About 3+1/2 weeks ago, Jean-François Quint gave a very nice talk (with same title as this post) during Paris 6 and 7 “Journées de dynamique” about his joint work with Yves Benoist on the regularity properties of stationary measures.

In what follows, I’m reproducing my notes for Jean-François Quint’s lecture. (As usual, all errors/mistakes in the sequel are my responsibility.)

**1. Introduction **

** 1.1. Limit sets of semigroups of matrices **

Let be a semigroup of invertible real matrices.

Recall that:

- is
*irreducible*if there are no non-trivial -invariant subspaces, i.e., and imply or ; - is
*proximal*if it contains a proximal element , i.e., has an unique eigenvalue with maximal modulus which has multiplicity one in the characteristic polynomial of ; equivalently, , , and has spectral radius or, in other terms, the action of on the projective space has an attracting fixed point.

Proposition 1Let be a irreducible and proximal semigroup. Then, the action of on admits a smallest non-empty invariant closed subset called the limit set of .

*Proof:* Let . It is clear that is non-empty, closed and invariant. Moreover, is the smallest subset with these properties thanks to the following argument. Let be a proximal element. If , then converges to as . If , we use the irreducibility of to find an element such that and, *a fortiori*, converges to as .

** 1.2. Stationary measures **

Suppose that is a probability measure on a semigroup acting on a space . We say that a probability measure on is –*stationary* if it is -invariant on average, i.e.,

is equal to .

In the case of irreducible and proximal semigroups of matrices, the following theorem of Furstenberg and Kesten ensures the existence and uniqueness of stationary measures for the corresponding projective actions:

Theorem 2 (Furstenberg-Kesten)Let be a Borel probability measure on and denote by the subsemigroup generated by the elements in the support of . Suppose that is irreducible and proximal. Then, has an unique -stationary measure on and .

In what follows, we shall also assume that is *strongly irreducible*, i.e., for all non-trivial proper subspaces , and we will be interested in the nature of in Furstenberg-Kesten theorem.

It is possible to show that if is absolutely continuous with respect to the Lebesgue (Haar) measure (on ), then is absolutely continuous with respect to the Lebesgue measure (on ).

For this reason, we shall focus in the sequel on the following question:

Can be absolutely continuous when is *finitely supported*?

It was shown by Kaimanovich and Le Prince that the answer to this question is *not* always positive:

Theorem 3 (Kaimanovich-Le Prince)There exists finite (actually, ) such that spans a Zariski dense subsemigroup of , but is the support of a probability measure such that the associated stationary measure on is singular with respect to the Lebesgue measure.

On the other hand, Bárány-Pollicott-Simon and Bourgain showed that the answer to this question is sometimes positive:

Theorem 4 (Bárány-Pollicott-Simon, Bourgain)There exists finite supporting a probability measure such that the corresponding stationary measure is absolutely continuous with respect to Lebesgue.

Remark 1As it was pointed out by Quint, the examples produced by Bourgain are explicit, but it would be desirable to get simpler explicit examples of sets satisfying the previous theorem. In this direction, he asked the following question. Denote by , and , , and consider the probability measures

Is it true that, for each fixed , if is small enough (and typical?), then the stationary measure associated to is absolutely continuous with respect to the Lebesgue measure? (Note that if is very large, then we are in the regime described by Kaimanovich-Le Prince theorem 3.)

** 1.3. Statement of the main result **

In a recent paper, Benoist and Quint extended Theorem 4 to higher dimensions:

Theorem 5 (Benoist-Quint)For any , there exists finite and a probability measure with and proximal and strongly irreducible such that the corresponding stationary measure on is absolutely continuous with respect to Lebesgue.

The remainder of this post is dedicated to the proof of this result.

**2. Proof of the main theorem **

** 2.1. Spectral theory of quasi-compact operators **

Let be a Banach space and denote by the space of bounded linear operators on .

Given , recall that the compact non-empty set

is the *spectrum* of , and the quantity

is the *spectral radius* of .

The space of compact operators is an ideal and the quotient comes equipped with a natural norm .

Recall that the *essential spectrum* of is

and the *essential spectral radius* of is

Note that and . Moreover, these objects are the same for and its adjoint :

Proposition 6One has the following identities: , , and .

The next proposition explains that the spectrum and the essential spectrum morally differ only by eigenvalues of finite multiplicity:

Proposition 7If , then there exists a decomposition into closed subspaces such that is finite-dimensional, , , and .

** 2.2. Spectral criterion for absolute continuity **

Let be a Borel probability measure on and consider the natural action of on .

Given a function , let .

Remark 2when is finitely supported.

We equip with the round measure induced from the natural Lebesgue measure on the sphere .

If has compact support, then is a bounded operator on .

One can infer the absolute continuity of the stationary measure of from the spectral properties of thanks to the following proposition:

Proposition 8If is proximal and strongly irreducible, and , then the -stationary measure on is absolutely continuous with respect to , i.e., .

*Proof:* Note that (where is the constant function with value one), so that .

By hypothesis, . Thus, there exists with . By definition, this means that the absolutely continuous measure is the -stationary measure.

** 2.3. Application of the spectral criterion **

The result in Theorem 5 (i.e., the case ) is easier to derive than Theorem 4 (i.e., the case ) because has elements equidistributing very quickly. Here, the word equidistribution means the following: if is a probability measure on , is Zariski dense on , then we say that the elements in the support of equidistribute whenever for all (with standing for the Haar measure).

This equidistribution property holds in presence of *spectral gap*, i.e., (where is the subspace of -functions with zero average [for Haar measure]). In particular, the works of Drinfeld and Margulis provide examples of elements of equidistributing very quickly:

Theorem 9 (Margulis and Drinfeld)There exists finite and a probability measure with and .

At this point, the proof of Theorem 5 is almost over. Indeed, if we take a proximal element and we denote by the probability measure provided by Margulis and Drinfeld, then the sequence of measures

have the property that converges to a rank one operator. Therefore,

In particular, by Proposition 8, it follows that satisfies the conclusions of Benoist-Quint theorem 5 for any sufficiently large.

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