aa r X i v : . [ m a t h . K T ] A p r On the Schr¨odinger group
Guy Roger Biyogmam .Department of Mathematics,Southwestern Oklahoma State University,Weatherford, OK 73096, USA
Email:[email protected]
Abstract
In this paper, we compute the Leibniz homology of the Schr¨odinger algebra. Weshow that it is a graded vector space generated by tensors in dimensions 2 n − n . The Leibniz homology of the full Galilei algebra is also calculated. Mathematics Subject Classifications(2000):
Key Words : Leibniz homology, Galilei algebra, Schr¨odinger algebra.
Leibniz homology was introduced by Jean-Louis Loday (see [7, 10.6]) as a non commutativeversion of Lie algebra homology. In this paper, we calculate this homology for one of the mostimportant non semisimple Lie algebra of mathematical physics. Thanks to its semidirect sumstructure, the Schr¨odinger algebra is presented as an abelian extension of a semisimple Liealgebra. This enables us to calculate its Leibniz (co)homology using techniques previouslyapplied on the Lie algebra of the euclidean group [1], the affine symplectic Lie algebra [9]and the Poincar´e algebra [2]. Recall that the (non centrally extended or massless) full Galileigroup ] GAL ( n ) of n -dimensional space consists of real ( n + 2) × ( n + 2) matrices X v a A n B n C n D n ( M. X ∈ O ( n ; R ) , a, v ∈ R n and A n , B n , C n , D n ∈ R with A n D n − B n C n = 0 . Its groupstructure is the semidirect product ] GAL ( n ) = ( O ( n ; R ) × GL (2; R )) ⋉ ( R n × R n ) . With the condition A n D n − B n C n = 1 , the matrices ( M.
1) above constitute the Schr¨odingergroup
Sch ( n ) . Its Lie algebra sch n is an abelian extension of the Lie algebra¯ h n = so ( n ; R ) ⊕ sl (2; R ) .
1e calculate its Leibniz homology and obtain the isomorphism of graded vector spaces HL ∗ ( sch n ; R ) ∼ = (cid:0) R ⊕ h ˜ ζ n i ⊕ h ˜ α n i (cid:1) ⊗ T ∗ (˜ γ n ) , where h ˜ ζ n i denotes a 1-dimensional vector space in dimension 2 n − sch n -invariant˜ ζ n = 1(2 n )! X σ ∈ S n − sgn( σ ) y σ (1) ⊗ . . . d y σ ( i ) . . . ⊗ y σ ( n ) ⊗ y σ ( n +1) ⊗ . . . \ y σ ( n + i ) . . . ⊗ y σ (2 n ) , h ˜ α n i denotes a 1-dimensional vector space in dimension 2 n generated by the sch n -invariant˜ α n = 1(2 n )! X σ ∈ S n sgn( σ ) y σ (1) ⊗ . . . ⊗ y σ ( n ) ⊗ y σ ( n +1) ⊗ . . . ⊗ y σ (2 n ) and T ∗ (˜ γ n ) denotes the tensor algebra on a (2 n − γ n = X ≤ i Let −→ I i −→ g e π −→ g −→ be an abelian extension of a (semi)-simple Lie algebra g over R . Then the following arenatural vector space isomorphisms H Lie ∗ ( g e ; R ) ∼ = H Lie ∗ ( g ; R ) ⊗ [ ∧ ∗ ( I )] g ,H Lie ∗ ( g e ; g e ) ∼ = H Lie ∗ ( g ; R ) ⊗ [ H Lie ∗ ( I ; g e )] g . Proof. The proof consists of applying the Hochschild-Serre spectral sequence to the (semi)-simple Lie algebra g , subalgebra of g e . See [10, lemma 2.1] for details when g is simple. Remark 2.2. The (co)homology groups of sl (2; R ) and so ( n ; R ) are known (see [5, p.1742]).So by lemma 2.1, to determine H Lie ∗ ( sch n ; R ) and H Lie ∗ ( sch n ; sch n ) , it is enough to determinethe appropriate modules of g -invariants. Recall also that for a Leibniz algebra ( Lie algebra in particular) g , the Leibniz homologyof g with coefficients in R denoted HL ∗ ( g , R ) , is the homology of the Loday complex T ∗ ( g ) , namely R ←− g [ , ] ←− g ⊗ d ←− . . . d ←− g ⊗ n − d ←− g ⊗ n ← . . . where g ⊗ n is the n th tensor power of g over R , and where d ( g ⊗ g ⊗ . . . ⊗ g n ) = X ≤ i Assume that R n is given the coordinates ( x , x , ..., x n ) , and let ∂∂x i be the unit vector fieldsparallel to the x i axes respectively. It is easy to show that the Lie algebra generated by thefamily ¯ B below of vector fields (endowed with the bracket of vector fields) is isomorphic to¯ h n : ¯ B = { X ij , a n , b n , c n } X ij := − x i ∂∂x j + x j ∂∂x i ≤ i < j ≤ n,a n := − x n +1 ∂∂x n +1 + x n +2 ∂∂x n +2 ,b n := x n +1 ∂∂x n +2 ,c n := − x n +2 ∂∂x n +1 . The brackets relations of the Lie algebra ¯ h n are:[ X ij , X ik ] = X jk , [ X ij , a n ] = 0 , [ X ij , b n ] = 0 , [ X ij , c n ] = 0 , [ a n , b n ] = − b n , [ a n , c n ] = 2 c n , [ b n , c n ] = a n . Remark 3.1. The brackets above yield the following Lie algebra isomorphisms sl (2; R ) ∼ = span { a n , b n , c n } , so ( n ; R ) ∼ = span { X ij , ≤ i < j ≤ n } . Denote by I n the Lie algebra of R n × R n . The following is a vector space basis of I n .B = { x i ∂∂x n +1 , x i ∂∂x n +2 ≤ i ≤ n } . Then the Schr¨odinger algebra sch n has an R -vector space basis ¯ B ∪ B and there is a shortexact sequence of Lie algebras [6, p.203]0 −→ I n i −→ sch n π −→ ¯ h n −→ i is the inclusion map and π is the projection sch n −→ ( sch n / I n ) ∼ = ¯ h n . Note thathere, I n is the standard representation of ¯ h n i.e ¯ h n acts on I n via matrix multiplicationon vectors. More precisely, I n is an abelian ideal of sch n acting on sch n via the followingbrackets of vector fields:[ X ij , x i ∂∂x n +1 ] = x j ∂∂x n +1 , [ X ij , x i ∂∂x n +2 ] = x j ∂∂x n +2 , [ a n , x i ∂∂x n +1 ] = x i ∂∂x n +1 , [ a n , x i ∂∂x n +2 ] = − x i ∂∂x n +2 , [ b n , x i ∂∂x n +1 ] = − x i ∂∂x n +2 , [ b n , x i ∂∂x n +2 ] = 0 , [ c n , x i ∂∂x n +1 ] = 0 , [ c n , x i ∂∂x n +2 ] = x i ∂∂x n +1 , [ x i ∂∂x n +1 , x j ∂∂x n +2 ] = 0 . In this framework, X ij , a n , b n , c n , x i ∂∂x n +1 and x i ∂∂x n +2 represent respectively the gener-ators of the rotations, dilation, time translation (Hamiltonian), conformal transformation,Galilean boosts and space translations (momentum operators).Also the Lie algebra ¯ h n acts on I n and sch n via the bracket of vector fields. This actionis extended to I ∧ kn by[ α ∧ α ∧ . . . ∧ α k , X ] = k X i =1 α ∧ α ∧ . . . ∧ [ α i , X ] ∧ . . . ∧ α k α i ∈ I n , X ∈ ¯ h n , and the action of ¯ h n on sch n ⊗ I ∧ kn is given by[ g ⊗ α ∧ α ∧ . . . ∧ α k , X ] = [ g, X ] ⊗ α ∧ . . . ∧ α k + k X i =1 g ⊗ α ∧ α ∧ . . . ∧ [ α i , X ] ∧ . . . ∧ α k for g ∈ sch n . For the remaining of the paper, we write so ( n ) , gl (2) and sl (2) for so ( n ; R ) , gl (2; R ) and sl (2; R ) respectively. Lemma 3.2. There is a graded vector space isomorphism: [ ∧ ∗ ( I n )] ¯ h n ∼ = R ⊕ h β n i ⊕ h ζ n i ⊕ h α n i where α n = y ∧ . . . ∧ y n ∧ y n +1 ∧ . . . ∧ y n , β n = n X i =1 y i ∧ y n + i and ζ n = n X i =1 y ∧ . . . b y i . . . ∧ y n ∧ y n +1 . . . d y n + i . . . ∧ y n with y i := x i ∂∂x n +1 and y n + i := x i ∂∂x n +2 for ≤ i ≤ n. Proof. Clearly [ R ] ¯ h n = R . Now let ω ∈ I n . Then ω = X ≤ i ≤ n c i x i ∂∂x n +1 + X ≤ i ≤ n c ′ i x i ∂∂x n +2 forreal constants c i , c ′ i . Assume without loss of generality that c i = 0 for some i = n. Then[ ω, X i n ] = − c i x n ∂∂x n +1 + c n x i ∂∂x n +1 = 0 . So ω / ∈ [ I n ] ¯ h n and thus [ I n ] ¯ h n = 0 . Now write I n = I n ⊕ I n with I n = (cid:10) y , . . . , y n (cid:11) and I n = (cid:10) y n +1 , . . . , y n (cid:11) . Then since so ( n ) is a Liesubalgebra of ¯ h n , it follows by [1, lemma 4.1] that (cid:2) ∧ k ( I in ) (cid:3) ¯ h n ⊆ (cid:2) ∧ k ( I in ) (cid:3) so ( n ) = R if k = 0 (cid:10) x ∂∂x n + i ∧ . . . ∧ x n ∂∂x n + i (cid:11) if k = n i = 1 , . However, x ∂∂x n + i ∧ . . . ∧ x n ∂∂x n + i / ∈ [ ∧ k ( I in )] ¯ h because (cid:2) x ∂∂x n + i ∧ . . . ∧ x n ∂∂x n + i , a n (cid:3) = ( − i n x ∂∂x n + i ∧ . . . ∧ x n ∂∂x n + i = 0 . Now let ω = X A ∗ ,B ∗ c ∗∗ A ∗ ∧ B ∗ ∈ (cid:2) ( I n ) ∧ r ∧ ( I n ) ∧ s (cid:3) ¯ h n with A ∗ ∈ ( I n ) ∧ r and B ∗ ∈ ( I n ) ∧ s . If r = s, we have [ ω, a n ] = (cid:2) X A ∗ ,B ∗ c ∗∗ A ∗ ∧ B ∗ , a n (cid:3) = ( s − r ) ω = 0 . This is a contradiction.If r = s = 1 , one easily shows that [( I n ) ∧ ( I n )] ¯ h n = h β n i . r = s = n − , one also shows that[( I n − ) n − ∧ ( I n − ) n − ] ¯ h n = h ζ n i . If r = s = n, a straightforward calculation shows that[( I n ) ∧ n ∧ ( I n ) ∧ n ] ¯ h n = h α n i . For 1 < r = s < n − , we show that [( I n ) ∧ s ∧ ( I n ) ∧ s ] ¯ h n = 0 by showing by induction on n that [( I n ) ∧ s ∧ ( I n ) ∧ s ] so ( n ) = 0 . Indeed, it is easy to check the result for n = 4 and r = s = 2 . By the inductive hypothesis, suppose [( I n − ) ∧ s ∧ ( I n − ) ∧ s ] so ( n − = 0 for s = 0 , , n − , n − z ∈ [( I n ) ∧ s ∧ ( I n ) ∧ s ] so ( n ) with s = 0 , , n − , n fixed. Then z = c A ∧ B + c A ∧ B ∧ y n + c A ∧ B ∧ y n + c A ∧ B ∧ y n ∧ y n where A , A ∈ ( I n − ) ∧ s , B , B ∈ ( I n − ) ∧ s , A , A ∈ ( I n − ) ∧ s − , B , B ∈ ( I n − ) ∧ s − , and c , c , c , c ∈ R . Let X ∈ so ( n − ⊆ so ( n ) as a Lie subalgebra, we have0 = [ z, X ] = c [ A ∧ B , X ]+ c [ A ∧ B , X ] ∧ y n + c [ A ∧ B , X ] ∧ y n + c [ A ∧ B , X ] ∧ y n ∧ y n . If the coefficients are non-zero, then the terms [ A ∧ B , X ] , [ A ∧ B , X ] , [ A ∧ B , X ]and [ A ∧ B , X ] are all zero By linear independence. This implies that A ∧ B ∈ [( I n − ) ∧ s − ∧ ( I n − ) ∧ s ] so ( n − = 0 , A ∧ B ∈ [( I n − ) ∧ s ∧ ( I n − ) ∧ s − ] so ( n − = 0by the case r = s above, and A ∧ B ∈ [( I n − ) ∧ s ∧ ( I n − ) ∧ s ] so ( n − = 0 , A ∧ B ∈ [( I n − ) ∧ s − ∧ ( I n − ) ∧ s − ] so ( n − = 0by inductive hypothesis. Hence z = 0. Lemma 3.3. For all integer k, [ sl (2) ⊗ I ∧ kn ] ¯ h n = 0 Proof. Let k and k with 0 ≤ k , k ≤ n and let ω ∈ [ sl (2) ⊗ ( I n ) ∧ k ∧ ( I n ) ∧ k ] ¯ h . Then ω = X A ∗ ,B ∗ c ∗∗ a a n ⊗ A ∗ ∧ B ∗ + X A ∗ ,B ∗ c ∗∗ b b n ⊗ A ∗ ∧ B ∗ + X A ∗ ,B ∗ c ∗∗ c c n ⊗ A ∗ ∧ B ∗ where A ∗ ∈ ( I n ) ∧ k , B ∗ ∈ ( I n ) ∧ k , and c ∗∗ a , c ∗∗ b , c ∗∗ c are real coefficients.0 = [ ω, a n ] =( k − k ) X A ∗ ,B ∗ c ∗∗ a a n ⊗ A ∗ ∧ B ∗ + ( k − k + 2) X A ∗ ,B ∗ c ∗∗ b b n ⊗ A ∗ ∧ B ∗ + ( k − k − X A ∗ ,B ∗ c ∗∗ c c n ⊗ A ∗ ∧ B ∗ . If k = k , k − , k +2 , this implies by linear independence that all the coefficients c ∗∗ a , c ∗∗ b , c ∗∗ c are zero and thus ω = 0 . However if k = k , only the coefficients c ∗∗ b , c ∗∗ c are zero by linearindependence. So ω = X A ∗ ,B ∗ c ∗∗ a a n ⊗ A ∗ ∧ B ∗ . Now0 = [ ω, c n ] = 2 X A ∗ ,B ∗ c ∗∗ a c n ⊗ A ∗ ∧ B ∗ + X A ∗ ,B ∗ c ∗∗ a a n ⊗ A ∗ ∧ [ B ∗ , c n ] . c ∗∗ a are zero by linear independence. Hence ω = 0 . If k = k − , only the coefficients c ∗∗ a , c ∗∗ c are zero by linear independence.So ω = X A ∗ ,B ∗ c ∗∗ b b n ⊗ A ∗ ∧ B ∗ . Now0 = [ ω, c n ] = X A ∗ ,B ∗ c ∗∗ b a n ⊗ A ∗ ∧ B ∗ + X A ∗ ,B ∗ c ∗∗ b b n ⊗ A ∗ ∧ [ B ∗ , c n ] . This implies that the coefficients c ∗∗ b are zero by linear independence. Hence ω = 0 . Similarly, if k = k +2 , only the coefficients c ∗∗ a , c ∗∗ b are zero by linear independence. Now thecondition [ ω, c n ] = 0 annihilates the coefficients c ∗∗ c by linear independence. Hence ω = 0 . Lemma 3.4. The following are verctor space isomorphisms [ so ( n ) ⊗ I ∧ kn ] ¯ h n = , if k = 2 , n − h ρ n i , if k = 2 h γ n i , if k = n − where ρ n = X ≤ i Clearly, [ so ( n )] ¯ h n = 0 since [ X ij , X ik ] = X jk = 0 . Also since so ( n ) is a subalgebra of¯ h n , it follows by [1, lemma 4.2] that [ so ( n ) ⊗ ∧ k ( I n )] ¯ h n is a submodule of[ so ( n ) ⊗∧ k ( I n )] so ( n ) = , if k = 2 , n − (cid:10) P ≤ i 7s [ A ∗ , a n ] = − rA ∗ and [ B ∗ , a n ] = sB ∗ . So all the c ∗∗ ij s are zero by linear independence. So ω = 0 . For the case r = s, first notice that since so ( n ) is a Lie subalgebra of ¯ h n , it followsthat [ so ( n ) ⊗ ∧ k ( I n )] ¯ h ⊆ [ so ( n ) ⊗ ∧ k ( I n )] so ( n ) . Now following again the proof of [1, lemma4.2], we have that dim [ so ( n ) ⊗ I ∧ kn ] so ( n ) = dimHom so ( n ) ( I ∧ n , I ∧ kn ) = ( , if k = 2 , n − , else . So two cases remain here for possible ¯ h n -invariants: Firstly, if r = s = 1 we have ω = X ≤ i For all integer k, [ I n ⊗ I ∧ kn ] ¯ h n = 0 . Proof. Write I n = I n ⊕ I n as in the proof of lemma 3.2. Note that since so ( n ) is a Liesubalgebra of ¯ h n , it follows by [1, lemma 4.3] that [ I in ⊗ ∧ k ( I in )] ¯ h n ⊆ [ I in ⊗ ∧ k ( I in )] so ( n ) = 0for all k = 1 , n − , for i = 1 , . Now by [1, lemma 4.4],[ I n ⊗ I n ] ¯ h n ⊆ [ I n ⊗ I n ] so ( n ) = (cid:10) n X i =1 y i ⊗ y i (cid:11) and by [1, lemma 4.5],[ I n ⊗ ∧ n − ( I n )] ¯ h n ⊆ [ I n ⊗ ∧ n − ( I n )] so ( n ) = (cid:10) n X m =1 ( − m − y m ⊗ y ∧ y . . . c y m . . . ∧ y n (cid:11) . However [ n X i =1 y i ⊗ y i , b n ] = − n X i =1 (cid:0) y i ⊗ y n + i + y n + i ⊗ y i (cid:1) = 0and[ P nm =1 ( − m − y m ⊗ y ∧ . . . c y m . . . ∧ y n , b n ] = − P nm =1 ( − m y n + m ⊗ y ∧ . . . c y m . . . ∧ y n + P ni =1 P nm =1 ( − m − y m ⊗ y ∧ . . . ∧ y n + i ∧ . . . c y m . . . ∧ y n = 0 . So[ I n ⊗ ∧ n − ( I n )] ¯ h n = 0 and [ I n ⊗ I n ] ¯ h n = 0 . Similarly, [ I n ⊗ ∧ n − ( I n )] ¯ h n = 0 and [ I n ⊗ I n ] ¯ h n = 0 . k and k with 1 ≤ k , k ≤ n and let ω ∈ [ I n ⊗ ( I n ) ∧ k ∧ ( I n ) ∧ k ] ¯ h n . Then ω = X ≤ i ≤ nA ∗ ,B ∗ c ∗∗ i y i ⊗ A ∗ ∧ B ∗ + X ≤ i ≤ nA ∗ ,B ∗ c ∗∗ n + i y n + i ⊗ A ∗ ∧ B ∗ with A ∗ ∈ ( I n ) ∧ k , B ∗ ∈ ( I n ) ∧ k and c ∗∗ i , c ∗∗ n + i real coefficients. We have0 = [ ω, a n ] =( k − k − X A ∗ ,B ∗ c ∗∗ i y i ⊗ A ∗ ∧ B ∗ + ( k − k + 1) X A ∗ ,B ∗ c ∗∗ n + i y n + i ⊗ A ∗ ∧ B ∗ . Two cases occur: Firstly, if k = k + 1 , all the coefficients c ∗∗ n + i are zero. Now since0 = [ ω, b n ] = X ≤ i ≤ nA ∗ ,B ∗ c ∗∗ i y n + i ⊗ A ∗ ∧ B ∗ + X ≤ i ≤ nA ∗ ,B ∗ c ∗∗ i y i ⊗ [ A ∗ , b n ] ∧ B ∗ , it follows that all the coefficients c ∗∗ i are zero by linear independence. So ω = 0 . Secondly, if k = k − , all the coefficients c ∗∗ i are zero. Similarly, the condition 0 = [ ω, c n ] annihilatesall the coefficients c ∗∗ n + i by linear independence. So ω = 0 . As a consequence of these lemmas we have the following: Theorem 3.6. There are graded vector space isomorphisms H Lie ∗ ( sch n ; R ) ∼ = H Lie ∗ ( sl (2); R ) ⊗ H Lie ∗ ( so ( n ); R ) ⊗ (cid:0) R ⊕ h ζ n i ⊕ h α n i (cid:1) , and HL ∗ ( sch ; R ) ∼ = (cid:0) R ⊕ h ˜ ζ n i ⊕ h ˜ α n i (cid:1) ⊗ T ∗ (˜ γ n ) , where ˜ α n = 1(2 n )! X σ ∈ S n sgn ( σ ) y σ (1) ⊗ . . . ⊗ y σ ( n ) ⊗ y σ ( n +1) ⊗ . . . ⊗ y σ (2 n ) is the antisymmetriza-tion of α n , ˜ ζ n = 1(2 n )! X σ ∈ S n − sgn ( σ ) y σ (1) ⊗ . . . d y σ ( i ) . . . ⊗ y σ ( n ) ⊗ y σ ( n +1) ⊗ . . . \ y σ ( n + i ) . . . ⊗ y σ (2 n ) is the antisymmetrization of ζ n and ˜ γ n is the ¯ h n -invariant cycle in sch ⊗ ( n − n representing γ n . Proof. The first isomorphism follows by lemma 2.1, lemma 3.2 and the K¨unneth formula forLie algebra homology. Note that the ¯ h n - invariants β n is zero in H Lie ∗ ( sch n ; R ) since d ( ¯ ρ n ) = − n − β n where d is the Chevalley-Eilenberg boundary map and¯ ρ n = X ≤ i There is a graded vector space isomorphism HL ∗ ( f gal n ; R ) ∼ = (cid:0)(cid:0) R ⊕ h ˜ ζ n i ⊕ h ˜ α n i (cid:1) ⊗ T ∗ (˜ γ n ) (cid:1) ∗ T ∗ ( R ) where ∗ is the non-commutative tensor product of N -graded modules.Proof. Since f gal n ∼ = sch n ⊕h d n i and H Lie ∗ ( h d n i ; R ) ∼ = T ∗ ( R ) , the result follows by [8, theorem3] and theorem 3.6. References [1] Biyogmam, G. R., On the Leibniz (Co)homology of the Lie Algebra of the EuclideanGroup, Journal of Pure and Applied Algebra, (2011), 1889 -1901.[2] Biyogmam, G. R., Leibniz Homology of the Affine Indefinite Orthogonal Lie Algebra, Cohomology theory of Lie groups and Lie algebras , Trans.Amer. Math. Soc, , 1 (1948), 85-124.104] Hilton, P. J., Stammbach, U., A course in homological algebra , Springer-Verlag, NewYork, 1971.[5] Ito, K., Nihon, S., Encyclopedic Dictionary of Mathematics , Cambridge, Mass: MITPress, 1987.[6] Kostrikin, A. I., Manin, I.,“ Linear Algebra and Geometry, Algebra, logic, and appli-cations”, Gordon and Breach Science Publishers, Vol. 1, New York, 1989.[7] Loday, J.-L., “Cyclic Homology, Springer-Verlag”, Berlin, Heidelberg, New York,1992.[8] Loday, J.-L., K¨unneth-style formula for the homology of Leibniz algebras , Mathema-tische Zeitschrift, (1996), 41-47.[9] Lodder, J. M., Lie algebras of Hamiltonian vector fields and symplectic manifold , Journal of Lie Theory , , 4 (2008), 897–914.[10] Lodder, J., A Structure Theorem for Leibniz Cohomology , Journal of Algebra, , 1(2012), 93 - 110.[11] Pirasvili, T., On Leibniz Homology , Annales de l’institut Fourrier, Grenoble44