![Page 1: The Heisenberg Representation and the Fast Fourier TransformThe Heisenberg Representation and the Fast Fourier Transform Shamgar Gurevich UW Madison July 31, 2014 Shamgar Gurevich](https://reader033.vdocuments.site/reader033/viewer/2022042709/5f458f5625607d16b9794c63/html5/thumbnails/1.jpg)
The Heisenberg Representation and the Fast FourierTransform
Shamgar Gurevich
UW Madison
July 31, 2014
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 1 / 25
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Motivation:
Discrete Fourier Transform
DFT =1√N
(e2πiN τ·ω
)0≤τ,ω≤N−1
Compute:
f̂ = DFT [f ]; f =
f (0)...
f (N − 1)
; Fast!!
Cooley—Tukey (1965): O(N · log(N)) operations!
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 2 / 25
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Motivation:
Discrete Fourier Transform
DFT =1√N
(e2πiN τ·ω
)0≤τ,ω≤N−1
Compute:
f̂ = DFT [f ]; f =
f (0)...
f (N − 1)
; Fast!!
Cooley—Tukey (1965): O(N · log(N)) operations!
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 2 / 25
![Page 4: The Heisenberg Representation and the Fast Fourier TransformThe Heisenberg Representation and the Fast Fourier Transform Shamgar Gurevich UW Madison July 31, 2014 Shamgar Gurevich](https://reader033.vdocuments.site/reader033/viewer/2022042709/5f458f5625607d16b9794c63/html5/thumbnails/4.jpg)
Motivation:
Discrete Fourier Transform
DFT =1√N
(e2πiN τ·ω
)0≤τ,ω≤N−1
Compute:
f̂ = DFT [f ]; f =
f (0)...
f (N − 1)
; Fast!!
Cooley—Tukey (1965): O(N · log(N)) operations!Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 2 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.
f : {0, ....,N − 1} → C.
Basic operations
Time shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.f : {0, ....,N − 1} → C.
Basic operations
Time shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.f : {0, ....,N − 1} → C.
Basic operations
Time shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.f : {0, ....,N − 1} → C.
Basic operationsTime shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.f : {0, ....,N − 1} → C.
Basic operationsTime shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Solution (Auslander—Tolimieri)(I) Heisenberg Group Representation
H = C(ZN ) – Hilbert space of digital signals.f : {0, ....,N − 1} → C.
Basic operationsTime shift: τ ∈ ZN ,
Lτ : H → H,Lτ [f ](t) = f (t + τ), t ∈ ZN .
Frequency shift: ω ∈ ZN ,
Mω : H → H,Mω [f ](t) = e
2πiN ωt f (t).
Note:
Mω ◦ Lτ = e2πiN ωτ · Lτ ◦Mω – Heisenberg commutation relations
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 3 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).
Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);
(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Combine: τ,ω, z ∈ ZN
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Identity:
π(τ,ω, z) ◦ π(τ′,ω′, z ′) = π(τ + τ′,ω+ω′, z + z ′ +12
∣∣∣∣ τ ωτ′ ω′
∣∣∣∣).Question. How to think on this?
Answer. Heisenberg group:
H = ZN ×ZN︸ ︷︷ ︸V
× ZN︸︷︷︸Z
;
(v , z) · (v ′, z ′) = (v + v ′, z + z ′ + 12
∣∣∣∣ vv ′∣∣∣∣);
(0, 0) · (v , z) = (v , z) · (0, 0) = (v , z);(v , z) · (−v ,−z) = (−v ,−z) · (v , z) = (0, 0).
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 4 / 25
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Heisenberg Rep’n, cont.
Summary: Heisenberg Rep’nπ : H → GL(H);
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ;
π(h · h′) = π(h) ◦ π(h′) – homomorphism.
DefinitionA representation of a group H on a complex vector space H is ahomomorphism
π : H → GL(H),π(h · h′) = π(h) ◦ π(h′).
Question. DFT ?
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Heisenberg Rep’n, cont.
Summary: Heisenberg Rep’nπ : H → GL(H);
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ;
π(h · h′) = π(h) ◦ π(h′) – homomorphism.
DefinitionA representation of a group H on a complex vector space H is ahomomorphism
π : H → GL(H),π(h · h′) = π(h) ◦ π(h′).
Question. DFT ?
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Heisenberg Rep’n, cont.
Summary: Heisenberg Rep’nπ : H → GL(H);
π(τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ;
π(h · h′) = π(h) ◦ π(h′) – homomorphism.
DefinitionA representation of a group H on a complex vector space H is ahomomorphism
π : H → GL(H),π(h · h′) = π(h) ◦ π(h′).
Question. DFT ?
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(II) Representation Theory
DefinitionsWe say that (π1,H,H1), (π2,H,H2) are equivalent, π1 ' π2, if
∃ α : H1→̃H2 − Intertwiner,
such that for every h ∈ H
H1π1(h)−−−→ H1yα
yα
H2π2(h)−−−→ H2
i.e., α ◦ π1(h) = π2(h) ◦ α.
Example: DFT is an intertwiner!
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(II) Representation Theory
DefinitionsWe say that (π1,H,H1), (π2,H,H2) are equivalent, π1 ' π2, if
∃ α : H1→̃H2 − Intertwiner,
such that for every h ∈ H
H1π1(h)−−−→ H1yα
yα
H2π2(h)−−−→ H2
i.e., α ◦ π1(h) = π2(h) ◦ α.
Example: DFT is an intertwiner!
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Rep’n Theory, cont.
H = V × Z = ZN︸︷︷︸T
× ZN︸︷︷︸W
× ZN︸︷︷︸Z
Time representation: HT = C(T ){πT : H → GL(HT );
πT (τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Frequency representation: HW = C(W ){πW : H → GL(HW );
πW (τ,ω, z) = e2πiN {−
12ωτ+z} ·Mτ ◦ L−ω.
Fact:DFT ◦ πT (h) = πW (h) ◦DFT ,
whereDFT [f ](w) =
1√N
∑t∈ZN
f (t)e2πiN wt .
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Rep’n Theory, cont.
H = V × Z = ZN︸︷︷︸T
× ZN︸︷︷︸W
× ZN︸︷︷︸Z
Time representation: HT = C(T ){πT : H → GL(HT );
πT (τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Frequency representation: HW = C(W ){πW : H → GL(HW );
πW (τ,ω, z) = e2πiN {−
12ωτ+z} ·Mτ ◦ L−ω.
Fact:DFT ◦ πT (h) = πW (h) ◦DFT ,
whereDFT [f ](w) =
1√N
∑t∈ZN
f (t)e2πiN wt .
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Rep’n Theory, cont.
H = V × Z = ZN︸︷︷︸T
× ZN︸︷︷︸W
× ZN︸︷︷︸Z
Time representation: HT = C(T ){πT : H → GL(HT );
πT (τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Frequency representation: HW = C(W ){πW : H → GL(HW );
πW (τ,ω, z) = e2πiN {−
12ωτ+z} ·Mτ ◦ L−ω.
Fact:DFT ◦ πT (h) = πW (h) ◦DFT ,
whereDFT [f ](w) =
1√N
∑t∈ZN
f (t)e2πiN wt .
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Rep’n Theory, cont.
H = V × Z = ZN︸︷︷︸T
× ZN︸︷︷︸W
× ZN︸︷︷︸Z
Time representation: HT = C(T ){πT : H → GL(HT );
πT (τ,ω, z) = e2πiN {
12ωτ+z} ·Mω ◦ Lτ.
Frequency representation: HW = C(W ){πW : H → GL(HW );
πW (τ,ω, z) = e2πiN {−
12ωτ+z} ·Mτ ◦ L−ω.
Fact:DFT ◦ πT (h) = πW (h) ◦DFT ,
whereDFT [f ](w) =
1√N
∑t∈ZN
f (t)e2πiN wt .
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(III) FFT Algorithm
Idea:HT
fast−−−→ HΛ
slow
yDFT ∥∥∥HW ←−−−
fastHΛ
More representation theory:
DefinitionA rep’n (π,H,H) is irreducible if
@ 0 6= H′ H
such thatπ(h) · H′ ⊂ H′, ∀h ∈ H.
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(III) FFT Algorithm
Idea:HT
fast−−−→ HΛ
slow
yDFT ∥∥∥HW ←−−−
fastHΛ
More representation theory:
DefinitionA rep’n (π,H,H) is irreducible if
@ 0 6= H′ H
such thatπ(h) · H′ ⊂ H′, ∀h ∈ H.
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(III) FFT Algorithm
Idea:HT
fast−−−→ HΛ
slow
yDFT ∥∥∥HW ←−−−
fastHΛ
More representation theory:
DefinitionA rep’n (π,H,H) is irreducible if
@ 0 6= H′ H
such thatπ(h) · H′ ⊂ H′, ∀h ∈ H.
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FFT Algorithm: more RT
Theorem (Stone—von Neumann)
If (πj ,H,Hj ), j = 1, 2, are irreducible representations of the Heisenberggroup H = V × Z , with
πj (z) = e2πiN z · IdHj , ∀z ∈ Z ,
then π1 ' π2 are equivalent, i.e., ∃ α : H1→̃H2 such that
α ◦ π1(h) = π2(h) ◦ α, ∀h ∈ H. (1)
Theorem (Schur’s lemma)
If π1 ' π2 equivalent irreducible representations, and if α, α′ satisfyequation (1), then
α = c · α′, for some scalar c .
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FFT Algorithm: more RT
Theorem (Stone—von Neumann)
If (πj ,H,Hj ), j = 1, 2, are irreducible representations of the Heisenberggroup H = V × Z , with
πj (z) = e2πiN z · IdHj , ∀z ∈ Z ,
then π1 ' π2 are equivalent, i.e., ∃ α : H1→̃H2 such that
α ◦ π1(h) = π2(h) ◦ α, ∀h ∈ H. (1)
Theorem (Schur’s lemma)
If π1 ' π2 equivalent irreducible representations, and if α, α′ satisfyequation (1), then
α = c · α′, for some scalar c .
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FFT Algorithm: Models of Heisenberg Rep’n
Examples
(1) Time model: V ⊃ T = {(t, 0); t ∈ ZN}
πT : H → GL(HT ).
(2) Frequency model: V ⊃ W = {(0,w); w ∈ ZN}
πW : H → GL(HW ).
CorollaryDFT : HT → HW – unique (up to a scalar) intertwiner between πT andπW .
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FFT Algorithm: Models of Heisenberg Rep’n
Examples(3) W—invariant model:Space:
HW =
f : H → C,
f (w · h) = f (h), ∀ w ∈ W , h ∈ H,f (z · h) = e 2πi
N z · f (h), ∀ z ∈ Z , h ∈ H.
Action: {πW : H → GL(HW ),
[πW (h′)f ](h) = f (h · h′).
Remark: We have a natural identification HT = HW , given by
f 7→ f̃ (wtz) = e2πiN z · f (t).
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Think on W-invariant functions as functions on T
A function f (t) on T = {(t, 0, 0); t ∈ Z/52}
0
5
10
15
20
25
0
5
10
15
20
250. 2
0. 15
0. 1
0. 05
0
0.05
0.1
0.15
0.2
TW
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FFT Algorithm: Models of Heisenberg Rep’n
Examples(4) T—invariant model:Space:
HT =
g : H → C,
g(t · h) = g(h), ∀ t ∈ T , h ∈ H,g(z · h) = e 2πi
N z · g(h), ∀ z ∈ Z , h ∈ H.
Action: {πT : H → GL(HT ),
[πT (h′)g ](h) = g(h · h′).
Remark: We have a natural identification HW = HT , given by
g 7→ g̃(twz) = e2πiN z · g(w).
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Think on T-invariant functions as functions on W
A function g(w) on W = {(0,w , 0); w ∈ Z/52}
0
5
10
15
20
25
0
5
10
15
20
250 .2
0 .15
0 .1
0 .05
0
0.05
0 .1
0 .15
0 .2
TW
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FFT Algorithm: The Arithmetic Model
Examples
(5) Arithmetic model: N = p2
Lagrangian: V = Z/p2 ×Z/p2 ⊃ Λ = {(p · a, p · b)}Space:
HΛ =
F : H → C,
F (λ · h) = F (h), ∀ λ ∈ Λ, h ∈ H,F (z · h) = e 2πi
N z · F (h), ∀ z ∈ Z , h ∈ H.
Action:πΛ : H → GL(HΛ), [πΛ(h′)F ](h) = F (h · h′).
Remark: We have a natural identification of HΛ with functionsF (x , y) on {0, ..., p − 1} × {0, ..., p − 1}
F 7→ F̃ (λ · (x , y , 0) · z) = e 2πiN z · F (x , y), λ ∈ Λ.
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Think on Lambda-invariant functions as functions on{0,...,p-1} x {0,...,p-1}
A function F (x , y) on (x , y) ∈ {0, ..., 4} × {0, ..., 4}, p = 5,
05
1015
2025
05
1015
20251
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
Y X
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FFT: The Algorithm
Algorithm: Case N = p2
HTDFT−−−→p4
HW
saw
y xsawHW HT
∑λ∈Λ f (λh)
y? ?
x∑t∈T F (tw )
HΛ HΛ
Observation (the saving!): For f ∈ HW , if λ,λ′ ∈ Λ differ by anelement of W , i.e., λ = wλ′, then
f (λh) = f (wλ′h) = f (λ′h), ∀h ∈ H.
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FFT: The Algorithm
Algorithm: Case N = p2
HTDFT−−−→p4
HW
saw
y xsawHW HT
∑λ∈Λ f (λh)
y? ?
x∑t∈T F (tw )
HΛ HΛ
Observation (the saving!): For f ∈ HW , if λ,λ′ ∈ Λ differ by anelement of W , i.e., λ = wλ′, then
f (λh) = f (wλ′h) = f (λ′h), ∀h ∈ H.
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FFT Algorithms: The Summands
So for a fixed h it is enough to know the summands f (λh), in thetransform, only for λ ∈ Λ/[Λ ∩W ] = {(0, 0), (p, 0), ..., ((p − 1)p, 0)}.
Example for p = 5
0
5
10
15
20
25
0
5
10
15
20
25
0. 2
0
0.2
TW
Enough to know summands here
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FFT Algorithm: The Saving
So we have
∑λ∈Λ
f (λh) = ∑λ∈Λ/[Λ∩W ]
f (λh) ·#(Λ ∩W ).
Only p = p2/p = #(Λ/[Λ ∩W ]) summands !!!.
Complexity of the algorithm
p2︸︷︷︸values of h
· p︸︷︷︸summands
+ p2 · p︸ ︷︷ ︸second operator
= p ·p2 · (1+ 1) = p︸︷︷︸constant
·N · log(N).
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FFT Algorithm: The Saving
So we have
∑λ∈Λ
f (λh) = ∑λ∈Λ/[Λ∩W ]
f (λh) ·#(Λ ∩W ).
Only p = p2/p = #(Λ/[Λ ∩W ]) summands !!!.Complexity of the algorithm
p2︸︷︷︸values of h
· p︸︷︷︸summands
+ p2 · p︸ ︷︷ ︸second operator
= p ·p2 · (1+ 1) = p︸︷︷︸constant
·N · log(N).
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FFT Algorithm: Schur’s lemma
Denote the intertwiners by
FFTΛ,W : HW → HΛ, and FFTT ,Λ : HΛ → HT .
WhyDFT = FFTT ,Λ ◦ FFTΛ,W ? (2)
In fact by Schur’s lemma: Both sides of (2) intertwine πW and πT ,so
DFT = c · FFTT ,Λ ◦ FFTΛ,W
for some scalar c . Easy to compute c = 1p .
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FFT Algorithm: Schur’s lemma
Denote the intertwiners by
FFTΛ,W : HW → HΛ, and FFTT ,Λ : HΛ → HT .
WhyDFT = FFTT ,Λ ◦ FFTΛ,W ? (2)
In fact by Schur’s lemma: Both sides of (2) intertwine πW and πT ,so
DFT = c · FFTT ,Λ ◦ FFTΛ,W
for some scalar c . Easy to compute c = 1p .
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Numerics for n=25
Start with the exponent function f (t) = e2πi25 t/5 on T = Z/25
0 5 10 15 20 250. 2
0. 15
0. 1
0. 05
0
0. 05
0. 1
0. 15
0. 2
T
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Numerics for n=25 : FFT(f)
Apply FFT and get g(w) = FFT [f ](w) on W = Z/25
0 5 10 15 20 250. 2
0
0.2
0.4
0.6
0.8
1
1.2
W
Indeed this is g(w) = δ1(w).
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FFT Algorithm: General
N = pk , k ≥ 1
V = Z/pk ×Z/pk
Good Lagrangian sequence ("interpolating" W and T )
Λ• : Λj = {(pk−j · a, pj · b)}, j = 0, ..., k ,
with LARGE intersections
# Λj ∩Λj+1 = pk−1.
FFT Algorithm
HW FFT Λ1 ,W→ HΛ1 → · · · → HΛk−1 FFTT ,Λk→ HT
DFT =1√pk· FFTT ,Λk−1 ◦ · · · ◦ FFTΛ1,W ,
with complexity
pk︸︷︷︸values of h
· { k︸︷︷︸k operators
· # Λj+1/(Λj ∩Λj+1)︸ ︷︷ ︸p summands
} = p︸︷︷︸constant
·pk︸︷︷︸N
· log(pk )︸ ︷︷ ︸log(N )
.
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FFT Algorithm: General
N = pk , k ≥ 1V = Z/pk ×Z/pk
Good Lagrangian sequence ("interpolating" W and T )
Λ• : Λj = {(pk−j · a, pj · b)}, j = 0, ..., k ,
with LARGE intersections
# Λj ∩Λj+1 = pk−1.
FFT Algorithm
HW FFT Λ1 ,W→ HΛ1 → · · · → HΛk−1 FFTT ,Λk→ HT
DFT =1√pk· FFTT ,Λk−1 ◦ · · · ◦ FFTΛ1,W ,
with complexity
pk︸︷︷︸values of h
· { k︸︷︷︸k operators
· # Λj+1/(Λj ∩Λj+1)︸ ︷︷ ︸p summands
} = p︸︷︷︸constant
·pk︸︷︷︸N
· log(pk )︸ ︷︷ ︸log(N )
.
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FFT Algorithm: General
N = pk , k ≥ 1V = Z/pk ×Z/pk
Good Lagrangian sequence ("interpolating" W and T )
Λ• : Λj = {(pk−j · a, pj · b)}, j = 0, ..., k ,
with LARGE intersections
# Λj ∩Λj+1 = pk−1.
FFT Algorithm
HW FFT Λ1 ,W→ HΛ1 → · · · → HΛk−1 FFTT ,Λk→ HT
DFT =1√pk· FFTT ,Λk−1 ◦ · · · ◦ FFTΛ1,W ,
with complexity
pk︸︷︷︸values of h
· { k︸︷︷︸k operators
· # Λj+1/(Λj ∩Λj+1)︸ ︷︷ ︸p summands
} = p︸︷︷︸constant
·pk︸︷︷︸N
· log(pk )︸ ︷︷ ︸log(N )
.
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FFT Algorithm: General
N = pk , k ≥ 1V = Z/pk ×Z/pk
Good Lagrangian sequence ("interpolating" W and T )
Λ• : Λj = {(pk−j · a, pj · b)}, j = 0, ..., k ,
with LARGE intersections
# Λj ∩Λj+1 = pk−1.
FFT Algorithm
HW FFT Λ1 ,W→ HΛ1 → · · · → HΛk−1 FFTT ,Λk→ HT
DFT =1√pk· FFTT ,Λk−1 ◦ · · · ◦ FFTΛ1,W ,
with complexity
pk︸︷︷︸values of h
· { k︸︷︷︸k operators
· # Λj+1/(Λj ∩Λj+1)︸ ︷︷ ︸p summands
} = p︸︷︷︸constant
·pk︸︷︷︸N
· log(pk )︸ ︷︷ ︸log(N )
.
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Conclusion
Conclusion:
Λ• =⇒ Cooley—Tukey FFT
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THANK YOU
Thank You!
Shamgar Gurevich (UW Madison) Heisenberg Repn and FFT July 31, 2014 25 / 25