# Given a compact manifold M , we denote by Diff

## Full text

(1)

MMANUEL

### M

ILITON

LetS be a compact orientable surface andf be an element of the group Homeo0.S/ of homeomorphisms ofS isotopic to the identity. Denote byfza lift of f to the universal coverSzofS. In this article, the following result is proved: If there exists a fundamental domainD of the coveringSz!S such that

n!C1lim 1

ndnlog.dn/D0;

where dn is the diameter of fzn.D/, then the homeomorphism f is a distortion element of the group Homeo0.S/.

37C85

r0

r

r0

r0

r0

### . M / . It will provide some very partial answers to these questions. Here is the definition. Remember that a group G is

finitely generated

G

G

11

22

n

i

G

i

G

G

G

G

G

G

n

n0

G

n

n

distorted

### or a

distortion element

G

(2)

BS

1

p

n

n

pn

G

G

pn

BS

nilpotent

n

n0

0

nC1

n

n

n

n1

1

n

n

G

n2

n

n

fa;bg

n2

n

n

(3)

### (6) In mapping class groups (see Farb, Lubotzky and Minsky

[6]) and in the group

### of interval exchange transformations (see Novak

[22]), any distorted element is

area

1

area

1

### –diffeomorphisms of the surface S that preserve the measure

area. Then we have the following statement:

### Theorem 1 (Polterovich

[23], Franks and Handel[9])

1

area

n

Z

1

area

Sd

[5])

0

d

[2]

10

S1

1

### –topology (such an element will be said to be

recurrent) is distorted in the group Diff10

1

[20]):

10

1

1

### –diffeomorphisms of this manifold. Notice that, thanks to the Anosov–Katok method (see Herman

[13], and Fathi and Herman[7]), we can build

1

BS

10

(4)

S1

R

1

2

BS

BS

### . 1 ; 2 / on

R. This last action can be made

1

1

BS

Rd

Rd

10

0

0

(5)

isotopy

t

t0;1

0

1

R2

H2

1

H2

R2

lift

t

t0;1

t

t0;1

t

t0;1

t

t

0

0

D2

1

D2

0

2

0

S1

[11]): If

0

0

0

(6)

0

0

0

0

0

1

2

i

i

1

2

1

2

1

0

t

1

t

t 1

### Definition 2.3 We call a

fundamental domain

1

1

[9, Lemma 6.1]):

1

0

0

0

0

n!C1

n

T2

S1

[21]

R2

12

12

(7)

T2

n

12

12

n2

12

0

0

G

1

2

p

0

0

G.

G

n

n1

n

i1

i2

iln

n

G

n

n

i1

i2

iln

1ip;zx2 zS

i

1

i

i

n

n

n

n

n

n

n

n

n

n0

0

0

n!C1

n

n

0

0

(8)

n!C1

n

n

n

n

0

Theorem 3

### satisfy the hypothesis of

Theorem 2.6. However, we already knew

Theorem 3.

0

2

n

n

### is bounded (see

[14, Proposition 2.1]). His examples are skew-products over a

### Denjoy counterexample. By

Theorem 2.6, such homeomorphisms are distorted

0

T2

### (4) In

Section 8, for any sequence

n

n

n!C1

n

n

n

n

Theorem 2.6

n!C1

n

n

(9)

1

T2

n!C1

n

n

n

### is unbounded (see

[17, Theorem 3]). The idea

R2

R

### is a continuous map which sends 0 and 1 to 0. Of course, . r / has to converges sufficiently fast to 0 when r tends to 1 to ensure that the homeomorphism f is well-defined. If . r / converges sufficiently fast to 0 when r tends to 1 , one can check that such a homeomorphism satisfies the hypothesis of

Theorem 2.6. In the same article,

R2

[17,

Theorem 2.6

Rd

Rd

d

1

2

d

RN

1

d

d

d

d

U

[4]

### or Fisher

[8]), there exists

i

1in

0

U

1

2

n

U

U

U

1

### . S / . We now describe the two steps of the proof of

Theorem 2.6. The first step of the proof consists of checking that the quantity FragU

(10)

0

0

0

U

0

0

0

0

U

0

0

0

0

0

U

0

U

U

U

### a finite family of closed balls or half-balls whose interiors cover M . In

Section 4, we will prove the following theorem. It asserts that, for a homeomorphism

0

U

n

n!C1

U

n

U

n

0

(11)

U

0

0

[2]).

Theorem 2.6

Proposition 2.4

n

n1

n!C1

n

0

R

Z

R

Z

n

n

0

R

Z

R

Z

R=Z

2.9

### and

2.10. More precisely, we

5,6

7.

U

0

0

U

D

0

0

0

(12)

0

U

0

0

0

0

U

D

0

0

0

R

quasi-isometric

0

0

0

0

0

R

0

1

Figure 1):

0

0

### is a closed disc bounded by a polygon with geodesic edges. We require that any edge of this polygon that is not contained in @ S z connects two edges contained in @ S z .

D0 D0 D0

@Sz

@Sz

@Sz @Sz

Case of the torus Case of the torus with one hole Case of the genus 2 closed surface Figure 1: The fundamental domainD0

D

0

1

0

D, we denote

D

0

(13)

0

D

D

G

1

0

D

0

0

G

1

G

1

1

R;

G

1

1

1

G

1

D

0

0

D

1

G

G

D

0

0

### Given a compact subset A of S z , we call the

discrete diameter

D

D

0

D

0

D

∅;

0

∅g

1

D

éloignement

1

D1

D

1

D

∅g

1

∅, we have

D1

D

D1

Proposition 3.1:

U

0

R

U

D

0

U

U0

U

U0

(14)

0

0

0

0

U

0

S0

0

0

S0

R

U

D

0

2.9

2.10.

Proposition 3.4

0

0

1

2

p

1

0

∅g:

0Sp

iD1

i

0

p

[

iD1

i

0

0

0

i

1ip

Sp

iD1

i

x2X;y2X

1ip

i

(15)

Z

i

1ikC1

1

kC1

i

iC1

.i/

k

X

iD1

i

iC1

k

X

iD1

.i/

1ip

i

U

U0

### as in the statement of

Proposition 3.4, the maps FragU

U0

Rd

V

0

0

V

1

2

p

V

0

0

0

V

1

N0

0

0

i

j

1

2

N0

Lemma 3.8

[3, Lemma 5.2]

Lemma 3.9

### a consequence of the proof of Theorem 1.2.3 in

[4]. These two lemmas imply that, for

D2

0

2

2

U

U0

U0

U0

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U

N

U

U

U0

N

U

U

U0

1

S

0

0

D

0

0

0

1

0

G

[12]

D

1

G

S

D

∅g:

1

10

1

10

D

1

10

0

U

### as in the statement of

Proposition 3.4, there exist

0

0

D

0

0

U

U

(17)

U

D

D

U

1

2

k

i

U

1

2

k

Lemma 3.11

Z;

D

j 1

11

0

D

0

i

D

k1

11

0

D

0

U

D

0

Lemma 3.11

U

1

D

U

h

U

0

D

0

D

0

0

1

h

h

h

h

D

0

0

0

D

0

D

0

D

D

### Thus, to complete the proof of

Proposition 3.4, it suffices to proveProposition 3.1.

(18)

U

3.1

3.2

3.1

### and

3.2, in each of these cases, consists in putting back

0

0

U

U

0

U

0

0

U

0

U

i

0

U

.1/

.2/

.l/

0

n!C1

U

n

0

0

0

U

0

0

0

U

(19)

0

0

n!C1

U

n

0

U

U

U

Theorem 2.11.

4.1

4.2

G

G

U

G0

n

n

G0

n

G

n

n!C1

n

U

n

G0

n

n!C1

U

n

### In the case of

Proposition 4.2, there is only one new difficulty: the elements of G

U

0

0

G

U

1

G

4.1

4.2

Theorem 2.12

### (construction of the example). In order to prove

Theorem 2.6, we just used Theorem 2.11, which is weaker.

U

U

0

Sd

[5]). This

### is a consequence of the annulus theorem by Kirby (see

[16]) and Quinn (see[24]).

### Thus, the following theorem by Calegari and Freedman (see

[5]) is a consequence of Theorem 2.11:

[5])

0

d

(20)

Proposition 4.1

[2]):

n

n1

Rd

d

d

G

Rd

d

n

G

G

n

r

n

n

n

n

1

Y

nD1

n

n

n1

1

n

G

n

G

lC1

G

### Before proving

Lemma 4.5, let us see why this lemma implies Propositions4.1

4.2.

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4.1

4.2

n!C1

U

n

N

N

n!C1

U

.n/

U

1

2

p

i

Rd

i

i

d

d

i

i

n

n

U

.n/

n

n

0

1;n

2;n

kn;n

U

.n/

n

N

N

N

.n/

Pn

iD1

.i/

n!C1

n

n

n!C1

n

Pm

iD1

.i/

n

.mC1/

PmC1

iD1

.i/

N

NN

.j/

N

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