Kotani–Last problem and Hardy spaces on surfaces of Widom type

Abstract

This is a small theory of non almost periodic ergodic families of Jacobi matrices with purely (however) absolutely continuous spectrum. The reason why this effect may happen is that under our “axioms” we found an analytic condition on the resolvent set that is responsible for (exactly equivalent to) this effect.

Appendix: the resolvent representation for the transfer matrix and the Christoffel-Darboux identity

For ${\mathcal {K}}=\hat{H}^2(\alpha )\ominus \check{H}^2(\alpha )$ we define $T:{\mathcal {K}}\oplus \{\check{e}_0\}\rightarrow \{\hat{e}_{-1}\}\oplus {\mathcal {K}}$ by

$$\begin{aligned} T=P_{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}\,{\mathfrak {z}}\cdot \,| {\mathcal {K}}\oplus \{\check{e}_0\}. \end{aligned}$$

(9.1)

We define $\hat{\mathcal {E}}$ and $\check{\mathcal {E}}$ acting in ${\mathbb {C}}^2$ by

$$\begin{aligned} \check{\mathcal {E}}\begin{bmatrix} c_{-1}\\ c_0 \end{bmatrix} =c_{-1}\check{p}_0\check{e}_{-1}+c_0\check{e}_0 ,\quad \hat{\mathcal {E}}\begin{bmatrix} c_{-1}\\ c_0 \end{bmatrix} =c_{-1}\hat{p}_0\hat{e}_{-1}+c_0\hat{e}_0. \end{aligned}$$

(9.2)

Proposition 9.1

For $T$, $\hat{\mathcal {E}}$, $\check{\mathcal {E}}$ defined in (9.1), (9.2), the transfer matrix (5.1) is of the form

$$\begin{aligned} {\mathfrak {A}}(z)= \begin{bmatrix} 0&\quad 0\\ 0&\quad \langle \hat{e}_0,\check{e}_0\rangle \end{bmatrix} + j \check{\mathcal {E}}^*P_{{\mathcal {K}}\oplus \{\check{e}_0\}}(T-z P_{\mathcal {K}})^{-1}P_{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}\hat{\mathcal {E}}. \end{aligned}$$

(9.3)

Proof

With necessity the vector $(T-z_0 P_{\mathcal {K}})^{-1}\hat{e}_{-1}$ is of the form

$$\begin{aligned} x\oplus c \check{e}_0=\frac{\hat{e}_{-1}-A \check{e}_{-1}-B\check{e}_{0}}{{\mathfrak {z}}-z_0}\oplus c \check{e}_0. \end{aligned}$$

Since $x\in H^2$ we have

$$\begin{aligned} \hat{e}_{-1}(\zeta _0)= \begin{bmatrix} \check{e}_{-1}(\zeta _0)&\check{e}_{0}(\zeta _0) \end{bmatrix} \begin{bmatrix} A\\ B \end{bmatrix} \end{aligned}$$

(9.4)

Since $x\in \hat{H}^2(\alpha )\ominus \check{H}^2(\alpha )$ the dual vector $\tilde{x}(\zeta )$, such that $b(\zeta )\tilde{x}(\zeta )=\Delta (\zeta ) x(\bar{\zeta })$ for $\zeta \in {\mathbb {T}}$, belongs to $\hat{H}^2(\alpha ^{-1}\mu ^{-1}\nu )\ominus \check{H}^2(\alpha ^{-1}\mu ^{-1}\nu )$. Thus, in notations (2.20) we have

$$\begin{aligned} \widetilde{\hat{e}}_{0} (\zeta _0)= \begin{bmatrix} \widetilde{\check{e}}_{0}(\zeta _0)&\widetilde{\check{e}}_{-1}(\zeta _0) \end{bmatrix} \begin{bmatrix} A\\ B \end{bmatrix} \end{aligned}$$

(9.5)

In this case, indeed,

$$\begin{aligned} (T-z_0P_{\mathcal {K}}) x&= P_{\mathcal {K}}({\mathfrak {z}}-z_0)x+P_{\{\hat{e}_{-1}\}}{\mathfrak {z}}x\\&= P_{\mathcal {K}}(\hat{e}_{-1}-A \check{e}_{-1}-B \check{e}_{0})+\hat{e}_{-1}\langle x, \hat{e}_0\rangle \hat{p}_0\\&= -A P_{\mathcal {K}}\check{e}_{-1}+\hat{p}_0\frac{(b\hat{e}_{-1})(0)-A(b\check{e}_{-1})(0)}{(b{\mathfrak {z}})(0)\hat{e}_0(0)}\hat{e}_{-1} \end{aligned}$$

and $T\check{e}_0=P_{{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}} \check{p}_0 \check{e}_{-1}=\check{p}_0 P_{{\mathcal {K}}} \check{e}_{-1}+ \check{p}_0\frac{(b\check{e}_{-1})(0)}{(b\hat{e}_{-1})(0)}\hat{e}_{-1}$. That is, for $c=A/\check{p}_0$ we have

$$\begin{aligned} (T\!-\!zP_{\mathcal {K}})(x+c \check{e}_0)\!=\!\left( A\frac{(b\check{e}_{-1})(0)}{(b\hat{e}_{-1})(0)} \!+\!\frac{(b\hat{e}_{-1})(0)-A(b\check{e}_{-1})(0)}{(b \hat{e}_{-1})(0)}\right) \hat{e}_{-1}\!= \!\hat{e}_{-1} \end{aligned}$$

Thus

$$\begin{aligned} (T-zP_{\mathcal {K}})^{-1} \hat{e}_{-1}= \frac{\hat{e}_{-1}-A \check{e}_{-1}-B\check{e}_{0}}{{\mathfrak {z}}-z_0}\oplus \frac{A}{\check{p}_0} \check{e}_0, \end{aligned}$$

(9.6)

Similarly we get

$$\begin{aligned} (T-zP_{\mathcal {K}})^{-1} P_{\mathcal {K}}\hat{e}_{0}= \frac{\hat{e}_{0}-C \check{e}_{-1}-D\check{e}_{0}}{{\mathfrak {z}}-z_0}\oplus \frac{C}{\check{p}_0} \check{e}_0, \end{aligned}$$

(9.7)

where

$$\begin{aligned} \begin{bmatrix} \hat{e}_{0}(\zeta _0)\\ \widetilde{\hat{e}}_{-1} (\zeta _0) \end{bmatrix} = \begin{bmatrix} \check{e}_{-1}(\zeta _0)&\quad \check{e}_{0}(\zeta _0)\\ \widetilde{\check{e}}_{0}(\zeta _0)&\quad \widetilde{\check{e}}_{-1}(\zeta _0) \end{bmatrix} \begin{bmatrix} C\\ D \end{bmatrix}. \end{aligned}$$

(9.8)

We see that all four entries $A,B,C,D$ given in (9.4), (9.5), (9.8) actually form the transfer matrix ${\mathfrak {A}}$. Using (9.6), (9.7) and the reproducing properties of the vectors $\check{e}_{-1}, \check{e}_0$ we get (9.3). $\square$

Lemma 9.2

The following Christoffel-Darboux type identity holds

$$\begin{aligned} \frac{{\mathfrak {A}}^*(z)j{\mathfrak {A}}(z)-j}{z-\bar{z}}=\hat{\mathcal {E}}^*P_{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}(T^*-\bar{z} P_{\mathcal {K}})^{-1} P_{\mathcal {K}}(T-z P_{\mathcal {K}})^{-1}P_{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}\hat{\mathcal {E}}.\nonumber \\ \end{aligned}$$

(9.9)

Proof

We start with an easy identity

$$\begin{aligned} P_{\mathcal {K}}T-T^*P_{\mathcal {K}}=\check{\mathcal {E}}\begin{bmatrix} 0&\quad \! 1\\ -1&\quad \!0 \end{bmatrix} \check{\mathcal {E}}^*. \end{aligned}$$

(9.10)

Indeed, for $f=x+ c\check{e}_{0}$ we have $\langle P_{\mathcal {K}}T (x+c\check{e}_{0}),x+c\check{e}_{0}\rangle = \langle {\mathfrak {z}}(x+c\check{e}_{0}),x\rangle = \langle {\mathfrak {z}}x,x\rangle +c \langle \check{p}_0 \check{e}_{-1},x\rangle =\langle {\mathfrak {z}}x,x\rangle +\check{p}_0 \langle f,\check{e}_{0}\rangle \langle \check{e}_{-1},f\rangle$. We can rewrite (9.10) in a more sophisticated form

$$\begin{aligned} (z-\bar{z}) P_{{\mathcal {K}}}=\check{\mathcal {E}}\begin{bmatrix} 0&\quad \! 1\\ -1&\quad \! 0 \end{bmatrix} \check{\mathcal {E}}^* - P_{\mathcal {K}}(T-zP_k)+(T^*-\bar{z} P_k)P_{\mathcal {K}}. \end{aligned}$$

(9.11)

Now we multiply (9.11) by $P_{{\mathcal {K}}\oplus \{\check{e}_0\}}(T-z P_{\mathcal {K}})^{-1}P_{\{\hat{e}_{-1}\}\oplus {\mathcal {K}}}\hat{\mathcal {E}}$ from the right and by the conjugated expression from the left. We use (9.6), (9.7) and the reproducing properties of $\hat{e}_{-1}, \hat{e}_0$ to obtain (9.9). $\square$