Simplicity of a vertex operator algebra whose Griess algebra is the Jordan algebra of symmetric matrices

0.3 MB·English

by  Hidekazu Niibori

Simplicity of a vertex operator algebra whose Griess
algebra is the Jordan algebra of symmetric matrices
Hidekazu Niibori
Graduate School of Pure and Applied Sciences, University of Tsukuba,
Tsukuba, Ibaraki 305-8571, Japan (e-mail: [email protected])
9 and
0
0
Daisuke Sagaki
2

n Institute of Mathematics, University of Tsukuba,
a
J Tsukuba, Ibaraki 305-8571, Japan (e-mail: [email protected])

7

]
A Abstract
Q
Let r ∈ C be a complex number, and d ∈ Z a positive integer greater than
. ≥2
h or equal to 2. Ashihara and Miyamoto introduced a vertex operator algebra V of
t J
a central charge dr, whose Griess algebra is isomorphic to the simple Jordan algebra
m
of symmetric matrices of size d. In this paper, we prove that the vertex operator
[ algebra VJ is simple if and only if r is not an integer. Further, in the case that
1 r is an integer (i.e., V is not simple), we give a generator system of the maximal
J
v
proper ideal I of the VOA V explicitly.
1 r J
4
8
0 1 Introduction.
.
1
0 Let V = V be a vertex operator algebra (VOA for short) over the field C of
9 n∈Z≥0 n
complex numbers with Y(·, z) : V → (EndV)[[z,z−1]] the vertex operator. As usual,
0 L
:
v for each v ∈ V, we define v ∈ EndV, m ∈ Z, by: Y(v, z) = v z−m−1. It
m m∈Z m
i
X follows from the axiom of a VOA that V becomes a C-algebra with the product given
2 P
r
a by u·v := u v ∈ V for u, v ∈ V . In addition, we know (see [FLM, §10.3] and also [M,
1 2 2
§5]) that if dimV = 1 and dimV = 0, then the C-algebra V is commutative (but not
0 1 2
necessarily associative), which we call the Griess algebra of V.
Various kinds of commutative C-algebras appear as the Griess algebras of VOAs. The
Griess algebra of the moonshine VOA V♮ is isomorphic to the 196884-dimensional, com-
mutative C-algebra introduced by Griess [G], whose automorphism group is isomorphic
to the Monster sporadic simple group (see [FLM, p.319]); the name “Griess algebra” is
derived from this fact. Also, for a given associative, commutative C-algebra A equipped
1
with an A-invariant bilinear form, Lam [L1] constructed a VOA whose Griess algebra
is isomorphic to A. Further, in [L2], [AM], [As], they constructed some VOAs whose
Griess algebras are (simple) Jordan algebras; for the definition of a Jordan algebra, see
§2.2 below. In this paper, we will mainly treat VOAs introduced in [L2, §4.1] and [AM],
whose Griess algebras are isomorphic to the simple Jordan algebra Sym (C) of symmetric
d
matrices of size d ∈ Z with entries in C. Because the Jordan algebra Sym (C) has a
≥2 d
strong connection to symmetric cones and zeta functional equations (see [FK]), Ashihara,
Miyamoto (see [AM, Introduction]), and the authors of this paper expect that the results
in [AM] and this paper contribute a VOA theoretical approach to the theory of symmetric
cones and zeta functional equations.
An essential difference between the VOA introduced in [L2, §4.1] and the one intro-
duced in [AM] (which we denote by V ) is their central charges. The central charge of the
J
former is equal to the (fixed) positive integer d ∈ Z , the size of symmetric matrices. On
≥2
the other hand, the central charge of the later is equal to dr, where r is an arbitrary com-
plex number (see Theorem 2.5 below). In general, the structure and the representation
theory of a VOA deeply depends on its central charge. For example, it is well-known that
the simplicity of a Virasoro VOA M /hL 1i (with notation in [W]) and the rationality
c,0 −1
of a simple Virasoro VOA V (with notation in [W]) depend on their central charges c (see
c
[W] and also [DMZ]). Moreover, rational Virasoro VOAs (i.e., V of special central charge
c
c, such as V ) and their irreducible modules play very important roles in the theory of
1/2
VOAs. So, as in the case of Virasoro VOAs, it is quite natural and important to study
how the VOA V introduced in [AM] depends on its central charge dr. In this paper,
J
we study the condition of r ∈ C for the VOA V to be simple. The main result of this
J
paper is the following theorem, which means that the simplicity of V also depends on its
J
central charge.
Theorem. Keep the notation above. The VOA V is simple if and only if r is not an
J
integer.
Many important VOAs are obtained as the nontrivial simple quotients of nonsimple
VOAs (e.g., rational Virasoro VOAs, and VOAs associated to integrable highest weight
modules over affine Lie algebras). So we are interested in the simple quotient V /I
J r
with r ∈ Z rather than the VOA V . When we study the structure of V /I and its
J J r
representation theory, it is very important to determine some relations in V /I induced
J r
by the maximal proper ideal I . In this paper, as a first step for studying the simple VOA
r
V /I , we will give a generator system of the maximal proper ideal I of the VOA V
J r r J
(1)
explicitly, whose elements are singular vectors for a certain Lie algebra L , and have a
r
high symmetry (see §6 below).
2
Thispaperisorganizedasfollows. In§2,werecallthedefinitionoftheGriessalgebraof
aVOA,thedefinition ofa Jordanalgebra, andtheconstruction oftheVOAV introduced
J
by Ashihara and Miyamoto [AM]. Then we state our main theorem (Theorem 2.6), and
the plan how we prove the theorem. In §3–§6, following the plan, we will prove some key
propositions (Propositions 3.1, 3.4, 4.1, 5.1, 6.1); our main theorem follows immediately
from these propositions. In §6, we also give a generator system of the maximal ideal I
r
of the VOA V explicitly when r is an integer, that is, when V is not simple.
J J
Acknowledgments. TheauthorswouldliketoexpresstheirsinceregratitudetoProfes-
sor Masahiko Miyamoto, Professor Toshiyuki Abe, Professor Hiroki Shimakura, Professor
Hiroshi Yamauchi, and Dr. Takahiro Ashihara for valuable discussions.
2 Vertex operator algebra whose Griess algebra is a Jordan al-
gebra.
2.1 Griess algebras. Let V = V , Y(·,z), 1, ω be a vertex operator
n∈Z≥0 n
algebra (VOA for short), with Y(·,z) : V → (EndV)[[z,z−1]] the vertex operator, 1 ∈ V
(cid:0) L (cid:1) 0
the vacuum element, and ω ∈ V the Virasoro element (for the details about VOAs, see,
2
e.g., [LL]). As usual, for each v ∈ V, we define v ∈ EndV, m ∈ Z, by: Y(v, z) =
m
v z−m−1. For a,b ∈ V , we define a·b := a b. Then it follows from the axiom of a
m∈Z m 2 1
VOA that a·b ∈ V for every a,b ∈ V , i.e., V becomes a C-algebra with · the product.
P 2 2 2
In addition, if V = C1 and V = {0}, then the C-algebra V is commutative (see [FLM,
0 1 2
§10.3] and also [M, §5]). In this case, we call V the Griess algebra of V. Note that the
2
Griess algebra of a VOA is not necessarily associative.
2.2 Jordan algebras. Let us recall the definition of a Jordan algebra. For the details
about Jordan algebras, see, e.g., [Al1], [Al2], and [J].
Definition 2.1. Let J be a C-algebra with the product a·b (a,b ∈ J). The C-algebra J
is called a Jordan algebra if a·b = b·a and a2·(b·a) = (a2·b)·a hold for every a,b ∈ J.
Let Sym (C) be the set of symmetric matrices of size d ∈ Z with entries in C. It is
d ≥2
well-known that Sym (C) becomes a (simple) Jordan algebra, where the product is given
d
by: A·B = 1(AB +BA) for A, B ∈ Sym (C) (see also [L2, Theorem 6B]).
2 d
2.3 VOA V . Let (and fix) d ∈ Z . In this subsection, we recall a VOA V intro-
J ≥2 J
duced by Ashihara and Miyamoto [AM], whose Griess algebra is isomorphic to the Jordan
algebra Sym (C) of symmetric matrices.
d
3
Let h be an (infinite-dimensional) vector space over C with a linear basis vi(m) | 1 ≤
i ≤ d, m ∈ Z ∪{c}, and define a Lie bracket on h by:
(cid:8)
b
[v(cid:9)i(m), vj(n)] = δ δ mc for 1 ≤ i, j ≤ d and m,n ∈ Z,
m+n,0 i,j b
(2.1)
[c, h] = {0}.
Denote by U(h) the universal enveloping algebra of the Lie algebra h, and let U(h)/hc−1i
b
be the quotient algebra with respect to the two-sided ideal hc−1i of U(h) generated by
b b b
c−1 ∈ U(h). We define a subspace L of U(h)/hc−1i as follows: For 1 ≤ i, j ≤ d and
b
m, n ∈ Z, we set
b b
vij(m, n) := vi(m)vj(n) mod hc−1i. (2.2)
Let
B := vii(m, n) | 1 ≤ i ≤ d and m, n ∈ Z with m ≤ n ∪
(cid:8) vij(m, n) | 1 ≤ i < j ≤ d and m(cid:9), n ∈ Z .
Then it follows from the Poincar´e-(cid:8)Birkhoff-Witt theorem that B ∪{1 ∈ U((cid:9)h)/hc−1i} is
a linearly independent subset of U(h)/hc−1i. We set
b
L := Span B ⊕C ⊂ U(h)/hc−1i.
b C
Remark 2.2. Let 1 ≤ i, j ≤ d,(cid:0)and m, n(cid:1)∈ Z. It can be easily seen from the definition
b
(2.1) of the Lie bracket on h that
vij(m, n) = vji(n, m) if i 6= j, or if i = j and m 6= −n,
b
(2.3)
vii(m, −m) = vii(−m, m)+m.
In particular, vij(m, n) ∈ L for all 1 ≤ i, j ≤ d and m, n ∈ Z.
Wesee by direct computationthat[x,y] = xy−yxiscontained inLforevery x, y ∈ L,
and hence L becomes a Lie algebra with respect to the natural Lie bracket. Now, let (and
fix) r ∈ C be an (arbitrary) complex number. For each x, y ∈ L, we define
[x, y] := π ([x,y])+rπ ([x,y]), (2.4)
r 1 2
where π : L ։ Span B and π : L ։ C denote projections from L onto Span B and
1 C 2 C
C, respectively. Then we know from [AM, §2.1] that [·, ·] is a Lie bracket on L. Let us
r
denote by L the Lie algebra L with the new Lie bracket [·, ·] .
r r
Example 2.3. As an example, let us compute [vii(m, n), vii(−n, −m)] for 1 ≤ i ≤ d and
r
m, n ∈ Z with m ≤ n. By direct computation and (2.3), we see that in L,
>0
[vii(m, n), vii(−n, −m)]
= n(1+δ )vii(−m, m)+m(1+δ )vii(−n, n)+mn(1+δ ).
m,n m,n m,n
4
Hence, by the definition (2.4) of the Lie bracket [·, ·] , we obtain
r
[vii(m, n), vii(−n, −m)]
r
= π ([vii(m, n), vii(−n, −m)])+rπ ([vii(m, n), vii(−n, −m)])
1 2
= n(1+δ )vii(−m, m)+m(1+δ )vii(−n, n)+rmn(1+δ ). (2.5)
m,n m,n m,n
We now set
B := vij(m, n) ∈ B | m ∈ Z or n ∈ Z ,
+ ≥0 ≥0
B− := (cid:8)vij(m, n) ∈ B | m, n ∈ Z<0 , (cid:9)
(cid:8) (cid:9)
and L+ := Span B ⊕C. It is easily seen that L+ is a Lie subalgebra of L . Let C1 be
r C + r r
a one-dimensional L+-module such that x·1 = 0 for all x ∈ B , and s·1 = s1 for each
(cid:0) r(cid:1) +
s ∈ C ⊂ L+. Denote by M the L -module induced from the L+-module C1, that is,
r r r r
M := U(L )⊗ C1.
r r U(L+r)
Here we give a linear basis B of M by using the Poincar´e-Birkhoff-Witt theorem.
r
Take (and fix) a total ordering ≻ on the set B . Denote by S the set of finite sequences
−
of elements of B that is weakly decreasing with respect to the total ordering ≻. For
−
x = (x (cid:23) x (cid:23) ··· (cid:23) x ) ∈ S with x ∈ B for 1 ≤ q ≤ p, we set
p p−1 1 q −
w(x) := x x ···x 1 ∈ M .
p p−1 1 r
In view of the Poincar´e-Birkhoff-Witt theorem, B := {w(x) | x ∈ S} is a linear basis of
the L -module M .
r r
Remark 2.4. We see from the definition (2.4) of the Lie bracket on L that xy = yx for
r
all x, y ∈ B . Therefore, if y , y , ..., y ∈ B , then y y ···y 1 = w(x) ∈ B, where
− 1 2 p − 1 2 p
x ∈ S is the sequence of length p obtained by arranging y , y , ..., y in the weakly
1 2 p
decreasing order with respect to the total ordering ≻. Also, we note that if m, n ∈ Z ,
<0
then vij(m, n) = vji(n, m) for every 1 ≤ i, j ≤ d (see Remark 2.2).
If x = (x (cid:23) x (cid:23) ··· (cid:23) x ) ∈ S with x = viqjq(m , n ) ∈ B for 1 ≤ q ≤ p, then
p p−1 1 q q q −
we define the degree of w(x) ∈ B by:
p
deg(w(x)) = − (m +n ) ∈ Z .
q q ≥0
q=1
X
Then the L -module M admits the degree space decomposition as follows:
r r
M = (M ) , where (M ) := Span b ∈ B | degb = n .
r r n r n C
nM∈Z≥0 (cid:8) (cid:9)
5
Note that
(M ) = C1, and (M ) = {0}. (2.6)
r 0 r 1
Define an operator Lij(m) ∈ End(M ) for 1 ≤ i, j ≤ d and m ∈ Z by:
r r
1
vij(m−h, h) if i 6= j or m 6= 0,
2
Lij(m) =  Xh∈Z (2.7)
r 1
 vii(0, 0)+ vii(−h, h) if i = j and m = 0,
2
 hX∈Z>0



and set 
d
ωij := Lij(−2)1 ∈ (M ) , and ω := ωii ∈ (M ) .
r r r 2 r r 2
i=1
X
Remark that Lij(m) = Lji(m) for every 1 ≤ i, j ≤ d and m ∈ Z (see Remark 2.2),
r r
and hence that ωij = ωji for every 1 ≤ i, j ≤ d. Let J := ωij, 1 | 1 ≤ i, j ≤
r r r
d ⊂ M , and let V be the subspace of M spanned by all elements of the form:
r J r (cid:8)
L(cid:9)ri1j1(m1)Lri2j2(m2)···Lripjp(mp)1 with p ≥ 0, and 1 ≤ iq, jq ≤ d, mq ∈ Z for 1 ≤ q ≤ p.
Then, V also admits the degree space decomposition induced from that of M , i.e.,
J r
V = (V ) with (V ) := V ∩ (M ) for n ∈ Z . We should remark that
J n∈Z≥0 J n J n J r n ≥0
(V ) = C1 and (V ) = {0} by (2.6).
J 0 L J 1
Define a map Y (·, z) : J → End(V )[[z,z−1]] by:
0 J
Y (ωij, z) = Lij(m)z−m−2 for 1 ≤ i, j ≤ d,
0 r r
m∈Z (2.8)
X
Y (1, z) = id .
0 VJ
The following theorem is the main result of [AM].
Theorem 2.5. Keep the notation above. The map Y (·,z) : J → End(V )[[z,z−1]] can
0 J
be uniquely extended to a linear map Y(·,z) : V → End(V )[[z,z−1]] in such a way that
J J
the quadruple V = (V ) , Y(·,z), 1, ω becomes a VOA of central charge dr,
J n∈Z≥0 J n
with 1 the vacu(cid:16)um element, and ω the Virasoro ele(cid:17)ment. Furthermore, the Griess algebra
L
of V is isomorphic to the Jordan algebra Sym (C) of symmetric matrices.
J d
The purpose of this paper is to determine the condition of r ∈ C for the VOA V to
J
be simple. The following theorem is the main result of this paper.
Theorem 2.6. Keep the notation above. The VOA V is simple if and only if r ∈ C is
J
not an integer, that is, r ∈ C\Z.
6
We will prove Theorem 2.6 as follows. First, in §3, we will show that V ⊂ M is,
J r
in fact, identical to the whole of M (Proposition 3.1), and then prove that the VOA V
r J
(= M ) is simple if and only if M is irreducible as an L -module (Proposition 3.4). Let
r r r
L(1) be a Lie subalgebra of L generated by v11(m, n) | m, n ∈ Z with m ≤ n ⊂ B, and
r r
set M(1) = U(L(1))1 ⊂ M . In §4, it will be shown that the L -module M is irreducible
r r r (cid:8) r r (cid:9)
(1) (1)
if and only if M is irreducible as an L -module (Proposition 4.1). In §5, we will prove
r r
that if r ∈ C \ Z, then M(1) is an irreducible L(1)-module, and hence V is a simple
r r J
(1)
VOA (Proposition 5.1). Finally, in §6, we will give some singular vectors of the L -
r
module M(1) explicitly in the case that r ∈ Z (Proposition 6.1), which implies that M(1)
r r
is reducible, and hence V is not simple.
J
3 Simplicity of V and irreducibility of M .
J r
3.1 Relation between V and M . As in the previous section, we fix d ∈ Z and
J r ≥2
r ∈ C. This subsection is devoted to proving the following proposition.
Proposition 3.1. The subspace V ⊂ M is identical to the whole of M , that is, V = M
J r r J r
holds.
In order to prove Proposition 3.1, we need some technical lemmas.
Lemma 3.2. (1) For each 1 ≤ i, j ≤ d, we have
vij(−1, −1)1 = 2Lij(−2)1. (3.1)
r
(2) Let 1 ≤ i, j ≤ d with i 6= j, and m, n ∈ Z . Then,
<0
1
vij(m−1, n)1 = − Lii(−1)vij(m, n)1. (3.2)
m r
(3) Let 1 ≤ i, j ≤ d with i 6= j, and m, n ∈ Z . Then,
<0
2
vii(m−1, n)1 = Lii(0)Lij(−1)vij(n, m)1. (3.3)
m(m−n+1) r r
Proof. (1) By the definition (2.7) of Lij(m),
r
2Lij(−2)1 = vij(−2−h, h)1.
r
h∈Z
X
Note that vij(−2 − h, h) = vji(h, −2 − h) for all h ∈ Z (see Remark 2.2). Since
vij(m, n)1 = 0 if vij(m, n) ∈ B , it follows that vij(−2 − h, h)1 = 0 unless h = −1.
+
Thus we obtain 2Lij(−2)1 = vij(−1, −1)1, and hence (3.1).
r
7
(2) As in the proof of part (1), we can easily show that
1
Lij(−1)1 = vij(−1−h, h)1 = 0 (3.4)
r 2
h∈Z
X
for all 1 ≤ i, j ≤ d. Hence,
Lii(−1)vij(m, n)1 = [Lii(−1), vij(m, n)]1 since Lii(−1)1 = 0 by (3.4)
r r r
1
= [vii(−1−h, h), vij(m, n)]1.
2
h∈Z
X
By direct computation (as in Example 2.3), we see that
[vii(−1−h, h), vij(m, n)] =
r
δ (−1−h)vij(h, n)+δ hvij(−1−h, n).
−1−h+m,0 h+m,0
Therefore,
1
[vii(−1−h, h), vij(m, n)]1
2
h∈Z
X
1
= δ (−1−h)vij(h, n)+δ hvij(−1−h, n) 1
−1−h+m,0 h+m,0
2
h∈Z
X(cid:8) (cid:9)
1
= (−m)vij(m−1, n)+(−m)vij(m−1, n) 1
2
= (−(cid:8)m)vij(m−1, n)1. (cid:9)
Thus we obtain Lii(−1)vij(m, n)1 = (−m)vij(m−1, n), and hence (3.2).
r
(3) We have
Lij(−1)vij(n, m)1 = [Lij(−1), vij(n, m)]1 since Lij(−1)1 = 0 by (3.4)
r r r
1
= [vij(−1−h, h), vij(n, m)]1.
2
h∈Z
X
As in Example 2.3, we see that
[vij(−1−h, h), vij(n, m)] =
r
δ hvii(n, −1−h)+δ (−1−h)vjj(h, m).
h+m,0 −1−h+n,0
Therefore we get
1
[vij(−1−h, h), vij(n, m)]1
2
h∈Z
X
1
= δ hvii(n, −1−h)+δ (−1−h)vjj(h, m) 1
h+m,0 −1−h+n,0
2
h∈Z
X(cid:8) (cid:9)
1
= (−m)vii(n, m−1)1+(−n)vjj(n−1, m)1 ,
2
(cid:8) (cid:9)
8
and hence
m n
Lij(−1)vij(n, m)1 = − vii(n, m−1)1− vjj(n−1, m)1.
r 2 2
Now, since Lii(0)1 = 0, it follows that
r
Lii(0)Lij(−1)vij(n, m)1
r r
m n
= − Lii(0)vii(n, m−1)1− Lii(0)vjj(n−1, m)1
2 r 2 r
m n
= − [Lii(0), vii(n, m−1)]1− [Lii(0), vjj(n−1, m)]1.
2 r 2 r
It can be easily checked that [Lii(0), vjj(n−1, m)] = 0 since i 6= j. Thus,
r
m n
− [Lii(0), vii(n, m−1)]1− [Lii(0), vjj(n−1, m)]1
2 r 2 r
m
= − [Lii(0), vii(n, m−1)]1
2 r
m m
= − [vii(0, 0), vii(n, m−1)]− [vii(−h, h), vii(n, m−1)] 1.
4 2
(cid:26) =0 hX∈Z>0 (cid:27)
Since n, m−1 ∈ Z| , it follow{zs that }
<0
[vii(−h, h), vii(n, m−1)] = δ hvii(−h, m−1)+δ hvii(n, −h)
r h+n,0 h+m−1,0
for h ∈ Z . Therefore,
≥1
m
− [vii(−h, h), vii(n, m−1)]1
2
hX∈Z>0
m
= − δ hvii(−h, m−1)+δ hvii(n, −h) 1
h+n,0 h+m−1,0
2
hX∈Z>0(cid:8) (cid:9)
m
= − (−n)vii(n, m−1)+(−m+1)vii(n, m−1) 1
2
(cid:8) (cid:9)
m(m+n−1)
= vii(m−1, n)1.
2
Thus we obtain
m(m+n−1)
Lii(0)Lij(−1)vij(n, m)1 = vii(m−1, n)1,
r r 2
and hence (3.3). This completes the proof of the lemma.
Lemma 3.3. (1) Let 1 ≤ i, j ≤ d with i 6= j, and m, n ∈ Z . Then,
<0
vij(m, n)1 = αLii(−1)−m−1Ljj(−1)−n−1Lij(−2)1 (3.5)
r r r
9
for some α ∈ C\{0}.
(2) Let 1 ≤ i ≤ d, and m, n ∈ Z with m ≤ −2. Take 1 ≤ j ≤ d with j 6= i arbitrarily.
<0
Then,
vii(m, n)1 = βLii(0)Lij(−1)Lii(−1)−n−1Ljj(−1)−m−2Lij(−2)1 (3.6)
r r r r r
for some β ∈ C\{0}.
Proof. (1) We prove (3.5) by induction on −m−n (notethat−m−n ≥ 2). If −m−n = 2,
that is, m = n = −1, then (3.5) followsimmediately from(3.1). Assume that−m−n > 2.
By Remark 2.2, we may assume that m < −1. Then, by (3.2), we have
1
vij(m, n)1 = − Lii(−1)vij(m+1, n)1.
m+1 r
Applying theinductive assumption totheright-handsideoftheequationabove, we obtain
1
vij(m, n)1 = − Lii(−1)vij(m+1, n)1
m+1 r
1
= − Lii(−1) αLii(−1)−m−2Ljj(−1)−n−1Lij(−2)1
m+1 r r r r
α (cid:8) (cid:9)
= − Lii(−1)−m−1Ljj(−1)−n−1Lij(−2)1,
m+1 r r r
where α ∈ C\{0}. Thus we have proved part (1).
(2) Using (3.3) and (3.5), we have
2
vii(m, n)1 = Lii(0)Lij(−1)vij(n, m+1)1 by (3.3)
(m+1)(m+n) r r
2β
= Lii(0)Lij(−1)Lii(−1)−n−1Ljj(−1)−m−2Lij(−2)1 by (3.6),
(m+1)(m+n) r r r r r
where β ∈ C \ {0}. Thus we have proved part (2), thereby completing the proof of
Lemma 3.3.
Proof of Proposition 3.1. We set
p ≥ 0, and 1 ≤ i , j ≤ d,
q q
U := Span Li1j1(m )Li2j2(m )···Lipjp(m )1 .
C r 1 r 2 r p
( (cid:12) mq ∈ {−2,−1,0} for 1 ≤ q ≤ p )
(cid:12)
(cid:12)
(cid:12)
In order to prove Proposition 3.1, it suffices to show that U = M . Indeed, it is obvious
(cid:12) r
fromthedefinitionofV thatU ⊂ V . Hence, ifU = M holds, thenM = U ⊂ V ⊂ M ,
J J r r J r
which implies that V = M .
J r
Claim. We have xU ⊂ U for every x ∈ B.
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