Fourier analysis on the symmetric group Risi Kondor . Fourier - PowerPoint PPT Presentation

Fourier analysis on the symmetric group Risi Kondor

. Fourier transform on S n Forward transform: ∑ � f ( λ ) = f ( σ ) ρ λ ( σ ) λ ⊢ n σ ∈ S n Inverse transform: ∑ [ � ] f ( σ ) = 1 f ( λ ) ρ λ ( σ − 1 ) d λ tr . n ! λ ⊢ n 2 / 16 2/16 .

. Integer partitions λ ⊢ n means that λ = ( λ 1 , . . . , λ k ) is an integer partition of n , i.e., k ∑ λ 1 ≥ λ 2 ≥ . . . ≥ λ k and λ i = n. i =1 Graphically represented by so-called Young diagrams (Ferrers diagrams), such as for λ = (5 , 3 , 2) ⊢ 10 . Fact: The irreps of S n are indexed by { λ ⊢ n } . 3 / 16 3/16 .

. Young tableaux Filling in the rows/columns of a Young diagram with 1 , . . . , n gives a so-called Young tableau . A standard tableau is a Young tableau in which the numbers strictly increase from left to right in each row, and top to bottom in each column. Example: 1 3 5 6 2 4 7 The set of standard tableaux of shape λ we’ll denote T λ . Fact: The rows/columns of ρ λ ( σ ) are in bijection with T λ . 4 / 16 4/16 .

. Hook rule Given any box i in a Young diagram, a hook is that box, plus the boxes to the right, plus the boxes below. The length ℓ ( i ) of the hook is the total number of boxes involved. Theorem. n ! ∏ | T λ | = d λ = i ℓ ( i ) . Corollary. ∑ d 2 λ = n ! λ ⊢ n 5 / 16 5/16 .

. λ d λ ( n ) 1 ( n − 1 , 1) n − 1 n ( n − 3) ( n − 2 , 2) 2 ( n − 1)( n − 2) ( n − 2 , 1 , 1) 2 n ( n − 1)( n − 5) ( n − 3 , 3) 6 n ( n − 2)( n − 4) ( n − 3 , 2 , 1) 3 ( n − 1)( n − 2)( n − 3) ( n − 3 , 1 , 1 , 1) 6 n ( n − 1)( n − 2)( n − 7) ( n − 4 , 4) 24 n ( n − 1)( n − 3)( n − 6) ( n − 4 , 3 , 1) 8 n ( n − 1)( n − 4)( n − 5) ( n − 4 , 2 , 2) 12 n ( n − 2)( n − 3)( n − 5) ( n − 4 , 2 , 1 , 1) 8 6 / 16 6/16 .

. Young’s Orthogonal Representation Let τ k be the adjacent transposition ( k, k +1) . The action of τ k on a tableau t is to swap the numbers k and k +1 . 1 3 5 6 2 4 7 Define c t ( i ) as the column index minus the row index of the cell in t where i is found. Define the signed distance d t ( i, j ) = c t ( j ) − c t ( i ) . In YOR the matrix ρ λ ( τ k ) is defined as follows. Take any t ∈ T λ . In row t : • The diagonal element is [ ρ λ ( τ k )] t,t = 1 /d t ( k, k +1) • If τ k ( t ) is a standard tableau, then we also have the off-diagonal element √ 1 − 1 /d t ( k, k + 1) 2 . [ ρ λ ( τ k )] t,τ k ( t ) = • All other elements are zero. 7 / 16 7/16 .

. Young’s Orthogonal Representation • Adapted to the subgroup chain S n > S n − 1 > S n − 2 > . . . > S 1 . • The matrices of adjacent transpositions are very sparse ( ≤ 2 nonzeros in each row). Any contiguous cycle � i, j � = ( i, i +1 , . . . , j ) can be written as a product of j − i adjacent transpositions: � i, j � = τ i τ i +1 . . . τ j − 1 . Any σ ∈ S n can be written as a product of at most n − 1 contiguous cycles (insertion sort). 8 / 16 8/16 .

Clausen’s FFT on S n

. Separation of variables Let G be a finite group, f : G → C and H < G . • For each g ∈ G/H , define f g ( x ) = f ( gx ) . • Compute each sub-FT { � f g ( ρ ′ ) } ρ ′ ∈R H . • Assemble { � f g ( ρ ′ ) } ρ ′ ∈R G . • ∑ � ρ ( g ) � f ( ρ ) = f g ( ρ ) ρ ∈ R G . g ∈ G/H 10 / 16 10/16 .

. Clausen’s FFT (1989) In a representation adapted to S n > . . . > S 1 , define f i ( τ ) = f ( � i, n � τ ) . n ∑ ∑ ∑ � f ( ρ ) = f ( σ ) ρ λ ( σ ) = f ( � i, n � τ ) ρ λ ( � i, n � τ ) = σ ∈ S n i =1 τ ∈ S n − 1 n n ∑ ∑ ∑ ⊕ ∑ ρ λ ( � i, n � ) f i ( τ ) ρ λ ( τ ) = ρ λ ( � i, n � ) f i ( τ ) ρ λ ′ ( τ ) = i =1 τ ∈ S n − 1 i =1 τ ∈ S n − 1 λ ′ ∈ λ ↓ n − 1 n ∑ ⊕ � f i ( λ ′ ) = ρ λ ( � i, n � ) i =1 λ ′ ∈ λ ↓ n − 1 where λ ↓ n − 1 := { λ ′ ⊢ n − 1 | λ ′ ≤ λ } . 11 / 16 11/16 .

. The Bratelli diagram 12 / 16 12/16 .

. Complexity • The multiplication ρ λ ( σ ) · M takes only 2 d 2 λ operations. • Therefore, computing ρ λ ( � i, k � ) ⊕ λ ′ � f i ( ρ λ ′ ) takes 2( k − i ) d 2 λ ops. • ∏ λk d 2 λ = k ! , but at level k need n ! /k ! FT’s. Total: n k n ∑ ∑ ∑ k ( k − 1) = ( n +1) n ( n − 1) 2 n ! ( k − i ) = 2 n ! n ! , 2 3 k =1 i =1 k =1 • Complexity of inverse transform the same (by unitarity of each level). • May be possible to improve [Maslen, 1998][Maslen & Rockmore, 2000] 13 / 16 13/16 .

. 14 / 16 14/16 .

. FFT on homogeneous spaces f ( ρ ) := � Recall: if f : G/H → C , then � ( f ↑ G )( ρ ) where f ↑ G ( g ) := f ( gx 0 ) . This gives rise to a column sparse structure in the � f ( ρ ) matrices determined by the multiplicity of the trivial irrep in ρ ↓ H . The columns in � f ( ρ λ ) are indexed by the paths from λ 1 = (1) to λ in the Bratelli diagram. 15 / 16 15/16 .

. Example: S n / S n − 2 Proposition. The Fourier transform of a function on S n / S n − 2 has the following structure: • The one dimensional matrix ρ ( n ) • Two columns in ρ ( n − 1 , 1) • One column in ρ ( n − 2 , 2) • One column in ρ ( n − 2 , 1 , 1) . 16 / 16 16/16 .