linear independence of time-frequency translates · a lattice in r2 is the integer span of two...
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LINEAR INDEPENDENCE OF TIME-FREQUENCY TRANSLATES
Christopher Heil
Georgia Tech
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.