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Calculating Lyapunov exponents for random products of positive matrices Mark Pollicott Warwick University 7 July, 2020 Greetings from Kenilworth This is a 1830 painting of the ruin of Kenilworth Castle by J.M.W. Turner. Of course, this is 190
Mark Pollicott
Warwick University
7 July, 2020
This is a 1830 painting of the ruin of Kenilworth Castle by J.M.W. Turner. Of course, this is 190 years out of date: The second picture is a modern photograph of the castle.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 2 / 29
We want to discuss the (top) Lyapunov exponent for random products of matrices. We want to discuss three ways (not) to estimate the Lyapunov exponent; and a 4th way to compute it - in the particular case that the matrices are positive Why positive matrices? - because the method works. The new ingredient is the improved estimate on the error in the approximation. Why is it interesting?- There are other different approaches (e.g., Bandtlow-Slipantshuk, Bahsoun, Braviera-Duarte, Galatolo-Monge-Nisoli, ... ) but I like this approach because it uses some interesting underlying mathematics and “Tudo vale a pena quando a alma n˜ ao ´ e pequena” - Fernando Pessoa
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 3 / 29
Let A be a single k × k matrix (k 2) with entries in R. Some comforting concepts: The eigenvalues λ1 · · · , λk of the matrix A; The determinant det A =
i λi ∈ R
The spectral radius spr(A) = maxi |λi| “Computational mathematics is mainly based on two ideas: Taylor series, and linear algebra” - Nick Trefethen We are going to use both.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 4 / 29
The spectral radius also comes from
Theorem (The spectral radius formula)
spr(A) = lim
n→+∞ An1/n
where we can define the norm of a matrix A = (aij) by A = maxi,j |aij|. (The specific choice of norm isn’t so important.) I.M. Gelfand Surprisingly, there doesn’t seem to be any reference to this result before Gelfand’s paper in 1941. [N.B., After the lecture Michael Benedicks pointed
years earlier than that of Gelfand.]
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 5 / 29
Let us begin with two k × k matrices A1, A2 ( k 2) with entries in R. Consider weights 0 < p1, p2 < 1 (p1 + p2 = 1). We can consider: the 2n possible products Ai1Ai2 · · · Ain (i1, · · · , in ∈ {1, 2}); their norms Ai1Ai2 · · · Ain; and the weights pi1pi2 · · · pin (i1, · · · , in ∈ {1, 2}). We can define the (top) Lyapunov exponent by λ := lim
n→+∞
1 n
(pi1 · · · pin) log Ai1 · · · Ain.
Trivial case: A = A1 = A2 then λ = log spr(A).
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 6 / 29
Everything generalizes in the obvious way to more matrices. Consider k × k matrices A1, . . . , Ad with entries in R (d 2, k 2). Consider weights 0 < p1, . . . , pd < 1 (p1 + · · · + pd = 1). We can consider: the dn possible products Ai1Ai2 · · · Ain (i1, · · · , in ∈ {1, . . . , d}); their norms Ai1Ai2 · · · Ain; and the weights pi1pi2 · · · pin (i1, · · · , in ∈ {1, . . . , d}). We can define the (top) Lyapunov exponent by λ := lim
n→+∞
1 n
(pi1 · · · pin) log Ai1 · · · Ain. However, henceforth I will usually restrict to the case of two 2 × 2 matrices (mainly to avoid the problem of mixing up k and d) and take p1 = p2 = 1
2 (mainly
to cut down on notation).
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 7 / 29
Question: Given matrices A1, A2 and 0 < p1, p2 < 1, how easy is it to compute λ? Sir John Kingman (1973): “Pride of place among the unsolved problems of subadditive ergodic theory must go to calculation of the value λ ... and indeed this usually seems to be a problem of some depth.” Yuval Peres (1992): “We turn now to the excruciating problem of the subject: Devise reasonably general and effective algorithms for explicit calculation (or at least approximation) of Lyapunov exponents” At least we have the definition to work from...
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 8 / 29
We can try this out with a simple example. Consider A1 = 2 1 1 1
3 1 2 1
2.
Then working from the definition:
1 2 log Ai = 1.77767 . . . 1 2
1 4 log AiAj = 1.45723 . . . 1 3
8 log AiAjAk = 1.35236 . . . 1 4
1 16 log AiAjAkAl = 1.30008 . . . 1 5
1 32 log AiAjAkAlAm = 1.26872 . . .
Unfortunately, this doesn’t converge particularly quickly (typically O( 1
n)).
“In general, things either work out or they don’t, and if they don’t, you figure out something else, a plan B. There’s nothing wrong with plan B” - Dick Van Dyke
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 9 / 29
We first need to introduce the following notation. Let Σ = {1, 2}N then we can write x = (xn)∞
n=1 ∈ Σ. Let µ = (p1, p2)N be the usual Bernoulli measure on Σ.
Theorem (Furstenberg-Kesten, Sir John Kingman)
For almost all (µ) x ∈ Σ, 1 n log Ax1 · · · Axn → λ, as n → +∞. This is a (subadditive) ergodic theorem.
(2020 Abel prize winner)
Sir John Kingman
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 10 / 29
Let us try a little experiment. We can (again) let A1 =
1 1 1
1 2 1
2.
We can compute (for example 15) values of
1 1000 log Ax1 · · · Ax1000 for products of
“random” choices of 1000 matrices: 1.14649 . . . 1.14777 . . . 1.14924 . . . 1.15448 . . . 1.15181 . . . 1.14341 . . . 1.14569 . . . 1.15094 . . . 1.14975 . . . 1.14683 . . . 1.15213 . . . 1.14924 . . . 1.13802 . . . 1.15244 . . . 1.14983 . . . This suggests the value of the Lyapunov exponent to a couple of decimal places, but it is not clear how to get a rigorous estimate, so let us turn to a third approach.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 11 / 29
Henceforth we will restrict to positive matrices. We say that the matrices A1, A2 are positive if all the entries are strictly larger than zero. When k = 2 then we can write A1 =
11
a(1)
12
a(1)
21
a(1)
22
11
a(2)
12
a(2)
21
a(2)
22
ij
> 0 and a(2)
ij
> 0 for 1 i, j 2. Let Ai : R2 → R2 (i = 1, 2) be the usual linear action of the matrix Ai. We can consider the positive quadrant R2
+ = {(x, y) ∈ R2 : x, y 0} ⊂ R2.
Clearly positivity implies Ai(R2
+) ⊂ R2 + for i = 1, 2
Ai(R2
+)
R2
+
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 12 / 29
We can consider the “restriction” to the simplex ∆ = {(x, y) ∈ R2
+ : x + y = 1}.
Let Ai : ∆ → ∆ (i = 1, 2) be the projective action defined by (x, y) → Ai ( x
y ) →
Ai ( x
y )
Ai ( x
y )1
→ Ai(x, y) ( x
y )
Ai ( x
y )
∆ Since A1, A2 are positive we have Ai(∆) ⊂ int(∆). ( x
y )
Ai ( x
y )
∆ x ξ We can just take the first coordinate of Ai(x, y) = (ξ, η) to get a map Ti : [0, 1] → [0, 1] defined by Ti : x → ξ.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 13 / 29
Of course we can explicitly write expressions for Ti in terms of the entries of Ai (i = 1, 2). More precisely, we associate to the positive matrices A1 =
11
a(1)
12
a(1)
21
a(1)
22
11
a(2)
12
a(2)
21
a(2)
22
calculation are M¨
Ti(x) = (a(i)
11 − a(i) 12 )x + a(i) 12
(a(i)
11 + a(i) 21 − a(i) 12 − a(i) 22 )x + a(i) 12 + a(i) 22
(i = 1, 2). The positivity of the matrices ensures that T1 and T2 are both (eventually) contractions of the interval.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 14 / 29
Enter a deus ex-machina: We can define a linear operator L : C 1([0, 1]) → C 1([0, 1]) on C 1 functions by: Lw(z) = p1w(T1z) + p2w(T2z) where w ∈ C 1([0, 1]) and z ∈ [0, 1]. We can write the nth iterate (for n 2) as Lnw(z) =
piw(Tiz) where
1
i = (i1, · · · , in) ∈ {1, 2}n and |i| = n;
2
Ti = Ti1 ◦ · · · ◦ Tin : [0, 1] → [0, 1]; and
3
pi = pi1 · · · pin. 1 λ p1(Lnf1) + p2(Lnf2)
Theorem (Y. Peres)
One can associate functions f1, f2 ∈ C 1([0, 1]) and then p1(Lnf1) + p2(Lnf2) → λ, as n → +∞ In fact, there exists 0 < θ < 1 such that p1(Lnf1) + p2(Lnf2) = λ + O(θn).
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 15 / 29
Although p1Lnf1 + p2Lnf2 is exponentially close to λ there is an exponential amount of work needed to compute these approximations. However, this approach is good for something: The Lyapunov exponent λ depends analytically on the entries in the matrices (Ruelle) The Lyapunov exponent λ depends analytically on the weights p1, p2 (Peres) Of course, the positivity of the matrices can be replaced with “preserving a cone”,
different assumptions.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 16 / 29
Of course, for a single matrix A the eigenvalues come from zeros of z → det(I − zA). Can we get the Lypapunov exponents from a complex function? To define a suitable complex function we recall some notation: i = (i1, · · · , in) ∈ {1, 2}n and |i| = n; Ti = Ti1 ◦ · · · ◦ Tin : [0, 1] → [0, 1]; xi = Ti(xi) is the unique fixed point for the map Ti;and pi = pi1 · · · pin. Then d(z, s) := exp −
∞
zn n
pi |(Ti)′(xi)|s 1 − (Ti)′(xi) which converges for |z| and |s| sufficiently small.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 17 / 29
The definition looks a little complicated, but let us concentrate on the properties
1 z d(z, s) z(s) d(z, s) is analytic; For each s > 0 we have d(0, s) = 1; and the smallest zero z(s) > 0 (i.e., d(z(s), s) = 0) is simple. These follow (essentially) from results of Ruelle.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 18 / 29
We can write the Lyapunov exponent λ in terms of the zero z(s) of the function(s) z → d(z, s) (i.e., d(z(s), s) = 0).
Lemma (N. Jurga and I. Morris)
The Lyapunov exponent is given by λ = 1
2z ′(0).
(This improved on a clunkier formula by M.P.) By differentiating the identity d(z(s), s) = 0 we can now write λ in terms of d(z, s): λ = −1 2 ∂d(1, s) ∂s |s=0 ∂d(z, 0) ∂z |z=1 −1 , since z(0) = 1.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 19 / 29
In case we have lost the plot, the basic idea is the following. We want to estimate the Lypapunov exponent λ. We write λ in terms of a function d(z, s) d(x, s) was defined in terms of fixed points
T1, T2 : [0, 1] → [0, 1]. The contractions T1, T2 : [0, 1] → [0, 1] are written in terms of the matrices A1 and A2. And, of course, this generalizes for finitely many positive matrices in arbitrary dimensions. This looks a complicated piece of machinery, but it actually works reasonably effectively. A Heath Robinson machine for making pancakes.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 20 / 29
Of course z → d(z, s) depends on infinitely many computations, so taking pity on my computer we need to make an approximation. We can expand as a Taylor series d(z, s) = 1 +
∞
an(s)zn and approximate it by a polynomial d(N)(z, s) = 1 +
N
an(s)zn, for some N 1. We can then approximate λ = 1
2 ∂d(1,t) ∂t
|t=0
∂z
|z=1 −1 by λN = −1 2 ∂d(N)(1, t) ∂t |t=0 ∂d(N)(z, 0) ∂z |z=1 −1 .
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 21 / 29
Let A1 =
1 1 1
1 2 1
2.
We can then consider the associated two contractions T1(x) = x + 1 x + 2 and T2(x) = 2x + 1 3x + 2. Letting N = 9, say, we can use this approach to estimate λ ≈ λ9 = 1.143311035041828694244 . . . . Question: To how many decimal places is this accurate? Answer: It is accurate to 20 decimal places
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 22 / 29
Let A1 = 11
10
1
1 10
1
1
1 10
1
11 10
2.
We can then consider the associated maps T1(x) =
1 10x + 1
− 4
5x + 2 and T2(x) = 9 10x + 1 10 4 5x + 6 5
. Letting N = 9, say, we can use this approach to estimate λ ≈ λ9 = 0.4660 . . . . Question: To how many decimal places is this accurate? Answer: It is accurate to 2 decimal places
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 23 / 29
The main approximation we made was replacing the function d(z, s) by dN(z, s) and so we need to bound the difference |d(z, s) − dN(z, s)|.
1
It is relatively easy to show there exists 0 < θ < 1 such that |d(z, s) − dN(z, s)| = O
an(s)zn
leading to |λ − λN| = O
.
2
Recall that previously one had that |λ − λN| = O
(by Plan C = Peres’ result). So we seem to be doing better. However, realistically N cannot be too large (unless you have a really big computer) so we would also like to get the best possible bounds for |an(s)| for n N + 1 ≈ 10.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 24 / 29
Recall that to estimate λ we expanded d(z, s) = 1 + ∞
n=1 an(s)zn and we want
to get the best bounds on |an| for n N + 1 ≈ 10. We can use the classical approach of Grothendieck, Ruelle, Fried, Jurga, Morris, etc. to get bounds on the an(s) (called Euler bounds). But often we can do better by using improved bounds on the an(s) (called Computed bounds) using:
◮ More operator theory (Composition Operators, Approximation Numbers,
Weyl’s Inequality); and then
◮ A bit more computation to bound an for 10 ≈ N + 1 n M ≈ 500
This is a variant on using ideas in recent(-ish) work of Jenkinson-P. (for computing Hausdorff Dimension of certain sets) and Jenkinson-P.-Vytnova (for computing Lypapunov exponents - the other type: for expanding maps of the interval). This improvement is more marked when there is less hyperbolicity. This is best illustrated by examples.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 25 / 29
Consider A1 = 2 1 1 1
3 1 2 1
2.
Recalling d(z, s) = d9(z, s) + ∞
n=10 an(s)zn we can compare the new and old
bounds on an(s) (e.g. when s = 0): n Old bound on an(0) New bound on an(s) 10 1.088971 . . . × 10−21 7.0265930 . . . × 10−23 11 1.956063 . . . × 10−26 6.3590078 . . . × 10−28 12 1.171186 . . . × 10−31 1.8728592 . . . × 10−33 13 2.337478 . . . × 10−37 1.8001868 . . . × 10−39 14 1.555061 . . . × 10−43 5.6589790 . . . × 10−46 15 3.448469 . . . × 10−50 5.8273585 . . . × 10−53 In fact we can show that λ = 1.143311035041828694244 ± 3 × 10−21 Conclusion: In this example there is no significant improvement (perhaps only an
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 26 / 29
Let A1 = 11
10
1
1 10
1
1 10
1
11 10
2.
Recalling d(z, s) = d9(z, s) + ∞
n=10 an(s)zn we can compare the old and new
bounds on an(s) (e.g., when s = 0): n Old bound on an(0) New bound on an(0) 10 10.0321212 . . . 0.0000161 . . . 11 2.8224811 . . . 9.2280733 . . . × 10−7 12 0.6450222 . . . 4.1184631 . . . × 10−8 13 0.1203062 . . . 1.4468302 . . . × 10−9 14 0.0183832 . . . 4.0200884 . . . × 10−11 15 0.0023083 . . . 8.8694309 . . . × 10−13 In fact we can show that λ = 0.4660 ± 0.003 Conclusion: In this example the old error bounds are still bigger than the estimate (for N ≈ 9) whereas the newer bounds lead to meaningful result.
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 27 / 29
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 28 / 29
Ao Cai (Universidade de Lisboa), Pedro Duarte (Universidade de Lisboa), Jos´ e Pedro Gaiv˜ ao (Universidade de Lisboa), Silvius Klein (Pontif´ ıcia Universidade Cat´
Jo˜ ao Lopes Dias (Universidade de Lisboa), Telmo Peixe (Universidade de Lisboa) Jaqueline Siqueira (Universidade Federal do Rio de Janeiro)
Mark Pollicott (Warwick University) Calculating Lyapunov exponents for random products of positive matrices 7 July, 2020 29 / 29