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The bounded isometry conjecture for the kodaira-thurston manifold and the 4-torus

The purpose of this note is to study the bounded isometry conjecture proposed by Lalonde and Polterovich. In particular, we show that the conjecture holds for the Kodaira-Thurston manifold with the standard symplectic form and for the 4-torus with all linear symplectic forms.


Introduction and main results
Let (M, ω) be a closed symplectic manifold. There is a natural bi-invariant norm, called the Hofer norm ρ, defined on the Hamiltonian diffeomorphism group Ham(M, ω). That is, ρ(f ) is the Hofer distance between the identity map id and f for all f ∈ Ham(M, ω), see Section 4 for details. Lalonde and Polterovich [11] have studied the full symplectomorphism group Symp(M, ω) within the framework of Hofer's geometry. We first recall the notion of bounded and unbounded symplectomorphisms. Namely, for each φ ∈ Symp(M, ω), define Denote by BI 0 (M, ω) the set of all bounded elements in the identity component Symp 0 (M, ω) of Symp(M, ω). Since ρ is bi-invariant, it follows from the inequality ρ([φ, f ]) 2ρ(φ) that Ham(M, ω) is a subgroup of BI 0 (M, ω). The converse is the following conjecture in [11]. This conjecture was proved in [11] for closed surfaces with area form and for arbitrary products of closed surfaces of genus greater than 0 with product symplectic form; Lalonde and Pestieau [12] confirmed it for product symplectic manifolds M = N × W with N being any product of closed surfaces and W being any closed symplectic manifold of first real Betti number equal to zero. In this note, we give a positive answer to this conjecture for the Kodaira-Thurston manifold with the standard symplectic form and for the 4-torus with all linear symplectic forms.
Let {φ t } ∈ Symp 0 (M, ω), i.e. φ t is a smooth isotopy in Symp 0 (M, ω). There exists a unique family of vector fields X t which generates the flow φ t , i.e. It is well known that this map is surjective, and its kernel is equal to Ham(M, ω).
In other words, we have the following exact sequence of groups We refer to [14] Chapter 10 for more details.
Since whether or not the flux is equal to 0 distinguishes a Hamiltonian diffeomorphism from a nonHamiltonian symplectomorphism, one main step in our applications is to understand the flux subgroup Γ.
For this, we denote as in [6] by C(M ) the space of continuous maps from M to M with the compact open topology. Given p ∈ M , we define the evaluation map ev c : C(M ) → M by ev c (f ) = f (p). Denote by ev s the restriction of ev c to Symp 0 (M, ω). We will use the same notation for the induced maps on the fundamental groups. By ev s we denote the homomorphism from π 1 (Symp 0 (M, ω)) to H 1 (M, Z), which is the composition of ev s with the Hurewitz map from π 1 (M ) to H 1 (M, Z).
The following commutative diagram due to Lalonde, McDuff and Polterovich [10] plays a crucial role in the calculation of the flux subgroup Γ.
Note that ρ mnkℓ preserves the symplectic form ω = ds ∧ dt + dx ∧ dy on R 4 . Hence the quotient (M := R 4 /G, ω) is a closed symplectic manifold, known as the Kodaira-Thurston manifold, see [19]. It was the first known example of a closed symplectic manifold which admits no kähler structure, since its first betti number b 1 = 3, see [14] Example 3.8.
The manifold M = R 4 /G can also be described as a torus bundle over a torus, that is M = R 2 × Z 2 T 2 . Here Z 2 acts on R 2 in the usual way, and it acts on T 2 via (m, n) → A mn : x y → 1 m 0 1 x y .
Our first task is to understand the flux subgroup Γ of the Kodaira-Thurston manifold described above. In particular, we have Theorem 3.1. The flux subgroup Γ ⊂ H 1 (M, R) of the Kodaira-Thurston manifold with the standard symplectic form ω = ds ∧ dt + dx ∧ dy has rank 2 over Z. Namely, Γ = Z ds, dy .
To prove Theorem 3.1, we need the following result on the cohomology groups of the Kodaira-Thurston manifold.
Proof. This follows from an easy calculation.

The Hofer norm
Let (M, ω) be a closed symplectic manifold of dimension 2n. Denote by A the space of all normalized smooth functions on M with respect to the volume form ω n , i.e.
It is well known that A can be identified with the space of Hamiltonian vector fields, which is the Lie algebra 1 of the ∞-dimensional Lie group Ham(M, ω). The x) is the time-dependent Hamiltonian function generating the path α. The Hofer distance d between two Hamiltonian diffeomorphisms f and g is defined by where the infimum is taken over all Hamiltonian paths α connecting f and g. The Hofer norm ρ(f ) is the Hofer distance between the identity map id and f , i.e.
It is easy to check that d is bi-invariant in the sense that for all f, g, h ∈ Ham(M, ω). The fact that d is nondegenerate is highly nontrivial. This was proved by Hofer [5] for the case of R 2n , then generalized by Polterovich [17] to some larger class of symplectic manifolds, and finally proved in the full generality by Lalonde and McDuff [8] using the following energy-capacity inequality for a subset S of M . Here the capacity of S is equal to πr 2 when S is a symplectically embedded ball of radius r, and is defined in general as the supremum of the capacities of all symplectically embedded balls in S. The displacement energy e(S) is defined to be the infimum of the Hofer norms of all f ∈ Ham(M, ω) such that Note that the energy-capacity inequality provides a lower bound for the Hofer norm. Namely, we have This fact will be crucial in our proof of Theorem 1.3.
Recall in Definition 1.
1 As a vector space, the Lie algebra is by definition the tangent space to the Lie group at the identity. The tangent spaces to the Lie group at other points are identified with the Lie algebra with the help of right shifts of the group.
Note that all Hamiltonian diffeomorphisms are bounded since r(g) 2ρ(g) < ∞ for all g ∈ Ham(M, ω), where ρ(g) is the Hofer norm of g. According to Proposition 1.2.A in [11], r satisfies the triangle inequality r(φψ) r(φ) + r(ψ). Since Ham(M, ω) is the kernel of the flux homomorphism, two symplectomorphisms φ and ψ have the same flux if and only if they differ by a Hamiltonian diffeomorphism. Combining these facts, we have the following Observation A. [11] In order to prove BI 0 (M, ω) = Ham(M, ω), it suffices to show that for each nonzero value v ∈ H 1 (M, R)/Γ, there exists some unbounded element φ ∈ Symp 0 (M, ω) with flux(φ) = v.

The admissible lift
To prove an element φ ∈ Symp 0 (M, ω) is unbounded, one has to show that ρ([ φ, f ]) can be arbitrarily large by choosing different f ∈ Ham(M, ω). Hence the energy-capacity inequality will not work directly for closed manifolds since the capacity of the manifold itself is finite. To go around this difficulty, we recall the notion of admissible lifts which was first introduced by Lalonde and Polterovich [11]. We shall point out that our definition is slightly different from theirs, but the two definitions are equivalent.
Let π : ( M , ω) → (M, ω) be a symplectic covering map, i.e. a covering map π between two symplectic manifolds such that ω = π * ω.  Proof. The existence follows from the definition. For the uniqueness, it suffices to show that the admissible lift g of g is independent of the choice of the Hamiltonian function H t .
Note that the choice of H t is equivalent to the choice of the Hamiltonian isotopy g t connecting id to g. For every point p ∈ M , let be the map induced by the evaluation map ev p : Ham(M, ω) → M which takes g to g(p). It follows from Floer theory that for all symplectic manifolds (M, ω), the induced map ev p is trivial, see Chapter 11 [14] for instance. This deep result implies that for any two different paths g 1 t and g 2 t in Ham(M, ω) connecting id to g, g 1 t (p) and g 2 t (p) must be homotopic paths in M . Therefore, for every point p ∈ M , the image g( p) of p under g, being the endpoint of the lift of the path g t (p), is independent of the choice of the Hamiltonian isotopy g t . This proves the uniqueness of admissible lifts.
For our purposes, we consider the universal cover M of M . Note that M is not necessarily compact, and the admissible lift g of g ∈ Ham(M, ω) is not necessarily compactly supported in M . Instead, it belongs to Ham b ( M , ω) of time-1 maps of bounded Hamiltonians M × [0, 1] → R. The Hofer norm is still well defined and the same energy-capacity inequality still holds for this setting. This idea is due to Lalonde and Polterovich [11]. We shall spell out some details here for the sake of clarity.
Denote by (N, σ) a noncompact symplectic manifold without boundary. We do not often consider the group Ham(N, σ) of all Hamiltonian diffeomorphisms with arbitrary support. One reason in our context is that it would not be possible to define the Hofer norm on Ham(N, σ) using the L ∞ norm on the space A of all Hamiltonian functions with arbitrary support, since not all elements in A have finite L ∞ norms.
One may consider the group Ham c (N, σ) of Hamiltonian diffeomorphisms with compact support. The Hofer norm ρ is well defined on Ham c (N, σ), and the energycapacity inequality To prove the other inequality, note that if f ∈ Ham b (N, σ) displaces a compact subset S from itself, one can easily construct some cut-off f cut ∈ Ham c (N, σ) of f which still displaces S from itself, and the Hofer norm satisfies ρ(f ) ρ(f cut ). Taking the infimum implies e b (S) e c (S).
It is easy to see that φ αβc preserves ω, and the obvious isotopy from id to φ αβc implies that φ αβc ∈ Symp 0 (M, ω). The calculation for flux(φ αβc ) = v is straightforward using (1) in Section 2.
The following theorem due to Lalonde and Polterovich [11] is an important criteria for unbounded symplectomorphisms. For the proof, one passes to the universal cover M of M . The hypothesis implies that the lift of a neighbourhood U of L has infinite capacity. One then constructs a Hamiltonian isotopy f τ supported in U so that the admissible lift [φ, f τ ] of the commutator [φ, f τ ] will displace a symplectic ball of arbitrarily large capacity as τ goes to infinity. This implies φ is unbounded according to Observation B. See [11] for details.
Proof of Theorem 1.3. In view of Observation A, it suffices to show that the symplectomorphisms φ αβc constructed in Lemma 6.1 are unbounded in all cases, as long as the flux v = αds + βdy + cdt does not vanish. We argue case by case. In the first two cases, this is a direct consequence of Theorem 6.2. Case 1. α = 0 ∈ R/Z.
Let L ⊂ M be the subset of M defined by It is easy to check that L is a Lagrangian torus satisfying the hypothesis of Theorem 6.2, and φ αβc displaces L from itself. Thus φ αβc is unbounded.
In this case, φ β := φ αβc maps (s, t, x, y) to (s, t, x + β, y). As in the first case, φ β displaces from itself a Lagrangian torus L of M defined by We again get φ β is unbounded in view of Theorem 6.2.
Consider two different situations, one of which is simple, while the other is more complicated.
In contrast to all previous cases where we used the same argument, here we are facing a difficulty. The trouble is that in this case we are unable to find a Lagrangian torus of M which is disjoined from itself by the map φ βc . Thus the above argument breaks down.
To resolve this difficulty, we take f τ to be the Hamiltonian isotopy whose support is in the subset U := {(s, t, x, y) ∈ M | |s| < ǫ, |x| < ǫ} of M . We require f τ to flow only along y and t direction in U and its restriction to is defined by f τ (s, t, x, y) = (s, t, x, y − τ ).
In the discussion below, [f, g] := f gf −1 g −1 stands for the commutator of f and g. Our goal is to show that the unique admissible lift [φ βc , f τ ] of [φ βc , f τ ] still displaces from itself a subset of R 4 of arbitrarily large capacity when τ goes to infinity. For this, we need the following Proof. Note that f τ is Hamiltonian implies [φ, f τ ] is Hamiltonian. So both admissible lifts [φ, f τ ] and f τ make sense. To simplify notation, denote Now back to the proof of Theorem 1.3. To prove φ βc is unbounded, we need to show that the commutator [φ βc , f τ ] has arbitrarily large Hofer norm when τ goes to infinity. Let V 0 ⊂ R 4 be the subset of R 4 defined by Since V 0 has arbitrarily large capacity as τ goes to infinity, according to Observation B, it suffices to show that the admissible lift For this, denote by φ βc : R 4 → R 4 the preferred lift of the map φ βc such that φ βc (s, t, x, y) = (s, t, x + β − cy, y).
Thus for sufficiently large τ , these two sets do not share the same values in x coordinates. Since the flow f −1 τ only changes the y and t-coordinates when restricted As we have already mentioned above, this implies φ βc is unbounded in case 3B, which completes the proof of Theorem 1.3.

Proof of Theorem 1.4
We have already mentioned in Section 1 that the bounded isometry conjecture holds for the torus with the standard symplectic form. In this section we prove Theorem 1.4 which states that the conjecture holds for the 4-torus with any linear symplectic form. We begin with a remark on the linear symplectic form ω on T 4 .
Here we take the convention that a ij = −a ji . In particular, a ii = 0. Proof. According to Lemma 2.1, we have the following commutative diagram for the manifold (T 4 , ω).
Note that ev s is surjective, and ∧[ω] : Note also that vol(T 4 ) = a 12 a 34 −a 13 a 24 +a 14 a 23 . It follows from a similar argument as in the proof of Theorem 3.1 that ξ i (1 i 4) span the flux subgroup Γ over Z.
where α i ∈ R/Z for 1 i 4. Then Recall that in view of Observation A in Section 4, to prove Theorem 1.4, it suffices to show φ is unbounded as long as at least one α i ∈ R/Z is nonzero. One may attempt to apply Theorem 6.2 by showing φ disjoins some Lagrangian torus L ⊂ T 4 from itself. For a general symplectic form ω, however, there may not exist any such Lagrangian torus in T 4 . Nevertheless, we can still prove φ is unbounded using the following This result can be found in Oh [16], Schwarz [18] and Kerman-Lalonde [7]. The main idea of the argument is that the Hofer norm is bounded from below by the spectral norm, while the spectral norm of such f τ grows linearly with respect to τ .
As discussed above, it suffices to show φ is unbounded when at least one α i ∈ R/Z is nonzero. Assume α 1 = 0 without loss of generality.
for sufficiently small ǫ.
Let H be a time-independent Hamiltonian function of T 4 supported in U . Denote by f τ the (autonomous) Hamiltonian flow generated by H. Since φ(U ) ∩ U = ∅, we know that [φ, f τ ] := φf τ φ −1 f −1 τ is also an autonomous Hamiltonian flow supported in the union of two disjoint sets U ∪ φ(U ). If we further require that H depend only on the first coordinate x 1 , using the fact that ω is a linear symplectic form, we conclude that [φ, f τ ] has no nonconstant contractible orbits. Thus it follows from Lemma 7.3 that the Hofer norm ρ([φ, f τ ]) goes to infinity as τ goes to infinity. Hence φ is unbounded in the sense of Definition 1.1.

The Kodaira-Thurston manifold with linear symplectic forms
So far we have studied bounded isometries for the Kodaira-Thurston manifold with the standard symplectic form and for the 4-torus with all linear symplectic forms. In particular, we have shown that the bounded isometry conjecture holds in both cases. In this section we will study the same question for the Kodaira-Thurston manifold with all linear symplectic forms. We expect the answer to be positive. Although we are not able to give a complete proof yet at this time, we shall provide some partial results below. We begin by describing the linear symplectic forms on the Kodaira-Thurston manifold M . Recall that it follows from Lemma 3.2 that H 2 (M, R) is of rank 4, generated by γ ∧ ds, γ ∧ dy, ds ∧ dt, and dy ∧ dt where γ = dx − sdy. We consider linear 2-forms ω abef := aγ ∧ ds + bγ ∧ dy + eds ∧ dt + f dy ∧ dt.
It remains to show that flux(φ αβc ) = v. Note that φ αβc is the time-1 map of the flow generated by the time-independent symplectic vector field Using (1) in Section 2, we have To answer Question 8.1, one has to check whether φ αβc constructed in Lemma 8.3 is always unbounded whenever its flux v is nonzero in H 1 (M, R)/Γ. This is in general a very hard question. In the remaining of this section, we will give a proof for some known cases. For the unknown cases, we will try to point out what difficulty is involved.
Case 1: α = 0 ∈ R/Z. In this case we will prove φ αβc is always unbounded. Note that φ αβc (U ) ∩ U = ∅ where U ⊂ M is defined by for sufficiently small ǫ. We will apply Lemma 7.3 as in the proof of Theorem 1.4. Recall that the only thing we need to do is to construct time-independent Hamiltonian H supported in U whose flow has no nonconstant contractible orbits. This follows from a tedious but straightforward calculation which asserts that Note that this is actually a special case of the construction in Lemma 8.3. And the fact that X is a well defined vector field on M follows from the equivalence relation (s, t, x, y) ∼ (s + 1, t, x + y, y). Since a and b can not be both zero, if we further require H to depend only on the t-coordinates, we know that the Hamiltonian flow generated by H will have no nonconstant contractible orbits. Therefore φ αβc is always unbounded in this case.
The problem here is that f τ does not descend to a Hamiltonian isotopy on M . Note that in proving φ βc itself is unbounded, it is crucial to have such a Hamiltonian isotopy on M , not just on the universal cover R 4 . Hence this case is still unsolved.
We do not know in general how to prove φ β is unbounded for this seemingly easy case. The difficulty in applying Theorem 6.2 is that the obvious torus L := {(s, t, x, y) ∈ M | s = 0, x = 0} displaced by φ β is not necessarily Lagrangian with respect to all symplectic forms ω abef . If we assume f = 0, then L is actually a Lagrangian torus, and φ β will be unbounded in view of Theorem 6.2.
Note also that Lemma 7.3 does not work here either since our situation here is different from Case 1 above. The main reason is that U := {(s, t, x, y) ∈ M | |x| < ǫ} is not a well defined set in M . Thus one can no longer apply Lemma 7.3 by constructing a time-independent Hamiltonian H supported in U whose flow has no nonconstant contractible orbits.