the collinear vector. But this is impossible since dr e = − r2 e 6= 0modπ. Thus this vector turns by angle which is not a multiple of π.

§ 3. Existence of H-complementary transversal if there exists a compact transversal on M. Let (M,F ) be a foliation of codimension 1 generated by locally free action of the commutative group H. Let P be a transversal of the foliation F and π0 : B → P be a map deﬁned in paragraph 1.4.1.

Theorem 4. If the transversal P is compact then π0 : B → P is covering.

Proof. a) Assume that H is compact then leaves of the foliation F are also compact. Then for any point x ∈ P the set Hx is ﬁnite. Let us recall that the map R : P × H → M,(x, h) → hx is local diﬀeomorphism. Now leaves of the trivial bundle π : P ×H → P pass into leaves of the foliation F under the map R and P is transverse to the foliation F . Consider (x, h) ∈ B = R−1(P ). Then B ⊂ P × H is transverse to the leaves of π. Hence there exist a neighbourhood Vh(x) and a mapping sTh : Vh(x) → B such that π0sh = IdV (x) and sh(x) = (x, h). Consider V (x) = Vh(x). Then for any two points x, y ∈ P consider the path h∈Hx γ : [0, 1] → P such that γ(0) = x, γ(1) = y. There exists a division t0 = 0 < t1 < . . . tN = 1 such that γ([ti, ti+1]) ⊂ V (γ(ti)). Then for any h ∈ Hx we have sh(γ(t1) = (γ(t1), h1) and

f sh(V (γt0 ) ∩ sh1 (V (γ(t1)) 6= .

Thus there exists a path (necessarily unique) γe :[t0 = 0, t1] → B such that γe(0) = (x, h) π0γe(t) = γ(t), t ∈ [0, t1]. Similarly we construct a unique path γe : [0, 1] → B such that γe(0) = (x, h) π0γe(t) = γ(t), t ∈ [0, 1]. Thus π0 : B → P is a covering. b) Assume that H is not compact. 68 Pyotr N. Ivanshin

Let V (x0) be a neighbourhood of he point x0 in P . Let us prove that for any h0 ∈ Hx0 there exists a map sh0 : V (x0) → B such that π0sh0 = IdV (x0) and sh0 (x0) = (x0, h0).

Note that repeating considerations of a) we show that for h0 ∈ Hx0 there exists sh0 deﬁned in a neighbourhood U(x0) ⊂ V (x0) depending on the choice of h0. Consider

U(x) = (x1, x2). Let us show that there exists lim sh0 (x). x→x1

Let sh0 (x) = (x, h(x)). By compactness of P the angle of the inclination of the 0 leaves of F to P is bounded from above. Hence |h (x)| ≤ K, x ∈ U(x0). Then 0 |h(x)| ≤ |h(x0)| + K|x − x0|. Since |h | is bounded on U(x) h is a function with ﬁnite change. Then by Corollary 2 of the theorem 6 from the ﬁfth chapter of [14] there exist a limit lim h(x). x→x1

Let lim sh0 (x) = (x1, h1). Then there exists sh1 (x) deﬁned in a neighbourhood x→x1

U(x1) such that π0sh1 = IdU(x1) and sh1 (x1) = (x1, h1). It seems clear that for any x ∈ U(x0) ∩ U(x1) sh1 (x) = sh0 (x). Then we proceed as in the case of the compact group H.

3.1 Properties of the set of translations.

Let π0 : B → P be a covering. Consider x ∈ P , g ∈ Hx, γ : [0, 1] → P , γ(0) = x, −1 γ(1) = gx. Then for any (x, h) ∈ π0 (x) there exists a unique path γe : [0, 1] → B, γe(0) = (x, h), γe(1) = (gx, h0) covering γ. Let us deﬁne a mapping

0 (3.4) Πγ,g : Hx → Hx, Πγ,g(h) = h g.

−1 Let us call Πγ,g a translation of the leaf Hx = π0 (x) along γ connecting x to gx ∈ P .

Statement 4. For any point x ∈ P the set of translations Φx = {Πγ,h}, here Πγ,h are deﬁned by the formula (3.4) is a group of transformations of the set Hx. This group Φx acts transitively on Hx.

Proof. Let us prove ﬁrst that the set Φx is closed under composition. Consider γ1 : [0, 1] → P , γ2 : [0, 1] → P , γ1(0) = γ2(0) = x, and g1, g2 ∈ Hx such that g1x = γ1(1), g2x = γ2(1). Let us show that there exist γ0 : [0, 1] → P , γ0(0) = x, and g0 ∈ Hx (here g0x = γ0(1)) such that

(3.5) Πγ2,g2 (Πγ1,g1 (h)) = Πγ0,g0 (h), ∀h ∈ Hx.

0 Let γe1(t) = (γ1(t), h1(t)) be the lift of the path γ1, γe1(0) = (x, h) and γe2(t) = 0 00 (γ2(t), h2(t)) be the lift of γ2, γe2(0) = (x, g1 + h1(1)), then γe2 (t) = (γ2(t), g(t)) is the 00 lift of the curve γ2, γe2 (0) = (x, g1). Consider

δ : [0, 1] → P, δ(t) = π1(γe2(t)) = g(t)γ2(t).

Now put γ0 = δ · γ1, here · is the composition of the paths. Put also g0 = g(1)g2. Let us show that then (3.5) holds true. e e e Let δ : [0, 1] → B, δ(t) = (δ(t), h0(t)), δ(0) = (g1x, h1(1)) be the lift of the curve δ. Ehresmann connection on foliations 69

Now in the equality (3.5) we have Πγ2,g2 (Πγ1,g1 (h)) = g2 + h2(1) on one side; and

Πγ0,g0 (h) = h0(1) + g(1) + g2 on the other. Let us show that these elements coincide.

It suﬃces to prove that h0(1) + g(1) = h2(1). Note ﬁrst that h0(t) + g(t) ∈ Hγ2(t) since g(t) ∈ Hγ2(t), moreover g(t)γ2(t) = δ(t) h0(t) ∈ Hδ(t). Hence the path Γ(t) = (γ2(t), h0(t) + g(t)) belongs to B and Γ(0) = (x, g1 + h1(1)), i.e. Γ = γe2. So, by uniqueness of the covering path we get h0(1) + g(1) = Γ(1) = γe2(1) = h2(1). Let us now show that there exists an inverse element for any Πγ,h ∈ Φx Φx. To prove this consider a path γ−1 : [0, 1] → P , γ−1(0) = gx, γ−1(1) = x. Let −1 −1 γe1(t) = (γ (t), h1(t)) be the lift of the curve γ , γe1(0) = (hx, −h). Now consider 0 −1 0 γ (t) = h1(t)γ (t) and h = h1(1). Let us show that Πγ0,h0 is inverse to Πγ,h. The identity (3.5) implies that

0 0 −1 −1 Πγ,hΠγ ,h = Πγ·γ ,h+h1(0) = Πγ·γ ,0.

Since the loop γ00 = γ ·γ−1 is contractible the lift γe00 of the loop γ00, starting into (x, g) also ends in (x, g) for each g ∈ Hx. So by the deﬁnition of Πγ,h we get Πγ00,0 = Id. Thus the set Φx is transformation group of Hx. Note now that since P is connected 0 ∈ Hx can be mapped into any h ∈ Hx by element from Φx. Then action of Φx on Hx is transitive.

Statement 5. If Πγ,g = Id then g = 0 and γ is the loop in P .

Proof. Since there exists a zero section B0 ⊂ B Πγ,g(0) = g + 0 = 0, i.e. g = 0. ∼ Statement 6. If π0 : B → P is locally trivial bundle and H = R then π0 : B → P is trivial.

Proof. If P is diﬀeomorphic to R then the covering π0 : B → P is naturally trivial. 1 Now let P be diﬀeomorphic to S . Recall that the bundle π0 : B → P possess global section σ : P → B, σ(x) = (x, 0). Now consider a loop δ : [0, 1] → P , δ(0) = δ(1) = x. Let (δ(t), h(t)) be the lift of the path δ into B such that (δ(0), h(0)) = (x, h). Since the group R is orientable h1 = h(1) has the same sign as h. Assume that h > 0. If h1 < h we consider the loop δ · δ and h2 which is generated similarly to h1. Again we have h2 < h1. Let us continue the construction of hn. Since the sequence (hn)n∈N decreases and is bounded from below (∀n ∈ N, 0 < hn since B0 is connected component; then if the lift of the path starts into the point which does not belong to B0 this lift does not intersect B0), there exists an element h0 ∈ H such that hn → h0, n → ∞. Since action of H is continuous h0 ∈ Hx. But Hx is discrete. −1 This is a contradiction. Assume that h1 > h. Consider a set h1, δ , h instead of the set h, δ, h1 of the previous considerations. Then we again arrive to the contradiction. Hence for any loop δ ⊂ P and h ∈ Hx and the lift (δ(t), h(t)) into B of the curve δ such that (δ(0), h(0)) = (x, h) we get h(1) = h.

Corollary 1. Let a foliation F on 2-dimensional manifold be generated by R. Let P be a total connected compact transversal to F . Then P is homotopic to H-complementary.

Proof. By statement 6 π0 : B → P is globally trivial. Now apply theorem 2. 70 Pyotr N. Ivanshin

0 0 Lemma 2. Assume that for x ∈ P , h ∈ Hx the paths γ, γ ⊂ P ,(γ(0) = γ (0) = x, 0 γ(1) = γ (1) = hx) are homotopic to each other then Πγ,h = Πγ0,h.

Proof. If paths are homotopic then their lifts into B are also homotopic to each other and end at the same point.

Corollary 2. The set Φx is countable for any x ∈ P .

Proof. There are no more than countable number of not homotopic toT each other paths between any pair of points of S1. The set of intersection points L P is also at most countable for any L ∈ F . Hence the set Φx is countable. ∼ Statement 7. For any pair of points x, y ∈ P , Φx = Φy.

Proof. Let a pair x, y ∈ P be ﬁxed. Let us construct a map ψ :Φx → Φy. Let us ﬁx a leaf δ : [0, 1] → P such that δ(0) = x, δ(1) = y. Consider Πγ,g ∈ Φx. There exists the lift δg of the path δ δg = (δ(t), g(t)) into B starting at (x, g). Consider a path δ1 : [0, 1] → P , δ1(t) = δ(t)g(t). Put −1 ψ(Πγ,g) = Πδ ·γ·δ1,g(1).

Let us show that ψ(Πγ1,g1 ◦ Πγ2,g2 ) = ψ(Πγ1,g1 ) ◦ ψ(Πγ2,g2 ). Let (γ2(t), g1(t)) be 0 the lift of the path γ2 into B,(γ2(0), g1(0)) = (x, g1). Consider γ2 : [0, 1] → P , 0 0 0 γ2(t) = g1(t)γ2(t). Let (δ(t), g (t)) be the lift of δ into B such that (δ(0), g (0)) = 0 0 0 (x, g1(1) + g2). Consider δ : [0, 1] → P , δ (t) = δ(t)g (t). Then ψ(Πγ1,g1 ◦ Πγ2,g2 ) =

ψ(Π 0 ) = Π −1 0 0 0 . γ1·γ2,g1(1)+g2 δ ·γ1·γ2·δ ,g (1) 0 0 0 0 On the other hand ψ(Πγ1,g1 ) = Π −1 , ψ(Πγ2,g2 ) = Π −1 . Here δ ·γ1·δ1,g1(1) δ ·γ2·δ2,g2(1) 0 0 (y, gi(1)) is the endpoint of the lift (δ(t), gi(t)) of the path δ starting at (x, gi), i = 1, 2. 0 0 0 0 0 0 δ (t) = δ(t)g (t). Then ψ(Πγ1,g1 ) ◦ ψ(Πγ2,g2 ) = Π −1 ◦ Π −1 = i i δ ·γ1·δ1,g1(1) δ ·γ2·δ2,g2(1) −1 0 0 0 −1 Π −1 0 0 0 since the lift δ ·γ2 ·δ2 into B starting at (y, g1(1)) is (δ(t), g1(t)) · δ ·γ1·γ2·δ ,g (1) 0 (γ2(t), g1(t)) · (δ(t), g (t)). To prove that ψ is isomorphism we ﬁrst show that Ker(ψ) = {e}. Let ψ(Πγ,g) = 0 Πδ−1·γ·δ0,g0 = Id. Hence g = 0 and by construction of ψ we get g(t) = 0. Hence γ is a loop and δ0 = δ−1. But δ0 · γδ−1 is contractible. Hence the loop γ is also contractible.

Then Πγ,g = Πconstx,0 = Id. Thus ψ is a monomorphism. Let us show first that there exists an element Πγ,g ∈ Φx such that Πγ0,g0 = ψ(Πγ,g) −1 for arbitrary Πγ0,g0 ∈ Φy. To do this consider the lift ((δ) (t), g(t)) of the path (δ)−1 into B such that ((δ)−1(0), g(0)) = (y, g0). Assume that δ0 : [0, 1] → P is defined by the equation δ0(t) = (δ)−1(t)g(t). Put γ = δ · γ · δ0 and g = g(1). Then 0 ψ(Πγ,g) = Πδ−1·δ·γ·δ0·δ0−1,g0 since the lift (δ(t), g (t)) of the curve δ into B starting at (x, g(1)) is uniquely defined. At the same time (δ(0), g0(0)) = ((δ)−1(1), g(1)). Note 0 now that δ−1 · δ · γ · δ0 · δ −1 = γ. Then ψ is an epimorphism.

Statement 8. If the action of the group Φx on Hx is free then the bundle π0 : B → P is trivial. If π0 : B → P is trivial and for any x, y ∈ P , Hx = Hy then the action of Φx on Hx is free.

Proof. Since action of Φx is free then Πγ,0(0) = 0 implies that ∀h ∈ Hx Πγ,0(h) = h. Thus lift of any loop is again a loop. Hence, π0 : B → P is trivial. Ehresmann connection on foliations 71

Assume now that for some h ∈ Hx Πγ,g(h) = h. By deﬁnition Πγ,g(h) = g + h(1), here h(1) is the endpoint of the lift (γ(t), h(t)) of the path γ into B starting at (x, h). Since the bundle π0 : B → P is trivial and Hx = Hγ(t) for any t, we get h(1) = h. Henceforth, g = 0, i.e. γ is a loop on P . Then the triviality of a bundle implies 0 that the lift of any loop from B is again a loop, i.e. for any h ∈ Hx the equality 0 0 Πγ,g(h ) = h holds true.

3.2 Existence of the invariant transversal for crystallographic group of translations.

Assume that the action of Φx on Hx is free for any x ∈ P . Hence, statement 8 implies triviality of the covering π0 : B → P . Assume also the following: A) There exist a diﬀeomorphism α : P × H → P × Rn and a subset ∆ ⊂ Rn such that the diagram (vertical arrows are the projections on the ﬁrst factor)

P × H −−−−→α P × Rn     y y

P −−−−→Id P is commutative and α(B) = P × ∆. then α(x, h) = (x, αx(h)) and αx(Hx) = ∆. B) There exists a subgroup Φ of the isometry group of Euclidean space Rn such that i) Φ is a semi-direct product of ﬁnite subgroup G of the group O(Rn) and ﬁnitely- generated subgroup L of the group of shifts of Rn (in this case Φ is a simmorphic crystallographic group [21]); ii) ∆ is the orbit of the action of Φ on Rn; iii) for any x ∈ P −1 Φx = {αx φαx | φ ∈ Φ}.

Conditions A) and B) imply that the action of Φx on Hx can be extended from ∼ n the set Hx ⊂ H = R onto the whole H. Let Πg ∈ Φx. Let h, g ∈ H and α(x, h) = 0

(x, h0), α(x, g) = (x, g0) and for g ∈ Hx α(Πg) = Πg0 |∆. By deﬁnition of α we −1 0 0 00 0 n have αx (x, Πg0 (h0)) = (x, Πg(h)) for h ∈ Hx. Put Πg(h ) = h for h ∈ R , here 00 −1 0 0 0 (x, h ) = αx (x, Πg0 (h0)) and α(x, h ) = (x, h0). n Note 3. Consider lift of the path γ : [0, 1] → P into P × R starting at (x, h0), n h0 ∈ R . By triviality of the bundle B and independence of Πg0 (h0) of x ∈ P we get 0 −1 γ(t) = (x(t), h0). Then γ = α (γ) is the lift of the path γ : [0, 1] → P into P × H, 0 −1 γ (t) = αγ(t)(γ(t), h0). Note that action of the group does not preserve under the trivialisation α, i.e. Πg0 (h0)x does not equal (h0 + g0)x. But under the action of α 0 lifts (x(t), Πg(t)h(t)) of the path γ starting at (x, g + h ) map to (x(t), Πg0 (h0)). Statement 9. Let there exist a closed compact transversal on the orientable manifold M with foliation F of codimension 1, generated by the locally free action of a commutative Lie group H. Assume also that the conditions A and B are true for (M,F ). Then there exists a transversal P 0 on (M,F ) such that for any x ∈ P 0 the group Φx is a free Abel group. 72 Pyotr N. Ivanshin

n Proof. Let y0 ∈ R be a fixed point of action of the elements g ∈ G (definition of G is given in the condition B i)). Then since by assumption the action of Φx on Hx is free the action of the group Φ on ∆ is also free. Hence, y0 does not belong to ∆. −1 Let (x, h(x)) = α (x, y0). Consider a map ρ : P → M, ρ(x) = h(x)x. Let P 0 = ρ(P ). Let us prove first that P 0 = ρ(P ) is a manifold. Note first that P 0 is factor of P by the free action of finite group. Let h(x)x = h(y)y and γ : [0, 1] → P , γ(0) = x, γ(1) = y. Since (y, z) is the fixed point of the transformation groupG and the mapping −1 α is isomorphism the lift γ0 of γ γ0(t) = αγ(t)(γ(t), z) starting at (x, h(x)) ends in (y, h(y)). Then Πh(x)−h(y)(h(x)) = h(y)+h(x)−h(y) = h(x). Hence, Πh(x)−h(y) ∈ G. Note now that by triviality of Π0 : B → P each map Πg ∈ Φx defines homeomorphism 0 0 Πg : P → P , x(t) 7→ g(x(t))x(t). That is, h(x)x = h(y)y implies y = u (x), u ∈ G. Assume that y = u0(x), u ∈ G. Then since there exists only one fixed point of the set of transformations G h(x)x = h(y)y. Hence, xh(x) = yh(y) if and only if y = u0(x), here u ∈ G. Consider now the following group of diffeomorphisms of the manifold PG0 = 0 0 {Πg|Πg ∈ G}. Since xh(x) = yh(y) if and only if y = u (x), u ∈ G we obtain P 0 =∼ P/G0. As M is an orientable manifold homeomorphisms u0 ∈ G0 preserve orientation on P . Now the finiteness of G0 for any u0 ∈ G0 implies the existence of a 0n number n ∈ N such that u = Id. Hence, there exists a diffeomorphism ψu0 : P → P such that ψ◦u0 ◦ψ−1(x) is a turn onto angle 2π/n. Then there exists a neighbourhood 0n Vu0 (x) such that for each pair x1, x2 ∈ Vu0 (x) and any n ∈ Z, x1 6= u (x2) for any T 0 x ∈ P . Consider V (x) = Vu0 (x). Hence, P/G is a manifold. u0∈G0 Consider a natural projection π : P → P/G0. Next consider the mapping β : P/G0 → P 0 such that β(t) ◦ π = ρ. This mapping β is correctly defined since if ρ(x) = ρ(y) then x = u0(y), u0 ∈ G0, so π(x) = π(y). The mapping β is a local homeomorphism since for any t ∈ P/G0 there exist a neighbourhood U(t) ⊂ π(V (π−1(t))) and a homeomorphism ρ of U(t) onto π−1(U(t)). The set P 0 is transversal to the 0 −1 foliation F since P = π0(α (P × {y0})), α is diffeomorphism that maps x × H in n x × R (i.e. it conserves leaves and maps transversal to transversal) and π0 is a local dρ diffeomorphism. The equality ρ(x) = π (α−1((x, z))) implies that 6= 0, hence dρ 0 dx 0 is an isomorphism of TxP onto Tρ(x)P . 0 0 0 0 0 Let π0 : B ⊂ P × H → P be the canonical covering for the transversal P (cf 1.4.1.). Consider now the mapping ρ∗ : B → B0, ρ∗(x, g) = (ρ(x), h(gx) + g − h(x)). This mapping maps each path γ : [0, 1] → B into the path ρ∗(γ) ⊂ B0. Let us show that ρ∗ is a covering. Let γ0 : [0, 1] → B0 be th path in B0 starting in 0 0 0 0 (h(x0)x0). Here x0 is defined up to some element of G . Let γ (t) = (x (t), g (t)). 0 Consider a neighbourhood V (x0) ⊂ P such that the mapping h(x), h(x)x ∈ P is uniquely defined in it. Hence we get h(x(t)) (we extend this function considering it in the neighbourhoods V (y) and V (z) in V (x0), here V (x) = (y, z)). Note now 0 0 0 that g (0)x (0) = h(x1)x1, here x1 is again defined up to element of G . Then there exists the element g1 ∈ Hx such that x1 = g1x0. Here g1 is defined up to addition with the element of the isotropy group of the leaf passing through x0, x1, but since L = H/HL we can assume that g1 is uniquely defined. Then in the neighbourhood 0 0 V (x1) we get the mapping g1(t) such that g1(t)x(t) = x1(t). Hence g (t)x (t) = 0 0 h(x1(t))x1(t) = (g1(t)+h(g1(t)x(t)))x(t) = (g1(t)+h(g1(t)x(t))−h(t))x (t), so g (t) = Ehresmann connection on foliations 73

0 0 ∗ g1(t) + h(g1(t)x(t)) − h(t). Thus (x (t), g (t)) = ρ (x(t), g1(t)). Note that the right- hand side of the last equality does not depend on u0 ∈ G0. Note also that if there exists a point x ∈ P such that u0(x) = x then for any other point x0 ∈ P , u0(x0) = x0 0 (otherwise there does not exist n ∈ N such that u n = Id). Hence power of the set ρ∗−1((x0, g0)) equals that of G. Note now that the diagram

ρ∗ B −−−−→ B0     0 π0 , y yπ0 ρ P −−−−→ P 0 0 0 0 0 is commutative, here π0(x , g ) = x . 0 0 0 0 Let us show that π0 : B → P is a trivial covering. Consider a loop γ : [0, 1] → P , 0 0 0 0 0 0 0 0 0 γ(0) = γ(1) = x , a point (x0, g ) ∈ B and a lift γ : [0, 1] → B , γ (0) = (x0, g ). −1 0 Then the curve γ0 ∈ ρ (γ), starting at x0, here h(x0)x0 = x0, finishes at the point u(x0). Then we have the uniquely defined (since the diagram given above is ∗ 0 commutative) curve γ1 ⊂ B, such that π0(γ1) = γ0 and ρ (γ1) = γ . At the same ∗ time γ1(1) = (u(x0), gu(x0)). Then ρ (γ1(1)) = (ρ(u(x0)), h(gu(x0)u(x0)) + gu(x0) − 0 0 h(u(x0))) = (ρ(x0), h(gx0) + g − h(x0)) = γ (0). This means that γ is a loop. ∗ 0 The mapping ρ naturally defines the mapping ρ :Φx → Φρ(x). Let Πg ∈ Φx.

Then Πg1 (g2) = g2(1)+g1, here γ1(t) = (x(t), g2(t)) is a lift of the path γ0, connecting ∗ 0 ∗ 0 0 x with g1x. Consider ρ (x, g2) = (ρ(x), g2) and ρ (γ1(1)) = (g1ρ(x), g2(1)). Then 0 0 0 0 0 0 0 0 0 g ρ(x) = ρ(gix). Put ρ (Πg1 )(g2) = g2(1) + g1 = Π (g2) or ρ (Πg1 ) = Π . i g1 g1 0 Let us show that if Πg ∈ G then ρ (Πg) = Id. Indeed the former means that ∗ 0 ∗ ρ(gx) = ρ(x), thus ρ (γ0) is a loop. Now triviality of B implies triviality of Πρ (γ0),0 ∈ 0 Φρ(x). Assume that Πg 6∈ G, thus gx 6= ux, u ∈ G. Hence ρ(gx) 6= ρ(x), this implies 0 0 0 0 that g x = ρ(gx) 6= ρ(x) = x . Hence Πg0 (0) = g 6= 0 and Πg0 6= Id. Thus Ker(ρ0) = G. Let us show now that ρ0 is a group homomorphism. To do this ﬁrst note that 0 0 0 ρ (Πg1 ◦ Πg2 ) = ρ (Πg1 ) ◦ ρ (Πg2 ). Then by deﬁnition of the group operation on

Φx we have Πg1 ◦ Πg2 = Πg1+g2(1), here (x(t), g2(t)) is the lift of the path connecting x with g2x to B starting at the point g1. Thus it suﬃces to show that

Πh((g1+g2(1))x)+g1+g2(1)−h(x) = Πh(g1x)+g1−h(x) ◦ Πh(g2x)+g2−h(x). This follows from commutativity of the last diagram. Since ρ∗ is a covering ρ0 is an epimorphism. 0 0 Thus we get an epimorphism ρ :Φx → Φρ(x), such that Ker(ρ ) = G. Hence 0 Φρ(x) = Φx/Ker(ρ ). Now we combine the last two statements with theorem 3 to get the Corollary 3. Let a foliation F of codimension 1 on the manifold M be generated by locally free action of a commutative Lie group H. Let there exist a compact connected transversal P on (M,F ). Assume that the triple (M,F,P ) meet conditions A and B. Then there exists an H-invariant Ehresmann connection on (M,F ). Let us give examples of discrete subsets of Rn, which possess a free and transitive action of the discrete group Φ though there is no diﬀeomorphism α : Rn → Rn such that α ◦ Φ ◦ α−1 becomes a crystallographic group. 74 Pyotr N. Ivanshin

S S S Example 4. Consider Z = ({(k − 1, k)} {(k, k)} {(k + 1, k)}) ⊂ R2. k∈Z Consider then the transformations φ1, φ2 : Z → Z,

φ1(x, y) = (x + 1, y + 1);   (x + 1, y) x = y = 2k;   (x + 1, y + 1) x = 2k + 1, y = 2k;  (x, y + 1) x = 2k, y = 2k − 1; φ = 2  (x + 1, y + 1) x = 2k − 1, y = 2k;   (x + 1, y) x = 2k, y = 2k + 1;  (x, y + 1) x = y = 2k + 1

2 3 Here k ∈ Z. Then φ1 = φ2. At the same time φ1 ◦ φ2 6= φ2 ◦ φ1. n n −1 Let there exist a diffeomorphism α : R → R such that α ◦ φi ◦ α turn to be −1 iα −1 iγ elements of a crystallographic group. Then α◦φ1 ◦α = e z+β, α◦φ2 ◦α = e z+ 2 3 3γi 2γi γi 2αi αi δ. Since φ1 = φ2 we have an equality e z + (e + e + 1)δ = e z + (e + 1)β for ∼ 2 any z ∈ C = R . Hence 3γ = 2α. Since by assumption φi belongs to crystallographic group, there are four different cases: 2π 1) α = π, γ = ± 3 ; 2π 2) α = 0, γ = ± 3 ; 3) α = π, γ = 0; 4) α = γ = 0. 4/3πi 2/3πi iπ −1 In 1) we have (e + e + 1)δ = (e + 1)β, or 0 = 0, hence α ◦ φ1 ◦ α (z) = iπ iπ iπ −1 e z + β. But e (e z + β) + β = z. Then the orbit of the action of α ◦ φ1 ◦ α is finite. This is impossible since orbit of any point x ∈ Z with respect to the action of φ1 is infinite by construction. Consider the second case. The equality e3γiz+(e2γi +eγi +1)δ = e2αiz+(eαi +1)β −1 implies β = 0. Thus again the orbit of the action of the group generated by α◦φ1 ◦α is finite. The third case provides us with the equality δ = 0. Hence again the orbit α ◦ φ2 ◦ α−1 is finite. The fourth case is impossible since φ1 ◦ φ2 6= φ2 ◦ φ1. Statement 10. Let a foliation F on two-dimensional manifold M be generated by locally free action of R, orientable [15] and possess complete transversal P . Then there exists a transversal homotopic to H-complementary on (M,F ).

Proof. We consider several cases. T Let there exist a compact leaf L ∈ F such that the set P L consits of at least two points. Then we can apply construction from the proof of lemma 1 of [15] and get a compact transversal P 0. Now apply to P 0 corollary 1 of the section 2.1. Now assume that all leaves of F are non-compact. Let there exist a leaf L ∈ F , such T 0 that L P = {x1, x2,... }. Consider a Poincare map φ : U(x1) = (x0, x0) ⊂ P → P , φ(x1) = x2. Then construct a continuous mapping f : U(x1) → R, f(t) = ht, here htt ∈ P , t ∈ U(x1) and f(x1)x1 = x2. Let us investigate the behaviour of f at the ends 0 of the interval (y0, y0). If there exists a ﬁnite limit lim f(x) then the mapping φ can x→y0 be extended continuously into y0 and apply similar procedure in the neighbourhood 0 U(y0) = (y1, y1). Now let lim f(x) = ∞. Consider z = lim φ(x). If z and y0 belong x→y0 x→y0 Ehresmann connection on foliations 75

S to the same leaf consider new transversal P 0 = (P \ (φ(x0), ∞)) f(x0)(x0, ∞), here ∞ 0 S 0 (x , ∞) ⊂ P . Thus let y0, z belong to different leaves. Then the set U(x) = (yi, yi) i=0 possess two maximal elements with respect to the order of [15]. Now let P intersect each leaf in one point. Then for any x ∈ P the set Hx is isotropy group of the leaf Lx. Now constructions similar to that of statement 7 tell us that these groups are isomorphic to each other for different x, x0 ∈ P . Let there ∼ ∼ exist x ∈ P , Hx = Z, otherwise the statement is obvious. Note first that if Hx = Z ∼ ∼ then there exists a neighbourhoodS U(X) ⊂ P , such that ∀y ∈ U(X), Hy = Hx = Z. Assume that S = 6= P , i.e. there exists a limit point z ∈ P of the open set ∼ x∈P,Hx=Z T S. Let the set R = {hx|x ∈ S U(z), < hx >= Hx} be bounded. Then there exists a limit of the sequence hz = lim x → zhx. Assume that at least one of these limits does ∼ not vanish. Then continuity of the action of R says us that Hz = Z. If there exists only one limit equal to 0 consider the sequence of numbers (hn)n∈N and the sequence of elements (xn) ⊂ P such that hnxk≥n = x. Then for each n ∈ N hnxn = xn → z but by assumptionT hnz 6= z, this contradicts continuity of the action of R. Now let R = {hx|x ∈ S U(z), < hx >= Hx} be unbounded then the action of R does not generate foliation in the point z, since for any h ∈ R, hx → x as x → z. Now apply construction of theorem 2 to the set B ⊂ P × R.

§ 4. Existence of an Ehresmann connection orientability of the foliation F .

4.1 Elimination of non-triviality of B. Our aim here is to construct a transversal P˜ such that the corresponding bundle B˜ → P˜ is trivial. Assume that the fibration B → P is non-trivial. Then there exist x ∈ P , a, b ∈ Hx, ψ ∈ Φx such that ψ(a) = a, ψ(b) 6= b. Then the path [a, b] under translation ψ passes to path [a, ψ(b)]. This means that there exists a natural mapping f : P → P of i degree greater than one but evidently less than ∞. Then there exist points ψS(b), i = 2,...,K. Then there exists a domain [x, y) ⊂ P bijective to P , here x, y ∈ L P . i Consider now union of all the curves γi, i = 1,...K, here γi connects ψ (b) with i+1 K ψ (b) for i = 1,...,K − 1 and γK connects ψ (b) with b. Then union of all these curves translated along transversal P defines a torus T2 ⊂ M. Now the curve nS−1 γ = φi(γ) intersects P in finite number of points. Then there exists a transversal i=0 P˜ intersecting γ in exactly one point.

4.2 Compact transversal Here we assume that the transversal P is compact. Then the mapping π : B → P is a fibration and the set Φx for any x ∈ P possess a natural group structure. Our first goal is to find the other transversal P1 such that the set Φy is a commutative group for any y ∈ P1. 76 Pyotr N. Ivanshin

4.2.1 Elimination of ﬁnite cyclic subgroups Here we construct the transversal P 0 whose group Φ0 does not contain ﬁnite cyclic subgroups. The main idea coincides with one given in the section 3.2.

Statement 11. Assume that there exists the element φ ∈ Φx, φ : P → P and the number n ∈ N such that φn = Id. Then there exists a transversal P 0 such that the generators of Φ0 are those of Φ with the exception of φ. Proof. Consider a curve γ : [0, 1] → L, γ(0) = x, γ(1) = φ(x) and define up to homotopy its translation along P into the curve starting at φ(x) and ending at φ2(x). Let us denote this curve by φ(γ). Locally — in some neighbourhoods of the points x and φ(x) on the transversal P — this translation is correctly defined sine the mapping π : B → P is a fibration. Now, compactness of the transversal P gives us the possibility to extend this translation on all P . The question is whether the image of γ under φn is homotopic to γ. This holds true since the endpoints of γ and φn(γ) coincide by assumption and any loop in Rk = H is contractible. Next we deform the translation so that φn(γ) = γ. This can be done since the set of points x, φ(x), . . . , φn−1(x) is finite subset of P , hence each of them has a neighbourhood that does not intersect one of adjacent point. So we have a set of curves γ, φ(γ), . . . φ−1(γ), whose union on H bounds some 2-dimensional domain (a set homeomorphic to the disk D2). Now we are able to use the method introduced for crystallographic groups. Let us first divide the transversal P into intervals isomorphic to P/ < φ >. Consider a continuous map ρ : P → Rk such that ρ(x) = 0, ρ(φ(x)) = h−1, here φ(x) = hx, etc. It is possible again since the set x, φ(x), . . . , φn−1(x) is finite one, or the parts representing continuous sections of P/ < φ >→ P are nontrivial intervals. This gives us a desired transversal P 0 = ρ(P ). The rest of the considerations coincide with that of the main statement proof of section 3.2.

4.2.2 Construction of the transversal P 00 with Abel group Φ00. In this section we assume that the transversal P is such that the translation group Φ does not contain ﬁnite cyclic (hence just ﬁnite) subgroups. QN βi Consider the relation aαi = 1, here for j1, . . . , jk ∈ 1,N i=1 X βj = nj 6= 0, i = 1, k

αji =αi

Let us simplify further expressions denoting elements aαi by bi, i = 1, k. Consider a group Φ1 ⊂ Φ generated by bi, i = 1, k and the inﬁnite cyclic group < b1 >⊂ Φ1, generated by the element b1. hen the set S = Φ1/ < b1 > is ﬁnite. 0 Consider a set of representatives of the equivalence classes S = Φ / < b1 > S = {e, c1, . . . , cm}, here e is the neutral element of Φ. Now we can construct a disk D with the boundary consisting of curves connecting points x, c1x, . . . , cmx, x. Let us construct the new transversal as follows: Fix a point x0 ∈ D, x0 6= x, c1x, . . . , cmx, x and the set of its images under translations of D by elements of Ehresmann connection on foliations 77

< b1 >. Let us construct a continuous mapping φ : P → M. First put φ(x) = φ(c1x) = ... = φ(cmx) = x0. Then we extend this mapping to the whole transversal P . Consider now P 0 = φ(P ). This again can be done since Sx ⊂ P is ﬁnite. The resulting curve P 0 is a transversal since we can assume that the diameter of the translated disk D is bounded from above. The action of the image of transformations c1, . . . , cm ∈ Φ under mapping φ will all coinside with that of the image of x0 → b1x0. Recall now that the elementT a ∈ Φx is characterised by two components, namely, 1) the image a(x) ∈ L P of a ﬁxed point x ∈ P ; 2) the curve on P connecting x with a(x) ∈ P . Thus any edge of the disk D gives rise to some (not one) curve on P . Hence the boundary of D generates a loop in P . It is clear that the image of this loop under the mapping φ is also a loop on P 0 simply because image of all the vertices of D is one point. After all the commutation relations satisfying 4.2.2 are thus eliminated we must QN βi introduce the relations of the type aαi = 1, here for any j = 1,N i=1 X βj = 0.

αj =αi

−1 −1 So we must either introduce relations aba b = 1 or show that the group Φx does not contain free noncyclic subgroups. The latter of the statements appears to be better suited for the solution of our problem.

Statement 12. The group Φ does not contain free noncyclic subgroups.

Proof. Assume the contrary, i.e. that the group under consideration contains at least one such subgroup A. Let us denote basic elements of this group as a, b. Consider any Riemannian metric d0 on M. n n As in theorem 2 construct a homeomorphism φ : R → R , such that ∀α, β, a ∈ Φx d(φ(aαx), φ(aβx)) =T c(α, β). Since the set L P ⊂ M is discrete there exist the leaf L0, the point x ∈ L0 and the element a ∈ Hx such that d(x, axT) = min{d(y, by)|y ∈ P, b ∈ Hy}. Consider these points x, ax ∈ L P . We can deform action of H on M so that in some neighbourhood U(x) = (x1, x2) ⊂ P the distance between points y(t), a(t)y(t), y(t) ∈ U(x) in the leaf Ly(t) becomes constant. Recall that B → P is a fibration. So this procedure can be extended to the whole transversal P . Next consider the leaf L1, the point x1 ∈ L and the element a1 ∈ Hx such that d(x1, a1x1) = min{d(y, by)|y ∈ P, b ∈ Hy, b does not belong to the connected component of a in the fibration B }. Repeat deformation of the group action with these elements as instead of x, ax. And so on. Note that no one of these processes nor any combination of them produce singular action of the group H. At thesame time we have an equality d(x, b−nabnx) = d(bnx, abnx) = d(x, ax) by construction of the new group action. Since group A is free the set of points b−nabnx is infinite. Then the set Ax is not discrete. This contradiction completes the proof. 78 Pyotr N. Ivanshin

4.2.3 Finiteness of elimination processes.

According to the results of the previous sections we can say that the group Φx does not contain either commutative ﬁnite or free noncyclic subgroups for any x ∈ P . Let us show now that this group is also ﬁnitely generated.

Statement 13. The commutation relations elimination process of section 4.2.2 is ﬁnite.

Proof. Assume the contrary. Then the sequence of coverings of the previous sections P → P1 → P2 → ... is infinite, here each transversal Pi is obviously diffeomorphic to S1. Hence, any neighbourhod of arbitrary point of M contains closed transversal. Contradiction with the definition of foliated manifold.

Corollary 4. Group Φ is ﬁnitely generated.

Proof. First note that statement 13 implies that there is finite number of commutation relations. Thus the only problem is finiteness of basicT set of Φ. Again assume the contrary. Then for any leaf L and any point x ∈ L P there must exist a sequence xn ⊂ L, d(xn, xm) → 0, (n, m → ∞). Indeed, recall first that dL is Φ-invariant. Then the basic set of Φ provides us with infinite number of distinct points in L. Consider k S S j now the diameter of the net Σk = ai x. Note that diameter dn = diam(Σn) i=1 j∈Z is monotone decreasing as n → ∞. Consider R = infn(dn). Then by definition of ∞ infimum there is infinite number of points (ank x)k=1 such that R < d(x, ank x) < 2R. ∞ Then the set (ank x)k=1 has an accumulation point, hence R = 0. A contradiction with the assumption of closedness of P ⊂ M.

4.2.4 Modiﬁcation of the existing commutation relations Now the translation group Φ of the transversal P is ﬁnitely generated and does not possess commutation relations of the type 4.2.2. Let us denote this transversal by P0. This transversal is such that the commutation relations of its group Φ0 are only lij lji lji lij of the type ai aj = aj ai . N Now put li = LCM(li,j)j=1, here N equals number of basic elements of Φ0. li Consider again a polygon in L with vertices x, aix, . . . , a x ⊂ L. We must apply to it the procedure of section 4.2.2. Let us denote the result transversal by P 0. Note that it is li-covered by P . Thus we can assume that ∀a, b ∈ Φ0, ab = ba.

4.2.5 Existence of the Ehresmann connection The main construction idea is as in note 1.

Statement 14.T If for some x ∈ P group Φx is commutative then its action on Hx{x} = Lx P is free. Ehresmann connection on foliations 79

Proof. Assume the contrary, i.e. ∃h ∈ Hx and a ∈ Φx, a(x) 6= x, but a(hx) = hx. But then by transitivity of the action of Φx ∃ψ ∈ Φx, such that ψ(x) = hx. Hence −1 −1 ψ◦a◦ψ (x) = x. But the group Φx is commutative so ψ◦a◦ψ = a, thus a(x) = x. Contradiction with the assumption. Hence a = id ∈ Φx.

1) By statement 8, B → P0 is trivial bundle and the group Φ can be identiﬁed with Hx. 2) Φ is a commutative free group without ﬁnite subgroups by 4.2.2. Thus Φ =∼ Zk. 3) The proof of statement 12 implies that Φ =∼ Zk is isometrically embedded intoH =∼ Rn. Hence k ≤ n. 1)-3) combined tell us that the group Φ meets conditions of theorem 3. Theorem 5. Let a manifold M with foliation F meet the following conditions: 1) codimF = 1. 2) The foliation F is generated by locally free action of Rn. 3) There exists a compact connected transversal on (M,F ). Then there exist a transversal homotopic to H-complementary one and Ehresmann connection on (M,F ).

4.3 Noncompact transversal on transversally orientable foliation Our aim is to prove statement similar to the last one of the previous section under slightly diﬀerent conditions, namely 1) codimF = 1. 2) The foliation F is generated by locally free action of the group Rn. 3) (M,F ) is transversally orientable. Main idea of solution is to reduce the problem to the one considered in section 4.2. Statement 15. Let (M,F ) be transversally orientable. Then there exists a transversal P on M such that B → P is a ﬁbre bundle. Proof. Construction of this transversal is similar to one of the proof of statement 10.

Now consider two cases. 1) There exist only noncompact transversals on (M,F ). 2) There exists at least one compact transversal on (M,F ). It holds true for compact manifold M [15]. The second case clearly coincides with one considered in the previous section. Statement 16. Let the foliation F on a manifold M be transversally orientable. Assume that there exist only noncompact transversals on (M,F ). Then the following holds true. ∀(pn)n∈N, pn → ±∞ (n → ∞), 6 ∃p0, pn → p0 as n → ∞. Proof. Assume the contrary. Then application of the procedure from the proof of lemma 1 [15] in the coordinate chart adapted to the foliation of the neighbourhood of p0 ∈ P produces a compact transversal of (M,F ). 80 Pyotr N. Ivanshin

S Now let us compactify a transversal P =∼ R in the usual way to get S1 =∼ P {∞}. This can be done on M by statement 16. 1) Deform the action of the group H on M so that ∀h ∈ H lim hp = lim hp. p→∞ p→−∞ 2) Assume that there exists a limit lim h(p) and |hp| & lim h(p) for any p→±∞ p→±∞ element of Hx, x ∈ P , here h(p) ∈ Hp is an element of the connected component of H ∈ Hp × p ⊂ B. 3) Deﬁne action of H in the {∞} ∈ M as follows:

h{∞} = lim hp p→∞

Note that then ∀h(p) ∈ Hp, lim h(p){∞} = {∞}. p→∞ S Thus we get a manifold M 0 = M H{∞} such that the action of H on it is 0 0 0 locally free, hence this action generatesT a foliation F on it. At the same time (M ,F ) possess a compact transversal P {∞}. The coordinate chart in the neighbourhood 0 of {∞} × 0 ∈ M is a cartesian product (P \ K) × Bmin{|h|/2|h∈H∞}(0) with the natural coordinates. The coordinate chart of any point {∞} × h is then (P \ K) ×

0 0 Bmin{|h−h |/2|h ∈H∞}(h) Thus again M,F,P meets conditions of the main statement of the previous section. Hence we get a corollary of theorem 6.

Theorem 6. Let a manifold M with foliation F meet following conditions: 1) codimF = 1. 2) Foliation F is generated by action of Rn. 3) (M,F ) is transversally orientable. Then there exist a transversal homotopic to H-complimentary and an Ehresmann connection on (M,F ).

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