Representations of trigonometric Cherednik algebras of rank 1 in positive characteristic

0.21 MB·English

by  Frédéric Latour

Representations of trigonometric Cherednik algebras of rank 1 in
positive characteristic
Fr´ed´eric Latour
4
0
0 February 1, 2008
2

n
a
1 Introduction
J

7
2 Cherednik’s double affine Hecke algebras are an important class of algebrasattached to root systems. They
were introduced in [Che95] as a tool of proving Macdonald’s conjectures, but are also interesting by them-

] selves,sincetheyprovideuniversaldeformationsoftwistedgroupalgebrasofdoubleaffineWeylgroups. One
T
may distinguish rational, trigonometric,and elliptic Cherednik algebras,which contain 0, 1, and 2 copies of
R
the root lattice, respectively (rational and trigonometric algebras are degenerations of the elliptic ones; see
.
h [EG02]).
t Development of representation theory of Cherednik algebras (in particular, description of all irreducible
a
m finite dimensionalrepresentations)isanimportantopenproblem. Inthe characteristiczerocase,itissolved
completely only for type A, while in other types only partial results are available (see [EG02],[BEG03], and
[
[CO] for the rank 1 case). In positive characteristic,the rank 1 case (in the more generalsetting of complex

1
reflectiongroups)issettledbytheauthorin[Lat],afterwhichthehigherrankcase(oftypeA)wasconsidered
v
in [FG].
7
8 The goal of this paper is to extend the results of [Lat] to the trigonometric case. That is, we study the
3 representationtheory oftrigonometricCherednik algebrasin positive characteristicp in the simplestcase of
1 rank 1. Our main result is a complete description of irreducible representations of such algebras.
0
The paper is organized as follows.
4
In Section 2, we state the main results.
0
/ In Section 3, we prove the results for the “classical” case, i.e. the case when the “Planck’s constant” t
h
is zero. In this case, generic irreducible representations have dimension 2; one-dimensional representations
t
a exist when the “coupling constant” k is zero.
m
In Section 4, we prove the results for the “quantum” case, i.e. the case when the “Planck’s constant” t
: is nonzero. In this case, generic irreducible representations have dimension 2p; smaller representations exist
v
when the “coupling constant” k is an element of F ⊂k; namely, if k is an integer with 0≤k ≤p−1, then
i p
X there exist irreducible representations of dimensions p−k and p+k.
r Acknowledgements. The author thanks his adviser Pavel Etingof for posing the problem and useful
a
discussions, as well as for helping to write an introduction. The work of the author was partially supported
by the National Science Foundation (NSF) grant DMS-9988796and by a Natural Sciences and Engineering
Research Council of Canada (NSERC) Julie Payette researchscholarship.
2 Statement of Results
Let k be an algebraically closed field of characteristic p, where p 6= 2. Let t,k ∈ k, and let H(t,k) be the
algebra (over k) generated by X,X−1,s and y, subject to the following relations:
1
sX = X−1s (1)
s2 = 1 (2)
sy+ys = −k (3)
XyX−1 = y−t+ks. (4)
We will classify the irreducible representations of H(t,k). Now, for t 6= 0, H(t,k) is clearly isomorphic
to H(1,k) under the map
t
1
X7→X,s7→s,y7→ y.
t
Thus itis sufficientto classify irreduciblerepresentationsofH(0,k)andH(1,k). For brevitywe willuse the
def def
notation H = H(0,k) and H = H(1,k), assuming that k has been fixed once and for all.
0 1
2.1 Irreducible representations of H
0
Proposition 2.1. Let k6=0. Then the irreducible representations of H are the following:
0
• For a,β ∈ k, a,β 6= 0, we have a two-dimensional representation Vβ,a with basis {v ,v }, defined by
0,1 0 1
the following:
yv = βv ,
0 0
yv = −βv ,
1 1
k2
Xv = av − v ,
0 0 4β2 1
1 k2
Xv = v + − v ,
1 0 a 4aβ2 1
(cid:18) (cid:19)
k k3−4kβ2
sv = − v + v ,
0 2β 0 8aβ3 1
2aβ k
sv = − v + v ;
1 0 1
k 2β
• For a = ±1,b ∈ k, we have a two-dimensional representation Va,b with basis {v ,v }, defined by the
0,2 0 1
following:
yv = 0,
0
yv = v ,
1 0
sv = v −kv ,
0 0 1
sv = −v ,
1 1
Xv = a(v −kv ),
0 0 1
Xv = bv +(a−kb)v .
1 0 1
Vβ,a and Vβ′,a′ are isomorphic if and only if β′ = β,a′ = a or β′ = −β,a′ = 4β2−k2. Va,b and Va′,b′ are
0,1 0,1 4aβ2 0,2 0,2
isomorphic if and only if a=a′ and b=b′. Furthermore, representations with different subscripts are never
isomorphic.
Proposition 2.2. Let k=0. Then the irreducible representations of H are the following:
0
2
• For a,β ∈ k, a,β 6= 0, we have a two-dimensional representation Vβ,a with basis {v ,v }, defined by
0,3 0 1
the following:
yv = βv ,
0 0
yv = −βv ,
1 1
Xv = av ,
0 0
1
Xv = v ,
1 1
a
sv = v ,
0 1
sv = v ;
1 0
• For a ∈ k, a ∈/ {0,±1}, we have a two-dimensional representation Va with basis {v ,v }, defined by
0,4 0 1
the following:
yv = 0,
0
yv = 0,
1
Xv = av ,
0 0
1
Xv = v ,
1 1
a
sv = v ,
0 1
sv = v ;
1 0
• For a = ±1,b = ±1, we have a one-dimensional representation Va,b on which y,X and s act as 0,a
0,5
and b respectively.
Vβ,a and Vβ′,a′ are isomorphic if and only if β′ =β,a′ =a or β′ =−β,a′ = 1. Va and Va′ are isomorphic
0,3 0,3 a 0,4 0,4
if and only if a′ = a or a′ = 1. Va,b and Va′,b′ are isomorphic if and only if a′ = a,b′ = b. Furthermore,
a 0,5 0,5
representations with different subscripts are never isomorphic.
2.2 Irreducible representations of H
1
Proposition 2.3. Let k∈/ F . Then the irreducible representations of H are the following:
p 1
• For µ,d∈k,d6=0,b=(µp−µ)2 with k not a root of f(y)=(yp−y)2−b, and also for µ=±k,d6=0,
2 2
we have a 2p-dimensional representation V1µ,1,d with basis {vµ+j,v−µ+j,j = 0,1,...,p−1}, defined by
the following:
yv = βv , β =±µ,±µ+1,...,±µ+p−1; (5)
β β
1 k
sv−µ−j = − vµ+j + v−µ−j, j =1,2,...,p−1; (6)
µ+j 2(µ+j)
k2 k
svµ+j = −(µ+j) v−µ−j − vµ+j, j =1,2,...,p−1; (7)
4(µ+j) 2(µ+j)
(cid:18) (cid:19)
k d
sv−µ = v−µ− vµ; (8)
2µ µ
k2 µ k
svµ = − v−µ− vµ; (9)
4dµ d 2µ
(cid:18) (cid:19)
Xvβ = sv−β−1, β =±µ,±µ+1,...,±µ+p−1; (10)
3
• Forθ =±1,wehavea2p-dimensionalrepresentationVθ withbasis {v ,w ,j =0,1,...,p−1},defined
1,2 j j
by the following:
yv = jv , j =0,1,...,p−1; (11)
j j
yw = jw +v , j =0,1,...,p−1; (12)
j j j
sv = −kw ; (13)
0 0
1
sw = − v ; (14)
0 0
k
1 k p−1
sv−j = vj + v−j, j =1,2,..., ; (15)
j 2j 2
k2 k p−1
svj = j− v−j − vj, j =1,2,..., ; (16)
4j 2j 2
(cid:18) (cid:19)
1 k 1 k p−1
sw−j = j2vj + 2j2v−j − jwj + 2jw−j j =1,2,..., 2 ; (17)
k2 k k k2 p−1
swj = 1+ 4j2 v−j + 2j2vj − 2jwj + 4j −j w−j, j =1,2,..., 2 ; (18)
(cid:18) (cid:19) (cid:18) (cid:19)
p−1
Xvj = −sv−j−1, j 6= ; (19)
2
Xvp−1 = θsvp−1; (20)
2 2
p−1
Xwj = sw−j−1, j 6= ; (21)
2
Xwp−1 = −θswp−1. (22)
2 2
Vµ,d and Vµ′,d′ are isomorphic if and only if
1,1 1,1
k2
(µ′−µ∈F and d′ =d) or (µ′+µ∈F and dd′ = −(µ+c)2 ).
p p
4
cY∈Fp(cid:18) (cid:19)
Vθ and Vθ′ are isomorphic if and only if θ =θ′ Furthermore, representations with different subscripts are
1,2 1,2
never isomorphic.
Now, in the case where k ∈F , note that there is an isomorphism between H(1,k) and H(1,−k), given
p
by
y7→y,s7→−s,X7→X,k 7→−k.
So we may assume that k is an even integer with 0≤k ≤p−1.
Proposition 2.4. Let k be even with 2 ≤ k ≤ p−1. Then the irreducible representations of H are the
1
following:
• For µ,d∈k,d6=0, we have Vµ,d, defined as in Proposition 2.3.
1,1
• For θ = ±1, we have a (p − k)-dimensional representation Vθ with basis {v ,v ,...,v },
1,3 k k+1 −k−1
2 2 2
defined by
k k k
yv = jv , j = , +1...,− −1; (23)
j j
2 2 2
k 1 k p−1
sv−j = v−j − vj, j = +1,..., ; (24)
2j j 2 2
k2 k k p−1
svj = −jv−j + v−j − vj, j = +1,..., ; (25)
4j 2j 2 2
sv = −v ; (26)
k k
2 2
4
p−1
Xvj = −sv−j−1, j 6= ; (27)
2
Xvp−1 = θsvp−1. (28)
2 2
• For θ =±1, we have a (p+k)-dimensional representation Vθ with basis {v ,w ,j =0,...,p−1,i=
1,4 j i
−k,...,k −1}, defined by
2 2
yv = jv , j =0,1,...,p−1; (29)
j j
k k k
yw = jw +v , j =− ,− +1,..., −1; (30)
j j j
2 2 2
sv = −kw ; (31)
0 0
1
sw = − v ; (32)
0 0
k
1 k k k p−1
sv−j = vj + v−j, j =1,..., −1, +1, ; (33)
j 2j 2 2 2
k2 k k k p−1
svj = j− v−j − vj, j =1,..., −1, +1, ; (34)
4j 2j 2 2 2
(cid:18) (cid:19)
sv = v ; (35)
−k −k
2 2
sv = 2v −v ; (36)
k −k k
2 2 2
1 k 1 k k
sw−j = j2vj + 2j2v−j − jwj + 2jw−j j =1,..., 2 −1; (37)
k2 k k k2 k
swj = 1+ 4j2 v−j + 2j2vj − 2jwj + 4j −j w−j, j =1,..., 2 −1; (38)
(cid:18) (cid:19) (cid:18) (cid:19)
2 2
sw = − v + v +w ; (39)
−k k −k −k
2 k 2 k 2 2
p−1
Xvj = −sv−j−1, j 6= ; (40)
2
Xvp−1 = θsvp−1; (41)
2 2
k k
Xwj = =sw−j−1, j =− ,..., −1. (42)
2 2
• For c ∈ k, we have a 2p-dimensional representation Vc with basis {v ,w ,u ,j = 0,...,p−1,i =
1,5 j i l
−k,...,k −1,l= k,...,−k −1}, defined by
2 2 2 2
yv = jv , j =0,1,...,p−1; (43)
j j
k k k
yw = jw +v , j =− ,− +1,..., −1; (44)
j j j
2 2 2
sv = w ; (45)
0 0
sw = v ; (46)
0 0
1 k k k p−1
sv−j = vj + v−j, j =1,..., −1, +1, ; (47)
j 2j 2 2 2
k2 k k k p−1
svj = j− v−j − vj, j =1,..., −1, +1, ; (48)
4j 2j 2 2 2
(cid:18) (cid:19)
sv = v ; (49)
−k −k
2 2
sv = 2v −v ; (50)
k −k k
2 2 2
5
1 k 1 k k
sw−j = j2vj + 2j2v−j − jwj + 2jw−j j =1,..., 2 −1; (51)
k2 k k k2 k
swj = 1+ 4j2 v−j + 2j2vj − 2jwj + 4j −j w−j, j =1,..., 2 −1; (52)
(cid:18) (cid:19) (cid:18) (cid:19)
2 2
sw = − v + v +w ; (53)
−k k −k −k
2 k 2 k 2 2
1 k k p−1
suj = − u−j − uj, j = +1,..., ; (54)
j 2j 2 2
k2 k k p−1
su−j = −j uj + u−j, j = +1,..., ; (55)
4j 2j 2 2
(cid:18) (cid:19)
2c
su = v −u ; (56)
k −k k
2 k 2 2
p−1
Xvj = −sv−j−1, j 6= ; (57)
2
Xvp−1 = sup−1; (58)
2 2
k p−3 p+1 k
Xuj = −su−j−1, j = ,..., , ,...,− −1; (59)
2 2 2 2
Xup−1 = svp−1. (60)
2 2
Vµ,d and Vµ′,d′ are isomorphic if and only if
1,1 1,1
k2
(µ′−µ∈F and d′ =d) or (µ′+µ∈F and dd′ = −(µ+c)2 ).
p p
4
cY∈Fp(cid:18) (cid:19)
Vθ and Vθ′ are isomorphic if and only if θ =θ′. Vc and Vc′ areisomorphic if and only if c=c′. Vc and
1,3 1,3 1,4 1,4 1,5
Vc′ are isomorphic if and only if c = c′. Furthermore, representations with different subscripts are never
1,5
isomorphic.
Proposition 2.5. Let k=0. Then the representations of H are the following:
1
• For µ,d∈k,d6=0,b=(µp−µ)2, we have Vµ,d, defined as in Proposition 2.3.
1,1
• For c,θ =±1, we have a p-dimensional representation Vc,θ with basis {v ,j =0,1,...,p−1}, defined
1,6 j
by
yv = jv , j =0,1,...,p−1; (61)
j j
sv = cv ; (62)
0 0
p−1
svj = −jv−j, j =1,..., ; (63)
2
1 p−1
sv−j = − vj, j =1,..., ; (64)
j 2
p−1
Xvj = sv−j−1, j 6= ; (65)
2
Xvp−1 = θsvp−1. (66)
2 2
• For c=±1,a∈k, we have a 2p-dimensional representation Vc,a with basis {v ,u ,j =0,1,...,p−1},
1,7 j j
defined by
yv = jv , j =0,1,...,p−1; (67)
j j
yu = ju , j =0,1,...,p−1; (68)
j j
sv = v ; (69)
0 0
su = av −u ; (70)
0 0 0
6
1 p−1
sv−j = − vj j =1,2,..., ; (71)
j 2
p−1
svj = −jv−j j =1,2,..., ; (72)
2
1 p−1
suj = u−j, j =1,2,..., ; (73)
j 2
p−1
su−j = juj, j =1,2,..., ; (74)
2
p−1
Xvj = sv−j−1, j 6= ; (75)
2
Xvp−1 = sup−1; (76)
2 2
p−1
Xuj = su−j−1, j 6= ; (77)
2
Xup−1 = svp−1. (78)
2 2
Vµ,d and Vµ′,d′ are isomorphic if and only if
1,1 1,1
k2
(µ′−µ∈F and d′ =d) or (µ′+µ∈F and dd′ = −(µ+c)2 ).
p p
4
cY∈Fp(cid:18) (cid:19)
Vc,θ and Vc′,θ′ are isomorphic if and only if θ = θ′. Va and Va′ are isomorphic if and only if a = a′.
1,6 1,6 1,7 1,7
Furthermore, representations with different subscripts are never isomorphic.
3 Proof of Propositions 2.1 and 2.2
Lemma 3.1 (PBW for H , easy direction). The elements
0
siXjyl, j,l ∈Z,l ≥0,i∈{0,1}
span H over k.
0
Proof. Given a product of X,y,s,X−1 in any order,one can ensure that the y’s are to the rightof all the X’s
by using yX=Xy−ksX repeatedly, and one can also ensure that the s’s are to the left of all the X’s and y’s
by using Xs=sX−1 and ys=−k−sy repeatedly.
Lemma 3.2. X+X−1,y2 and Xy−yX−1 belong to the center Z(H ) of H .
0 0
Proof. First, let us show that y2 ∈Z(H ). We have
0
Xy2 = (yX+ksX)y
= yXy+ksXy
= y(yX+ksX)+ks(yX+ksX)
= y2X+k(ys+sy)X+k2s2X
= y2X+−k2X+k2X
= y2X;
thus [X,y2]=0. We also have
sy2 = (−ys−k)y=−ysy−ky=−y(−ys−k)−ky=y2s;
7
thus [s,y2]=0. It follows that y2 ∈Z(H ). Next, we show that X+X−1 ∈Z(H ). We have
0 0
y(X+X−1)=Xy−ksX+X−1y+kX−1s=(X+X−1)y,
and
s(X+X−1)=X−1s+Xs=(X+X−1)s.
Thus [y,X+X−1]=[s,X+X−1]=0, and so X+X−1 ∈Z(H ). Finally, we show that Xy−yX−1 ∈Z(H ).
0 0
First we note that
yX−X−1y=yX+Xy−(X+X−1)y=yX+Xy−y(X+X−1))=Xy−yX−1,
and thus
X(Xy−yX−1)=X(yX−X−1y)=(Xy−yX−1)X,
y(Xy−yX−1)=y(yX−X−1y)=y2X−yX−1y=Xy2−yX−1y=(Xy−yX−1)y,
and
s(Xy−yX−1)=s(yX−X−1y)=−(ys+k)X−Xsy=−yX−1s−kX+X(ys+k)=(Xy−yX−1)s.
Thus Xy−yX−1 ∈Z(H ).
0
Corollary 3.3. H is finitely generated as a module over its center.
0
Proof. From Lemmas 3.1 and 3.2, we see that H is generated over its center by
0
siXjyl, i,j,l∈{0,1}.
Corollary 3.4. Every irreducible H -module is finite-dimensional over k.
0
Proof. Standard.
Thus Schur’s lemma implies that central elements of H act as scalars in any irreducible H -module.
0 0
Fromthis point,wewillusethe followingnotation: theeigenspaceofy witheigenvalueβ willbe denoted
V[β].
Corollary 3.5. Let V be an irreducible H -module, and let β be an eigenvalue of y. Suppose β 6=0. Then,
0
V =V[β]⊕V[−β],
and dimV[β]=dimV[−β]=1.
Proof. Suppose V[β]6= 0, and let v ∈V[β] be nonzero. From the proof of corollary 3.3, we know that V is
spanned by
{v,Xv,sv,sXv}.
Now let w =sXv; then
yw =ysXv =−syXv−kXv =−sXyv+ks2Xv−kXv =−βsXv =−βw.
Thus, w ∈V[−β]. Clearly, w 6=0, and thus V[−β]6=0. Now let v′ =2βXv−kw. Then,
yv′ =2βyXv+βkw =2βXyv−2kβsXv+βkw =2β2Xv−βkw =βv′.
Hence, v′ ∈V[β]. Also, if w′ =kv+2βsv, then
yw′ =kyv+2βysv =βkv−2βsyv−2βkv =−βkv−2βsyv =−βw′,
8
Hence, w′ ∈V[−β]. From this it follows that
V =V[β]⊕V[−β].
Now let H be the subalgebra of H generated by Z(H ) and 2βX−ksX. It is clear that V[β] = Hv.
0 0 0
Since this is true for all nonzero v ∈V[β], it follows that V[β] is an irreducible representation of H . Since
0
H is commutative, we see that V[β] is one dimensional. The same holds for V[−β], and the corollary is
0
proved.
Corollary 3.6. Assume k 6= 0. Let V be an irreducible H -module, and suppose 0 is an eigenvalue of y.
0
Then, V =V [0], the generalized eigenspace of 0. We also have dimV =2 and dimV[0]=1.
gen
Proof. Let v ∈V[0] be nonzero. From the proof of corollary 3.3, we know that V is spanned by
{v,Xv,sv,sXv}.
Let w=−sv; then,
yw =−ysv =syv+kv =kv.
Let v′ =sXv =X−1sv; then,
yv′ =yX−1sv =X−1ysv+kX−1s2v =−X−1syv =0.
Let w′ =−Xv =−sv′; then, as above, we have yw′ =v′.
So we have
yv =0, yw =kv, yv′ =0, yw′ =kv′;
therefore, V = V [0] and V[0] is spanned by v and v′. Now let H be the subalgebra of H generated by
gen 0 0
Z(H ) and sX. It is clear that V[0] = Hv. Since this is true for all nonzero v ∈ V[0], it follows that V[0]
0
is an irreducible representation of H . Since H is commutative, we see that V[0] is one dimensional. The
0 0
corollary follows from this.
Corollary 3.7. Assume k = 0. Let V be an irreducible H -module, and suppose 0 is an eigenvalue of y.
0
Then,
V =V[0],
the eigenspace of 0. We also have
1 if 1 or −1 is an eigenvalue of X
dimV =
(2 otherwise.
Proof. From the proofof corollary3.6, we see that y acts on V as the zero operator. Let λ be an eigenvalue
of X, let VX[λ] denote the associated eigenspace and let v ∈ VX[λ] be nonzero. From the proof of corollary
3.3, we know that V is spanned by {v,sv}. Now
Xsv =sX−1v =λ−1sv,
so sv ∈VX[λ−1]. Clearly, sv 6=0; thus, if λ6=±1, then
V =VX[λ]⊕VX[λ−1] and dimV =2.
If λ = ±1, it follows that X and s commute as operators on V; since V is irreducible, this implies that
dimV =1.
9
Proof of Proposition 2.1. Letβ 6=0,andletV beatwo-dimensionalrepresentationofH inwhichV[β]and
0
V[−β] both have dimension 1. Let v ∈ V[β], v ∈ V[−β] be nonzero. Let the matrices representing s and
0 1
X with respect to the basis {v ,v } be as follows:
0 1
γ δ θ ω
s7→ 0 0 , X7→ 0 0 .
γ δ θ ω
1 1 1 1
(cid:18) (cid:19) (cid:18) (cid:19)
First, we note that X and y cannot have a common eigenvector; for if Xw = γw and yw = β′w, then
ksw = XyX−1w−yw = 0, and combining this with s2 = 1 gives w = 0. Hence, by scaling, we can assume
that ω =1.
0
Now the central element Xy−yX−1 acts on V as a scalar. The matrix representation of Xy−yX−1 is
β (θ detX−ω ) − β (detX−1)
detX 0 1 detX .
βθ1 (detX−1) − β (θ detX−θ )
(cid:18) detX detX 0 0 (cid:19)
Thus, 0=− β (detX−1), which means that detX=1. Hence,
detX
θ =θ ω −1. (79)
1 0 1
Using (4), we see that XyX−1 − y − ks = 0. Using (79), we see that the matrix yrepresentation of
XyX−1−y−ks is
2βθ ω −2β−kγ −2βθ −kδ
0 1 0 0 0 .
2βθ ω2−2βω −kγ −2βθ ω +2β−kδ
(cid:18) 0 1 1 1 0 1 1 (cid:19)
Hence,
2β
γ = (θ ω −1), (80)
0 0 1
k
2β
γ = θ ω2−ω , (81)
1 k 0 1 1
2β(cid:0) (cid:1)
δ = − θ (82)
0 0
k
2β
δ = (1−θ ω ). (83)
1 0 1
k
Using (2), we see that s2 =1. Using (80)–(83), we see that the matrix representation of s2 is
4β2(1−θ ω ) 0
k2 0 1 .
0 4kβ22(1−θ0ω1) !
Thus,
1 k2
ω = 1− . (84)
1 θ 4β2
0 (cid:18) (cid:19)
Using (79)–(84), we see that V is isomorphic to Vβ,θ0. Furthermore, it is easy to see that for all a,β ∈
0,1
k\{0}, Vβ,a is a representation of H ; furthermore, each eigenvector of y clearly generates H , and thus
0,1 0 0
Vβ,θ0 is irreducible.
0,1
Now the eigenvalues of y in Vβ,a are β and −β, and 2β(Xy−yX−1)−2β2(X+X−1) acts on Vβ,a as
0,1 0,1
k2−4β2 Id, while for β = k, X+X−1 acts on Vβ,a as aId. From this it follows that Vβ,a and Vβ′,a′ are
a 2 0,1 0,1 0,1
isomorphic if and only if β′ =β,a′ =a or β′ =−β,a′ = 4β2−k2.
4aβ2
Now let V be a two-dimensional representation of H in which V[0] has dimension 1 and V [0] has
0 gen
dimension 2. Let v ,v ∈V be nonzero elements such that yv =0,yv =v . Let the matrices representing
0 1 0 1 0
s and X with respect to the basis {v ,v } be as follows:
0 1
γ δ θ ω
s7→ 0 0 , X7→ 0 0 .
γ δ θ ω
1 1 1 1
(cid:18) (cid:19) (cid:18) (cid:19)
10