LINEAR INDEPENDENCE OF TIME-FREQUENCY TRANSLATES

Christopher Heil

Georgia Tech

heil@math.gatech.edu

http://www.math.gatech.edu/∼heil

NOTATION

Time-frequency shift: gλ(x) = ga,b(x) = e2πibx g(x − a), λ = (a, b) ∈ R2

Gabor system: G(g, Λ) = {gλ}λ∈Λ

STATEMENT OF THE CONJECTURE

Time-frequency shift: gλ(x) = ga,b(x) = e2πibx g(x − a), λ = (a, b) ∈ R2

Gabor system: G(g, Λ) = {gλ}λ∈Λ

HRT Conjecture (1996)

If:

(a) 0 <

∫ ∞

−∞|g(x)|2 dx < ∞,

(b) Λ = {(ak, bk)}Nk=1 is a finite set of distinct points in R2,

then G(g, Λ) is linearly independent. That is,

N∑

k=1

ck gak,bk = 0 a.e. ⇐⇒ c1 = · · · = cN = 0.

(PUBLISHED?) REFERENCES

1. H/R/T, PAMS, 1996.

2. Linnell, Von Neumann algebras . . . , PAMS, 1999.

3. Rzeszotnik, Four points, 2004.

4. Balan, A noncommutative Wiener Lemma . . . , TAMS, 2008.

5. Balan/Krishtal, almost periodic . . . Wiener’s lemma, JMAA, 2010.

6. Bownik/Speegle, . . . Parseval wavelets, Ill. J. Math., 2010.

7. Demeter, . . . special configurations, MRL, 2010.

8. Demeter/Gautam, . . . lattice Gabor systems, PAMS, 2012.

9. Demeter/Zaharescu, . . . (2, 2) configurations, JMAA, 2012.

10. Bownik/Speegle, . . . faster than exponential decay, Bull. LMS, 2013.

11. Benedetto/Bourouihiya, . . . behaviors at infinity, J. Geom. Anal., 2014

12. Grochenig, . . . independence of time-frequency shifts?, preprint, 2014.

Talk by Eric Weber: “My failed attempts at the HRT Conjecture.”

FALSE FOR THE AFFINE GROUP

Time-scale shifts of any compactly supported scaling function are

dependent.

0.5 1 1.5 2 2.5 3

-0.25

0.25

0.5

0.75

1

1.25

D4(x) = 1+√

34

D4(2x) + 3+√

34

D4(2x − 1) + 3−√

34

D4(2x − 2) + 1−√

34

D4(2x − 3).

0.5 1.0 1.5 2.0 2.5 3.0

-1.0

-0.5

0.5

1.0

1.5

2.0

The Devil’s Staircase (Cantor–Lebesgue Function)

1 2

1

ϕ(x) = 12ϕ(3x) + 1

2ϕ(3x − 1) + ϕ(3x − 2) + 1

2ϕ(3x − 3) + 1

2ϕ(3x − 4).

ZERO DIVISOR CONJECTURE

The group algebra of a group G is

CG =

{

∑

g∈G

cgg : cg ∈ C with only finitely many cg 6= 0

}

.

Finite order elements yield zero divisors: If gn = e, then

(g − e)(gn−1 + · · · + g + e) = 0.

ZERO DIVISOR CONJECTURE

The group algebra of a group G is

CG =

{

∑

g∈G

cgg : cg ∈ C with only finitely many cg 6= 0

}

.

Finite order elements yield zero divisors: If gn = e, then

(g − e)(gn−1 + · · · + g + e) = 0.

Zero Divisor Conjecture (c. 1940, still open)

If G is a torsion-free group and α, β ∈ CG, then

α 6= 0 and β 6= 0 =⇒ αβ 6= 0.

Attributed to Higman, Kaplansky, . . .

Higman (1940): True if G is a locally indicable group

HEISENBERG GROUP

H ={

zMbTa : z ∈ T, a, b ∈ R}

,

CH =

{ N∑

k=1

ckMbkTak: N > 0, ck ∈ C, ak ∈ R, bk ∈ R

}

.

Theorem (Speegle)

CH has no zero divisors.

Proof idea: If

α =

M∑

j=1

zjMbjTaj, β =

N∑

k=1

wkMdkTck,

then

αβ =

M∑

j=1

N∑

k=1

tjkMbj+dkTaj+ck.

Order the translations and then the modulations, obtain a “highest-

order term” . . .

A similar proof works for the affine group. This is essentially the

“locally indicable group” argument (assume every finitely generated

subgroup can be mapped homomorphically onto Z)

Corollary: The set of time-frequency shift operators

{MbTa : a, b ∈ R}

is a finitely linearly independent set of operators:

N∑

k=1

ckMbkTak= 0 =⇒ c1 = · · · = cN = 0.

Similarly, the set of time-scale shifts is finitely linearly independent.

But HRT asks: Is there a g 6= 0 such that

N∑

k=1

ckMbkTakg = 0?

STATEMENT OF THE CONJECTURE

Time-frequency shift: gλ(x) = ga,b(x) = e2πibx g(x − a), λ = (a, b) ∈ R2

Gabor system: G(g, Λ) = {gλ}λ∈Λ

HRT Conjecture (1996)

If:

(a) 0 <

∫ ∞

−∞|g(x)|2 dx < ∞,

(b) Λ = {(ak, bk)}Nk=1 is a finite set of distinct points in R2,

then G(g, Λ) is linearly independent. That is,

N∑

k=1

ck gak,bk = 0 a.e. ⇐⇒ c1 = · · · = cN = 0.

Notation:

HRT(Λ) = HRT for that Λ (and all g)

HRT(g) = HRT for that g (and all Λ; implicitly g 6= 0)

METAPLECTIC TRANSFORMS

If T is a 2 × 2 matrix with | det(T )| = 1, then there exists a unitary

metaplectic transform U = UT on L2(R) such that

G(Ug, TΛ) = G(g, Λ).

Corollary

If T is an area-preserving linear transformation on R2, then

HRT(Λ) is true ⇐⇒ HRT(TΛ) is true.

(Similar result if T is a translation operator.)

This result does not generalize to linear transformations in higher

dimensions!

LINES

Λ ⊂ {0} × R is easy:

N∑

k=1

ck e2πibkxg(x) = 0 a.e. ⇐⇒ c1 = · · · = cN = 0.

(Exponentials are independent on any set of positive measure.)

Corollary

HRT(Λ) is true if Λ ⊂ line.

Proof. Rotation is an area-preserving linear transformation.

LATTICES

A lattice in R2 is the integer span of two independent vectors.

Lemma: HRT for Λ ⊂ lattice ⇐⇒ HRT for Λ ⊂ aZ × bZ.

Proof. Every (full-rank) lattice in R2 is the image of aZ × bZ under

an area-preserving linear transformation.

Theorem

HRT is true for Λ ⊂ aZ × bZ.

ab = 1 is easy via Zak transform.

ab 6= 1 is not easy. Proofs by:

Linnell (von Neumann algebras)

Bownik/Speegle (shift-invariant spaces)

Demeter/Gautam (spectral theory/random Schrodinger operators)

These results do not generalize to higher dimensions!

FOUR POINTS

Four points in R2 need not lie on a translate of a lattice.

Theorem (Rzeszotnik)

HRT(Λ) is true for Λ ={

(0, 0), (1, 0), (0, 1), (√

2, 0)}

.

HRT Subconjecture (currently open)

HRT(Λ) is true for Λ ={

(0, 0), (1, 0), (0, 1), (√

2,√

2)}

.

(Open even if we assume g is continuous—or Schwartz!)

Some special four point cases

Demeter/Zahrescu: HRT true if Λ = vertices of a trapezoid.

Demeter: HRT true if Λ ⊂ two parallel lines.

Benedetto/Bourouihiya: Conditions on g

CONDITIONS ON g

HRT(g) is true if:

Theorem (H/R/T, 1996)

(a) supp(g) is contained in [a,∞) or (−∞, a], or

(b) g(x) = p(x)e−x2

, where p is a polynomial.

Theorem (Bownik/Speegle, 2013)

(a) limx→∞

|g(x)| ecx2

= 0 for all c > 0, or

(b) limx→∞

|g(x)| ecx log x = 0 for all c > 0.

Still open: limx→∞

|g(x)| ecx = 0 for all c > 0.

Theorem (Benedetto/Bourouihiya)

(a) g is ultimately positive and b1, . . . , bN are independent over Q, or

(b) |Λ| = 4, g is ultimately positive, g(x) and g(−x) are ultimately

decreasing.

PERTURBATIONS

Theorem (H/R/T)

(a) If HRT(g) is true, then there exists an ε = εg > 0 such that

‖h − g‖2 < ε =⇒ HRT(h) is true.

(b) If HRT(Λ) is true, then there exists an ε = εΛ > 0 such that

‖Λ′ − Λ‖∞ < ε =⇒ HRT(Λ′) is true.

Corollary

HRT(g) is true for an open, dense set of g ∈ L2(R).

Proof. HRT(g) is true for all compactly supported g;

perturb g by εg.

Theorem

If HRT(g) is true, then there exists an εg > 0 such that

‖h − g‖2 < εg =⇒ HRT(h) is true.

Key ingredient: If HRT(g) is true, then G(g, Λ) is a Riesz basis for its

span. The value of εg is determined by the lower frame bound.

Fundamental Problem

Determine the lower frame bound of G(g, Λ) in terms of properties of

g or Λ.

(Some results by Christensen et al. exist, but they are not strong

enough.)

SPECTRUM

HRT(Λ) is true if and only if the point spectrum of

T =N

∑

k=1

ckMbkTak

is empty (for c1, . . . , cN not all zero; note I = M0T0).

Theorem (Balan, 2008)

T cannot have an isolated eigenvalue of finite multiplicity.

Therefore, if the point spectrum of T is not empty, then it can only

contain eigenvalues of infinite multiplicity, or eigenvalues that also

belong to the continuous spectrum of T.

ASYMPTOTICS

Modulation space condition:∫∫

|〈g, MbTag〉| v(a, b) da db < ∞.

The weight v is submultiplicative and satisfies GRS.

Theorem (Grochenig, 2014)

If G(g, Λ) is a frame but not a Riesz basis for L2(R), then the lower

Riesz bound An of G(g, Λn) decays like

An ≤ C sup|λ|>n

v(λ)−2,

where Λn = Λ ∩ Bn(0).

Remark: An = 0 ⇐⇒ G(g, Λn) is dependent.

Polynomial weight v(a, b) = (1 + |a| + |b|)s:

An = O(n−2s).

Subexponential weight v(z) = ec(|a|+|b|)s:

An = O(e−cns

).

“From a numerical point of view even small sets of time-frequency

shifts may look linearly dependent. [This] illustrates the spectacular

difference between a conjectured mathematical truth and a compu-

tationally observable truth.”

JUST THREE POINTS

We sketch a proof that

{

g(x), g(x − a), g(x)e2πix}

is linearly independent.

Suppose that

c1 g(x) + c2 g(x − a) + c3 g(x)e2πix = 0, all x ∈ R.

Rewriting,

g(x − a) = m(x) g(x),

where m(x) = −c1

c2− c3

c2e2πix is 1-periodic.

g(x − a) = m(x) g(x) for all x.

Iterate:

g(x − 2a) = g((x − a) − a)

= m(x − a) g(x − a)

= m(x − a) m(x) g(x).

g(x − ka) =

k−1∏

j=0

m(x − ja)

g(x).

|g(x − ka)| =

k−1∏

j=0

|m(x − ja)|

|g(x)|.

ln |g(x − ka)| =k−1∑

j=0

ln |m(x − ja)| + ln |g(x)|.

ln |g(x − ka)| =k−1∑

j=0

p(x − ja) + ln |g(x)|

where p is 1-periodic.

1

kln |g(x − ka)| =

1

k

k−1∑

j=0

p(x − ja) +1

kln |g(x)|.

The value1

kp(x− ja) is the area of the box with base centered at (the

fractional part of) x − ja and height p(x − ja) and width 1/k:

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

Since p is 1-periodic, these boxes are scattered around the interval

[0, 1].

1

k

k−1∑

j=0

p(x − ja)

The sum is the sum of the areas in the boxes:

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

1

k

k−1∑

j=0

p(x − ja)

The sum is the sum of the areas in the boxes:

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

It’s like a Riemann sum approximation to

∫ 1

0

p(x) dx, except the boxes

are “randomly distributed” — at least, if a is irrational.

Ergodic Theorem:1

k

k−1∑

j=0

p(x − ja) →∫ 1

0

p(x) dx as k → ∞.

1

k

k−1∑

j=0

p(x − ja) ≈∫ 1

0

p(x) dx = C

k−1∑

j=0

ln |m(x − ja)| ≈ Ck

k−1∏

j=0

|m(x − ja)| ≈ eCk

|g(x − ka)| =

(k−1∏

j=0

|m(x − ja)|)

|g(x)| ≈ eCk |g(x)|

If C > 0 then g is growing as x → −∞, contradicting

∫ ∞

−∞|g(x)|2 dx < ∞.

Symmetric argument if C < 0. C = 0 takes more work.

CHALLENGE

HRT Subconjecture (currently open . . . but . . . ?)

HRT(Λ) is true for Λ ={

(0, 0), (1, 0), (0, 1), (√

2,√

2)}

,

even if g is Schwartz.

A survey article is available

http://www.math.gatech.edu/∼heil/papers/hrtnotes.pdf

A newer survey by H. and Speegle will be available shortly.

THANK YOU.