Quantitative lower bounds on the Lyapunov exponent from multivariate matrix inequalities

Abstract

Lyapunov exponents characterize the asymptotic behavior of long matrix products. In this work we introduce a new technique that yields quantitative lower bounds on the top Lyapunov exponent in terms of an efficiently computable matrix sum in ergodic situations. Our approach rests on two results from matrix analysis—the n-matrix extension of the Golden–Thompson inequality and an effective version of the Avalanche Principle. While applications of this method are currently restricted to uniformly hyperbolic cocycles, we include specific examples of ergodic Schrödinger cocycles of polymer type for which outside of the spectrum our bounds are substantially stronger than the standard Combes–Thomas estimates. We also show that these techniques yield short proofs of quantitative stability results for the top Lyapunov exponent which are known from more dynamical approaches. We also discuss the problem of finding stable bounds on the Lyapunov exponent for almost-commuting matrices.

Appendices

Proof of the Avalanche principle, version 2

Proof of Theorem 2.6

The proof follows the general steps in [25]. We only summarize the necessary changes in the argument, which amount to various refined estimates and a different choice of the scaling parameter t (see below).

We define $\varepsilon _0=\frac{1}{5}$ and $c_0=\frac{1}{6}$. Using that $(\varepsilon _0+\sqrt{1-\varepsilon _0^2})\pi /2\le 1.86$, the estimate $\frac{\sqrt{2}\pi }{2}\kappa \varepsilon ^{-2}$ on the Lipschitz constant in Corollary 5.7(b) in [25] can be replaced by

$$\begin{aligned} 1.86*\kappa \varepsilon ^{-2}. \end{aligned}$$

(A.1)

This replacement then propagates through the argument in [25], specifically to Lemma 5.14, Corollary 5.15 and the proof of Theorem 5.5.

Another change is made starting with Lemma 5.12 in [25]. We define ${\tilde{\varepsilon }}=t\varepsilon$ with $t=\frac{72}{100}$. (This choice of t is close to the largest one allowed here which is important for having a good bound in (A.4) below.) Lemma 5.12 in [25] then holds true with this choice of t under our present assumptions on $\kappa$ and $\varepsilon$. To see this, we note that condition (5.7) in [25] can be rearranged as

$$\begin{aligned} t>c_0\sqrt{\frac{1+2c_0^2\varepsilon ^2}{1+c_0^2\varepsilon ^2}} , \end{aligned}$$

(A.2)

which is comfortably satisfied for all $\varepsilon \in (0,\varepsilon _0)$. To establish Lemma 5.12, it then remains to check that

$$\begin{aligned} f\left( \frac{c_0\varepsilon }{t}\sqrt{1-t^2\varepsilon ^2},\sqrt{1-\frac{\varepsilon ^2}{1+2c_0^2\varepsilon ^2}}\right) <\sqrt{1-t^2\varepsilon ^2} \end{aligned}$$

(A.3)

for $f(x,y)=x\sqrt{1-y^2}+y\sqrt{1-x^2}$, which is tedious, but elementary. (Here we avoid estimating the square roots by their linearization as was done in (5.8) of [25].)

The previously established estimates yield the following modification of Lemma 5.16 in [25]. The relevant Lipschitz constant is now

$$\begin{aligned} L=1.86*\kappa {\tilde{\varepsilon }}^{-2}\le 1.86*c_0 t^{-2}<\frac{3}{5}<1 , \end{aligned}$$

(A.4)

where the fact that this is strictly less than 1 is important for having a contraction. The relevant estimate after summing the geometric series in the proof of Lemma 5.16 then yields

$$\begin{aligned} \frac{\kappa {\tilde{\varepsilon }}^{-1}}{1-L} = \frac{1/t}{1-L}\frac{\kappa }{\varepsilon } <\frac{7}{2}\frac{\kappa }{\varepsilon } \end{aligned}$$

(A.5)

and then the right-hand side replaces the upper bound from Lemma 5.16 (which was $3\kappa /\varepsilon$ in [25]).

For the proof of Theorem 5.5(i), it suffices to note that

$$\begin{aligned} \frac{7}{2}\frac{\kappa }{\varepsilon } +\frac{1}{2}\le \frac{7}{2}c_0 \varepsilon _0+\frac{1}{2}=\frac{7}{30}+\frac{1}{2}<\sqrt{1-t^2\varepsilon _0^2}<\sqrt{1-{\tilde{\varepsilon }}^2}\, . \end{aligned}$$

(A.6)

In the conclusion of Theorem 5.5(i), the bound $3\frac{\kappa }{\varepsilon }$ is replaced by $\frac{7}{2}\frac{\kappa }{\varepsilon }$ due to our alternative version of Lemma 5.16 described above.

Finally, in the proof of Theorem 5.5(ii), we replace every instance of $3\frac{\kappa }{\varepsilon }$ by $\frac{7}{2}\frac{\kappa }{\varepsilon }$ in (5.17)–(5.19). This has the effect of replacing the lower bound in (5.19) by

$$\begin{aligned} \frac{\varepsilon }{\sqrt{1+2\frac{\kappa ^2}{\varepsilon ^2}}}-\frac{7}{2}\frac{\kappa }{\varepsilon } \ge \frac{\varepsilon }{\sqrt{1+2\frac{\kappa ^2}{\varepsilon ^2}}}-\frac{7}{12}\varepsilon > \frac{1}{3}\varepsilon \end{aligned}$$

(A.7)

and therefore the upper bound in (5.20) by $\frac{21}{2}\frac{\kappa }{\varepsilon ^2}$. Combining this with the bound (5.21) in [25] and using $\kappa \le c_0\varepsilon _0^2=\frac{1}{150}$ yields the lower bound on the main quantity

$$\begin{aligned} \frac{\rho (L_1,L_2,\ldots ,L_{n})}{\rho (L_1,L_2)\cdots \rho (L_{n-1}, L_{n})}\ge \mathrm {e}^{-11 n\kappa /{\varepsilon ^2}} \end{aligned}$$

(A.8)

so $c_l=11$ as desired.

It remains to prove the upper bound on the main quantity. We first note that the upper bound $\kappa '$ in Lemma 5.17 needs to be replaced by $\kappa _j'=\left( \frac{3}{5}\right) ^{j-1} \frac{6}{5}\kappa$. Indeed, Lemma 5.16 (with the constant $\frac{7}{2}\frac{\kappa }{\varepsilon }\le \frac{7}{12}\varepsilon =:r$) and Proposition 2.29 in [13] imply $\Vert (D\hat{g_0})_{\hat{\mathfrak {v}}(g^j)}\Vert \le \kappa \frac{r+\sqrt{1-r^2}}{1-r^2}\le \frac{60}{53}\kappa$. From there, applying the chain rule as in [25] implies

$$\begin{aligned} \mathrm {gr}(g^j)\le L^{j-1}\frac{60}{53}\kappa <\kappa ', \end{aligned}$$

(A.9)

with the Lipschitz constant $L=1.86*\kappa {\tilde{\varepsilon }}^{-2}<\frac{3}{5}$ defined in (A.4). Equipped with this new version of Lemma 5.17 and recalling (A.7), the relevant bound becomes

$$\begin{aligned} \log \sqrt{1+2\frac{\big (\kappa _j'\big )^2}{\varepsilon /3}} \le \frac{\big (\kappa _j'\big )^2}{\varepsilon /3} \le 7.2\left( \frac{3}{5}\right) ^{j}\frac{\kappa ^2}{\varepsilon }. \end{aligned}$$

(A.10)

Since the new upper bound in (5.20) is $\frac{21}{2}\frac{\kappa }{\varepsilon ^2}$ and we assume $n\ge 36$, we have

$$\begin{aligned} \frac{21}{2}n+7.2\sum _{j=1}^n \left( \frac{3}{5}\right) ^{j} \le \frac{21}{2}n+7.2\frac{5}{2} \le 11n \end{aligned}$$

(A.11)

and hence the upper bound on the main quantity

$$\begin{aligned} \frac{\rho (L_1,L_2,\ldots ,L_{n})}{\rho (L_1,L_2)\cdots \rho (L_{n-1}, L_{n})} \le \mathrm {e}^{11 n\kappa /{\varepsilon ^2}} \end{aligned}$$

(A.12)

as desired. This proves Theorem 2.6. $\square$

Proofs for the almost-commuting case

1.1 Proof of Proposition 5.2

Note that for each index $1\le k\le n$ it holds that $A_k^{1+\mathrm {i}t}=A_k A_k^{\mathrm {i}t}=A_k^{\mathrm {i}t} A_k$. We start by rewriting the difference as a telescopic sum

$$\begin{aligned}&A_1^{1+\mathrm {i}t}A_2^{1+\mathrm {i}t}\ldots A_n^{1+\mathrm {i}t}-A_1\ldots A_n A_1^{\mathrm {i}t}\ldots A_n^{\mathrm {i}t}\\&=\sum _{j=1}^{n-1} A_1^{1+\mathrm {i}t}\ldots A_{j-1}^{1+\mathrm {i}t} A_j [A_j^{\mathrm {i}t},A_{j+1}\ldots A_n]A_{j+1}^{\mathrm {i}t}\ldots A_n^{\mathrm {i}t} . \end{aligned}$$

We express the long commutator $[A_j^{\mathrm {i}t},A_{j+1}\ldots A_n]$ as a sum of individual commutators $[A_j^{\mathrm {i}t},A_k ]$ by iteratively applying the Leibniz rule for commutators,

$$\begin{aligned} {[}B,CD]=C[B,D]+[B,C]D . \end{aligned}$$

We find

$$\begin{aligned} {[}A_j^{\mathrm {i}t},A_{j+1}\ldots A_n] =\sum _{k=j+1}^n A_{j+1}\ldots A_{k-1} \left[ A_j^{\mathrm {i}t}, A_k \right] A_{k+1}\ldots A_{n} . \end{aligned}$$

The unitary invariance of the operator norm then gives for $\varepsilon _t:=\max _{j,k \in \mathbb {N}} \Vert [A_j^{\mathrm {i}t},A_k]\Vert$

$$\begin{aligned} X_{1,n}(t) \le X_{1,n}(0) + \varepsilon _t \sum _{j<k\le n} X_{1,j-1}(t) X_{j,j}(0) X_{j+1,k-1}(0) X_{k+1,n}(0) . \end{aligned}$$

Thus we obtain

$$\begin{aligned} \frac{X_{1,n}(t)}{X_{1,n}(0)}&\le 1 + \varepsilon _t \sum _{j<k \le n} \frac{X_{1,j-1}(t) X_{j,j}(0) X_{j+1,k-1}(0) X_{k+1,n}(0)}{X_{1,n}(0)} \\&\le 2\varepsilon _t \sum _{j<k \le n} \frac{X_{1,j-1}(t) X_{j,j}(0) X_{j+1,k-1}(0) X_{k+1,n}(0)}{X_{1,n}(0)} , \end{aligned}$$

where the final step is valid because we can assume without loss of generality that $1 \le \varepsilon _t \sum _{j<k \le n} \frac{X_{1,j-1}(t) X_{j,j}(0) X_{j+1,k-1}(0) X_{k+1,n}(0)}{X_{1,n}(0)}$, as otherwise the assertion of Proposition 5.2 holds trivially.

Assumption 5.1 then gives

$$\begin{aligned} \frac{X_{1,j-1}(t) X_{j,j}(0) X_{j+1,k-1}(0) X_{k+1,n}(0)}{X_{1,n}(0)}&\le \frac{\mathrm {e}^{(j-1) \gamma _1(t) + c} \mathrm {e}^{\gamma _1 + c} \mathrm {e}^{(k-j-1) \gamma _1 + c} \mathrm {e}^{(n-k) \gamma _1+ c}}{\mathrm {e}^{n \gamma _1 - c}} \\&\le \mathrm {e}^{(j-1) (\gamma _1(t) - \gamma _1)} \mathrm {e}^{5c} . \end{aligned}$$

We thus find

$$\begin{aligned} \frac{X_{1,n}(t)}{X_{1,n}(0)}&\le 1+2 \mathrm {e}^{5c} \varepsilon _t \sum _{j<k\le n} \mathrm {e}^{j (\gamma _1(t) - \gamma _1)} \\&\le 1+ 2\mathrm {e}^{5c} \varepsilon _t \sum _{j<k\le n} \mathrm {e}^{j |\gamma _1(t) - \gamma _1|} \\&\le 1+ 2\mathrm {e}^{5c} \varepsilon _t \sum _{j=0}^n (n-j) \mathrm {e}^{j |\gamma _1(t) - \gamma _1|} . \end{aligned}$$

We next bounding the sum with an integral, i.e.,

$$\begin{aligned} \sum _{j=0}^n (n-j) \mathrm {e}^{j |\gamma _1(t) - \gamma _1|} \le \int _{0}^n (n-x) \mathrm {e}^{x |\gamma _1(t) - \gamma _1|} \mathrm {d}x + \frac{ \mathrm {e}^{n |\gamma _1(t) - \gamma _1| }}{|\gamma _1(t) - \gamma _1|} , \end{aligned}$$

which is correct because the function $\phi _{n,c}: [0,n] \ni x \mapsto (n-x)\mathrm {e}^{x c}$ for $c>0$ is monotonically increasing in $[0,x^\star ]$ and monotonically decreasing in $[x^\star ,n]$ where $x^\star = n-1/c$. Furthermore we have $\phi _{n,c}(x^\star )=\frac{1}{c}\mathrm {e}^{nc -1} \le \frac{1}{c}\mathrm {e}^{nc}$. Hence we find

$$\begin{aligned} \frac{X_{1,n}(t)}{X_{1,n}(0)}&\le 1 +2\mathrm {e}^{5c} \varepsilon _t \left( \frac{\mathrm {e}^{n |\gamma _1(t) - \gamma _1|}}{|\gamma _1(t) - \gamma _1|^2} + \frac{ \mathrm {e}^{n |\gamma _1(t) - \gamma _1| }}{|\gamma _1(t) - \gamma _1|} \right) . \end{aligned}$$

Together with Assumption 5.1 this implies

$$\begin{aligned} |\gamma _1(t) - \gamma _1 |&\le \frac{1}{n} \log X_{1,n}(t) - \frac{1}{n} \log X_{1,n}(0) + \frac{2c}{n} \\&\le \frac{1}{n} \log \left( 1 +2\mathrm {e}^{5c} \varepsilon _t \left( \frac{\mathrm {e}^{n |\gamma _1(t) - \gamma _1|}}{|\gamma _1(t) - \gamma _1|^2} + \frac{ \mathrm {e}^{n |\gamma _1(t) - \gamma _1| }}{|\gamma _1(t) - \gamma _1|} \right) \right) + \frac{2c}{n} . \end{aligned}$$

As a result we obtain

$$\begin{aligned} \mathrm {e}^{n |\gamma _1(t) - \gamma _1| } \le \left( 1 +2\mathrm {e}^{5c} \varepsilon _t \left( \frac{\mathrm {e}^{n |\gamma _1(t) - \gamma _1|}}{|\gamma _1(t) - \gamma _1|^2} + \frac{ \mathrm {e}^{n |\gamma _1(t) - \gamma _1| }}{|\gamma _1(t) - \gamma _1|} \right) \right) \mathrm {e}^{2c} . \end{aligned}$$

Since this is true for all $n\in \mathbb {N}$ we can conclude that

$$\begin{aligned} |\gamma _1(t) - \gamma _1| \le \max \big \{\sqrt{ 4\mathrm {e}^{5c} \varepsilon _t}, 4\mathrm {e}^{5c} \varepsilon _t \big \} , \end{aligned}$$

which proves the assertion. $\square$

1.2 Proof of Corollary 5.3

The n-matrix Golden–Thompson inequality from Theorem 2.1 implies that

$$\begin{aligned} \int _{\mathbb {R}} f(t) \gamma _1(t) \mathrm {d}t&= \lim _{n \rightarrow \infty } \frac{1}{n} \int _{\mathbb {R}} f(t) \log \left\| \prod _{k=1}^n A_k^{1+\mathrm {i}t} \right\| \\&\ge \lim _{n \rightarrow \infty } \frac{1}{n} \log \left\| \exp \left( \sum _{k=1}^n \log A_k \right) \right\| \\&= \lim _{n \rightarrow \infty } \lambda _{\max } \left( \frac{1}{n} \sum _{k=1}^n \log A_k \right) \\&= \lambda _{\max }({\mathbb {E}}\log A_1) , \end{aligned}$$

where in the first step we swap the limit and the integral which is valid by the dominated convergence theorem. Proposition 5.2 implies

$$\begin{aligned} \gamma _1 \ge \int _{\mathbb {R}} f(t) \gamma _1(t) \mathrm {d}t - \int _{\mathbb {R}} f(t) \max \left\{ \sqrt{ 4 \mathrm {e}^{5c}\max _{j,k \in \mathbb {N}} \left\| \big [A_j^{\mathrm {i}t},A_k\big ] \right\| },4 \mathrm {e}^{5c}\max _{j,k \in \mathbb {N}} \left\| \big [A_j^{\mathrm {i}t},A_k\big ] \right\| \right\} \mathrm {d}t . \end{aligned}$$

We can use a well known bound [42, Lemma 3.8] to further simplify the bound⁴ by using

$$\begin{aligned} \int _{\mathbb {R}} f(t) \sqrt{\max _{j,k \in \mathbb {N}} \left\| \big [A_j^{\mathrm {i}t},A_k\big ] \right\| } \mathrm {d}t&\le \max _{j,k \in \mathbb {N}} \sqrt{\left\Vert [\log A_j, A_k]\right\Vert } \int _{\mathbb {R}} f(t) \sqrt{|t|} \mathrm {d}t \\&\le \max _{j,k \in \mathbb {N}} \sqrt{\left\Vert [\log A_j, A_k]\right\Vert } , \end{aligned}$$

which then proves the assertion. $\square$