violation of bulk-edge correspondence in a hydrodynamic model...violation of bulk-edge...

Violation of bulk-edge correspondence in ahydrodynamic model

Gian Michele GrafETH Zurich

PhD School: September 16-20, 2019@Universita degli Studi Roma Tre

based on joint work with Hansueli Jud, Clement Tauber

Outline

A hydrodynamic model

Topology by compactification

The Hatsugai relation

Violation

What goes wrong?

The Great Wave off Kanagawa

(by K. Hokusai, ∼1831)




Violation

What goes wrong?

The model (take it or leave it)

I The Earth is rotating.

Sure

I The Earth is flat. Well, locally yes

I The Sea covers the Earth. Don’t despair. We’ll sight land

I The Sea is shallow. Compared to wavelength

Incompressible, shallow water equations (preliminary):

∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥

I fields (dynamic): velocity v = v(x , y), height above averageη = η(x , y)

I parameters: gravity g , average depth h, angular velocity f /2


I The Earth is rotating. Sure





∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥





I The Earth is flat.

Well, locally yes




∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥









∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥






I The Sea covers the Earth.

Don’t despair. We’ll sight land



∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥









∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥







I The Sea is shallow.

Compared to wavelength


∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥









∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥



A quick derivationStarting point: Euler equations for an incompressible fluid in dimension 3.

~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v

Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction

(a) η � h, ~v · ~∇ � ∂/∂t. Hence D/Dt ≈ ∂/∂t(b) ~v = ~v(x , y , z) =: (v , v);

~g = (0,−g), ~f = (0, f ), hence ~f ∧ ~v = (f v⊥, ∗); to leading order

ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η

(c) Replace v by its average over 0 ≤ z ≤ h: ; v = v(x , y)

∂η

∂t= −h∇ · v , ρ

∂v

∂t= −ρf v⊥ − ρg∇η


~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v

I fields: velocity ~v = ~v(x , y , z) =: (v , v), pressure p = p(x , y , z)I parameters: density ρ; gravity in z-direction




ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η(c) Replace v by its average over 0 ≤ z ≤ h: ; v = v(x , y)

∂η

∂t= −h∇ · v , ρ

∂v



~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v


(a) η � h, ~v · ~∇ � ∂/∂t. Hence D/Dt ≈ ∂/∂t

(b) ~v = ~v(x , y , z) =: (v , v);~g = (0,−g), ~f = (0, f ), hence ~f ∧ ~v = (f v⊥, ∗); to leading order

ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η


∂η

∂t= −h∇ · v , ρ

∂v



~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v


(a) η � h, ~v · ~∇ � ∂/∂t. Hence D/Dt ≈ ∂/∂t(b) ~v = ~v(x , y , z) =: (v , v)

;~g = (0,−g), ~f = (0, f ), hence ~f ∧ ~v = (f v⊥, ∗); to leading order

ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η


∂η

∂t= −h∇ · v , ρ

∂v



~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v



~g = (0,−g), ~f = (0, f ), hence ~f ∧ ~v = (f v⊥, ∗)

; to leading order

ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η


∂η

∂t= −h∇ · v , ρ

∂v



~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v




ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η

(c) Replace v by its average over 0 ≤ z ≤ h:

; v = v(x , y)

∂η

∂t= −h∇ · v , ρ

∂v



~∇ · ~v = 0 , ρD~v

Dt= ρ~g − ρ~f ∧ ~v − ~∇p

p = 0 at z = η(x , y)

Dη

Dt= v




ρg + ∂p/∂z = 0 , p = ρg(η − z) , ∇p = −ρg∇η


∂η

∂t= −h∇ · v , ρ

∂v





Violation

What goes wrong?

A convenient extension

Momentum equations (in dimension 2):

ρDv

Dt= b +∇ · σ

body forces ~b, stress tensor σ.

To σij = −pδij (Euler) add either (vi ,j := ∂vi/∂xj):

I even viscosity (Navier-Stokes)

σ = −η(

2v1,1 v1,2+v2,1v1,2+v2,1 2v2,2

), ∇ · σ = η∆v

I odd viscosity (Avron)

σ = −η(−(v1,2+v2,1) v1,1−v2,2v1,1−v2,2 v1,2+v2,1

), ∇ · σ = −η∆v⊥

The model (final form)Equations of motion

∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥−ν∆v⊥

with ν = η/ρ.

∂η

∂t= −∇ · v

∂v

∂t= −∇η − (f + ν∆)v⊥

In Hamiltonian form (v =: (u, v), px := −i∂/∂x)

i∂ψ

∂t= Hψ

ψ =

ηuv

, H =

0 px pypx 0 i(f − νp2)

py −i(f − νp2) 0

= H∗

The model (final form)Equations of motion

∂η

∂t= −h∇ · v

∂v

∂t= −g∇η − f v⊥−ν∆v⊥

with ν = η/ρ. After rescaling (gh = 1)

∂η

∂t= −∇ · v

∂v

∂t= −∇η − (f + ν∆)v⊥


i∂ψ

∂t= Hψ

ψ =

ηuv

, H =


py −i(f − νp2) 0

= H∗

The model (final form)

Equations of motion

∂η

∂t= −∇ · v

∂v

∂t= −∇η − (f + ν∆)v⊥


i∂ψ

∂t= Hψ

ψ =

ηuv

, H =


py −i(f − νp2) 0

= H∗

The model as a spin 1 bundle

By translation invariance (momentum k ∈ R2), H reduces to fibers

H =

( 0 kx kykx 0 i(f−νk2)

ky −i(f−νk2) 0

)

= ~d · ~S , ~d(k) = (kx , ky , f − νk2)

where ~S is an irreducible spin 1 representation

S1 =(

0 1 01 0 00 0 0

), S2 =

(0 0 10 0 01 0 0

), S3 =

(0 0 00 0 i0 −i 0

)Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2



H =


ky −i(f−νk2) 0

)= ~d · ~S , ~d(k) = (kx , ky , f − νk2)


S1 =(

0 1 01 0 00 0 0

), S2 =

(0 0 10 0 01 0 0

), S3 =

(0 0 00 0 i0 −i 0

)

Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2



H =


ky −i(f−νk2) 0

)= ~d · ~S , ~d(k) = (kx , ky , f − νk2)


S1 =(

0 1 01 0 00 0 0

), S2 =

(0 0 10 0 01 0 0

), S3 =

(0 0 00 0 i0 −i 0

)Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2


H = ~d · ~S , ~d(k) = (kx , ky , f − νk2)

Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2

Eigenvectors (only ω+):Same as for ~e · ~S with ~e = ~d/|~d |, denoted

|~e, j = 1〉 , k 7→ ~e(k)

Remarks.I The compactification of R2 is S2.I ~e(k) 7→ (0, 0,− sgn ν) as k →∞ by ~d(k) = (kx , ky , f − νk2)I ~e : R2 → S2 extends to a continuous map S2 → S2

Lemma. Let f ν > 0. The line bundle P(1)+ = |~e, 1〉〈~e, 1| defined by ~e(k)

on S2 has Chern numberch(P

(1)+ ) = 2

Proof. If ~S were a spin-12 representation, then

ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1


H = ~d · ~S , ~d(k) = (kx , ky , f − νk2)

Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2

Left: ω+ as a function of kRight: projected along ky as a function of kx

Remark: Gap is f > 0


|~e, j = 1〉 , k 7→ ~e(k)




(1)+ ) = 2


ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1


H = ~d · ~S , ~d(k) = (kx , ky , f − νk2)

Eigenvalues

ω0(k) = 0 , ω±(k) = ±|~d(k)| = ±(k2 + (f − νk2)2)1/2


|~e, j = 1〉 , k 7→ ~e(k)




(1)+ ) = 2


ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1

The model as a spin 1 bundleEigenvectors (only ω+):Same as for ~e · ~S with ~e = ~d/|~d |, denoted

|~e, j = 1〉 , k 7→ ~e(k)

Remarks.

I The compactification of R2 is S2.

I ~e(k) 7→ (0, 0,− sgn ν) as k →∞ by ~d(k) = (kx , ky , f − νk2)

I ~e : R2 → S2 extends to a continuous map S2 → S2



(1)+ ) = 2


ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1


|~e, j = 1〉 , k 7→ ~e(k)

Remarks.






(1)+ ) = 2

(cf. Souslov et al.; Tauber et al.)

Proof. If ~S were a spin-12representation, then

ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1


|~e, j = 1〉 , k 7→ ~e(k)

Remarks.






(1)+ ) = 2


ch(P(1/2)+ ) = deg(~e) = +1

Now P(1)+ = P

(1/2)+ ⊗ P

(1/2)+ , so ch(P

(1)+ ) = 1 + 1

Topological phenomena at interfaces

f > 0 (< 0) on northern (southern) hemisphere

(Source: NASA)

The role of the coast

The figure illustrates the clockwise motion of both a particle in amagnetic field and of a wave in presence of a Coriolis force.

Boundary waves are gapless (Halperin 1982, Kelvin 1879).

Halperin’s work led to the far reaching bulk-edge correspondence.




Violation

What goes wrong?

The Hatsugai relation and bulk-edge correspondenceA (projected) band separated from the rest of the bulk spectrum; edgestates (aka evanescent states, bound states).

+

−k−π π

j-th band

ch(Pj) = n+j − n−j

n±j : signed number of eigenvalues crossing the fiducial line ±.

I Remark: n−j = n+j−1I Edge index: N ] := n+j for uppermost occupied band jI Bulk index: N :=

∑j ′≤j ch(Pj ′)

I Bulk-edge correspondence: N = N ]

I Proof: Telescoping sum.


+

−k−π π

j-th band

ch(Pj) = n+j − n−jn±j : signed number of eigenvalues crossing the fiducial line ±.Alternatively: merging with the band from above/below






+

−k−π π

j-th band



I Remark: n−j = n+j−1

I Edge index: N ] := n+j for uppermost occupied band jI Bulk index: N :=





+

−k−π π

j-th band










Violation

What goes wrong?

Bulk-edge correspondence?Sea restricted to upper half-space y > 0.Boundary condition at y = 0 (parametrized by real parameter a):

v = 0 , ∂xu + a∂yv = 0

(boundary condition defines self-adjoint operator Ha).

Bulk-edge correspondence predicts: The signed number of eigenstatesmerging with the band ω+(k) is +2.Remark. Merging with the band from below, but boundary is negativelyoriented. Spectra of Ha

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = −1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 3

I Kelvin waves are seen in all casesI Bulk-edge correspondence is violated!I There are edge states never merging with a bandI There are edge states “merging at infinity”


v = 0 , ∂xu + a∂yv = 0

(boundary condition defines self-adjoint operator Ha).Bulk-edge correspondence predicts: The signed number of eigenstatesmerging with the band ω+(k) is +2.

Remark. Merging with the band from below, but boundary is negativelyoriented. Spectra of Ha

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = −1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 3



v = 0 , ∂xu + a∂yv = 0

(boundary condition defines self-adjoint operator Ha).Bulk-edge correspondence predicts: The signed number of eigenstatesmerging with the band ω+(k) is +2.Remark. Merging with the band from below, but boundary is negativelyoriented.

Spectra of Ha

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = −1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 3


Bulk-edge correspondence?

Spectra of Ha

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = −1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 3

I Kelvin waves are seen in all cases

I Bulk-edge correspondence is violated!

I There are edge states never merging with a band

I There are edge states “merging at infinity”

Bulk-edge correspondence?

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = −1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 1.25

-5 0 5

-5

0

5

C = +2

C = −2

C = 0

kx

ω

a = 3

Theorem. (Violation of correspondence) As a function of the boundaryparameter a, the edge index takes the values

N ] =

2 (a < −

√2)

3 (−√

2 < a < 0)

1 (0 < a <√

2)

2 (a >√

2)

Recall: The bulk index is N = 2.

Back to the Hatsugai relation

+

−k−π π

j-th band

ch(P) = n+ − n−

Relation can be split in two (Porta, G.):

ch(P) = N (S+)−N (S−)

N (S±) = n± (Levinson theorem)

where

I S± = S±(k) = S(k ,E±(k)∓ 0), (k ∈ S1)

I N (f ) winding number of f : S1 → S1.


+

−k−π π

j-th band

ch(P) = n+ − n−

Relation to scattering from inside the bulk:

|in〉 |out〉

defines scattering mapS : |in〉 7→ |out〉

and scattering phase S(k,E ) = 〈in|out〉 (k : longitudinal momentum)


ch(P) = N (S+)−N (S−)


whereI S± = S±(k) = S(k ,E±(k)∓ 0), (k ∈ S1)I N (f ) winding number of f : S1 → S1.


+

−k−π π

j-th band

ch(P) = n+ − n−


ch(P) = N (S+)−N (S−)


where

I S± = S±(k) = S(k ,E±(k)∓ 0), (k ∈ S1)

I N (f ) winding number of f : S1 → S1.




Violation

What goes wrong?

What goes wrong?Is it?

ch(P) = N (S+)−N (S−)

Pictures of torus (Brillouin zone; kx , ky longitudinal/transversalmomentum)

kx

ky

kx

ky

kx

S+

S−

ky

That still holds for waves: On the compactified sphere (instead of torus)one hemisphere contains incoming states, one outgoing.

ch(P) = N (S)


ch(P) = N (S+)−N (S−)


kx

ky

|in〉

|in〉

|out〉

Regions of |out〉, |in〉 states

kx

ky

kx

ky

kx

S+

S−

ky


ch(P) = N (S)


ch(P) = N (S+)−N (S−)


kx

ky

kx

ky

kx

S+

S−

ky

Left: Region admitting (extended) section of states |in〉Middle: Region admitting (extended) section of states |out〉Right: The scattering phases S±(k) as transition functions


ch(P) = N (S)


ch(P) = N (S+)−N (S−)


kx

ky

kx

ky

kx

S+

S−

ky


ch(P) = N (S)

What goes wrong?

Is it Levinson’s theorem?N (S) = n

More precisely: Suppose H(k) depends on some parameter k ∈ R

��

��

k∗

E

kk1 k2

The scattering phase jumps when a bound state reaches threshold

limE→0

arg S(k ,E )∣∣∣k2k1

= ∓2π

The Levinson scenario

limE→0

arg S(kx ,E )∣∣∣k2k1

= ∓2π

Structure of scattering phase

S(kx ,E ) = −g(kx , ky )

g(kx , ky )

where

I ky and ky are the incoming/outgoing momenta withE (kx , ky ) = E (kx , ky ) = E

I ky = −ky if E is even

I g is analytic in ky

Bound states of H(kx) correspond to poles of S(kx ,E ) with Im ky < 0(“bound out-state without in state”); i.e. to g(kx , ky ) = 0


limE→0


= ∓2π


S(kx ,E ) = −g(kx , ky )

g(kx , ky )

where




Bound states of H(kx) correspond to poles of S(kx ,E ) with Im ky < 0(“bound out-state without in state”)

; i.e. to g(kx , ky ) = 0


limE→0


= ∓2π


S(kx ,E ) = −g(kx , ky )

g(kx , ky )

where




Bound states of H(kx) correspond to poles of S(kx ,E ) with Im ky < 0(“bound out-state without in state”); i.e. to g(kx , ky ) = 0


k∗

E

kxk1 k2

Bound states of H(kx) correspond to complex zeros ky of g(kx , ky )

Re ky

Im ky

Re ky

Im ky

(kx < k∗) (kx > k∗)

Fact 1: As kx crosses zero, a bound state disappears.

As for waves, this is the relevant scenario for (almost) all critical, finitemomenta kx .


k∗

E

kxk1 k2


Re ky

Im ky

Re ky

Im ky

(kx < k∗) (kx > k∗)

−ε−ε

Fact 2: As kx crosses zero, arg g(kx , ky = −ε) changes by −π (andarg g(kx , ε) by π), hence S winds by −2π.

As for waves, this is the relevant scenario for (almost) all critical, finitemomenta kx .


k∗

E

kxk1 k2


Re ky

Im ky

Re ky

Im ky

(kx < k∗) (kx > k∗)

−ε−ε

Fact 2: As kx crosses zero, arg g(kx , ky = −ε) changes by −π (andarg g(kx , ε) by π), hence S winds by −2π.As for waves, this is the relevant scenario for (almost) all critical, finitemomenta kx .

Waves at infinite momentum

A convenient, orientation preserving change of coordinates oncompactified momentum space S2 is

λx =kx

k2x + k2y, λy = − ky

k2x + k2y

The map k 7→ λ maps ∞→ 0. (Antipodal map in stereographiccoordinates.)

Not the Levinson scenarioλx = 0 is always critical (regardless of whether an edge state mergesthere).

Structure of g(λx , λy ) for λx fixed, small: Two sheets joined by slits.

Reλy

Reλy

Imλy

Imλy

It takes two zeros, both with Imλy < 0, to make a bound state

Not the Levinson scenario: Alternative I

It takes two zeros, both with Imλy < 0, to make a bound state. Atλx = 0 the slits touch.

Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: No bound state is created nor destroyed at transition.Fact 2: There is a jump of arg g by ±π, hence S winds by ±2π



Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: No bound state is created nor destroyed at transition.

Fact 2: There is a jump of arg g by ±π, hence S winds by ±2π



Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: No bound state is created nor destroyed at transition.Fact 2: There is a jump of arg g by ±π, hence S winds by ±2π

Not the Levinson scenario: Alternative II


Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: A bound state is destroyed at transitionFact 2: There is no jump of arg g and hence S does not wind.



Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: A bound state is destroyed at transition

Fact 2: There is no jump of arg g and hence S does not wind.



Reλy

Reλy

Reλy

Reλy

Imλy

Imλy Imλy

Imλy

(λx < 0) (λx > 0)

Fact 1: A bound state is destroyed at transitionFact 2: There is no jump of arg g and hence S does not wind.

Back to Theorem

Edge:

N ] =

2 (a < −

√2)

3 (−√

2 < a < 0)

1 (0 < a <√

2)

2 (a >√

2)

Bulk:N = 2

Back to Theorem, case by case

-5 0 5

-10

-5

0

5

10

kx

ω

a=-5

N ] = 2 , (a < −√

2)

Alternative II: Edge state merging at infinity; no winding of S there


-5 0 5

-10

-5

0

5

10

kx

ω

a=-1.25

N ] = 3 , (−√

2 < a < 0)

Alternative I: No edge state merging at infinity; winding of S by −1


-5 0 5

-10

-5

0

5

10

kx

ω

a=1.25

N ] = 1 , (0 < a <√

2)

Alternative I: No edge state merging at infinity; winding of S by +1


-5 0 5

-10

-5

0

5

10

kx

ω

a=3

N ] = 2 , (a >√

2)

Alternative II: Edge state merging at infinity; no winding of S there

The transition at a = 0

-5 0 5

-10

-5

0

5

10

kx

ω

a=-0.25

-5 0 5

-10

-5

0

5

10

kx

ω

a=0

-5 0 5

-10

-5

0

5

10

kx

ω

a=0.25

a = −0.25 a = 0 a = 0.25

I The transition occurs within Alternative 1.I Winding of S at infinity changes from −1 to +1I The fibers Ha(kx) of the edge Hamiltonian are self-adjoint for

almost all kx (as it must)

, but not for a = 0, kx = 0. In fact theboundary condition

ikxu + a∂yv = 0

becomes empty.


-5 0 5

-10

-5

0

5

10

kx

ω

a=-0.25

-5 0 5

-10

-5

0

5

10

kx

ω

a=0

-5 0 5

-10

-5

0

5

10

kx

ω

a=0.25

a = −0.25 a = 0 a = 0.25


almost all kx (as it must), but not for a = 0, kx = 0.

In fact theboundary condition

ikxu + a∂yv = 0

becomes empty.


-5 0 5

-10

-5

0

5

10

kx

ω

a=-0.25

-5 0 5

-10

-5

0

5

10

kx

ω

a=0

-5 0 5

-10

-5

0

5

10

kx

ω

a=0.25

a = −0.25 a = 0 a = 0.25


almost all kx (as it must), but not for a = 0, kx = 0. In fact theboundary condition

ikxu + a∂yv = 0

becomes empty.

Summary

I The shallow water model has edge states in presence of Coriolisforces.

I The model is topological if compactified by odd viscosity

I The model violates bulk-boundary correspondence

I Scattering theory (of waves hitting shore) clarifies the cause

I Levinson’s theorem does not apply in its usual form

violation of bulk-edge correspondence in a hydrodynamic model...violation of bulk-edge...

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