Time-Frequency Analysis and the Dark Side of Representation Theory - PowerPoint PPT Presentation

Time-Frequency Analysis and the Dark Side of Representation Theory Gerald B. Folland February 21, 2014

We consider time-frequency translations on L 2 ( R ): M y f ( t ) = e 2 πiyt f ( t ) T x f ( t ) = f ( t + x ) , We have T x M y = e 2 πixy M y T x , so the collection of operators � � e 2 πiz M y T x : x, y, z ∈ R forms a group, essentially the (real) Heisenberg group. More precisely, the real Heisenberg group H R is R 3 equipped with the group law ( x, y, z )( x ′ , y ′ , z ′ ) = ( x + x ′ , y + y ′ , z + z ′ + xy ′ ) .

Given τ, ω > 0, consider the subgroup generated by the T jτ and M kω with j, k ∈ Z , namely, � � e 2 πiτωl M kω T jτ : j, k, l ∈ Z . There is a large literature on the use of families { M kω T jτ φ } as building blocks to synthesize more general functions.

Given τ, ω > 0, consider the subgroup generated by the T jτ and M kω with j, k ∈ Z , namely, � � e 2 πiτωl M kω T jτ : j, k, l ∈ Z . There is a large literature on the use of families { M kω T jτ φ } as building blocks to synthesize more general functions. By rescaling, we can and shall take τ = 1. This is a unitary representation of the discrete Heisenberg group H , whose underlying set is Z 3 and whose group law is ( j, k, l )( j ′ , k ′ , l ′ ) = ( j + j ′ , k + k ′ , l + l ′ + jk ′ ) . That is, the representation in question is defined by ρ ω ( j, k, l ) f ( t ) = e 2 πiωl e 2 πiωkt f ( t + j ) ( f ∈ L 2 ( R )) .

Given τ, ω > 0, consider the subgroup generated by the T jτ and M kω with j, k ∈ Z , namely, � � e 2 πiτωl M kω T jτ : j, k, l ∈ Z . There is a large literature on the use of families { M kω T jτ φ } as building blocks to synthesize more general functions. By rescaling, we can and shall take τ = 1. This is a unitary representation of the discrete Heisenberg group H , whose underlying set is Z 3 and whose group law is ( j, k, l )( j ′ , k ′ , l ′ ) = ( j + j ′ , k + k ′ , l + l ′ + jk ′ ) . That is, the representation in question is defined by ρ ω ( j, k, l ) f ( t ) = e 2 πiωl e 2 πiωkt f ( t + j ) ( f ∈ L 2 ( R )) . How does this representation decompose into irreducible representations?

Some Background ◮ A (unitary) representation of a locally compact group G is a continuous homomorphism ρ : G → U ( H ) where H is a Hilbert space. ◮ ρ is irreducible if there are no nontrivial closed subspaces of H that are invariant under the operators ρ ( g ), g ∈ G . ◮ ρ : G → U ( H ) and ρ ′ : G → U ( H ′ ) are (unitarily) equivalent if there is a unitary map V : H → H ′ such that V ρ ( g ) = ρ ′ ( g ) V for all g ∈ G . ◮ The set of equivalence classes of irreducible unitary representations of G is denoted by � G . If G is compact, every unitary representation of G is a direct sum of irreducible representations. The equivalence classes (elements of � G ) occurring in it and the multiplicities with which they occur are uniquely determined. If G is noncompact, there are “continuous families” of irreducible representations, and in general one must employ direct integrals instead.

Direct Integrals � � Suppose we have a family π α : α ∈ A of representations of G parametrized by a measure space ( A, µ ), where π α acts on H α . The direct integral of the Hilbert spaces H α is the Hilbert space � ⊕ H = H α dµ ( α ) � � � � � f ( α ) � 2 = f : A → H α : f ( α ) ∈ H α ∀ α, H α dµ ( α ) < ∞ . (Some issues of measurability are being swept under the rug, but note that if the H α are all the same, say H α = K for all α , then H is just L 2 ( A, K ).) The direct integral of the representations π α is the representation � ⊕ π = π α dµ ( α ) on H defined by [ π ( g ) f ]( α ) = π α ( g )[ f ( α )] .

Example If G = R , the irreducible representations are all one-dimensional and are parametrized by ξ ∈ R : π ξ ( x ) = e 2 πiξx . � ⊕ π ξ dξ acts on L 2 ( R ) by The direct integral π = R π ( x ) f ( ξ ) = e 2 πiξx f ( ξ ) . Conjugation by the Fourier transform � e − 2 πitξ f ( t ) dt F f ( ξ ) = turns this into the regular representation of R on L 2 ( R ): F − 1 π ( x ) F f ( t ) = f ( t + x ) , F − 1 π ( x ) F = T x . i.e.,

What Should Happen: ◮ � G is a geometrically “reasonable” object, equipped with a natural σ -algebra of measurable sets, and we can choose a representative π α from each equivalence class α in � G in a “reasonable” way. ◮ Given a representation ρ , there is a measure µ on � G and disjoint measurable sets E 1 , E 2 , . . . , E ∞ (some of which may be empty) such that � ⊕ � ⊕ � ⊕ ρ ∼ π α dµ ( α ) ⊕ 2 π α dµ ( α ) ⊕ · · · ⊕ ∞ π α dµ ( α ) . E 1 E 2 E ∞ (The coefficients in front of the integrals denote multiplicities.) µ is determined up to equivalence (mutual absolute continuity), and the E j are determined up to sets of µ -measure zero.

What Actually Happens: There is a sharp dichotomy in the class of locally compact groups: ◮ For “good” (type I) groups, this all works as advertised.

What Actually Happens: There is a sharp dichotomy in the class of locally compact groups: ◮ For “good” (type I) groups, this all works as advertised. ◮ For “bad” groups, it all fails. ◮ � G is horrible. ◮ Representations can be decomposed into direct integrals of irreducibles, but usually not with � G as the parameter space. ◮ There is usually no uniqueness in such decompositions!

What Actually Happens: There is a sharp dichotomy in the class of locally compact groups: ◮ For “good” (type I) groups, this all works as advertised. ◮ For “bad” groups, it all fails. ◮ � G is horrible. ◮ Representations can be decomposed into direct integrals of irreducibles, but usually not with � G as the parameter space. ◮ There is usually no uniqueness in such decompositions! ◮ Some type I groups: Abelian groups; compact groups; connected Lie groups that are nilpotent, semisimple, or algebraic; discrete groups with an Abelian normal subgroup of finite index. ◮ Some non-type I groups: some solvable Lie groups, all other discrete groups.

Now back to the discrete Heisenberg group H with group law ( j, k, l )( j ′ , k ′ , l ′ ) = ( j + j ′ , k + k ′ , l + l ′ + jk ′ ) , and our representation ρ ω of H , ρ ω ( j, k, l ) f ( t ) = e 2 πiωl e 2 πiωkt f ( t + j ) ( f ∈ L 2 ( R )) . Note that the center of H (also its commutator subgroup) is � � Z = (0 , 0 , l ) : l ∈ Z , and it acts by scalars: ρ ω (0 , 0 , l ) = e 2 πiωl I. The representation l �→ e 2 πiωl of Z is called the central character of ρ ω . Only those irreducible representations having the same central character will occur in ρ ω .

Case 1: ω is rational, say ω = p/q ( p, q ∈ Z + , gcd( p, q ) = 1). Here the central character is trivial on multiples of (0 , 0 , q ), so ρ ω factors through the group H q = Z × Z × Z q ( Z q = Z /q Z ) , — same group law, with arithmetic mod q in the last factor.

Case 1: ω is rational, say ω = p/q ( p, q ∈ Z + , gcd( p, q ) = 1). Here the central character is trivial on multiples of (0 , 0 , q ), so ρ ω factors through the group H q = Z × Z × Z q ( Z q = Z /q Z ) , — same group law, with arithmetic mod q in the last factor. Subcase 1a: ω ∈ Z , i.e., q = 1. Here H 1 = Z 2 with the standard Abelian group structure. Its irreducible representations are one-dimensional; they are the characters χ u,v ( j, k ) = e 2 πi ( ju + kv ) , u, v ∈ R / Z . Claim: � ⊕ If ω = p ∈ Z , then ρ ω ∼ p ( R / Z ) 2 χ u,v du dv .

The intertwining operator that gives this equivalence is the Zak transform . This is a map from (reasonable) functions on R to functions on R 2 defined by � e 2 πinu f ( v + n ) . Z f ( u, v ) = n ∈ Z

The intertwining operator that gives this equivalence is the Zak transform . This is a map from (reasonable) functions on R to functions on R 2 defined by � e 2 πinu f ( v + n ) . Z f ( u, v ) = n ∈ Z Note that Z f ( u, v + m ) = e − 2 πimu Z f ( u, v ) , Z f ( u + m, v ) = Z f ( u, v ) , so Z f is determined by its values on [0 , 1) × [0 , 1). Moreover, by the Parseval identity, � 1 � 1 � 1 � � | Z f ( u, v ) | 2 du dv = | f ( v + n ) | 2 dv = | f ( t ) | 2 dt, 0 0 0 R n so Z is an isometry from L 2 ( R ) to L 2 ([0 , 1) 2 ) which is easily seen to be surjective, hence unitary.

Moreover, since ρ p ( j, k, l ) f ( t ) = e 2 πipkt f ( t + j ), we have � e 2 πinu e 2 πipk ( v + j ) f ( v + j + n ) Z ρ p ( j, k, l ) f ( u, v ) = n � e 2 πi ( n − j ) u e 2 πipkv f ( v + n ) = n = e − 2 πiju e 2 πipkv Z f ( u, v ) = χ − u,pv ( j, k ) Z f ( u, v ) . Thus Z intertwines ρ p with � ⊕ � [0 , 1) 2 χ − u,pv du dv ∼ p ( R / Z ) 2 χ u,v du dv.