Analytic Cocycles with negative infinite Lyapunov exponents

Singular Analytic Linear Cocycles with negative infinite Lyapunov exponents


We show that linear analytic cocycles where all Lyapunov exponents are negative infinite are nilpotent. For such one-frequency cocycles we show that they can be analytically conjugated to an upper triangular cocycle or a Jordan normal form. As a consequence, an arbitrarily small analytic perturbation leads to distinct Lyapunov exponents.

Moreover, in the one-frequency case where the -th Lyapunov exponent is finite and the st negative infinite, we obtain a simple criterion for domination in which case there is a splitting into a nilpotent part and an invertible part.

Key words and phrases:
complex analytic cocycles, quasi-periodic cocycles, nilpotent cocycles, normal forms
2010 Mathematics Subject Classification:
Primary 37C55, Secondary 34C20, 37A20, 37F99, 37G05

1. Introduction

Let be a compact space, a probability measure on the Borel -algebra of and a measure preserving transformation, for all Borel sets . Iterations of the map define a dynamical system on , the so called base dynamics. By we denote the set of matrices with complex entries. For a measurable map one obtains the linear cocycle denoting the map

Some examples of linear cocycles are the derivative cocycle of a map of torus, the random products of matrices, Schrödinger cocycles, etc.

In general we want to consider analytic cocycles:

Definition 1.

We call an analytic cocycle over a compact, connected measure space if the following three assumptions hold:

  • is a compact, connected, real analytic manifold

  • For any analytic chart (bi-analytic map) the push-forward measure on has a continuous density with respect to the Lebesgue measure on .

  • and are (real) analytic, i.e. and .

Note, if would not be connected then using compactness one finds that a certain iterative power of would leave the connected components invariant and one could consider the corresponding powers of inducing cocycles on these components.

The prime example we are thinking about are cocycles over the rotation on a torus, i.e. , is the canonical Haar measure (or Lebesgue measure), with and . Then we may denote the cocycle also by and call it an -frequency cocycle, because the base dynamics is determined by the -frequency vector .

If is a rational vector, then is a periodic sequence, for irrational one calls it a quasi-periodic sequence and is a quasi-periodic cocycle. Such one-frequency quasi-periodic cocycles have been intensively studied in the past because they are very important for the theory of discrete quasi-periodic one-dimensional Schrödinger operators, see [A2] and references therein.

For analytic cocycles one often uses some inductive limit topology considering holomorphic extensions1 of and , in the one-frequency case see e.g. [AJS].

The main object of interest of linear cocycles is the asymptotic behavior of the products of along the orbits of , especially the Lyapunov exponents. Iterating a linear cocycle leads to or , where


Let denote the -th singular value of a matrix , i.e. and the squares, are the eigenvalues of . Then, the -th Lyapunov exponent is defined by


With we denote the linear operator on the anti-symmetric tensor product defined by . Then it is well known that giving


If we have an -frequency cocycle with , then we may also write .

Let where is the positive part of the logarithm, then Kingman’s Subadditive Ergodic Theorem shows that the Lyapunov exponents exist2 with . If is continuous and always invertible, then all Lyapunov exponents are finite, i.e. bigger than . But if can have a kernel, then one might end up with some Lyapunov exponents. We want to classify these situations for analytic cocycles.

Understanding the structure of cocycles is an important branch in the theory of dynamical systems. An important question is how frequent cocycles with simple Lyapunov spectrum occur (cf. [GM],[BV],[AV],[GR],[V],[FK],etc.). The Lyapunov spectrum is called simple if all Lyapunov exponents are different. Typically one would expect this to be true on a dense set of cocycles. This question, however, gets trickier the higher the considered regularity class. On the other hand, in low regularity (), failure of non-uniform hyperbolicity is a fairly robust phenomenon in the topological sense [Boc].

For -cocycles Avila showed that the set of cocycles with distinct (or positive) Lyapunov exponents is dense in all usual regularity classes [AV]. Distinctness of the largest and smallest Lyapunov exponent on a dense set of general symplectic or pseudo-unitary cocycles of matrices (in all regularity classes) was shown in [Xu]. It relies on Kotani theory and local averaging formulas combining ideas from [AV, AK, KS, S], but a certain real Lie-group structure is always very important. For complex analytic or cocycles the question is open. An approach to distinct Lyapunov exponents has been worked out by Duarte and Klein [DK, DK2] which is based on generalizations of the Avalanche principle and large deviation estimates. These tools had been used a lot for cocycles ([BJ, Bou, GS]).

Once there is some gap in the Lyapunov spectrum another important concept is that of domination (a generalisation of the notion of uniform hyperbolicity, a precise definition is given below). In [AJS] it was shown that within the set of complex, analytic one-frequency cocycles with a gap in the Lyapunov spectrum, the set of dominated cocycles is dense. However, for complex analytic cocycles it is not clear whether the set of cocycles where all Lyapunov exponents are equal has a non-empty interior.

We propose to attack this and further question for complex cocycles by looking for conjugated ’normal forms’ similar as Jordan normal forms or Hilbert-Schmidt decompositions for matrices. One should try to classify cocycles where all Lyapunov exponents are equal. In this work we consider cocycles where all Lyapunov exponents are negative infinite. Within the measurable, ergodic category, the Oseledets filtration gives some block upper-triangular normal form, cf. [O, R] which can be refined by looking at so called maximal invariant flags [ACO]. For invertible cocycles ( and invertible) one has an Oseledets splitting and a block diagonal normal form. Each block corresponds to a distinct Lyapunov exponent.

Before getting to the normal forms mathematically, we need a proper equivalence relation. Two cocycles and with the same base dynamics are dynamically conjugated, if

where is a measurable map into the general linear group. Then, and the cocycles are dynamically equivalent. However, if is only measurable and only almost surely defined, then one looses regularity features like e.g. analyticity of the cocycle and other certain fine distinctions such as non-uniform and uniform hyperbolicity or the notion of domination. Therefore, in terms of normal forms we are only interested at dynamical conjugation within the regularity class. Especially in this case we consider analytic cocycles and we also want (and hence ) to depend analytically on .

One way to create cocycles where all Lyapunov exponents are is by constructing cocycles such that after finitely many steps one arrives at the zero cocycle. We call such cocycles nilpotent:

Definition 2.

A linear cocycle is called nilpotent if for finite we have -almost surely. The minimal such natural number is called the nilpotency degree .

Clearly, for nilpotent cocycles all Lyapunov exponents are negative infinite. Our main result is that for analytic cocycles this is an equivalence. Let us note that in the regularity class it is wrong that implies nilpotency; even for cocycles! To show this let for , and continue periodically. Then, , and is clearly not nilpotent but

Nilpotency can be achieved by taking upper triangular matrices with zeroes along (and below) the diagonal. Our second main result is that in the analytic one-frequency case these are all possibilities up to analytic unitary dynamical conjugation. Particularly, an arbitrarily small analytic perturbation leads to simplicity of Lyapunov exponents.

If we have only some negative infinite Lyapunov exponents, but we can split of some nilpotent analytic invariant subspace corresponding to the negative infinite Lyapunov exponents. In this case we also get some simple criterion for a dominated splitting.

In the next section we state the precise theorems and give several remarks. In Section 3 we treat first the case when the rank of is at most one and show that implies nilpotency. Then, based on this result we can treat the case for general rank of in Section 4. Section 5 finally considers one-frequency cocycles where only some Lyapunov exponents are negative infinite. In the Appendix we give some important facts which are used multiple times.

Acknowledgement: This research has been funded by the People Programme (Marie Curie Actions) of the European Union’s Seventh Framework Programme FP7/2007-2013 under REA grant agreement number 291734. D.X. would like to thank his thesis advisor Artur Avila for the supervision and support, and this research was partially conducted during the period when D.X. visited the Institute of Science and Technology, Austria.

2. Results

Having only negative infinite Lyapunov exponents implies nilpotency in the analytic category:

Theorem 1.

Let be an analytic cocycle over a compact, connected measure space in the sense of Definition 1 and assume that . Then, is nilpotent, more precisely, for all , where is the maximal rank.

Very concrete normal forms can be found in the one-frequency case.

Theorem 2 (One frequency case).

Let and such that . Then the following hold:

  1. There exists a one-periodic analytic function with values in the unitary group , such that is upper triangular with zeroes on and below the diagonal. More precisely, if the nilpotency degree is then one can choose such that is divided into blocks (of different size) with upper-triangular block structure,

  2. Assume additionally that for all , is constant in . Then, there exists a one-periodic analytic function such that


    where and


Adding a diagonal perturbation , to and conjugating it back we obtain the following for as a corollary of the above theorem.

Theorem 3.

Let and such that , in which case for all . Then, there exists such that for any all Lyapunov exponents of are distinct. Hence, there are arbitrarily small analytic perturbations with simple Lyapunov spectrum.


In analogy to [ACO] we call and analytic Jordan normal forms of . As the form is much more restrictive, we may call it a completely reduced Jordan normal form. Let us make some remarks about the existence of such normal forms in the analytic category.

  1. The condition needed for Theorem 2 (ii) is satisfied on a dense set of cocycles with . For small enough one can define by analyticity and a local Taylor expansion. For any up to the nilpotency degree, there is only a finite set of within where the is not maximal (equal to which is independent of ). This follows from analyticity. Hence, for small enough , the cocycle satisfies the condition.

  2. For a completely reduced Jordan form as in Theorem 2 (ii) one may want to relax (2.3) and allow and to depend on , where still has only non-zero entries on the superdiagonal3 which may become for some . Then the condition that the ranks of are constant is not necessary for such a conjugation. However, alone is also not sufficient in this case. Examples where such a form can not be reached by (everywhere defined) analytic conjugations are

    In both cases the nilpotency degree is 3, . In the first scenario, is not constant, in the second one, is not constant.
    If one allows the conjugation to be not invertible in finitely many points, then one can always get such a conjugation so that (for almost all ). But as is then not defined for some (and of course not analytic) it is not an analytic conjugation of the cocycle.

  3. In the general analytic case with a higher-dimensional base one can not even necessarily get ’normal forms’ like above by everywhere analytic conjugations. One crucial ingredient missing in the general case is an analogue of Lemma A.1. Let us give an example of an analytic nilpotent 2-frequency cocycle that can not be conjugated to such a normal form. Let be the translation vector for the base dynamics , , and let

    Then we have with rank almost surely, and the direction of the kernel of (on projective space ) has no limit at or which contradicts with analyticity of to get

    One may choose for conjugating to such a normal form, however, the inverse of does not exist at or .

Next, we have a look at analytic one-frequency cocycles where some but not all Lyapunov exponents are . In this case there is an obvious gap after the last finite Lyapunov exponent and one can ask the question about domination. In general this was classified in [AJS], however, in this special case the classification is much simpler. For completeness let us repeat the definition of domination. Let denote the Grassmannian manifold of complex -dimensional subspaces of and the set of continuous functions from to .

Definition 3.

A continuous cocycle over is -dominated () if there is a continuous splitting of the space in a relatively stable and a relatively unstable invariant space, i.e. there exist , such that for all ,

and for some and all and all one has

Particularly, the kernel is always inside the relatively stable space, .

Theorem 4.
Let and such that and . Then the following hold. There exists such that is an upper triangular matrix with zeros on and below the diagonal, hence is nilpotent and is an almost surely invertible matrix. In particular, for all . Of course, the block can also be conjugated to a Jordan form by an analytic dynamical conjugation as described above. The cocycle is -dominated if and only if for all . It is also equivalent to as defined in (i) being invertible for all . In this case there is some analytic matrix such that with as in (i) we have This conjugation corresponds to the dominated splitting.

Without domination it is not always true that one can obtain this block-diagonal form with an analytic (or everywhere defined) conjugation. A counter-example is the following cocycle, with any frequency . A diagonal, analytic conjugated cocycle would necessarily be of the form . As one has or , i.e. for either or . But, for any . So there is a contradiction if is invertible for all .

3. Rank one case

In this section we will basically prove Theorem 1 and Theorem 2 in the rank one case by the following Proposition.

Proposition 3.1.

We have the following.

  1. Assume that is an analytic cocycle over a compact and connected space as defined in Definition 1. Assume further that has maximal rank and . Then, for all .

  2. Let be an analytic one-frequency cocycle, i.e. , , and let . Then, there is a one-periodic analytic function and a one-periodic analytic unitary function such that


We will first show (ii). The case for all is trivial, so assume for some . We find some column-vector of which is not always zero. By Lemma A.1 we find a one-periodic, real analytic function with such that is a complex multiple of . Doing the same with we obtain some one-periodic analytic function with . As (at most rank 1), we find some such that


where is a column-vector and a row vector and their product a matrix. As depends analytically on , has to be analytic. Thus,

which leads to


By Lemma A.3 this implies for all as is not the zero function. This gives . Moreover by Lemma A.1 (ii) one can extend to an orthonormal basis4 defining a unitary matrix such that  .

In the general case (i) we still find functions , with satisfying (3.1). However, we can only guarantee analyticity at points where . In general, there might be some union of sub-manifolds of where , (i.e. ) and where may not be analytic. But the functions


are always analytic. We assume again that is not identically zero, in which case is not identically zero. Then, similar to (3.2) we find

Using Lemma A.3 we find that for all , but this is equivalent to for all . ∎

4. General rank case

We start with the following simple observation:

Lemma 4.1.

Assume is an analytic cocycle over a compact, connected measure space . There is such that for all except a union of sub-manifolds of zero measure (w.r.t. ), and for all .


Let , such that for all . Then, for some and the equation is equivalent to . By analyticity and connectedness of , in any chart for , the equation defines a union of sub-manifolds of zero Lebesgue measure within the chart (see also Corollary A.4). Using a finite atlas for and Assumption (A2) in Definition 1 gives the claim. ∎

Note that in the one-dimensional case , this zero-measure set simply consists of finitely many points.

Another special point of analytic cocycles is the fact that the rank reduction has to take place in each step:

Lemma 4.2.

Let denote an analytic cocycle over a compact, connected measure space such that for some and all we have and . Then for all .


We claim if , then for all . The result then follows by backward induction. We let

Take some , then

As this means . Since and we find and hence



In summary, we prove that for all , either or , i.e. . This implies and by Corollary A.4, for all and hence . ∎

Now we can prove Theorem 1 and Theorem 2.

Proof of Theorem 1.

Let denote an analytic cocycle over a compact, connected measure space and . By Lemma 4.1 we know for any , there is such that for all except a -zero-measure set and for all . Then we have for all . Let , to prove the Lemma we need to establish . Suppose . Therefore, has maximal rank and

By Proposition 3.1 we have for all . As a result, which contradicts with our assumption of . Iterating Lemma 4.2 gives for all with . ∎

Proof of Theorem 2.

Now let be an analytic one-frequency cocycle and . By the proof above we know that is nilpotent. Let be the nilpotency degree. As a corollary of the lemma above we get that is strictly increasing, for and almost all , hence and the kernels have dimensions . Note that by Lemma A.2 the subspaces induce an analytic function from to . Using Lemma A.1 (ii) this means that we find analytic dependent matrices , such that:
(i) for all
(ii) is orthogonal to the kernel of for almost all5
(iii) the range of and the kernel of span the kernel of , for almost all .
Here, and so actually spans the kernel of .

As we get and hence defines an analytic unitary matrix. As for and almost all we obtain that is of the claimed form (2.1), at first, for almost all , but by analyticity for all . This shows part (i).

Now let us get to the completely reduced Jordan form, part (ii).

We assume that is constant for all . Recall that the nilpotency degree of was denoted by . By Lemma A.2 the subspaces

of fixed dimensions , are analytically dependent on , where we set . Clearly, and . Choosing some analytically dependent basis of it is clear that defined as mapping from to is analytic and by assumption of constant rank . Thus, by Lemma A.2, the subspaces and their orthogonal complements within , depend analytically on . Then, by constancy of the rank, the restriction of the map (or ) from to is analytic and invertible for all . Taking the inverse, we get some analytic function such that


We claim that for any , there exists analytic maps , such that


The values and depend on . Notice that to prove the existence of a Jordan form we only need to prove the claim for the case .

We prove the claim by induction: when , . By Lemma A.1 and Appendix of [AJS], there are analytic maps such that for any , is a basis of which proves the claim for the case .

Suppose the claim holds for , i.e. there are analytic maps satisfying (4.2), (4.3), (4.4). Then and using the analytic dependent maps as in (4.1) we can define the analytic vectors

By construction, where the inverse denotes the pre-image. By assumption, the latter pre-image is spanned by and . Hence, is spanned by , the vectors and some vectors in .

Now, by constancy of we get that the dimensions of are constant. Hence, the orthogonal complement of within has constant dimension and is an analytically dependent subspace. Using Lemma A.1 (ii) we find analytic functions such that for all ,

Moreover, by the considerations above, is spanned by , and for all . We claim that