Generic continuous spectrum for multi-dimensional quasi-periodic - - PowerPoint PPT Presentation

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Generic continuous spectrum for multi-dimensional quasi-periodic Schr¨

• dinger operators with rough

potentials

Joint work with Rui Han

Fan Yang University of California, Irvine

35th Annual Western States Mathematical Physics Meeting

February 13, 2017

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Main object

Multi-dimensional discrete quasi-periodic Schr¨

• dinger operators

(Hα,xu)n =

∑

|m−n|=1

um + f (T nx)un, f : Td → R is the potential function, T is the shift on Td with frequency vector α = (α1, α2, ..., αd) ∈ Td, x = (x1, x2, ..., xd) ∈ Td is the phase vector, m, n ∈ Zd and |m| = ∑d

i=1 mi.

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Main object

Multi-dimensional discrete quasi-periodic Schr¨

• dinger operators

(Hα,xu)n =

∑

|m−n|=1

um + f (T nx)un, f : Td → R is the potential function, T is the shift on Td with frequency vector α = (α1, α2, ..., αd) ∈ Td, x = (x1, x2, ..., xd) ∈ Td is the phase vector, m, n ∈ Zd and |m| = ∑d

i=1 mi.

d = 1 ⇒ many (sharp) results

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Main object

Multi-dimensional discrete quasi-periodic Schr¨

• dinger operators

(Hα,xu)n =

∑

|m−n|=1

um + f (T nx)un, f : Td → R is the potential function, T is the shift on Td with frequency vector α = (α1, α2, ..., αd) ∈ Td, x = (x1, x2, ..., xd) ∈ Td is the phase vector, m, n ∈ Zd and |m| = ∑d

i=1 mi.

d = 1 ⇒ many (sharp) results d > 1 ⇒ very few results

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Excluding point spectrum for one-dimensional operators

(Hα,xu)n = un+1 + un−1 + f (x + nα)un. Repetitions in potential leads to empty point spectrum: Developed by Gordon in 1976 H¨

• lder continuous potential: Avila-You-Zhou (sharp results for Almost

Mathieu operator) Generic continuous potential: Boshernitzan-Damanik (one-dimensional

• perator with multi-frequency)

Sturmian Hamiltonian: many authors Singular potential: Simon (Maryland model), Jitomirskaya-Liu (sharp results for Maryland model), Jitomirskaya-Y (meromorphic potential) Measurable potential: Gordon (one-dimensional operator with multi-frequency)

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Excluding point spectrum for multi-dimensional operators

General continuous potential: Simon’s Wonderland theorem implies empty point spectrum for generic frequencies. For a long time, there is no arithmetically quantitative result.

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Excluding point spectrum for multi-dimensional operators

General continuous potential: Simon’s Wonderland theorem implies empty point spectrum for generic frequencies. For a long time, there is no arithmetically quantitative result. H¨

• lder continuous potential:

Damanik: one frequency is sufficiently Liouville, the other frequencies are rational, for any x, Hα,x has no point spectrum Breakthrough by Gordon: when all the frequencies are sufficiently Liouville, for any x, Hα,x has no ℓ1(Zd) solution. Gordon-Nemirovski: when all the frequencies are sufficiently Liouville, for any x, Hα,x has no point spectrum.

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Theorem (Boshernitzan-Damanik, 2007)

Let (Hα,xu)n = un+1 + un−1 + f (x + nα)un, be a one-dimensional quasi-periodic operator with multi-dimensional frequency α ∈ Td. Then for any frequency, for generic continuous potential f , Hα,x has no point spectrum for a.e. x ∈ Td.

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Our main result for continuous potentials

Theorem

Assume there exists an infinite sequence Q = {τ (n) = (τ (n)

1 , τ (n) 2 , ..., τ (n) d )} such

that lim

n→∞

τ (n)

1

· · · τ (n)

d

τ (n)

i

∥τ (n)

i

αi∥T = 0 (1) for any i = 1, 2, ..., d. Then for generic continuous potential f , the multi-dimensional operator Hα,x has no point spectrum for a.e. x ∈ Td .

Remark

d = 2, (1) holds for a.e. (α1, α2). Actually (1) is equivalent to (α1, α2) are not both of bounded type. d ≥ 3, using Borel-Cantelli’s Lemma, (1) holds on a zero measure set of α.

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Our main result for measurable potentials

Theorem

Let the potential f be measurable, then for generic α ∈ Td, Hα,x has no point spectrum for a.e. x. Remark: This theorem extends a result of Gordon (2016) for d = 1 to the multi-dimensional setting.

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Rokhlin-Halmos Lemma

Let T be an invertible measure-preserving transformation on a measure space (X, M, µ) with µ(X) = 1. We assume T is aperiodic. Then for any ϵ > 0 and n ∈ N, there exists an E ∈ M such that: the sets E, TE, ..., T n−1E are pairwise disjoint; µ(∪n−1

j=0 T jE) > 1 − ϵ.

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Sketch of proof for continuous potentials

Notations: for 1 > δ > 0, let Γn = (2d + δ) ∏d

j=1 τ (n) j

. Denote x ⋆ y = (x1y1, x2y2, ..., xdyd), x, y ∈ Rd. Proof: By (1), for any k ∈ N, take nk so that Γnk τ (nk)

i

∥τ (nk)

i

αi∥ < 1 k2 for any i = 1, 2, ..., d .

Step 1

By Rokhlin-Halmos Lemma, there exists Onk so that {Onk + j ⋆ α}∥j∥∞≤(k+1)Γnk are pairwise disjoint, and |Onk| > 1 − 2−k−1 (2(k + 1)Γnk + 1)d .

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Step 2

Cut Onk into {Knk,l}l=1,2,...,snk , so that

snk

∑

l=1

|Knk,l| > 1 − 2−k (2(k + 1)Γnk + 1)d , and diam(Knk,l) < 1

k .

Let Ink = {m ∈ Zd : 0 ≤ mi ≤ τ (nk)

i

− 1 for any i = 1, 2, ..., d} and define Unk,l,m = ∪

|ji|≤kΓnk /τ

(nk ) i

Knk,l + m ⋆ α + j ⋆ τ (nk) ⋆ α.

Step 3

Unk,l,m satisfies diam(Unk,l,m) < 2

√ d+1 k

. for 1 ≤ l ≤ snk, m ∈ Ink, Unk,l,m are pairwise disjoint.

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Let Fnk = {f ∈ C(Td) : f is constant on each Unk,l,m}, and define Gnk = {g ∈ C(Td) : there exists f ∈ Fnk s.t. ∥g − f ∥0 < 1 2k− 2d+γ

2d+δ Γnk }.

Step 4

G = ∩

t≥1

∪

k≥t Gnk is a generic set of C(Td).

If f ∈ G, then there exists subsequence {˜ nk} of {nk} such that f ∈ G˜

• nk. For any

k > 4d − 1 + 2∥f ∥0, let T˜

nk =

∪

1≤l≤s˜

nk , ∥j∥∞≤(k−1)Γ˜ nk

(K˜

nk,l + j ⋆ α).

Step 5

For x ∈ T˜

nk, we have:

max

j∈Zd, |ji|≤Γ˜

nk /τ (˜ nk ) i

|f (x + j ⋆ τ (˜

nk) ⋆ α) − f (x)| < (4d − 1 + 2∥f ∥0)− 2d+γ

2d+δ Γ˜ nk .

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Step 6

T˜

nk satisfies:

|T˜

nk| ≥

((2k − 2)Γ˜

nk + 1

(2k + 2)Γ˜

nk + 1

)d

(1 − 2−k), thus

∑

k

|T c

˜ nk| =

∑

k

(1 − |T˜

nk|) < ∞.

Therefore, a.e. x ∈ Td belongs to infinitely many T˜

• nk. For such full

measure set of x, combining Step 5 and the following Theorem (Gordon-Nemirovski), Hα,x has no point spectrum. □

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Theorem (Gordon-Nemirovski, 2016)

LetV be a complex valued function on Zd, suppose there exists γ > δ > 0 and a infinite set P = {τ = (τ1, τ2, ..., τd)} ⊂ Nd which satisfy: lim

P∋τ→∞ τi = ∞,

i = 1, 2, ..., d. s.t. there exists τ-period function Vτ(·) satisfy: for some λ0 > 0, ρτ < (2d − 1 + Mτ + λ0)−(2d+γ)τ1τ2···τd, in which, ρτ = max

∥j∥∞≤(2d+δ)τ1τ2···τd |Vτ(j) − V (j)|,

Mτ = max

∥j∥∞≤(2d+δ)τ1τ2···τd |V (j)|.

Then for any |λ| ≤ λ0, Hu = λu has no non-trivial ℓ2(Zd) solution.

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Sketch of proof for measurable potentials

Notations: For x ∈ Rd, let ∥x∥Td = dist(x, Zd). For measurable f , let ∥f ∥∞ := ∥f ∥L∞. Set E(M) = {x ∈ Td : |f (x)| > M}. For τ ∈ Nd, let Mτ := inf{M : E(M) ≤ (τ1τ2 · · · τd)−d}. Also set F(y, ϵ) = {x ∈ Td : |f (x + y) − f (x)| ≥ ϵ}. The key to tackle measurable potential:

Proposition

For any ϵ, η > 0, there is κ(ϵ, η) > 0 such that if ∥y∥Td < κ(ϵ, η) then we have |F(y, ϵ)| < η.

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Key lemma

Lemma

Suppose there exists an infinite sequence τ ∈ Q with limQ∋τ→∞ τi = ∞ for i = 1, 2, ..., d, such that the frequencies satisfy ∥(τ1α1, τ2α2, ..., τdαd)∥Td < κ(M−(2d+γ)τ1τ2···τd

τ

, (τ1τ2 · · · τd)−2) for some γ > 0 and any τ ∈ Q, then Hα,x has no point spectrum for a.e. x ∈ Td.

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Proof of Lemma

We will only sketch the proof when ∥f ∥∞ = ∞. Let γ

4 > δ > 0. For any τ ∈ Q, let

X τ

j = F(j ⋆ τ ⋆ α, (|j1| + · · · + |jd|)M−(2d+γ)τ1···τd τ

), Y τ

j = {x ∈ Td : |f (x + j ⋆ τ ⋆ α)| > Mτ} = E(Mτ) − j ⋆ τ ⋆ α.

Bad phases

Let the following Bτ be the set of bad phases: Bτ = (∪Iτ X τ

j ) ∪ (∪Iτ Y τ j ),

where Iτ = {j ∈ Zd : |ji| ≤ (2d + δ)τ1 · · · τd/τi for i = 1, 2, ..., d}.

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We have |Y τ

j | ≤ 1 (τ1τ2···τd)d . By triangle inequalities, |X τ j | ≤ |j1|+···+|jd| τ 2

1 ···τ 2 d

.

Measure of bad phases

The measure of Bτ satisfies: |Bτ| ≤

∑

j∈Iτ

|j1| + · · · |jd| (τ1 · · · τd)2 + |Iτ| (τ1 · · · τd)d → 0 as Q ∋ τ → ∞. By Borel-Cantelli lemma, for a.e. x ∈ Td (good phase), there is an infinite sequence τ ∈ Px such that x / ∈ Bτ for any τ ∈ Px. Taking a good phase x ∈ Td and τ ∈ Px.

τ-periodic potential

Define a τ-periodic potential by setting Vτ(x + n ⋆ α) = f (x + m ⋆ α), where nj ≡ mj (mod τj) with 0 ≤ mj ≤ τj − 1.

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∥f ∥∞ = ∞ implies limPx∋τ→∞ Mτ = ∞, then for any λ0 > 0, for τ ∈ Px large, we have

Repetition of potential

We have max

∥n∥∞≤(2d+δ)τ1···τd

|Vτ(x+n⋆α)−f (x+n⋆α)| < (2d−1+Mτ+λ0)−(2d+ γ

4 )τ1···τd.

where Mτ ≥ max∥n∥∞≤(2d+δ)τ1···τd |f (x + n ⋆ α)|. The result then follows from Theorem (Gordon-Nemirovski). □

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Thank you!

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