On Irreducibility of Oseledets Subspaces
Abstract.
For a cocycle of invertible real by matrices, the Multiplicative Ergodic Theorem gives an Oseledets subspace decomposition of ; that is, above each point in the base space, is written as a direct sum of equivariant subspaces, one for each Lyapunov exponent of the cocycle. It is natural to ask if these summands may be further decomposed into equivariant subspaces; that is, if the Oseledets subspaces are reducible. We prove a theorem yielding sufficient conditions for irreducibility of the trivial equivariant subspaces and for valued cocycles and give explicit examples where the conditions are satisfied.
Keywords: Oseledets subspaces, multiplicative ergodic theorem, reducible matrix cocycles.
1. Introduction
The Multiplicative Ergodic Theorem (MET) has a rich history of generalizations and variations; there are versions of the theorem in many different situations (see Froyland, Lloyd, Quas [5] for a brief survey). The original theorem, by Oseledets [10], obtained a splitting of into equivariant subspaces, each subspace corresponding to a different Lyapunov exponent of the differentiable matrix cocycle; these splittings (and generalizations thereof) are now called Oseledets splittings.
These splittings play a very important role in the study of hyperbolic and nonuniformly hyperbolic dynamical systems, giving the tangent spaces for invariant submanifolds. Coming from differential equations, the exponential dichotomy, or SackerSell spectrum, gives a splitting of into equivariant subspaces with uniform gaps between exponential growth rates [12]. A natural question is that of reducibility of an equivariant family of subspaces: when can one find a lowerdimensional equivariant subspace inside a given one. One can check that the subspaces appearing in the SackerSell decomposition are a sum of subspaces appearing in the Oseledets decomposition. It can happen that the (nonuniform) Oseledets decomposition strictly refines the uniform SackerSell decomposition [9], even in cases where the SackerSell decomposition is trivial [8]. Bochi [3] has shown that generically for a matrix cocycle over a minimal base, the two decompositions coincide. Since these are results about reducibility of the SackerSell decomposition, it is natural to ask further about reducibility of the Oseledets decomposition.
One way to look at Oseledets splittings, for the case of real invertible matrix cocycles, is to use the notion of block diagonalization of the cocycle. From the MET, we get a splitting of which is equivariant; this equivariance property may be used to show that the cocycle is cohomologous to one which is block diagonal. The column vectors of the new cocycle exactly correspond to basis vectors of the subspaces in the splitting, and the blocks correspond to the subspaces. We then see that refinement of the Oseledets splitting, in the sense of further decomposing the subspaces, is equivalent to showing that the cocycle is cohomologous to a cocycle with a more refined block structure.
One of the key steps in the original proof of the Oseledets’ theorem ([10], in Russian, or [2], by Barreira and Pesin) is to construct a related triangular cocycle over an extended base space, thereby reducing the computation to the simpler case of triangular cocycles. Other parts of the theorem are obtained, however, using the original (nontriangularized) cocycle. A triangular cocycle is in some ways simpler than a nontriangularized cocycle, in that there is a flag of nested equivariant subspaces for the cocycle. In the twodimensional case, a cocycle is irreducible if and only if it is not triangularizable.
More recently, Arnold, Cong, and Oseledets showed in [1] that any real matrix cocycle over an invertible ergodic map, not just those satisfying the logintegrability condition of the MET, may be put into an equivariant form with blocks on the diagonals which are blockconformal, nothing below those large blocks, and arbitrary elements above those blocks; they call this is a Jordan normal form. With another condition, the elements above the diagonal may be removed, so that the form is blockdiagonal. In the setting of by real invertible matrix cocycles, Thieullen in [14] gave a classification of possible cocycles analogous to the classification of Möbius transformations.
In this paper, we consider valued cocycles, so that the only Lyapunov exponent is 0 and the decomposition arising from the Oseledets theorem is trivial. We exploit the nearcommutativity of to relate reducibility of the trivial equivariant subspaces and to the ergodic properties of a pair of skew product extensions of the base dynamical system, and give sufficient conditions for irreducibility of the subspaces, both over the reals and over the complex numbers.
The rationale for considering reducibility over the complex numbers is that even for a single matrix (the case where the underlying dynamics is a single fixed point), not every matrix can be triangularized over the reals, while it can always be triangularized over the complex numbers.
Using our criteria, we build two simple examples of valued cocycles: one with the property that the trivial complex equivariant subspace is reducible, but the real subspace is irreducible, and the other with the property that both and are irreducible.
We thank Kening Lu for bringing this problem to our attention.
2. Preliminaries and Formulation
Throughout this section, we use to refer to either or . Any measurability requirements for matrixvalued functions are with respect to the Borel algebra generated by the usual norm topology. Measurability of subspacevalued functions is with respect to the appropriate Grassmannian.
Definition 2.1 (Invertible Matrix Cocycle).
Let be a dynamical system, where is an invertible measurepreserving transformation, and be measurable. Define the cocycle by:
for all . One easily checks that for all . is the generator for the cocycle, and we often use the same letter for the cocycle and its generator.
Definition 2.2 (Equivariant Family of Subspaces).
Let be an invertible matrix cocycle over . The measurable function is called an equivariant family of subspaces for A if there exists a invariant set of full measure such that for all , .
We note that an equivariant family of subspaces is equivalent to an invariant measurable vector space bundle: the bundle is invariant under .
Note that when we speak of invariant vector bundles, for the remainder of the paper, we are referring to measurable invariant vector bundles.
Definition 2.3 (Reducible Bundle).
Let be an invertible matrix cocycle, and let be an invariant measurable vector bundle over . We say that is reducible if there exists an invariant vector bundle over such that for a.e. , .
We recall the statement of the MET.
Theorem 2.1 (Multiplicative Ergodic Theorem for Invertible Matrix Cocycles).
Let be an invertible and ergodic measurepreserving system, and let be an invertible matrix cocycle. Suppose that satisfies
Then there exist real numbers , positive integers with , a subset with and measurable families of subspaces such that:

Equivariance: For each , for ;

Growth: For each , and each nonzero , as .
That is, the MET states that the bundle may be decomposed as a sum of invariant bundles with different growth rates.
Definition 2.4 (Block Diagonalization).
Let be an ergodic dynamical system. We say that a measurable cocycle of by matrices over can be put into block diagonal form with block sizes over the field (or that is block diagonalizable, if the block sizes are understood), if there exist positive integers with and , a invariant set of full measure , and a measurable family of matrices such that is block diagonal, with block sizes , for all . We also say that is cohomologous to the resulting block diagonal matrix cocycle.
The previous definition may be extended to nonergodic systems; in this case, the quantities and can be functions on , but are now required to be constant on each ergodic component. We will only consider ergodic systems.
The cocycle is block diagonalizable with block sizes if and only if the bundle may be expressed as a sum of vector bundles of dimensions . To see the equivalence, notice that if is block diagonal, then the fibre, of the th bundle is the span of the st through th columns of . Conversely, if is expressed as the sum of bundles , ,…, with fibres of dimensions , then as in [4], we may measurably pick bases for each . Forming the matrix as described above yields the block diagonalization of the cocycle.
Definition 2.5 (Block Triangularization).
We say that a measurable cocycle can be put into block triangular form over the field , or is block triangularizable over , if is block diagonalizable over in the sense of Definition 2.4, and each of the blocks in the decomposition is triangular.
Remark 1.
We generally assume that any triangularization is uppertriangularization; it is equivalent to consider lower triangularization, because if we have triangularizing matrices , we may instead use matrices , where is the matrix with ones from bottomleft to topright and zeroes everywhere else. This obtains the opposite triangular form.
Remark 2.
In the case of a cocycle of matrices, by the argument outlined below Definition 2.4, we see that reducibility of the bundle is equivalent to the triangularizability over of the matrix cocycle.
We may now state precisely the question that we wish to address: Given an invertible matrix cocycle, , are the trivial bundles and irreducible under the action of ? In the matrix language, can necessarily be block triangularized?
The remainder of the paper proceeds as follows. In Section 3, we present the main theorem, Theorem 3.1, which gives sufficient conditions for the trivial bundle for a by orthogonal matrix cocycle to be irreducible over , to be irreducible over , and not to be cohomologous to a scalar multiple of the identity matrix. In Section 4, we present three explicit examples illustrating the application of Theorem 3.1, in both nontrivial situations. These examples provide a strong negative answer to the question posed above; namely, it is not always possible to block triangularize a matrix cocycle, even over the complex numbers. In the language of bundles, there exists cocycles for which and are both irreducible. In Section 5, we give the proof of Theorem 3.1. Proofs of some of the more technical details are left to the Appendix.
3. Sufficient conditions for irreducibility
Denote the collection of real by orthogonal matrices by . Let be an invertible and ergodic measurepreserving system on a probability space, and let be a measurable matrix cocycle over . For each , , as a map on , either rotates by some angle , or reflects in the line with angle ; let be where is a rotation, and let be where is a reflection.
We choose to restrict our study here to orthogonal matrices, because in the dimension case, the MET guarantees that if a cocycle has two different Lyapunov exponents, the cocycle is diagonalizable. Orthogonal matrices do not change the norm of any vectors, and so the only Lyapunov exponent for cocycles of orthogonal matrices is . Therefore, the MET yields the trivial decomposition for the cocycle.
Denote , and . For each , define the maps and by:
From these maps, define skew products given by , and given by . and are measurepreserving transformations on and , when each space is equipped with Haar measure, and , respectively ( is normalized counting measure on and is Lebesgue measure).
Theorem 3.1.
Let be an invertible ergodic measurepreserving system over a probability space. Let be a measurable cocycle over the map . Let and be as constructed above.

If is ergodic, is irreducible under the action of .

If both and are ergodic, then is irreducible under the action of .

If at least one of or is ergodic, then the cocycle is not cohomologous to a cocycle of the form , i.e. a scalar multiple of the identity.
Remark 3.
Notice that in the case of valued cocycles, if is a real invariant subbundle, then so is , where is just the orthogonal complement of ; the same holds for the Hermitian complement, in the case that is a complex vector bundle. Hence, for valued cocycles, if is a reducible bundle, then it may be decomposed as a sum of line bundles.
In matrix language, if an cocycle can be block triangularized, then it can be block diagonalized.
Remark 4.
The third part of Theorem 3.1 is the cocycle analogue of the linear algebra fact that given two eigenvectors for the same eigenvalue of a by matrix, either they are scalar multiples of each other, or the matrix is similar to a scalar multiple of the identity. If the cocycle is cohomologous to a scalar cocycle, then there is a continuum of proper subbundles of . This is ruled out by the ergodicity of either or .
Remark 5.
Theorem 3.1 gives sufficient conditions for to be an irreducible bundle for the cocycle , and for to be noncohomologous to a scalar multiple of the identity. None of these conditions are necessary; counterexamples are given in the Appendix.
4. Examples of irreducibility
In this section, we discuss three examples; one is a cocycle where the trivial bundle is reducible, but is irreducible, and the other two are cases where is irreducible (and hence so is ).
4.1. Example 1: rotation cocycle over a rotation
Let be the irrational rotation by over the unit interval with normalized Lebesgue measure, and consider the matrix cocycle generated by
Proposition 4.1.
The bundle is reducible under the action of , but is irreducible under the action of .
Proof.
To see that is irreducible, we first compute the map . Each matrix is a rotation in by , which allows us to compute the map as outlined above. We obtain , and then we compute the map on to be . It is well known (see, for example, [6], Theorem 2.1) that is an ergodic map with respect to Lebesgue measure on . Applying Theorem 3.1 yields that is irreducible.
On the other hand, observe that and Hence is reducible, as
∎
4.2. Example 2: rotation and flip cocycle over a rotation
Let the base dynamics space be the same as in the previous example: . This time, for an irrational number in , we define a matrix cocycle over , with
Proposition 4.2.
The trivial bundle is irreducible for the cocycle over as defined above.
We leave the proof to the Appendix. The proof uses a result by Schmidt stemming from his work on ergodic transformation groups [13]. We note that we need to have take values outside of the rotations , since the same proof as in the previous example shows that is reducible under the action of any measurable cocycle with values in .
4.3. Example 3: rotation and flip cocycle over a Bernoulli shift
In the next example, the base map will be a Bernoulli shift instead of a circle rotation. We again obtain irreducibility of the bundle, indicating that reducibility of the bundle is not primarily determined by the properties of the base system.
Let denote the left Bernoulli shift on two symbols, each with weight . Let be a matrix cocycle over , generated by the map , given by:
Proposition 4.3.
The trivial bundle is irreducible for the cocycle over as defined above.
Again, we leave the proof to the Appendix. This example has the feature that the proof is selfcontained, relying only on Fourier analysis.
5. Proof of Theorem 3.1
Proof.
We begin with the proof of irreducibility of the bundle . The proof will proceed in three steps.
Step 1: Let denote the complex Grassmannian of dimensional subspaces of . It is wellknown that is homeomorphic to , the onepoint compactification of . In particular, if we choose
then a homeomorphism is given by for , and ; this yields coordinates for the Grassmannian.
Consider the skew product of and the generator of , , on the space (as an invertible by matrix, acts naturally on ). In the coordinates above, we obtain a map on as follows: for ,
whereas for or , we have
The form of stems from the fact that and are eigenvectors for the rotation matrices, and are swapped and scaled by the reflection matrices.
To complete the setup, denote the complex unit circle by , let be given by
and let be given by . These maps are measurable, and will be used to relate with and .
Step 2: Assume, for the sake of contradiction, that is an invariant onedimensional subbundle of . Denote the coordinates of in by . The following lemma sets the stage for the remainder of the proof.
Lemma 5.1.
Let be the map described in Step 1, and let be described as above. Let be the union of two circles where if then , and . Then the graph of is invariant under , and is contained in for some . Moreover, is an invariant set under , for any .
Proof.
The invariance of and the definition of the map provide almosteverywhere invariance of the graph of under . For the containment of the graph inside for some , let , and observe that is an invariant measurable function with respect to the ergodic map , hence is almosteverywhere constant with respect to , with value . Then the graph of is contained in , up to a set of measure zero. The set is easily shown to be almosteverywhere invariant by applying to a point in the set. Finally, we may remove an invariant measurezero subset of on which equivariance of fails, and separately where containment inside fails. This completes the proof. ∎
Step 3: By the above lemma, and its complement are invariant sets for , and the graph of lies entirely in one of them. We split the remainder of the proof into two cases. First, consider the case where the graph of lies in . Using the definition of the map from Step 1, an explicit calculation shows that
Thus the graph of (a subset of ) is invariant under . Let be the distance on the torus . By using the explicit form of (as before Theorem 3.1), is shown to be an invariant function, and it is clearly measurable. Since is nonconstant, this contradicts the ergodicity of .
Then, consider the case that the graph of lies in the complement of . Using the map , we get that
so analogously to the previous case, the graph of is invariant under . We apply Fubini’s theorem to show that the graph of has measure , so since the graph is invariant, this contradicts the ergodicity of .
The two cases together show that cannot exist, and thus the onedimensional bundle, , cannot exist as assumed; therefore is irreducible.
Next, we show that if is ergodic, then is irreducible; this is part (2) of the theorem. Instead of the complex Grassmannian, we consider the real Grassmannian, . It is easy to show that it is homeomorphic to the unit circle equipped with the usual topology, so we have coordinates for . A similar argument to that in Step 1 for the complex case computes the skew product of and over , which is exactly the map . The exact same arguments in the remaining steps applied to the map go through, and hence is irreducible.
Finally, we prove part (3) of the theorem. Suppose, on the contrary, that is cohomologous to a scalar multiple of the identity, so that there exists a measurable matrix function and a measurable function such that on a set of full measure in ,
In the previous parts of the proof, we found that if there are equivariant subspaces which do not lie on the unit circle of the Grassmannian (in coordinates), then the map cannot be ergodic, and if there are equivariant subspaces which do not lie at the poles ( or in ), then the map cannot be ergodic. We will use the functions and to construct an equivariant family of subspaces which lie neither at the poles, nor on the unit circle. This will imply that neither nor is ergodic, which is the contrapositive of statement (3) in the theorem. Notice that for fixed and (not both 0), is a onedimensional invariant bundle; if we can find and such that the subspaces in this bundle live away from the unit circle or the poles, then we will be done.
Let be the homeomorphism from to , and let be given by
This function is measurable and nonnegative, with values in (similar to the function in Lemma 5.1). For fixed , is continuous, and for fixed not both zero, the function is measurable, and invariant. The latter is shown by an explicit computation, utilizing the relations between and , as well as the induced action of on the coordinates of the Grassmannian, similar to the proof of Lemma 5.1.
Hence, since is ergodic with respect to , we see that for any fixed , is almost everywhere constant. Note that if , then the span of is at one of the poles ( or ), and if , then the span of is on the unit circle. We will be done if we can find such that the almost everywhere value of is neither nor ; then the map
is exactly the equivariant family of subspaces we desire.
To do this, let be given by
then is exactly the almost everywhere value for . We will show that takes a value which is neither nor . For a fixed , let be a set of full measure such that for all , . Let , and for each , find such that has full measure, and for all , we have . Let
Then , and for any , we have:
using the continuity of for fixed . Hence is continuous.
Let be a countable dense subset of , which does not include . For each , find such that and for all , we have . Let
Then . Let . For any , find converging to , we get:
again using the continuity of for fixed , as well as the continuity of .
Finally, note that by definition of and , there exists a set of full measure in such that is a basis for . So for any , the map is surjective, and hence takes values which are neither nor . Now, has full measure, so find , and find such that . Since is also in , we have , and we have completed the proof. ∎
Appendix A Other proofs
Proof of Proposition 4.2.
is a rotation by for , and a reflection in the horizontal axis for . In our notation, , with a fixed rotation angle , and , with fixed reflection axis . Substituting this into our maps and gives us:
It is then sufficient to show that these maps are ergodic, as afterwards we simply apply Theorem 3.1.
The first step for both of these claims will be to induce the map on a subset of positive measure; ergodicity of the induced map will then imply ergodicity of the original map. We deal first with , then with .
In the case of on , we induce on the set . Since the set on which we are inducing is the product of and the whole space , the return time for only depends on the map and the first coordinate . The base map for this skew product is an irrational rotation, and inducing an irrational rotation on an interval of the same size as the rotation angle yields a map which is isomorphic to a different irrational rotation (see [11], noting that rotations are special cases of interval exchange transformations), with the rotation angle given by the fractional part of , denoted . Using this and using the fact that the only action in the component is when the map first leaves the set , we obtain that the induced map of on is isomorphic to the map , given by
Another wellknown result (Theorem 1.9 in [15]) says that because is compact, is a rotation, and is dense in , is ergodic; we are done in the case of .
In the case of on , we must do a bit more work. Similarly to before, we induce on the set , and obtain a new map on , which is given by
If we were to apply twice in a row, we would eliminate the flip in the coordinate. This idea is justified, because it is easy to show that if a map is measurepreserving and is ergodic, then is also ergodic.
Hence, we square to obtain the map , given by
We again induce, this time on the set , and after another coordinate change, we get a map acting on . Setting to be the fractional part of , we see that is given by
We now appeal to a result from the literature:
Proposition A.1 (Schmidt, [13]).
Consider the space as defined in the previous proposition. Define the map on by
Then is an ergodic measurepreserving transformation.
To use this proposition, we translate it from the finite case to the finite case by this proposition:
Proposition A.2.
Suppose is measurepreserving and ergodic with respect to Lebesgue measure, and let be a measurable function, with range , where is irrational. Let have the usual Lebesgue product measure and Borel sets, and let be the skew product extension of and to , so that:
Let be the skew product extension of and to with the (finite) product measure (Lebesgue and counting, with the discrete algebra for the counting measure), so that:
Then if is ergodic, so is .
Proof.
Let be a bounded measurable function invariant under , so . We shall show that must be a.e. constant; this will imply that is ergodic. For , define the measurable map
Then we see that . In addition, define , so that is a measurable function defined on . Since intertwines the dynamics on the two spaces, we get the following:
Thus is invariant under , and so is constant a.e. with respect to the product measure , since is ergodic.
We wish to use the fact that is a.e. constant for each to show that is constant a.e. To do this, we make an intermediate step. Define
Because is bounded, is finite, and by Fubini’s theorem, is measurable. Moreover, we have the following, since is a.e. constant on :
The map is ergodic on , thus we see that is a.e. constant on ; write for a.e. . Note that for all , is a.e. constant on , so we see that for a.e. , . Denote ; this set has full measure in . If , then for a.e. , . Computing the measure of the set of points where via Fubini’s Theorem yields the final statement: almost everywhere. Hence is ergodic. ∎
and are related in exactly the correct way for Proposition A.2 to be applied, and so is ergodic. This proves that the original map is ergodic, and so we are finished. ∎
Proof of Proposition 4.3.
The proof that is irreducible is similar, in outline, to the proof that the cocycle in Example 2 is not triangularizable. We give only a brief sketch and leave the details to [7].
The map , on , is shown to be ergodic by inducing on the product of with the set of binary strings with as the th component, and obtaining the product of a Bernoulli shift on countably many symbols (), which is strongly mixing, and the translation by on , which is ergodic. Their product is therefore ergodic, and so the map is ergodic.
The map , on , requires a similar process to that in Example 2. After inducing, squaring, and inducing again, the resulting map is a skew product over a Bernoulli shift on strings with symbols from , where the action on the torus is a rotation determined by the th component of the string. One then applies Fourier analysis techniques to show that this new map is strongly mixing, and hence the map is ergodic. ∎
Appendix B Counterexamples to necessity of conditions in Theorem 3.1
As mentioned in Remark 5, none of the sufficient conditions listed in Theorem 3.1 are necessary. We will illustrate with some simple counterexamples.
For condition (1), consider the system , where is the torus equipped with the usual Borel algebra and normalized Lebesgue measure and is an irrational rotation by , and let In this case, the map is given by . It is easy to see that this map is not ergodic: the set is invariant, and has measure . However, the cocycle is irreducible over . To see this, consider the projection , where is given by . This projection induces is a factor map for , given by . This map is ergodic, which prevents the existence of a real equivariant family of subspaces for , because the resulting invariant graph for factors to an invariant graph for , and this contradicts the fact that is ergodic.
The same example can be used to show that condition (3) is also not necessary; the associated map is just , which is also not ergodic. However, the cocycle is not cohomologous to a scalar multiple of the identity. Suppose it were; following the proof of condition (3) in Theorem 3.1, we obtain an equivariant family of subspaces which yields an invariant graph for . This yields an invariant graph for , which contradicts the ergodicity of .
For condition (2), modify the cocycle above as in Section 4.3. Let be the left Bernoulli shift on , where and are both given weight , and let be the cocycle over generated by
In this case, the map is still ergodic, but is not ergodic, because is invariant, with measure . Once again, using a similar construction to the above (taking care of the points separately), we obtain a similar factor map of , which is ergodic (by a similar proof to that of Proposition 4.3). Applying the arguments used in the proof of (2) in Theorem 3.1, but to and instead of and , we see that has no equivariant family of onedimensional subspaces over . Therefore, condition (2) is not necessary.
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