The Absolutely Continuous Spectrum of Finitely Differentiable Quasi-Periodic Schrödinger Operators

Abstract

We prove that the quasi-periodic Schrödinger operator with a finitely differentiable potential has purely absolutely continuous spectrum for all phases if the frequency is Diophantine and the potential is sufficiently small in the corresponding $C^k$ topology. This extends the work of Eliasson [19] and Avila–Jitomirskaya [5] from the analytic topology to the finitely differentiable one which is much broader, revealing the interesting phenomenon that small oscillation of the potential leads to both zero Lyapunov exponent in the whole spectrum and purely absolutely continuous spectrum. Our result is based on a refined quantitative $C^{k,k_0}$ almost reducibility theorem which only requires a quite low initial regularity “$k>14\tau +2$” and much of the regularity “$k_0\le k-2\tau -2$” is conserved in the end, where $\tau$ is the Diophantine constant of the frequency.

Appendix

Proof of Theorem 3.1

For readers who are quite familiar with the analytic KAM scheme, this proof can be skipped since the structure is similar to that in [13]. But the estimates here are sharp compared with those in [13] (see Remark 3.3), so we prefer to provide the detailed proof for self-containedness.

Recall that $sl(2,{{\mathbb {R}}})$ is isomorphic to su(1, 1), which consists of matrices of the form

$$\begin{aligned} \begin{pmatrix} it &{}\quad v\\ \bar{v} &{}\quad -it \end{pmatrix} \end{aligned}$$

with $t\in {{\mathbb {R}}}$, $v\in {{\mathbb {C}}}$. The isomorphism between them is given by $A\rightarrow MAM^{-1}$, where

$$\begin{aligned} M=\frac{1}{1+i}\begin{pmatrix} 1 &{}\quad -i\\ 1 &{}\quad i \end{pmatrix} \end{aligned}$$

and a simple calculation yields

$$\begin{aligned} M\begin{pmatrix} x &{}\quad y+z\\ y-z &{}\quad -x \end{pmatrix}M^{-1}=\begin{pmatrix} iz &{}\quad x-iy\\ x+iy &{}\quad -iz \end{pmatrix}, \end{aligned}$$

where $x,y,z\in {{\mathbb {R}}}$. SU(1, 1) is the corresponding Lie group of su(1, 1). We will prove this theorem in SU(1, 1), which is isomorphic to $SL(2,{{\mathbb {R}}})$. $\square$

We distinguish two cases:

Non-resonant case For $0<|n |\leqslant N=\frac{2}{r-r'} |\ln \epsilon |$, we have

$$\begin{aligned} |2\rho - \langle n,\alpha \rangle |\geqslant \epsilon ^{\sigma }; \end{aligned}$$

(5.1)

by (3.2) with $D>\frac{2}{\sigma }$, we have

$$\begin{aligned} \left|\langle n,\alpha \rangle \right|\geqslant \frac{\kappa }{\left|n \right|^{\tau }}\geqslant \frac{\kappa }{\left|N \right|^{\tau }}\geqslant \epsilon ^{\frac{\sigma }{2}}\geqslant \epsilon ^{\sigma }. \end{aligned}$$

(5.2)

It is well known that (5.1) and (5.2) are the conditions which are used to overcome the small denominator problem in KAM theory.

Define

$$\begin{aligned} \Lambda _N= \left\{ f\in C^{\omega }_{r}({{\mathbb {T}}}^{d},su(1,1))\mid f(\theta )=\sum _{k\in {{\mathbb {Z}}}^{d},0<|k |<N}\hat{f}(k)e^{i\langle k,\theta \rangle }\right\} . \end{aligned}$$

(5.3)

Our goal is to solve the cohomological equation

$$\begin{aligned} Y(\theta +\alpha )A-AY(\theta )=A(-\mathcal {T}_Nf(\theta )+\hat{f}(0)), \end{aligned}$$

i.e.,

$$\begin{aligned} A^{-1}Y(\theta +\alpha )A-Y(\theta )=-\mathcal {T}_Nf(\theta )+\hat{f}(0). \end{aligned}$$

(5.4)

Here $\mathcal {T}_N$ is the truncation operator such that

$$\begin{aligned} (\mathcal {T}_Nf)(\theta )=\sum _{k\in {{\mathbb {Z}}}^{d},|k |<N}\hat{f}(k)e^{i\langle k,\theta \rangle }. \end{aligned}$$

Take the Fourier transform for (5.4) and compare the corresponding Fourier coefficients of the two sides. By (5.1) (apply it twice to solve the off-diagonal) along with (5.2) (apply it once to solve the diagonal), we obtain that if $Y\in \Lambda _N$, then

$$\begin{aligned} |Y(\theta )|_r \leqslant \epsilon ^{-3\sigma }|\mathcal {T}_Nf(\theta )-\hat{f}(0)|_r, \end{aligned}$$

which gives

$$\begin{aligned} |A^{-1}Y(\theta +\alpha )A-Y(\theta )|_r\geqslant \epsilon ^{3\sigma }|Y(\theta )|_r. \end{aligned}$$

(5.5)

Moreover, we have $A^{-1}Y(\theta +\alpha )A \in \Lambda _N$ by (5.3). For $\eta =\epsilon ^{3\sigma }$, we define $\mathcal {B}_r^{nre}(\epsilon ^{3\sigma })$ by (3.1), then we have $\Lambda _N \subset \mathcal {B}_r^{nre}(\epsilon ^{3\sigma })$.

Since $\epsilon ^{3\sigma }\geqslant 13\Vert A\Vert ^2\epsilon ^{\frac{1}{2}}$ (it holds by $\sigma$ being smaller than $\frac{1}{6}$ and $\tilde{D}$ depending on $\sigma$), by Lemma 3.1 we have $Y\in \mathcal {B}_r$ and $f^{re}\in \mathcal {B}_r^{re}(\epsilon ^{3\sigma })$ such that

$$\begin{aligned} e^{Y(\theta +\alpha )}(Ae^{f(\theta )})e^{-Y(\theta )}=Ae^{f^{re}(\theta )}, \end{aligned}$$

with $|Y |_r\leqslant \epsilon ^{\frac{1}{2}}$ and

$$\begin{aligned} |f^{re}|_r\leqslant 2\epsilon . \end{aligned}$$

(5.6)

By (5.3)

$$\begin{aligned} (\mathcal {T}_N{f^{re}})(\theta )=\hat{f}^{re}(0), \ \ \Vert \hat{f}^{re}(0)\Vert \leqslant 2\epsilon , \end{aligned}$$

and

$$\begin{aligned} |(\mathcal {R}_N{f^{re}})(\theta )|_{r'}&= |\sum _{|n |>N}\hat{f}^{re}(n)e^{i\langle n,\theta \rangle }|_{r'}\nonumber \\&\leqslant 2\epsilon e^{-N(r-r')}(N)^{d}\nonumber \\&\leqslant 2\epsilon \cdot \epsilon ^{2}\cdot \frac{1}{4}\epsilon ^{-\sigma }\nonumber \\&=\frac{1}{2}\epsilon ^{3-\sigma }. \end{aligned}$$

(5.7)

Moreover, we can compute that

$$\begin{aligned} e^{\hat{f}^{re}(0)+\mathcal {R}_N{f^{re}}(\theta )}=e^{\hat{f}^{re}(0)}(Id+e^{-\hat{f}^{re}(0)}\mathcal {O}(\mathcal {R}_N{f^{re}}))=e^{\hat{f}^{re}(0)}e^{f_+(\theta )}, \end{aligned}$$

by (5.7), we have

$$\begin{aligned} |f_+(\theta )|_{r'}\leqslant 2|\mathcal {R}_N{f^{re}(\theta )}|_{r'} \leqslant \epsilon ^{3-\sigma }. \end{aligned}$$

Finally, if we denote

$$\begin{aligned} A_+=Ae^{\hat{f}^{re}(0)}, \end{aligned}$$

then we have

$$\begin{aligned} \Vert A_+-A\Vert \leqslant \Vert A\Vert \Vert Id-e^{\hat{f}^{re}(0)} \Vert \leqslant 2\Vert A\Vert \epsilon . \end{aligned}$$

Resonant case In fact, we only need to consider the case in which A is elliptic with eigenvalues $\{e^{i\rho },e^{-i\rho }\}$ for $\rho \in {{\mathbb {R}}}\backslash \{0\}$ since if $\rho \in i{{\mathbb {R}}}$, then the non-resonant condition is always satisfied due to the Diophantine condition on $\alpha$ and then it actually belongs to the non-resonant case.

Claim 2

$n_*$ is the unique resonant site with

$$\begin{aligned} 0<|n_*|\leqslant N=\frac{2}{r-r'} |\ln \epsilon |. \end{aligned}$$

Proof

Indeed, if there exists $n_{*}^{'}\ne n_*$ satisfying $|2\rho - \langle n_{*}^{'},\alpha \rangle |< \epsilon ^{\sigma }$, then by the Diophantine condition of $\alpha$, we have

$$\begin{aligned} \frac{\kappa }{|n_{*}^{'}-n_*|^{\tau }}\leqslant |\langle n_{*}^{'}-n_*,\alpha \rangle |< 2\epsilon ^{\sigma }, \end{aligned}$$

which implies that $|n_{*}^{'} |>2^{-\frac{1}{\tau }}\kappa ^{\frac{1}{\tau }}\epsilon ^{-\frac{\sigma }{\tau }}-N> 2N^2.$ $\square$

Since we have

$$\begin{aligned} |2\rho - \langle n_*,\alpha \rangle |< \epsilon ^{\sigma }, \end{aligned}$$

(5.8)

the smallness condition on $\epsilon$ implies that

$$\begin{aligned} |\ln \epsilon |^\tau \epsilon ^\sigma \leqslant \frac{\kappa (r-r')^\tau }{2^{\tau +1}}. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\kappa }{|n_*|^\tau } \leqslant |\langle n_*,\alpha \rangle |\leqslant \epsilon ^\sigma +2|\rho |\leqslant \frac{\kappa }{2|n_*|^\tau }+2|\rho |, \end{aligned}$$

which implies that

$$\begin{aligned} |\rho |\geqslant \frac{\kappa }{4|n_*|^{\tau }}. \end{aligned}$$

Then by Lemma 8.1 of Hou–You [22], one can find $P\in SU(1,1)$ with

$$\begin{aligned} \Vert P \Vert \leqslant 2 \left( \frac{\Vert A \Vert }{|\rho |}\right) ^{\frac{1}{2}}\leqslant 4 \left( \frac{\Vert A\Vert }{\kappa }\right) ^{\frac{1}{2}}|n_*|^{\frac{\tau }{2}}, \end{aligned}$$

such that

$$\begin{aligned} PAP^{-1}=\begin{pmatrix} e^{i\rho } &{}\quad 0\\ 0 &{}\quad e^{-i\rho } \end{pmatrix}=A'. \end{aligned}$$

Denote $g=PfP^{-1}$, by (3.2) we have:

$$\begin{aligned} \Vert P \Vert\leqslant & {} 4\left( \frac{\Vert A\Vert }{\kappa }\right) ^{\frac{1}{2}}|N |^{\frac{\tau }{2}}\leqslant 4 \left( \frac{\Vert A\Vert }{\kappa }\right) ^{\frac{1}{2}}\left( \frac{2}{r-r'} |\ln \epsilon |\right) ^{\frac{\tau }{2}}, \end{aligned}$$

(5.9)

$$\begin{aligned} |g |_r\leqslant & {} \Vert P \Vert ^2|f|_r \leqslant \frac{2^{4+\tau }\Vert A\Vert |\ln \epsilon |^{\tau }}{\kappa (r-r')^{\tau }}\times \epsilon :=\epsilon '. \end{aligned}$$

(5.10)

Now we define

$$\begin{aligned}&\Lambda _1(\epsilon ^{\sigma })=\{n\in {{\mathbb {Z}}}^{d}: |\langle n,\alpha \rangle |\geqslant \epsilon ^{\sigma }\},\\&\Lambda _2(\epsilon ^{\sigma })=\{n\in {{\mathbb {Z}}}^{d}: |2\rho -\langle n,\alpha \rangle |\geqslant \epsilon ^{\sigma }\}. \end{aligned}$$

For $\eta =\epsilon ^{\sigma }$, we define the decomposition $\mathcal {B}_r=\mathcal {B}_r^{nre}(\epsilon ^{\sigma }) \bigoplus \mathcal {B}_r^{re}(\epsilon ^{\sigma })$ as in (3.1) with A substituted by $A'$. Recall that su(1, 1) consists of matrices of the form

$$\begin{aligned} \begin{pmatrix} it &{} v\\ \bar{v} &{} -it \end{pmatrix} \end{aligned}$$

with $t\in {{\mathbb {R}}}$, $v\in {{\mathbb {C}}}$. Direct computation shows that any $Y\in \mathcal {B}_r^{nre}(\epsilon ^{\sigma })$ takes the precise form:

$$\begin{aligned} \begin{aligned} Y(\theta )&=\sum _{n\in \Lambda _1(\epsilon ^{\sigma })}\begin{pmatrix} i\hat{t}(n) &{} 0\\ 0 &{} -i\hat{t}(n) \end{pmatrix} e^{i\langle n,\theta \rangle }\\&\quad +\sum _{n\in \Lambda _2(\epsilon ^{\sigma })}\begin{pmatrix} 0 &{} \hat{v}(n)e^{i\langle n,\theta \rangle }\\ \overline{\hat{v}(n)}e^{-i\langle n,\theta \rangle } &{} 0 \end{pmatrix}. \end{aligned} \end{aligned}$$

(5.11)

Since $\epsilon ^{\sigma }\geqslant 13\Vert A'\Vert ^2 (\epsilon ') ^{\frac{1}{2}}$, we can apply Lemma 3.1 to remove all the non-resonant terms of g, which means there exist $Y\in \mathcal {B}_r$ and $g^{re}\in \mathcal {B}_r^{re}(\eta )$ such that

$$\begin{aligned} e^{Y(\theta +\alpha )}(A'e^{g(\theta )})e^{-Y(\theta )}=A'e^{g^{re}(\theta )}, \end{aligned}$$

with $|Y |_r\leqslant (\epsilon ')^{\frac{1}{2}}$ and $|g^{re}|_r\leqslant 2\epsilon '$.

Combining with the Diophantine condition on the frequency $\alpha$ and the Claim 2, we have:

$$\begin{aligned}&\{{{\mathbb {Z}}}^{d}\backslash \Lambda _1(\epsilon ^{\sigma })\}\cap \{n\in {{\mathbb {Z}}}^{d}:|n |\leqslant \kappa ^{\frac{1}{\tau }}\epsilon ^{-\frac{\sigma }{\tau }}\}=\{0\},\\&\{{{\mathbb {Z}}}^{d}\backslash \Lambda _2(\epsilon ^{\sigma })\}\cap \{n\in {{\mathbb {Z}}}^{d}:|n |\leqslant 2^{-\frac{1}{\tau }}\kappa ^{\frac{1}{\tau }}\epsilon ^{-\frac{\sigma }{\tau }}-N\}=\{n_*\}. \end{aligned}$$

Let $N':=2^{-\frac{1}{\tau }}\kappa ^{\frac{1}{\tau }}\epsilon ^{-\frac{\sigma }{\tau }}-N$, then we can rewrite $g^{re}(\theta )$ as

$$\begin{aligned} g^{re}(\theta )&=g^{re}_0+g^{re}_1(\theta )+g^{re}_2(\theta )\\&=\begin{pmatrix} i\hat{t}(0) &{}\quad 0 \\ 0 &{}\quad -i\hat{t}(0) \end{pmatrix}+\begin{pmatrix} 0 &{}\quad \hat{v}(n_*)e^{i\langle n_*,\theta \rangle } \\ \overline{\hat{v}(n_*)}e^{-i\langle n_*,\theta \rangle } &{}\quad 0 \end{pmatrix}\\&\quad +\, \sum _{|n |>N'}\hat{g}^{re}(n)e^{i\langle n,\theta \rangle }. \end{aligned}$$

Define the $4\pi {{\mathbb {Z}}}^d$-periodic rotation $Q(\theta )$ as below:

$$\begin{aligned} Q(\theta )=\begin{pmatrix} e^{-\frac{\langle n_*,\theta \rangle }{2}i} &{}\quad 0\\ 0 &{}\quad e^{\frac{\langle n_*,\theta \rangle }{2}i} \end{pmatrix}. \end{aligned}$$

So we have

$$\begin{aligned} |Q(\theta )|_{r'}\leqslant e^{\frac{1}{2}Nr'}\leqslant \epsilon ^{\frac{-r'}{r-r'}}. \end{aligned}$$

(5.12)

One can also show that

$$\begin{aligned} Q(\theta +\alpha )(A'e^{g^{re}(\theta )})Q^{-1}(\theta )=\tilde{A}e^{\tilde{g}(\theta )}, \end{aligned}$$

where

$$\begin{aligned} \tilde{A}=Q(\theta +\alpha )A'Q^{-1}(\theta )=\begin{pmatrix} e^{i(\rho -\frac{\langle n_*,\alpha \rangle }{2})} &{}\quad 0\\ 0 &{}\quad e^{-i(\rho -\frac{\langle n_*,\alpha \rangle }{2})} \end{pmatrix} \end{aligned}$$

(5.13)

and

$$\begin{aligned} \tilde{g}(\theta )=Qg^{re}(\theta )Q^{-1}=Qg^{re}_0Q^{-1}+Qg^{re}_1(\theta )Q^{-1}+Qg^{re}_2(\theta )Q^{-1}. \end{aligned}$$

Moreover,

$$\begin{aligned} Qg^{re}_0Q^{-1}&=g^{re}_0 = \begin{pmatrix} i\hat{t}(0) &{}\quad 0 \\ 0 &{}\quad -i\hat{t}(0) \end{pmatrix} \in su(1,1), \end{aligned}$$

(5.14)

$$\begin{aligned} Qg^{re}_1(\theta )Q^{-1}&=\begin{pmatrix} 0 &{}\quad \hat{v}(n_*) \\ \overline{\hat{v}(n_*)} &{}\quad 0 \end{pmatrix} \in su(1,1). \end{aligned}$$

(5.15)

Now we return back from su(1, 1) to $sl(2,{{\mathbb {R}}})$. Denote

$$\begin{aligned} L&=M^{-1}(Qg^{re}_0Q^{-1}+Qg^{re}_1(\theta )Q^{-1})M, \end{aligned}$$

(5.16)

$$\begin{aligned} F&=M^{-1}Qg^{re}_2(\theta )Q^{-1}M, \end{aligned}$$

(5.17)

$$\begin{aligned} B&=M^{-1}(Q\circ e^Y \circ P) M, \end{aligned}$$

(5.18)

$$\begin{aligned} \tilde{A}^{'}&=M^{-1}\tilde{A}M, \end{aligned}$$

(5.19)

then we have:

$$\begin{aligned} B(\theta +\alpha )(Ae^{f(\theta )})B^{-1}(\theta )=\tilde{A}^{'}e^{L+F(\theta )}. \end{aligned}$$

(5.20)

By (5.9) and (5.12), we have the following estimates:

$$\begin{aligned} \Vert B\Vert _0\leqslant & {} |e^{Y}|_r \Vert P\Vert \leqslant 8 \left( \frac{\Vert A\Vert }{\kappa }\right) ^{\frac{1}{2}}\left( \frac{2}{r-r'} |\ln \epsilon |\right) ^{\frac{\tau }{2}}, \end{aligned}$$

(5.21)

$$\begin{aligned} |B|_{r'}\leqslant & {} 8\left( \frac{\Vert A\Vert }{\kappa }\right) ^{\frac{1}{2}}\left( \frac{2}{r-r'} |\ln \epsilon |\right) ^{\frac{\tau }{2}}\times \epsilon ^{\frac{-r'}{r-r'}},\end{aligned}$$

(5.22)

$$\begin{aligned} \Vert L \Vert\leqslant & {} \Vert Qg^{re}_0Q^{-1}\Vert + \Vert Qg^{re}_1(\theta )Q^{-1}\Vert \leqslant \epsilon '+\epsilon ' e^{-|n_{*}|r},\end{aligned}$$

(5.23)

$$\begin{aligned} |F|_{r'}\leqslant & {} |Qg^{re}_2(\theta )Q^{-1}|_{r'} \leqslant \frac{2^{4+\tau }\Vert A\Vert |\ln \epsilon |^{\tau }}{\kappa (r-r')^{\tau }}\epsilon e^{-N'(r-r')}(N')^de^{Nr'}. \end{aligned}$$

(5.24)

By (5.23) and (5.24), direct computation shows that

$$\begin{aligned} e^{L+F(\theta )}=e^L+\mathcal {O}(F(\theta ))=e^L(Id+e^{-L}\mathcal {O}(F(\theta )))=e^L e^{f_+{(\theta )}}. \end{aligned}$$

(5.25)

It immediately implies that

$$\begin{aligned} |f_+{(\theta )}|_{r'}\leqslant 2 |F(\theta )|_{r'}\leqslant \frac{2^{5+\tau }\Vert A\Vert |\ln \epsilon |^{\tau }}{\kappa (r-r')^{\tau }}\epsilon e^{-N'(r-r')}(N')^de^{Nr'}\ll \epsilon ^{100}. \end{aligned}$$

Thus, we can rewrite (5.20) as

$$\begin{aligned} B(\theta +\alpha )(Ae^{f(\theta )})B^{-1}(\theta )=A_{+}e^{f_+(\theta )}, \end{aligned}$$

with

$$\begin{aligned} A_+=\tilde{A}^{'}e^L=e^{A''}, \ \ A''\in sl(2,{{\mathbb {R}}}). \end{aligned}$$

(5.26)

Now recall that Baker–Campbell–Hausdorff formula [28] says that

$$\begin{aligned} \ln (e^X e^Y)=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}([X,[X,Y]+[Y,[Y,X]])+\cdots ,\nonumber \\ \end{aligned}$$

(5.27)

where $[X,Y]=XY-YX$ denotes the Lie Bracket and $\cdots$ denotes the sum of higher order terms. Using this formula and by a simple calculation, (5.26) gives

$$\begin{aligned} MA''M^{-1}=\begin{pmatrix} it &{} v\\ \bar{v} &{} -it \end{pmatrix} \end{aligned}$$

where

$$\begin{aligned} t=\rho -\frac{\langle n_*,\alpha \rangle }{2}+\hat{t}(0)+ \mathrm{higher\, order\, terms} \end{aligned}$$

and

$$\begin{aligned} v=\hat{v}(n_*)+ \mathrm{higher\, order\, terms}. \end{aligned}$$

By (5.8) and (5.23), we obtain $|t|\leqslant \epsilon ^{\sigma }$ and

$$\begin{aligned} |v |\leqslant \frac{2^{4+\tau }\Vert A\Vert |\ln \epsilon |^{\tau }}{\kappa (r-r')^{\tau }}\epsilon e^{-|n_{*}|r}. \end{aligned}$$

Finally, the following estimate is straightforward:

$$\begin{aligned} \Vert A'' \Vert \leqslant 2(|\rho -\frac{\langle n_*,\alpha \rangle }{2}|+\Vert Qg^{re}_0Q^{-1}\Vert +\Vert Qg^{re}_1(\theta )Q^{-1}\Vert )\leqslant 2\epsilon ^{\sigma }. \end{aligned}$$

(5.28)

This finishes the proof of Theorem 3.1. $\square$