Spinorspaces in discrete Clifford analysis

0.19 MB·

by  Hilde De Ridder

Spinorspaces in discrete Clifford analysis
H. De Ridder∗, T. Raeymaekers†
7
1
0
2
Abstract
n
a In this paper we work in the ‘split’ discrete Clifford analysis setting, i.e. the m-
J dimensional function theory concerning null-functions, defined on the grid Zm, of the

6 discrete Dirac operator ∂, involving both forward and backward differences, which
2 factorizes the (discrete) Star-Laplacian (∆∗ = ∂2). We show how the space of
Mk
discrete homogeneous spherical monogenics of degree k, is decomposable into 22m−n
]
T isomorphicirreduciblerepresentationswithhighestweight k+ 1,1,...,1 intheodd-
2 2 2
R dimensional case and two times 22m−n isomorphic irreducible representations with
(cid:0) (cid:1)
. highest weight (k)′ = k+ 1,1,...,1,1 resp. (k)′ = k+ 1,1,...,1, 1 in the
h + 2 2 2 2 − 2 2 2 −2
even dimensional case.
t
a (cid:0) (cid:1) (cid:0) (cid:1)
m
MSC 2010: 17B15,47A67, 20G05, 15A66, 39A12
[
Keywords: discrete Clifford analysis, irreducible representation, orthogonal Lie al-

1 gebra, monogenic functions
v
1
4 1 Introduction
7
7
In classical Clifford analysis, the infinitesimal ‘rotations’ are given by the angular momen-
0
. tum operators L = x ∂ x ∂ . These operators satisfy the commutation relations
1 a,b a xb − b xa
0
7 [La,b,Lc,d] = δb,cLa,d δb,dLa,c δa,cLb,d+δa,dLb,c,
− −
1
: which are exactly the defining relations of the special orthogonal Lie algebra so(m) and
v
i they form endomorphisms of the space k(m,C) of scalar-valued harmonic homogeneous
X H
polynomials, thus transforming the latter in an (irreducible) so(m,C)-representation. To
ar establish k(m,S), i.e. thespinor-valuedhomogeneous monogenics of degreek, classically
M
as so(m,C)-representation, the following operators are considered
1
dR(e ) : (m,S) (m,S), M L + e e M .
a,b k k k a,b a b k
M → M 7→ 2
(cid:18) (cid:19)
∗Ghent University, Department of Mathematical Analysis, Building S22, Galglaan 2, 9000 Gent, Bel-
gium, email: [email protected]
†Ghent University, Department of Mathematical Analysis, Building S22, Galglaan 2, 9000 Gent, Bel-
gium, email: [email protected]
1
These operators are endomorphisms of the space of spinor-valued k-homogeneous polyno-
mials which also satisfy the defining relations of so(m,C):
[dR(e ),dR(e )] =δ dR(e ) δ dR(e ) δ dR(e )+δ dR(e ).
a,b c,d b,c a,d b,d a,c a,c b,d a,d b,c
− −
In [3], we developed similar operators in the discrete Clifford analysis setting: the angular
momentum operators are discrete operators L = ξ ∂ +ξ ∂ , a = b. For a = b, we define
a,b a b b a
6
L = 0. Then the operators Ω , acting on discrete functions f as Ω f = L fe e ,
aa a,b a,b a,b b a
satisfy the defining relations of the special lie algebra so(m):
[Ω ,Ω ] = δ Ω δ Ω δ Ω +δ Ω .
a,b c,d b,c a,d b,d a,c a,c b,d a,d b,c
− −
Furthermore, they are endomorphisms of the space of Clifford-algebra valued homoge-
k
n[4e]o,uwsehsahromwoednitchsaotf degirsetehek,susimncoefΩ22am,b icsoommmoruptheiscwHcoitphiesslo2f=the∆ir,rξed2,uEci+blem2rep,r∀es(ean,tba).tioInn
k
H (cid:8) (cid:9)
of so(m,C) with highest weight (k,0,...,0).
Thediscrete Dirac operator ∂ is however not invariant underthe operators Ω , hence
a,b k
M
cannotbeexpressed as so(m,C)-representation by meansof these operators. Therefore,we
1
consideredin[3]theoperatorsL andthefour-vectorV = e e e⊥e⊥ = e⊥e e⊥e .
a,b−2 a,b a b a b − a a b b
Let the operator dR(e ), a = b, act on discrete functions f as
a,b
6
1
dR(e )f = V L fe⊥e⊥.
a,b a,b a,b− 2 a b
(cid:18) (cid:19)
For a = b, we defined dR(e ) = 0. The operators dR(e ) satisfy the defining relations
a,a a,b
of the special lie algebra so(m):
[dR(e ),dR(e )] =δ dR(e ) δ dR(e ) δ dR(e )+δ dR(e ),
a,b c,d b,c a,d b,d a,c a,c b,d a,d b,c
− −
and commute with osp(12) = ∂,ξ,E+ m which makes them endomorphisms of the
| 2
space of k-homogeneous discrete monogenic polynomials. As such, the space of k-
k
(cid:8) (cid:9) M
homogeneousClifford-valuedmonogenicpolynomialsisareducibleso(m,C)-representation.
In [3], it was already suggested that can be decomposed into irreducible parts of high-
k
est weight k+ 1, 1,...,1 resp. kM+ 1,1,..., 1 , but this was left as open conjecture.
2 2 2 2 2 −2
In the following sections, we will show how this decomposition is done exactly.
(cid:0) (cid:1) (cid:0) (cid:1)
2 Preliminaries
Let Rm be the m-dimensional Euclidian space with orthonormal basis e , j = 1,...,m and
j
consider the Clifford algebra Rm,0 over Rm. Passing to the so-called ‘split’ discrete setting
[5,1], weimbedtheCliffordalgebraRm,0 intothebiggercomplexoneC2m,0,theunderlying
2
vector space of which has twice the dimension, and introduce forward and backward basis
elements e± satisfying the following anti-commutator rules:
j
e−,e− = e+,e+ = 0, e+,e− = δ , j, ℓ = 1,...,m.
j ℓ j ℓ j ℓ jℓ
n o n o n o
Theconnection to theoriginal basis e is given by e++e− = e , j = 1,...,m. This implies
j j j j
e2 = 1, in contrast to the usual Clifford setting where traditionally e2 = 1 is chosen. We
j j −
will often denote e+ e− = e+e− e−e+, j = 1,...,m.
j ∧ j j j − j j
Now consider the standard equidistant lattice Zm; the coordinates of a Clifford vector x
will thus only take integer values. We construct a discrete Dirac operator factorizing the
discrete Laplacian, using both forward and backward differences ∆±, j = 1,...,m, acting
j
on Clifford-valued functions f as follows:
∆+[f]() = f( +e ) f(), ∆−[f]() = f() f( e ).
j · · j − · j · · − ·− j
With respect to the Zm-grid, the usual definition of the discrete Laplacian in x Zm is
∈
m m
∆∗[f](x)= ∆+∆−[f]= (f(x+e )+f(x e )) 2mf(x).
j j j − j −
j=1 j=1
X X
This operator is also known as “Star Laplacian”; we will from now on simply write ∆. An
appropriate definition of a discrete Dirac operator ∂ factorizing ∆, i.e. satisfying ∂2 = ∆,
is obtained by combining the forward and backward basis elements with the corresponding
forward and backward differences, more precisely
m
∂ = e+∆++e−∆− .
j j j j
Xj=1(cid:16) (cid:17)
In order to receive an analogue of the classical Weyl relations ∂ x x ∂ = δ , the
co-ordinate vector variable operators ξ = e+X−+e−X+ are defixnjedk −by tkhexirj interjakction
j j j j j
with the corresponding co-ordinate operators ∂ = e+∆++e−∆−, according to the skew
j j j j j
Weyl relations, cf. [1]
∂ ξ ξ ∂ = 1, j = 1,...,m,
j j j j
−
which imply that ∂ ξk[1] = kξk−1[1]. The operators ξ and ∂ furthermore satisfy the
j j j j j
following anti-commutator relations:
ξ ,ξ = ∂ ,∂ = ∂ ,ξ = 0, j = k, j,k = 1,...,m
j k j k j k
{ } { } { } 6
implying that ∂ ξk[1] = 0, j = ℓ.
ℓ j 6
The natural powers ξk[1] of the operator ξ acting on the ground state 1 are the basic
j j
discrete k-homogeneous polynomials of degree k in the variable x , i.e. Eξk[1] = kξk[1],
j j j
3
where E = m ξ ∂ is the discrete Euler operator. They constitute a basis for all
j=1 j j
discretepolynomials. Explicitformulasforξk[1]aregivenforexamplein[1,2]; furthermore
P j
ξk[1](x ) = 0 if k >2 x +1.
j j | j|
Adiscretefunctionisdiscreteharmonic(resp. leftdiscretemonogenic)inadomainΩ Zm
⊂
if ∆f(x) = 0 (resp. ∂f(x) = 0), for all x Ω. The space of discrete harmonic (resp.
∈
monogenic) homogeneous polynomials of degree k is denoted (resp. ), while the
k k
H M
space of all discrete harmonic (resp. monogenic) homogeneous polynomials is denoted
H
(resp. ). It is clear that
M
∞ ∞
= , = .
k k
H H M M
k=0 k=0
M M
The respective dimensions over the discrete Clifford algebra are
k+m 1 k+m 3 k+m 2
dim( )= − − , dim( ) = − .
k k
H k − k M k
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
3 Orthogonal Lie algebras
We willstartby brieflyintroducingtheorthogonal Liealgebraso(m,C); adetailed descrip-
tion can be found for example in [6]. The orthogonal Lie algebra so(m,C) is generated in
even dimension m = 2n by m basis elements H , X , Y and Z (1 6 a,b 6 n) and
2 a a,b a,b a,b
in odd dimension m = 2n+1 these basis elements are extended to a full basis of so(m,C)
(cid:0) (cid:1)
by 2n extra elements U and V , 1 6 a 6 n:
a a
so(2n,C) = span H ,X ,Y ,Z ,1 6 a,b 6 n, a = b ,
C a a,b a,b a,b
{ 6 }
so(2n+1,C) = spanC Ha,Xa,b,Ya,b,Za,b,Ua,Va,1 6 a,b 6 n, a = b .
{ 6 }
The Cartan subalgebra can be chosen as
h = H ,1 6 a 6 n ,
a
{ }
independently of the parity of the dimension, i.e. so(2n,C) and so(2n+1,C) are both Lie
algebras of rank n. The roots of so(m,C) (see also [?]) are determined by considering the
adjoint representation (1 6 a,b,c,d 6 n):
[H ,Y ]= (δ +δ )Y = ((L +L )(H ))Y ,
c a,b ca cb a,b a b c a,b
[H ,X ]= (δ δ )X = ((L L )(H ))X ,
c a,b ca cb a,b a b c a,b
− −
[H ,Z ]= (δ +δ )Z = (( L L )(H ))Z ,
c a,b ca cb a,b a b c a,b
− − −
[H ,U ]= δ U = (L (H ))U ,
c a ca a a c a
[H ,V ]= δ U = ( L (H ))U .
c a ca a a c a
− −
4
Note in particular that the Cartan subalgebra elements H can be found by means of the
a
commutator of a positive root with a negative root of the same index:
[Y ,Z ]= H H , [X ,X ] = H H .
a,b a,b a b a,b b,a a b
− − −
We thus deduce the following roots and root vectors. Here L ,1 6 a 6 n is a basis of
a
{ }
the dual vector space h∗ of the Cartan subalgebra h, i.e. L (H )= δ .
a b a,b
m = 2n m = 2n+1
root root vector root root vector
L L X L L X
a b a,b a b a,b
− −
L +L Y L +L Y
a b a,b a b a,b
L L Z L L Z
a b a,b a b a,b
− − − −
L U
a a
L V
a a
−
By the usual convention, we choose the positive roots in even dimension to be
L +L :1 6 a = b 6 n L L : 16 a < b 6 n)
a b a b
{ 6 }∪{ −
and negative roots
L L : 16 a = b 6 n L L : 16 b < a 6 n).
a b a b
{− − 6 }∪{ −
In odd dimension, one finds positive roots
L +L : 16 a =b 6 n L L : 16 a < b 6 n) L : 16 a 6n
a b a b a
{ 6 }∪{ − ∪{ }
and negative roots
L L :1 6 a = b 6n L L : 16 b < a 6 n) L :1 6 a 6 n .
a b a b a
{− − 6 }∪{ − ∪{− }
In [3], we introduced the algebra so(m,C) (up to an isomorphism) in the discrete Clifford
analysis context. The generators of so(m,C) were not given in terms of the root vectors
andCartansubalgebra,butratherbythegenerators dR(e ) :1 6 a = b 6 m ,satisfying
a,b
{ 6 }
the defining relations of so(m,C):
[dR(e ),dR(e )] = δ dR(e )+δ dR(e ) δ dR(e ) δ dR(e ). (1)
a,b c,d a,d b,c b,c a,d a,c b,d b,d a,c
− −
In the following sections, we will re-establish the orthogonal Lie algebra in the discrete
Cliffordanalysissetting,butnowbydeterminingtheexplicitexpressionsoftherootvectors
and Cartan subalgebra.
5
4 Decomposition of in irreducible representations
k
M
4.1 Even dimension m = 2n
Definition 1. We define the operators H , X , Y and Z so(m,C):
a a,b a,b a,b
∈
Ha = idR(e2a−1,2a), 16 a 6 n,
1
Xa,b = (dR(e2a−1,2b−1)+idR(e2a−1,2b) idR(e2a,2b−1)+dR(e2a,2b)),
2 −
1
Ya,b = (dR(e2a−1,2b−1) idR(e2a−1,2b) idR(e2a,2b−1) dR(e2a,2b)),
2 − − −
1
Za,b = (dR(e2a−1,2b−1)+idR(e2a−1,2b)+idR(e2a,2b−1) dR(e2a,2b)), 1 6 a,b 6 n.
2 −
Note that, because dR(e ) = dR(e ), we find that Y = Y and Z = Z . For
a,b b,a b,a a,b b,a a,b
− − −
X , we find that X = X and that X = H , hence we will only consider couples
a,b b,a a,b a,a a
6
(a,b) with a = b.
6
We will now show that these operators indeed show the expected commutator relations:
Lemma 1. The operators H , X , Y and Z , 1 6 a,b,c,d 6 n, satisfy the commutator
c a,b a,b a,b
relations given in Lemma ??; in particular:
[H ,Y ] = (δ +δ )Y = (L +L )(H )Y ,
c a,b ca cb a,b a b c a,b
[H ,X ] = (δ δ )X = (L L )(H )X ,
c a,b ca cb a,b a b c a,b
− −
[H ,Z ] = (δ +δ )Z = (L +L )(H )Z ,
c a,b ca cb a,b a b c a,b
− −
[X ,Y ] = δ Y δ Y .
a,b c,d bc a,d bd a,c
−
In particular, X , a < b resp. Y are root vectors corresponding to the positive roots
a,b a,b
L L , resp. L +L . Furthermore, X with a > b and Z are root vectors corresponding
a b a b a,b a,b
−
to the negative roots L L resp. L L .
a b a b
− − −
Proof. Since the commutator relations between the operators dR(e ) are the same as
a,b
those between the operators Ω of the harmonics, the proof is completely similar as the
a,b
proof in [4].
We already established in [?] that although is a representation of so(2n,C), by means
k
M
of the operators dR(e ), acting on , this representation is not irreducible. The de-
a,b k
M
composition is done by splitting 1 into a sum of idempotents. We will now introduce the
appropriate idempotents for this situation. For a function P L to be an eigenfunction of
k
the maximal abelian subgroup h, it must certainly hold that Le⊥ e⊥ is again equal to
2a−1 2a
L up to a (complex) constant. Consider, for a = 1,...,n, the Clifford elements
L± = e+ e− ie+ , L± = e+e− e+ ,
2a−1 2a−1 2a−1 ± 2a−1 2a 2a 2a± 2a
M± = e− e+ ie− , M± = e−e+ e− ,
2a−1 (cid:0) 2a−1 2a−1 ± 2a−1(cid:1) 2a (cid:0) 2a 2a± 2a(cid:1)
(cid:0) (cid:1) (cid:0) (cid:1)
6
For the rest of this article, we will need the following notations. For a factor F
a
∈
L±,M± , a = 1,...,m, denote
a a
{ }
0, F =L+ or M−, 0, F = L±,
F = a a a and F = a a
| a| (1, Fa =L−a or Ma+. k ak (1, Fa = Ma±.
Furthermore, denote by F the idempotent
a
e F = L∓a, if Fa = L±a,
s
(Ma∓, if Fa = Ma±.
e
Then F = 1 F and F = F .
s s s s
| | −| | k k k k
Lemma 2. The multiplication from the right on the idempotent F L±,M± by e is
e e a ∈ { a a } a
given by
F2a−1e⊥2a−1 = (−1)|F2a|+1iF2a−1,
F2ae⊥2a = ( 1)|F2a|+1F2a.
−
As a result, for 16 a 6 n, we have that
e
F2a−1F2ae⊥2a−1e⊥2a = (−1)|F2a−1|+|F2a|+1iF2a−1F2a.
We also find that for 1 6 a < b 6 n and a general idempotent F = m F , with
s=1 s
F L±,M± , we get
s s s
∈{ } Q
V2a−1,2b−1F e⊥2a−1e⊥2b−1 = (−1)|F2a−1|+|F2b−1|+kF2a−1k+kF2b−1k+1F2a,2b−1,
V2a−1,2bF e⊥2a−1e⊥2b = (−1)|F2a−1|+|F2b|+kF2a−1k+kF2bkiF2a,2b−1,
V2a,2b−1F e⊥2ae⊥2b−1 = (−1)|F2a|+|F2b−1|+kF2ak+kF2b−1kiF2a,2b−1,
V2a,2bF e⊥2ae⊥2b = ( 1)|F2a|+|F2b|+kF2ak+kF2bkF2a,2b−1.
−
where we denote, for 1 6 s1 < s2 6 m:
Fs1,s2 = F1F2... Fs1−1Fs1Fs1+1... Fs2−1Fs2Fs2+1Fs2+2...Fm−1Fm.
Proof. Note that
e e e e
L± e⊥ = e+ ie+ e− = iL± , L± e⊥ = e+ e+e− = L∓,
2a−1 2a−1 2a−1∓ 2a−1 2a−1 ∓ 2a−1 2a 2a 2a∓ 2a 2a ∓ 2a
M2±a−1e⊥2a−1 =(cid:0)−e−2a−1±ie−2a−1e+2a−(cid:1)1 =±iM2±a−1, M2±ae⊥2a =(cid:0)−e−2a±e−2ae+2(cid:1)a =±M2∓a.
We may indeed(cid:0)summarize this as (cid:1) (cid:0) (cid:1)
F2a−1e⊥2a−1 = (−1)|F2a−1|+1iF2a−1, F2ae⊥2a = (−1)|F2a|+1F2a.
e
7
From this, it follows that
F2a−1e⊥2a−1 = (−1)|F2a−1|iF2a−1, F2ae⊥2a = (−1)|F2a|F2a.
Hence, for F L±,M± , we have
a ∈e{ a a } e e
F2a−1F2ae⊥2a−1e⊥2a = F2a−1e2a−1F2ae⊥2a = (−1)|F2a|+|F2a|+1iF2a−1F2a.
Also importantto noteis that e⊥e L± =L± ande⊥e M± = M± so for theidempotent
a a a ea a a a a
F = m F , we find that −
s=1 s
Q V F = e⊥e e⊥e F = ( 1)1+kFak+kFbkF.
a,b a a b b
− −
We thus get, for F = m F , that
s=1 s
V2a−1,2b−1F e⊥2a−1e⊥2b−Q1 = (−1)1+kF2a−1k+kF2b−1kF1F2...Fme⊥2a−1e⊥2b−1
= (−1)1+kF2a−1k+kF2b−1kF1...F2a−2F2a−1e⊥2a−1F2a ...F2b−1e⊥2b−1F2bF2b+1...Fm
= ( 1)|F2a−1|+|F2b−1|+kF2a−1k+kF2b−1ki2F1F2...F2a−2F2a−1F2a ...F2b−1F2bF2b+1...Fm
− e e
= ( 1)|F2a−1|+|F2b−1|+kF2a−1k+kF2b−1k+1F2a,2b−1.
− e e
Analogously, we find that
V2a−1,2bF e⊥2a−1e⊥2b = (−1)1+kF2a−1k+kF2bkF1F2...Fme⊥2a−1e⊥2b
= (−1)1+kF2a−1k+kF2bkF1...F2a−2F2a−1e⊥2a−1F2a ...F2be⊥2bF2b+1F2b+2...Fm
= ( 1)|F2a−1|+|F2b|+kF2a−1k+kF2bkiF1F2...F2a−2F2a−1F2a ...F2b−1F2bF2b+1...Fm
− e e
= ( 1)|F2a−1|+|F2b|+kF2a−1k+kF2bkiF2a,2b−1.
− e e
Also
V2a,2b−1F e⊥2ae⊥2b−1 = (−1)1+kF2ak+kF2b−1kF1F2...Fme⊥2ae⊥2b−1
= (−1)1+kF2ak+kF2b−1kF1...F2a−1F2ae⊥2aF2a+1 ...F2b−1e⊥2b−1F2bF2b+1...Fm
= ( 1)|F2a|+|F2b−1|+kF2ak+kF2b−1kiF1F2...F2a−1F2aF2a+1 ...F2b−1F2bF2b+1...Fm
− e e
= ( 1)|F2a|+|F2b−1|+kF2ak+kF2b−1kiF2a,2b−1.
− e e e
Finally
V2a,2bF e⊥2ae⊥2b = ( 1)1+kF2ak+kF2bkF1F2...Fme⊥2ae⊥2b
−
= ( 1)1+kF2ak+kF2bkF1...F2a−1F2ae⊥2aF2a+1 ...F2be⊥2bF2b+1...Fm
−
= ( 1)|F2a|+|F2b|+kF2ak+kF2bkF1F2...F2a−1F2aF2a+1 ...F2b−1F2bF2b+1...Fm
− e e
= ( 1)|F2a|+|F2b|+kF2ak+kF2bkF2a,2b−1.
− e e e
8
Consider the basic monogenic functions
g2k = ((ξ2 ξ1)(ξ2+ξ1))k, g2k+1 = (ξ2 ξ1)((ξ2+ξ1)(ξ2 ξ1))k.
− − −
From now on we denote (k)′ = k+ 1, 1,...,1 and (k)′ = k+ 1, 1,..., 1 . We will
show under which condition+s on(cid:0)the id2em2poten2t(cid:1)F = ns−=1Fs(cid:0), the2sp2ace sp−an2C(cid:1){gkF} is
a weight space of h with weight (k)′ resp. (k)′ .
+ − Q
Lemma 3. The polynomial g F , F = m F with F L±,M± , is a weight
k ∈ Mk s=1 s s ∈ { s s }
vector of so(m,C) with
Q
weight (k)′+ when k+ F1 + F2 + F1 + F2 is even and F2a−1 + F2a + F2a−1 +
• | | | | k k k k k k k k | |
F2a is even for 26 a 6n.
| |
• weight (k)′− when k+|F1|+|F2|+kF1k+kF2k iseven, kF2a−1k+kF2ak+|F2a−1|+|F2a|
is even for 2 6 a 6 n 1 and F2n−1 + F2n + F2n−1 + F2n is odd.
− k k k k | | | |
Proof. We consider the action of the Cartan subalgebra-elements H , 1 6 s 6 n, on the
s
gkF. Since gk only contains ξ1 and ξ2, we will first consider H1:
1
H1(gkF)= iV12 L12 gkF e⊥1 e⊥2.
− 2
(cid:18) (cid:19)
We will also denote
f2k = ((ξ2+ξ1)(ξ2 ξ1))k, f2k+1 = (ξ2+ξ1)((ξ2 ξ1)(ξ2+ξ1))k.
− −
In [] it was established that ∂jgk = ( 1)jkfk−1:
−
L12gk = (ξ1∂2+ξ2∂1)gk = k(ξ1 ξ2)fk−1 = k(ξ2 ξ1)fk−1 = kgk.
− − − −
We thus get that
1
H1(gkF) = i k V12gkF e⊥1 e⊥2.
− − 2
(cid:18) (cid:19)
Now we will show that V12gk = ( 1)kgkV12: consider again V12 = e⊥1e1e⊥2e2, then
− −
e⊥1e1ξ1 = e+1e−1 e−1e+1 X1+e−1 +X1−e+1 = X1+e−1 +X1−e+1
− −
= (cid:0)X1+e−1 +X1−e(cid:1)+1(cid:0) e+1e−1 +e−1e(cid:1)+1 =(cid:0) ξ1e⊥1e1, (cid:1)
− −
e⊥1e1ξ2 = ξ(cid:0)2e⊥1e1. (cid:1)(cid:0) (cid:1)
Hence
V12(ξ2 ξ1) = e⊥1e1e⊥2e2(ξ2 ξ1)= (ξ2 ξ1)e⊥1e1e⊥2e2 = (ξ2 ξ1)V12,
± − ± ± − ±
V12((ξ2 ξ1)(ξ2+ξ1)) = ((ξ2 ξ1)(ξ2+ξ1))V12
− −
9
and so V12gk = ( 1)kgkV12. Applying this, we find that
−
1
H1(gkF) = ( 1)k+1i k+ gkV12F e⊥1 e⊥2
− 2
(cid:18) (cid:19)
1
= ( 1)k+|F1|+|F2|+kF1k+kF2k k+ g F.
k
− 2
(cid:18) (cid:19)
To be a weight vector with weight k+ 1, it must hold that
2
k+ F1 + F2 + F1 + F2 is even.
| | | | k k k k
We thus find 8 possible combinations for F1F2:
k even:
•
F F L+L+, L−L−, L+M+, L−M−, M+L+, M+M+, M−L−, M−M− .
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
∈
(cid:8) (cid:9)
k odd:
•
F F L+L−, L−L+, L+M−, L−M+, M+L−, M+M−, M−L+, M−M+ .
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
∈
(cid:8) (cid:9)
Next, we consider Ha, 1 < a 6 n. Since the generator gk only contains ξ1 and ξ2, it
vanishes under the action of L2a−1,2a. Note that V2a−1,2agk = gkV2a−1,2a since gk contains
only e± and e±. Thus
1 2
i i
Ha(gkF) = −2V2a−1,2agkF e⊥2a−1e⊥2a = −2gkV2a−1,2aF e⊥2a−1e⊥2a
i
= ( 1)kF2a−1k+kF2ak+|F2a−1|+|F2a|+1i g F
k
− 2
1
= ( 1)kF2a−1k+kF2ak+|F2a−1|+|F2a| g F.
k
− 2
This equals +21 gkF when kF2a−1k + kF2ak + |F2a−1| + |F2a| is even and −12gkF when
F2a−1 + F2a + F2a−1 + F2a is odd. We may thus conclude that the statement holds.
k k k k | | | |
We find that F2a−1 + F2a + F2a−1 + F2a is even for F2a−1F2a in
k k k k | | | |
L+ L+,L− L−,L+ M+,L− M−,M+ L+,M+ M+,M− L−,M− M−
2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a
(cid:8) (cid:9)
and odd for F2a−1F2a in
L+ L−,L− L+,L+ M−,L− M+,M+ L−,M+ M−,M− L+,M− M+ .
2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a 2a−1 2a
(cid:8) (cid:9)
10