Ricardo Mañé. The Lyapunov exponents of generic area preserving diffeomorphisms

8/10/2019 Ricardo Ma. The Lyapunov exponents of generic area preserving diffeomorphisms.

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THE LYAPUNOV EXPONENTS OF GENERIC AREA

PRESERVING DIFFEOMORPHISMS

RICARDO MANE

Editors note. This article contains an outline of a proof of a result announced by

the author around 1980. The theorem and its generalization to higher dimensions are

reported in the authors address at the International Congress in 1983 [M]. The proofs

have not been published. For the 2-dimensional case, the closest to a written draft that

is available are some notes handed out by the author at a seminar in Paris. Except

for some very minor editing, this article is a direct transcription of those notes. We

stress that what is presented here is only the sketch of a proof, and that a substantial

amount of work is needed to fill in the details.

Let M be a compact connected boundaryless smooth 2-manifold and let be aprobability on its Borel -algebra generated by a smooth volume form. LetD(M)be the space of-preservingC1 diffeomorphisms endowed with the C1 topology.

0.1. Theorem. There exixts a residual subsetA D(M) such that everyf A iseither Anosov or

limn

1

n log (Dxf

n)= 0

for a.e. x.

In these notes we shall sketch the proof of this result. This sketch will consist of thestatements of all the lemmas conducting to the proof of the theorem. With few excep-tions these lemmas will be stated without proof. The sequence of lemmas is dividedin several steps. In the first step we introduce an entropy function h : D(M) Rthat is upper semicontinuous. Then its setA0 of points of continuity is residual. Inthe second step we shall prove the existence of a residual set A1 D(M) of diffeo-morphisms such that all its hyperbolic sets have measure 0 or 1. The third, fourthand fifth steps are devoted to show that diffeomorphisms in A0 do not exhibit certainpathological behaviours, because these behaviours would produce discontinuities ofh.The perturbations required to show this discontinuity are constructed with the lem-

mas given in section III. Using this absence of pathological behaviour we shall showthat a diffeomorphism f A0 A1 satisfies one of the alternatives of the theorem.

1. First step: the entropy function.

IffD(M) define (f) as the set of points x Msuch that the limits:

limn

1

|n|log (Dxf

n)

1


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2 RICARDO MANE

exist and coincide. By Oseledecs theorem:

((f))) = 1.

Define+ : (f) R by

+(x) = limn+

1

n log (Dxf

n)

and

H(f) ={ x (f)| +(x)= 0 } .

By Oseledecs theorem:

+(x) 0

for all x (f) and if+(x)= 0 there exists a splitting TxM=Es(x) Eu(x) such

that:

limn

1n

log (Dxfn)|Es(x)= +(x)

limn

1

nlog (Dxf

n)|Eu(x)= +(x)

Define the entropy functionh : D(M) R by

h(f) =

(f)

+ d.

1.1. Lemma. h is upper semicontinuous.

Proof. Define

cn=

(f)

log (Dxfn)|Eu(x) d

then:

cn+m cn+ cm

for alln >0, m >0. Hence

h(f) = limn+

1

n

(f)

log (Dxfn)|Eu(x) d= lim

n+

1

ncn

= infn

cn= infn

1

n(f)

log (Dxfn)|Eu(x) d.

1.2. Definition. A0= set of points of continuity ofh.

1.3. Corollary. A0 is residual.


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LYAPUNOV EXPONENTS 3

2. Second step: avoiding fat hyperbolic sets.

2.1. Proposition. There exists a residual subsetA1 D(M) such that if f A1all its hyperbolic sets have measure0 or1.

Proof. Let Bbe a basis of neighborhoods ofMand let Bbe the set of all finite subsetsofBsuch that BFB=M. IffD(M) and F Bdefine:

S(F, f) =nfn

BFB

.

If > 0 and F Bdefine D(, F) as the set of diffeomorphisms f D(M) suchthat one of the following properties is satisfies:

(1) There exists a neighborhood Uoffsuch thatS(F, g) is not hyperbolic for allg U.

(2) (S(F, f))< .

ObviouslyD(, F) is open. Moreover, it is dense because iff D(M) is C2 (that

is a dense property in D(M) as proved by Zhender) then every hyperbolic set offhas measure 0 or 1.

3. The fundamental perturbation lemma.

As we expalined in the introduction in the next section we shall have to produceperturbations of diffeomorphisms exhibiting certain behaviours. The key tool in theconstruction of these perturbations will be the next lemma.

3.1. Definition. Iff :M is a diffeomorphism andF Z+ we say that a sequenceof linear maps Lj : Tfj(x)M Tfj+1(x)M, x M, j = 0, 1, . . . , m is F-adapted if

Lj =Dfj(x)f for allj except for at mostF values ofj or if(LnLn1 L0)B1(0) = (Dxf

n)B1(0)

for all0 n m.

3.2. Fundamental Lemma. Iff D(M), F Z+ andU is a neighborhood off

there exists >0 such that if1 M is a Borel set, 0< 1 < andN : 1 Z+

is a measurable function with the property that for allx 1 and m > N(x) thereexists anF-adapted sequenceLj :Tfj(x)MTfj+1(x)M, 0 j m, satisfyingLj (Dfj(x)f) for all0 j m and m

j=0

Lj exp m1

then there existsg U such that

h(g) h(f) + 2 1

n0f

n(1)

+ d .

This lemma is obtained from the following sequence of properties:


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4 RICARDO MANE

3.3. Lemma. If f D(M), 0 < k < 1 and U is a neighborhood of f, there exist >0 andr0 > 0 such that ifx M, 0< r < r0, andL: TxMTf(x)M is a linearmap satisfying:

L (Dxf)<

then there existsg U such that:

g(y) =f(y)

wheny /Br(x) and

Dyg= L

ify Bkr(x).

3.4. Lemma. Givenf D(M), F Z+, 0 < k 0 and a neighborhoodU

off there exist >0 andr0 >0 such that for all0< r < r0 and every nonperiodicpoint x M, if Lj : Tfj(x)M Tfj+1(x)M, 0 j < n, is an F-adapted sequencesatisfying Lj (Dfj(x)f) for all 0 j < n then for all 0 < r < r0 there exists g U and a compact setKBr(x) with(K)> k (Br(x)) such that

f(y) =g(y)

ify / nm=0fm(Br(x)) and

Dygm =Lm1Lm2 L0

ify K and0 m < n.

3.5. Definition. Givenf D(M) we say that a family of Borel sets{U0, . . . , U m}is a tower if they are disjoint andUj =f

j(U) for all0 j m.

3.6. Lemma. Given f D(M), 0 < k < 1, F Z+ and a neighborhood U of f

there exists > 0 such that if {U(n)0 , . . . , U

(n)jn

}, n = 1, . . . , is a family of disjoint

towers and for allx n=1U(n)0 we have anF-adapted sequenceLj(x) : Tfj(x)M

Tfj+1(x)M, 0 j < jn, such thatLj(x) (Dfj(x)f)

for all0 j < jn, then there exists a compact setK

n=1U

(n)

0 andg U satisfying:(1) g(x) =f(x) forx S,

(2) (K) k

n=1U(n)0

,

(3) Ifx K there exist1 n andy U(n)0 such that:

Dxgjn =Ljn1(y) L0(y).

[Editors remark: Sapparently not defined.]


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3.7. Lemma. If f D(M), F Z+ and U is a neighborhood of f there exists

> 0 such that if 0 < 1 < and {U(n)0 , . . . , U

(n)jn

}, n = 1, . . . , are disjoint towers

whose union isf-invariant and for alln 1 andx U(n)0 there exists andF-adapted

sequenceLj(x) :Tfj(x)MTfj+1(x)M, 0 j < jn, such that(1)

Lj (Dfj(x)f) for all0 j < jn,(2) Ljn1 L0 exp n1.

[Editors remark: The author probably meant exp(jn1).]

then there existsg U satisfying

h(g) h(f) + 2 1

n,jU

(n)j

+ d.

The fundamental lemma follows from the last one and the following well knownresult about decompositions of invariant sets in unions of disjoint towers:

3.8. Lemma. If f D(M) and M is a Borel set such that the measure ofthe set of periodic points off contained in is0, then for everyN > 0, can bedecomposed in a countable union of disjoint towers with its basis contained in1 andheight larger thanN.

[Editors remark: The statemnet of this lemma in probably not intended as is.]

4. Third step: bounding the angles of the Oseledec splitting.

IffD(M) define 1 : H(f) R by:

i(x) = infn0

Es(fn(x)), Eu(fn(x))

where(, ) is the angle between the subspaces between the brackets. Let 1(f) ={ x H(f)| 1(f) = 0 }. Then f(1(f)) = 1(f).

4.1. Proposition. If f D(M) and (1(f)) = 0 then for every 0 > 0 and allneighborhoodU off there existsg U with

h(g) h(f) + 0

1(f)

+ d.

4.2. Corollary. f A0 = (1(f)) = 0.

Sketch of the proof. Given f D(M) with (1(f))> 0 and a neighborhood U off take >0 given by the fundamental lemma. Fix any c >0 and define:

c= {x 1(f)| +(x)> c}.

Then:

limn

1

n log

(Dxfn) | Es(x)

(Dxfn) | Eu(x) 2c


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6 RICARDO MANE

for allx c. [Editors remark: The author probably meant 2c.] Then there existsa >0 such that ifx c then

Es(fm(x)), Eu(fm(x))

a for some m >0. This

means that if we define:

a= {x c |

Es

(x), Eu

(x)

> a}

we have:

c = n>0fn(a).

Take A > 0 and 0 < 1 < min(, a) and a measurable function N : a Z+ such

that ifx a andn > N(x) then in every interval [j, j+ n1] [0, n] there exists msuch that:

Es(fm(x), Eu(fm(x))

1

(Dfm(x)fnm) | Es(fm(x))

exp(n m)(+(x) + A1)

(Dxfm

) exp m(+

(x) + A1)(Dfm(x)fnm) exp(n m)(+(x) + A1)(Dxf

m) | Es(x) exp m(+(x) + A1).

Choosem [(nn1)/2, (n+n1)/2] with all these properties and define a sequenceof linear mappings Lj : Tfj(x)M Tfj+1(x)M putting Lj = (Dfj(x)f) when 0 j m 2 or m j n 1 and

Lm1 = R(Dfm1(x)f)

whereR is the rotation that sends Eu(fm(x)) onEs(fm(x)). Then

n1j=0

Lj|Es(x)

exp 2(+(x) + A1)1 n n1j=0

Lj |Eu(x)

exp 2(+(x) + A1)1 n

Since (Es(x), Eu(x))a we can choose A = A(a) so small that these inequalitiesimply:

n1

j=0

Lj exp3C1nwhere:

C= supx(f)

+(x).

If1 > 0 is small enough, the rotation R is so near to the identity that:Lj (Dfj(x)f)<


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for all 0 j n 1. Then the hypothesis of the fundamental lemma are satisfiedand there exists g U with

h(g) h(f) + 6 C1 c

+ d

=h(f) + 6 C1

1(f)

+ d +

1(f)c

+ d

h(f) + 6 C1+ 6

1(f) c

1(f)

+ d.

Since1 is arbitrarily small the proposition is proved.

5. Fourth step: the domination inequalities.

Define 2 : H(f) R (i.e. R {+}) taking as 2(x) the supremum over alln >0, m >0 of(Dfn(x)fm) | Es(fn(x)) (Dfn+m(x)fm) | Eu(fn+m(x)) .Define 2(f) as the set of points x H(f) where 2(x) = + then f(2(f)) =2(f).

5.1. Proposition. IffD(M) and(2(f) 1(f)c)= 0 then for every neighbor-

hoodU off and every >0 there existsg U such that:

h(g) h(f) + 2(f)1(f)c

+ d.

5.2. Corollary. f A0 = (2(f)) = 0.

This proposition also follows from the fundamental lemma. To apply it take K >0,n0 > 0, k >0 and define as the set of points x 2(f) 1(f)

c such that

(Dxfn0) | Es(x)

(Dfn0 (x)fn0) | Eu(fn0(x))> K,1(x)> a.

[Editors question: k= a?]

Let

(K, n0, k) =n>0fn().

IfK and n0 are large and k is small, the measure of 2(f) 1(f)c (K, n0, k) is

small. Now we take 1 > 0 and a measurable function N : (K, n0, k) Z+ such

that ifx (K, n0, k) and n > N(x) in every interval [j, j+ n1] contains some msuch that fm(x) . We define the sequence of linear maps Lj as follows: choosem [(n n1)/2, (n+ n1)/2] with f

m(x) . This property implies that thereexists a subspace STfm(x)Mwith the properties described in the picture:


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8 RICARDO MANE

Define Lm1as the composition of (Dfm1(x)f) with a linear map near to Iand send-ing Es(fm(x)) on Sand leaving Eu(fm(x)) invariant. Define Lm+n01as (Dfm+n01(x)f)composed with a rotationRmapping (Dfm(x)f

n0)(S) onEu(fm+n0(x)). Finally define

Ln1 as the composition ofDfn1

(x)f with a linear map L near to the identity thatmaps (Dfm+n0(x)fnm)REu(fm+n0(x)) on Es(fn(x)) and leaves Eu(fn(x)) invariant.

The other linear maps Lj are defined as (Dfj(x)f). Thenn1

j=0Lj maps Es(x) on

Eu(fn(x)) and Eu(x) on Es(fn(x)) and the norm can be bounded by exp3Cn1 asin the previous section. In fact this estimate as well as the possibility of taking Lnear to the identity require some more precautions in the choice ofm that we didntexplain, hoping that this technical oversimplification may help to put in evidence theunderlying idea of the proof.

6. Fifth step: more domination inequalities.

Define3 : H(f) R and 3: H(f) R by

3(x) = sup{n >0 | (Dxfm)|Es(x) (Dfm(x)f

m)|Eu(fm(x))>1/2

for all 0 m n}

3(x) = supn>0

3(fn(x)).

Let 3(f) ={x | 3(x) = +}.

6.1. Proposition. If f D(M) and(3(f) 2(f)c 1(f)

c) > 0 [then] for all >0 and every neighborhoodU off there existsg U satisfying:

h(g) h(f) + 3(f)2(f)

c1(f)

c +

d.

6.2. Corollary. f A0 = (3(f)) = 0.

6.3. Lemma. IffD(M) then for allc >0 the closure of the set

c(f) ={x| 3(x)< c}

is hyperbolic ifc(f)=.


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Proof. The definition of c(f) implies the existence of constants K > 0, 0 < < 1such that:

(Dxfn)|Es(x)

(Dfn(x)fn)|Eu(fn(x))

K n

for alln 0, x c(f). This condition implies

infxc(f)

(Es(x), Eu(x))> 0.

Using the fact that f is area preserving and dim M= 2 it follows that there existK >0, 0< < 1 such that:

(Dxfn)|Es(x) K n(Dfn(x)fn)|Eu(x) K n

for allx c(f),n 0. From the uniformity of these estimates, the hyperbolicity of

c(f) follows easily.

Proof of the Theorem. Suppose thatf A0 A1 and(H(f))= 0. We shall provethatf is Anosov. By the corollary in section 1, the hypothesisf A0 implies:

(3(f)) = 0.

Hence(H(f) 3(f))> 0. But

H(f) 3(f) =c>0c(f).

Therefore(c(f)) > 0 for some c > 0. By the lemma the set c(f) is hyperbolicand since f A1 its measure must be 1. Then f is Anosov.

References

[M] R. Mane, Oseledecs theorem from the generic viewpoint, Proceedings of the International Con-

gress of Mathematicians, August 1983, Warzawa, Volume 2, p. 12591276 (Theorem C and itsCorollary in particular).

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Ricardo Mañé. The Lyapunov exponents of generic area preserving diffeomorphisms