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Mathematics 2019, 7(11), 1058; https://doi.org/10.3390/math7111058

Article
Smooth Counterexamples to Frame Bundle Freeness
School of Mathematics, University of Minnesota, Minneapolis, MN 55455, USA
Received: 11 September 2019 / Accepted: 24 October 2019 / Published: 5 November 2019

Abstract

:
We provide counterexamples to P. Olver’s freeness conjecture for C actions. In fact, we show that a counterexample exists for any connected real Lie group with noncompact center, as well as for the additive group of the integers.
Keywords:
prolongation; moving frame; dynamics
MSC:
57Sxx; 58A05; 58A20; 53A55

1. Introduction

P. Olver’s freeness conjecture (in his words) asserts: “If a Lie group acts effectively on a manifold, then, for some n < , the action is free on [a nonempty] open subset of the jet bundle of order n.” There is some ambiguity in this wording: No mention is made of connectedness of the group or manifold, the particular choice of jet bundle isn’t made precise and the smoothness of the action is left unspecified.
In this note, we provide counterexamples to one interpretation of the freeness conjecture for C actions and higher order frame bundles. Those who know Olver’s work will understand that there are a family of associated jet bundles to which he generally refers, and work in frame bundles then informs results in these jet bundles, through the associated bundle construction. In the C context, Olver has noted that, to avoid elementary counterexamples, “effective” must be strengthened to “fixpoint rare”, which we define in Section 2 below. In Theorem 1 and Lemma 2, we show that a counterexample exists for any connected real Lie group with a noncompact center, as well as for the additive group of the integers. We also prove [1] the validity of the conjecture for connected real Lie groups with compact center. Finally, in [2], we describe a certain “meager” modification of the C ω conjecture, and prove it holds for all connected real Lie groups.
For any fixed group, the C conjecture implies the C ω conjecture, so the C ω conjecture (on frame bundles) is now proved for connected real Lie groups with compact center. In [3], we offer a Z -action on a manifold with an infinitely generated fundamental group, which, after induction of actions, provides a C ω counterexample for any connected Lie group with a noncompact center. There is the possibility that the construction in Theorem 1 could be modified to make counterexamples to the C ω conjecture on a contractible manifold, e.g., R 4 . The main difficulty in such an extension appears to be technical, and revolves around developing a good understanding of convergence of sequences in C ω with respect to some well-chosen topology. For this, D. Morris’ unpublished note [4] may be useful.

2. Miscellaneous Notation and Terminology

A subset of a topological space is meager (a.k.a. of first category) if it is a countable union of nowhere dense sets. A subset of a topological space is nonmeager (a.k.a. of second category) if it is not meager. A subset of a topological space is comeager (a.k.a. residual) if its complement is meager.
Let N : = { 1 , 2 , 3 , } . For all d N , let Id d : R d R d be the identity map, defined by Id d ( σ ) = σ . For every subset S R , for every d N , we define S d : = S × S × × S R d .
Define V 0 : R 4 R 4 by: for all σ R 4 , V 0 ( σ ) = ( 0 , 0 , 0 , 1 ) .
Let d 1 be an integer. A function V : R d R d will be said to be complete if it is C and represents a complete vector field on R d . For any complete V : R d R d , we will use the notation Φ t V : R d R d to denote the time t flow of V, defined by the ODE ( d / d t ) ( Φ t V ( σ ) ) = V ( Φ t V ( σ ) ) and by the initial value condition Φ 0 V = Id d .
Let d 1 be an integer. Let V : R d R d be complete. For any A R , for any B R d , let Φ A V ( B ) : = { Φ a V ( b ) | a A , b B } . For any A R , for any b R d , let Φ A V ( b ) : = { Φ a V ( b ) | a A } . For any a R , for any B R d , let Φ a V ( B ) : = { Φ a V ( b ) | b B } .
Let d 1 be an integer. Let V : R d R d be complete. Let σ R d . We say that ( V , σ ) is periodic if there exists an integer n 0 such that Φ n V ( σ ) = σ . For any integer k 0 , we say that ( V , σ ) is periodic to order k if there is an integer n 0 such that the map Φ n V : R d R d agrees with the identity Id d : R d R d to order k at σ . We say that ( V , σ ) is periodic to all orders if there is an integer n 0 such that the map Φ n V : R d R d agrees with Id d : R d R d to all orders at σ .
An action of a group G on a topological space X is fixpoint rare if, for any nonempty open subset U of X, for all g G \ { 1 G } , there exists u U such that g u u . Any fixpoint rare action is effective. For a C ω action on a connected manifold, fixpoint rare and effective are equivalent. Any continuous transitive action of a real Lie group preserves a C ω structure, from which it follows that: By lemma 6.1 of [5], a continuous action of a connected real Lie group G on a topological space X is fixpoint rare iff, for every nonempty G-invariant open subset V of X, the G-action on V is effective.

3. Description of the Proof

To aid the reader, we give a broad description of how to construct a counterexample. Here, we focus only on flows in R 4 . We identify vector fields on R 4 with smooth maps R 4 R 4 .
Let V : R 4 R 4 be any complete vector field on R 4 . Then, PRP V will denote the set of all x R 4 such that t Φ t V ( x ) : R R 4 is proper. We say V is generically proper if PRP V is comeager in R 4 .
Note that PRP V 0 = R 4 , so V 0 is generically proper.
Let { ω 1 , ω 2 , ω 3 , } be a countable dense subset of R 4 . For all j N , let B j denote the ball about ( 0 , 0 , 0 , 0 ) of radius j in R 4 with its usual Euclidean metric. Let Id : = Id 4 , the identity map R 4 R 4 .
Let σ 1 : = ω 1 . Choose a complete vector field V 1 on R 4 such that, for some T 1 R \ { 0 } , Φ T 1 V 1 agrees with Id on a neighborhood of σ 1 . (For example, V 1 : R 4 R 4 could be the identically zero.) Note that V 1 is NOT generically proper because every point near σ 1 has a periodic (hence nonproper) orbit. Let Σ 1 : = Φ R V 1 ( σ 1 ) denote the orbit of σ 1 under the flow of V 1 . Then, Σ 1 is compact. Let C 1 denote an open ball in R 4 containing Σ 1 .
Let A 1 denote the set of all complete vector fields V : R 4 R 4 such that V agrees with V 1 to infinite order at every point of Σ 1 .
Exercise for the reader: Find a generically proper vector field V 2 in A 1 . We expect that many (and perhaps MOST) vector fields in A 1 are generically proper.
By density, choose σ 2 PRP V 2 such that | σ 2 ω 2 | < 1 / 2 .
Since σ 2 PRP V 2 , for all sufficiently large t > 0 , we have both Φ t V 2 ( σ 2 ) C 1 and Φ t V 2 ( σ 2 ) C 1 .
Choose a complete vector field V 2 on R 4 such that V 2 agrees with V 2 on C 1 and for some T 2 R \ { 0 } , Φ T 2 V 2 = Id on some neighborhood of σ 2 .
Note that V 2 is NOT generically proper because every point near σ 2 has a closed (hence nonproper) orbit. Let Σ 2 : = Φ R V 2 ( σ 2 ) denote the orbit of σ 2 under the flow of V 2 . Then, Σ 2 is compact. Let C 2 denote an open ball in R 4 containing B 1 C 1 Σ 1 .
Let A 2 denote the set of all complete vector fields V : R 4 R 4 such that V agrees with V 2 to infinite order at every point of Σ 2 and such that V agrees with V 2 on C 1 .
Exercise for the reader: Find a generically proper vector field V 3 in A 2 . We expect that many (and perhaps MOST) vector fields in A 2 are generically proper.
By density, choose σ 3 PRP V 3 such that | σ 3 ω 3 | < 1 / 3 .
Since σ 3 PRP V 3 , for all sufficiently large t > 0 , we have both Φ t V 3 ( σ 3 ) C 2 and Φ t V 3 ( σ 3 ) C 2 .
Choose a complete vector field V 3 on R 4 such that V 3 agrees with V 3 on C 2 and for some T 3 R \ { 0 } , Φ T 3 V 3 = Id on a neighborhood of σ 3 .
Note that V 3 is NOT generically proper because every point near σ 3 has a closed (hence nonproper) orbit. Let Σ 3 : = Φ R V 3 ( σ 2 ) denote the orbit of σ 3 under the flow of V 3 . Then, Σ 3 is compact. Let C 3 denote an open ball in R 4 containing B 2 C 2 Σ 3 .
Let A 3 denote the set of all complete vector fields V : R 4 R 4 such that V agrees with V 3 to infinite order at every point of Σ 3 and such that V agrees with V 2 on C 1 .
Exercise for the reader: Find a generically proper vector field V 3 in A 3 . We expect that many (and perhaps MOST) vector fields in A 3 are generically proper.
Continuing, we construct a sequence V 1 , V 2 , of vector fields on R 4 . By construction, there is a unique vector field W on R 4 such that:
j N , W = V j on C j .
The flow of W agrees to infinite order with Id on the set { σ 1 , σ 2 , } . Since { ω 1 , ω 2 , } is dense in R 4 , we see that { σ 1 , σ 2 , } is dense in R 4 as well. It remains to argue that the flow of W is fixpoint rare.
Given t R \ { 0 } and a nonempty open set U in R 4 , we want: u U s.t. Φ t W ( u ) u . Assume, for a contradiction that u U , Φ t W ( u ) = u .
For all j N , let Q j : = { z R 4 | Φ t W ( z ) = Φ t V j ( z ) } . By construction, j N Q j = R 4 , so j N ( Q j U ) = U . By the Baire Category Theorem, U is nonmeager in R 4 , so choose j N such that Q j U is nonmeager in R 4 . Since V j is generically proper, we get: PRP V j is comeager in R 4 . Then, Q j U PRP V j , and we fix a point z Q j U PRP V j . As z Q j , we see that Φ t W ( z ) = Φ t V j ( z ) . As z U , we see that Φ t W ( z ) = z . Then, Φ t V j ( z ) = z . As z PRP V j , we see that s Φ s V j ( z ) : R R 4 is proper. Since t 0 and Φ t V j ( z ) = z , we get: s Φ s V j ( z ) : R R 4 is nonproper. Contradiction.
The exercises for the reader (given above) have not been completed by the author. They motivate the following OPEN PROBLEM:
Let V : R 4 R 4 be a complete, generically proper vector field. Let C R 4 have compact closure in R 4 , and let p PRP V . Prove: there is a complete, generically proper vector field V ˜ : R 4 R 4 such that V ˜ agrees with V on C and for some T R \ { 0 } , Φ T V ˜ agrees with Id at p to infinite order.
To solve this open problem, one needs to take the V-orbit of p (which leaves compact sets, in both positive and negative time), and “bend” it, outside of C, so that it turns into an orbit that is periodic to infinite order. Moreover, this must be done in such a way as to preserve completeness and generic properness.
It is possible that this problem may admit a simple solution, in which case a relatively easy counterexample would be possible, as described above. In this paper, we replace “generically proper” by a stronger condition, “porous”. We provide the iterative step in Lemma 1. Its proof is complicated and appears in [6].

4. The Iteration

Let I : = { ( a , a ) R | a N } . For every I I , let a I : = sup I , so a I N and I = ( a I , a I ) . For every I I , for every integer n 1 , we define n I : = ( n a I , n a I ) I ; then, a n I = n a I . For every I I , let I ¯ : = [ a I , a I ] be the closure in R of I. We define I 0 : = ( 1 , 1 ) I ; then, a I 0 = 1 .
Let C be the set of C maps V : R 4 R 4 such that V ( R 4 ) I ¯ 0 4 . Then, V 0 C . For all V C , the vector field V : R 4 R 4 is complete.
Let V C . Let Π : R 4 R be projection onto the 4th coordinate, defined by Π ( x 1 , x 2 , x 3 , x 4 ) = x j . We say σ R 4 is undeterred by V if Π ( Φ R V ( σ ) ) = R . The undeterred set for V is the set U ( V ) of all σ R 4 such that σ is undeterred by V. This set U ( V ) is V-invariant. We say V is porous if U ( V ) is dense in R 4 . For example, U ( V 0 ) = R 4 , so V 0 is porous. By a deterrence system, we mean an element ( V , I ) C × I such that V = V 0 on ( R 4 ) \ ( I 4 ) . Let D be the set of all deterrence systems. Note, for all ( V , I ) D , that V V ( a I ) .
For any I I , we define C I : = { P C | P = V 0 on ( 3 I ¯ ) 4 } .
Let ( V , I ) D . For any ( W , J ) D , we say ( W , J ) is a modification of ( V , I ) if: both [ a I < a J ] and [ W = V on I ¯ 4 ].
We define
M ( V , I ) : = { ( W , J ) D | ( W , J ) is a modification of ( V , I ) } .
Let I I . We define D I × : = { ( P , K ) D | 4 I K and P C I } . We denote by P I the set of ( P , K ) D I × such that, for some integer m > 2 a I , we have:
  • Φ m P agrees with Id 4 to all orders at ( 0 , 0 , 0 , a I ) and
  • for all τ I 3 × { a I } , for all t ( 0 , m ) ,
    [ Φ t P ( τ ) I 4 ] [ t < 2 a I ] .
Lemma 1.
Let ( V , I ) D . Assume V is porous. Let σ U ( V ) . Then, there exists ( V , I ) M ( V , I ) such that V is porous and such that ( V , σ ) is periodic to all orders.
Proof. 
See Lemma 16.1 of [6]. □

5. A Result about Cyclic Groups

Lemma 2.
If G is a connected real Lie group whose center Z ( G ) is noncompact, then Z ( G ) has an infinite cyclic closed subgroup.
Proof. 
This is Lemma 17.3 of [6]. □

6. The Counterexamples

Note that Theorem 1 below applies when G is discrete and isomorphic to the additive group Z . Lemma 2 shows that Theorem 1 also applies when G is a connected real Lie group whose center Z ( G ) is noncompact. In addition, note by (ii) below that, if
either G is connected or G = Z ,
then M is connected.
For any C manifold M, for any integer k 0 , let π k M : F k M M denote the kth order frame bundle of M. Let Π : R 4 R be projection onto the 4th coordinate, defined by Π ( x 1 , x 2 , x 3 , x 4 ) = x j .
Theorem 1.
Let G be a real Lie group. Assume the center Z ( G ) of G admits an infinite cyclic closed subgroup Z. Then, there is a C manifold M and a fixpoint rare C action of G on M such that:
(i) 
for any integer k 0 , there is a dense subset D of F k M such that, for all δ D , the stabilizer Stab Z ( δ ) is infinite and
(ii) 
the number of connected components of M and G / Z are equal.
Proof. 
We have ( V 0 , I 0 ) D . In addition, U ( V 0 ) = R 4 , so V 0 is porous.
Let | | : R 4 [ 0 , ) be a norm on R 4 . Let { ω 1 , ω 2 , ω 3 , } be a countable dense subset of R 4 . Choose σ 1 R 4 such that | σ 1 ω 1 | < 1 . Then, σ 1 R 4 = U ( V 0 ) .
By Lemma 1, choose ( V 1 , I 1 ) M ( V 0 , I 0 ) such that V 1 is porous and such that ( V 1 , σ 1 ) is periodic to all orders. Because V 1 is porous, U ( V 1 ) is dense in R 4 , so fix σ 2 U ( V 1 ) such that | σ 2 ω 2 | < 1 / 2 .
By Lemma 1, choose ( V 2 , I 2 ) M ( V 1 , I 1 ) such that V 2 is porous and such that ( V 2 , σ 2 ) is periodic to all orders. Because V 2 is porous, U ( V 2 ) is dense in R 4 , so fix σ 3 U ( V 2 ) such that | σ 3 ω 3 | < 1 / 3 .
By Lemma 1, choose ( V 3 , I 3 ) M ( V 2 , I 2 ) such that V 3 is porous and such that ( V 3 , σ 3 ) is periodic to all orders. Because V 3 is porous, U ( V 3 ) is dense in R 4 , so fix σ 4 U ( V 3 ) such that | σ 4 ω 4 | < 1 / 4 .
Continuing yields a countable dense subset { σ 1 , σ 2 , σ 3 , } of R 4 , and a sequence ( V 1 , I 1 ) , ( V 2 , I 2 ) , ( V 3 , I 3 ) , in D . For each integer j 1 ,
  • a I j + 1 > a I j ,
  • V j + 1 = V j on I j 4 ,
  • V j is porous, and
  • ( V j , σ j ) is periodic to all orders.
We have a I 1 < a I 2 < a I 3 < and a I 1 , a I 2 , a I 3 , N . It follows both that I 1 I 2 I 3 and that I 1 I 2 I 3 = R . Then, I 1 4 I 2 4 I 3 4 and I 1 4 I 2 4 I 3 4 = R 4 . Define V C by the rule: For all integers j 1 , V = V j on I j 4 . Let X : = k = 1 ( U ( V k ) ) .
Claim 1: For all t R \ { 0 } , for all σ X , we have Φ t V ( σ ) σ .
Proof of Claim 1. 
Let t R \ { 0 } and let σ X . Assume, for a contradiction, that we have Φ t V ( σ ) = σ .
Since Φ R V ( σ ) = Φ [ 0 , t ] V ( σ ) , we see that Φ R V ( σ ) is compact. Then, since I 1 4 I 2 4 I 3 4 and I 1 4 I 2 4 I 3 4 = R 4 , fix an integer j 1 such that Φ R V ( σ ) I j 4 . We have V j = V on I j 4 . Then, by Lemma 6.1 of [6], Φ R V j ( σ ) = Φ R V ( σ ) .
Then, Π ( Φ R V j ( σ ) ) = Π ( Φ R V ( σ ) ) Π ( I j 4 ) = I j . However because σ X U ( V j ) , it follows that Π ( Φ R V j ( σ ) ) = R . Thus, we have R I j , contradiction. □
Claim 2: For all integers j 1 , ( V , σ j ) is periodic to all orders.
Proof of Claim 2. 
Fix an integer j 1 . We wish to show that ( V , σ j ) is periodic to all orders.
Because ( V j , I j ) D and ( V j , σ j ) is periodic, we conclude, from Lemma 8.13 of [6], that Φ R V j ( σ j ) I j 4 . For all τ I j 4 , because V and V j agree on I j 4 , which is an open neighborhood of τ , it follows that they agree to all orders at τ . Thus, for all t R , V and V j agree to all orders at Φ t V j ( σ j ) . Then, by Lemma 6.3 of [6], for all t R Φ t V : R 4 R 4 and Φ t V j : R 4 R 4 agree to all orders at σ j . Thus, as ( V j , σ j ) is periodic to all orders, it follows ( V , σ j ) is periodic to all orders as well. □
Because Z is infinite cyclic, it follows that Z is isomorphic to the additive discrete group Z . Let f : Z Z be an isomorphism. Define a Z-action on R 4 by: for all z Z , for all σ R 4 , z σ = Φ f ( z ) V ( σ ) . Let M : = G × Z R 4 . Because the Z-action on R 4 is C , it follows that the G-action on M is C as well. By construction, M is (the total space of) a fiber bundle over G / Z with fiber R 4 , so, because R 4 is connected, M has the same number of connected components as does G / Z .
By Corollary 8.4(i) of [6], U ( V ) is an open subset of R 4 . For all integers k 1 , V k is porous. Then, U ( V k ) is a dense open subset of R 4 . Then, because X = k = 1 ( U ( V k ) ) , we see, by the Baire Category Theorem, that X is dense in R 4 . By Claim 1, for all z Z \ { 1 Z } , for all σ X , we have Φ f ( z ) V ( σ ) σ , i.e., we have z σ σ . Thus, the Z-action on R 4 is fixpoint rare. Then, the G-action on M is also fixpoint rare.
Let p : G × R 4 G × Z R 4 = M be the canonical map. Define an injection ι : R 4 M by ι ( σ ) = p ( 1 G , σ ) . Let Σ : = ι ( { σ 1 , σ 2 , σ 3 , } ) . Because { σ 1 , σ 2 , σ 3 , } is dense in R 4 , A : = G Σ is dense in M.
Fix an integer k 0 , and let π : = π k M : F k M M be the kth order frame bundle of M. Let D : = π 1 ( A ) . Since A is dense in M, and since π : F k M M is open, it follows that D is dense in F k M . Fix δ D . We wish to show that Stab Z ( δ ) is infinite.
Let S : = Stab G ( δ ) . Then, Stab Z ( δ ) = S Z . We therefore wish to show that S Z is infinite.
Since π ( δ ) A = G Σ , fix g 0 G , τ 0 Σ such that π ( δ ) = g 0 τ 0 . Let δ 0 : = g 0 1 δ . Then, π ( δ 0 ) = τ 0 , i.e., δ 0 π 1 ( τ 0 ) . Let S 0 : = Stab G ( δ 0 ) . Then, S = g 0 S 0 g 0 1 . As Z Z ( G ) , we have Z = g 0 Z g 0 1 . Then, S Z = g 0 ( S 0 Z ) g 0 1 , so it suffices to show that S 0 Z is infinite.
Recall that Id 4 : R 4 R 4 is the identity map defined by Id 4 ( σ ) = σ . Let Id : M M be the identity map defined by Id ( ρ ) = ρ .
Since τ 0 Σ = ι ( { σ 1 , σ 2 , σ 3 , } ) , fix an integer j 1 such that τ 0 = ι ( σ j ) . By Claim 2, fix n 0 Z \ { 0 } such that Φ n 0 V : R 4 R 4 agrees with the identity Id 4 : R 4 R 4 to all orders at σ j .
Let z 0 : = f 1 ( n 0 ) . Then, z 0 Z \ { 1 Z } , and, for all σ R 4 , we have z 0 σ = Φ n 0 V ( σ ) . Then, the map σ z 0 σ : R 4 R 4 is equal to Φ n 0 V : R 4 R 4 , and therefore agrees with Id 4 : R 4 R 4 to all orders at σ j . Then, since Z Z ( G ) and since ι ( σ j ) = τ 0 , it follows that the map ρ z 0 ρ : M M agrees with Id : M M at τ 0 to all orders. In particular, ρ z 0 ρ : M M agrees with Id : M M at τ 0 to order k. Then, for all ρ π 1 ( τ 0 ) , we have z 0 ρ = ρ . Thus, since δ 0 π 1 ( τ 0 ) , we get z 0 δ 0 = δ 0 . That is, z 0 Stab Z ( δ 0 ) = S 0 Z . Let C 0 be the cyclic subgroup of S 0 Z generated by z 0 . Every nontrivial subgroup of an infinite cyclic group is infinite, so C 0 is infinite. Thus, because C 0 S 0 Z , it follows that S 0 Z is infinite, as desired. □

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

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