Aug 19, 1999 - destruction operators, together with a set fa+ i ji 2 Ig of creation operators, conjugate of the former by an anti-involution which we denote by J.
Classi cation of Quadratically Presented Number Operator Algebras of Finite Type Fabien Besnard August 19th, 1999 Abstract
Under some hypotheses, one can enumerate all algebras, generated by creation and destruction operators, in which number operators exist. We do so in the nitely generated case.
1 Introduction Following Dirac's ideas, D. Bennequin has proposed a new way of looking at quantization. In this point of view, one should quantize the equations of evolution rather than the canonical commutation relations, the latter being a consequence of the former. In the case, so crucial for eld theory, of a system of harmonic oscillators, we thus want to nd algebras (over C) of q-numbers, as Dirac should say, generated by a set fai ji 2 Ig of so-called destruction operators, together with a set fa+i ji 2 Ig of creation operators, conjugate of the former by an anti-involution which we denote by J . Our algebras will then have the structure of -algebras1 . It should be stressed that the word \operator" is just a convention here, since no Hilbert space is xed. We will want our algebras to contain elements Ni , for each i 2 I , so that the following equations hold : [Ni ; aj ] = ?ij ai
(1)
[Ni ; a+j ] = ij a+i
(2)
1 It is worth noticing that all we will say would be valid as well for algebras over any eld of characteristic 0, endowed with an involution (possibly the identity) taking place of the complex conjugation.
1
It is clear that if Ni0 is another element ful lling (1) and (2), then the difference Ni ? Ni0 belongs to the centre of the algebra. All this leads us to the following de nition : De nition 1 Let be a cardinal number, B a non-trivial algebra (that is B 6= 0, B 6= C), Z (B ) the centre of B . Take I a set of cardinal , and let XA = fai ji 2 Ig, XA+ = fa+i ji 2 Ig, N = fNi ji 2 Ig be 3 sets indexed by I , with the ai's and a+i 's in B , and Ni's in B=Z (B ). (B; XA; XA+ ; N ) is said to be a number operator algebra of type if, and only if : (i) B is generated by XA [ XA+ . (ii) One can uniquely de ne an anti-involution J on B by setting J (ai ) = a+i . (iii) (1) and (2) are ful lled. Remarks : We will sometimes say \B is a n.o.a." rather than (B; XA ; XA+ ; N ) is a n.o.a. It should be noted that I has no real signi cance : it is just a way of indexing the generators and number operators. We could do without it but it would soon become rather messy. If we want to avoid unneeded complications, we can x a representative I for , and say that all n.o.a. of type use I as their set of indices.
2 Categories of Number Operator Algebras We have to de ne morphisms between two n.o.a. : we will only need to do so for n.o.a. of the same type. De nition 2 Let B and B 0 be two n.o.a., fai ji 2 Ig, fNiji 2 Ig, J , Z (B ), fai 0ji 2 Ig, fNi0 ji 2 Ig, J 0, Z (B 0) be the corresponding generators, number operators, anti-involution, and centre. Let f be an algebra homomorphism from B to B 0 . We will say that f is a morphism of n.o.a. i : (i) f J = J 0 f (ii) f (Z (B )) Z (B 0 ) and there exists a bijection : I ! I such that : (iii) f (Ni ) 2 N 0 (i) + Z (B 0 ) 2
It is easy to see that n.o.a. of some type and their morphisms de ne a category. From now on, n will be an integer 1, and I will be the set f1; : : : ; ng. As we wish to put all the creation (and destruction) operators on an equal footing, we will de ne another category : De nition 3 Let B be a n.o.a. of type n and Sn be the group of permutations of I . If for all 2 Sn , the prescription ai 7! a(i) uniquely de nes an automorphism of B , then we will say that B is symmetric.
De nition 4 A n.o.a. morphism f between two symmetric n.o.a. of type n will be called symmetric i it commutes with , 8 2 Sn. Let Ln be the free algebra generated by the creation and destruction operators. We call the canonical projection from Ln onto B , and I the kernel of . If I is generated by elements of degree 2, we call B a quadratic n.o.a. We still have one hypothesis to state. It is of combinatorial nature. The terminology we use is that of [Berg]. We recall that given a presentation of an ideal I (i.e. a set of generators) and a monoid partial ordering > j > > > or > < (0;0) (0;0) If R = A3 : > (b): 9i0 ; j0 s.t. S = f(ai a+i ; ai0 a+i0 ); (a+j0 aj0 ; ? > > > 1 a a+ ? X a+ a )g > > > : i0 i0 j6=j0 j j i6=i0 X If R(0;0) = C : 9i0 s.t. S (0;0) = f(ai0 a+i0 ; ? aia+i )g ;
(0 0)
i6=i0
If R X
j 6=j0
;
(0 0)
aj aj )g +
IfXR j 6=j0
= C [D : 9i0 ; j0 s.t. S (0;0) = f(ai0 a+i0 ; ?
X i6=i0
aia+i ); (a+j0 aj0 ; ?
= A1 [ D : 9i0 ; j0 t.q. S (0;0) = f(ai a+i ; ai0 a+i0 ); (a+j0 aj0 ; ? a+j aj )gi6=i0
If R
;
(0 0)
;
= A2 [D : 9i0 t.q. S (0;0) = f(ai a+i ; ); (a+i0 ai0 ; ?
;
= E2 :
(0 0)
If R
(0 0)
X j 6=i0
a+j aj )g1in
8 (a) f(a a+ ; ? a+ a ? Xa+ a )g 1 i 2 > i i i j j 1in > > < j =i 6 or { If n 3 : S (0;0) = > X + + + > (b) f ( a a ; ? a a > 1 i i ? 2 aj aj )g1in i i : j 6=i
12
{ If n = 2 : S
;
(0 0)
8 (a) same as n 3 > > > > or > > > > (b) same as n 3 > > > > > or > 1 > + + + + > > > < (c) f(aj aj ; ( ? 1 ai ai ? ai ai )); (aj aj ;
=> 1 ) + ( 12 ? )a+ a + 1 a a+ )g; fi; j g = f1; 2g > (1 ? > i i i > 2 2 2 i or > 2 > > > > > > (d) f(a+i ai ; 1 ( ? 2 a+j aj ? ai a+i ); (aj a+j ; > > > 1 > > 2 > > > : (1 ? 2 ) + ( 2 ? 1 )a+ aj + 2 ai a+ )g; fi; j g = f1; 2g 2
j 1 1 1 i X+ (a)f(ai a+i ; ? 1 a+i ai + 1 n ? 1 aj aj )g1in
8 > > > > > > j 6=i > > > or X > > > (b) 9i0 s.t. S (0;0) = f(ai0 a+i0 ; ? aj a+j ); > > > > > j 6=i0 X > > + + + > ( a a ; ? + a a + 2 a a aj a+j )gi6=i0 > i i + i i i0 i0 < j 6=i;i0 If R(0;0) = E20 : > or X > > > (c) 9i0 s.t. S (0;0) = f(ai0 a+i0 ; ? aj a+j ); > > > > j 6=i0 > X + > > + + + > aj aj ); ( a a ; ? + a a + 2 a a + > i j j i j j 0 0 0 0 0 0 > > > j 6=i0 ;j0 > > > > + + + + > : (ai ai ; aj aj + aj aj ? ai ai )gi6=i0 ;j0 ; with = 1 ? n 0 0
0 0
8 X + + > (0;0) > (a) 9 i t.q. S = f ( a a ; ? ai ai 0 > i0 i0 > > > i =i0 6 > > > > X a+ a )g > ? > > < 1jn j j (0;0) If R = F : > or > ? X a a+ + (0;0) > (b) 9 j s.t. S = f ( a a ; > 0 i i j j0 0 > > > 1 i n > > X > > > ? a+j aj )g : j 6=j0
13
n 1
8 > > (a) 9i0 t.q. S (0;0) = f(ai a+i ; ai0 a+i0 ); (a+j aj ; ? > > > > > + > ai0 ai0 )gi6=i0 ;1jn > < (0;0) If R = A1 [B1[F : > or > > (0 ; 0) > (b) 9i0 s.t. S = f(a+i ai ; a+i0 ai0 ); (aj a+j ; ? > > > > a+ a )g > : i0 i0 i6=i0 ;1jn
If R ; = E [ C [ D : 9K; L [1; ::; n], K\L = ;, i 2= K[L, K[L[ fi g = [1::n] such that S ; = f(ai ai ; ? ai ai); (aj aj ; ? aj aj ); X X X X (ai0 ai0 ; 0 ? ak ak + ak ak ); (ai0 ai0 ; 0 ? ak ak + ak ak )g, (0 0)
1
0
+
(0 0)
0
+
+
+
+
+
+
+
+
+
L L 0 0 i 2 L; j 2 K, et = ?Card(L) , = ?Card(K) .
K
K
and also the forms corresponding to the cases B2 , A2 [ B1 , B3 , and E200 , symmetrical to the forms A2 , A1 [ B2 , A3 et E20 , by the exchange of ai and a+i . The next lemma will allow us to use symmetry between ai and a+i and x the value of some constant when it is useful.
Proposition 3 Let be one of the following automorphisms of Ln : ; : Ln ! Ln : L n ! Ln ai 7?! ai ai 7?! ai ai 7?! ai ai 7?! ai with 2 R n f0g. Then if I ful ls (P ),. . . ,(P ), so does (I ). +
+
+
0
+
4
We do it for ; , the even easier case of is left to the reader. For (P0 ), it is obvious. Let us show that ; (I ) ful ls (P1 ) : it is only needed to verify that the image under ; of a standard presentation of I , which is a presentation of ; (I ), is sent to ; (I ) by J . If x is a homogenous quadratic element of a standard presentation of I , ; (x) is proportional to x, so J ( ; (x))Pis proportional P +to ; (J (x)) and + therefore belongs to ; (I ). Now if x = i ai ai + i ai ai ? :1, with i , i , 2 R, is a generator of I , then ;(x) = (P iai a+i + P i a+i ai ) ? :1 is stable under J , since 2 R. It is clear by its de nition that ; commutes to the action of Sn . Thus, ( ; (I )) = ; ( (I )) = ; (I ) 14
Lastly, let S , whose elements we denote by (ms ; fs ), be a quadratically con uent reduction system, adapted to < and associated to I , and let W be the vector space spanned by the irreducible monomials relative to S . If we set ; (S ) = f(ms ; k1k ; (fs ))js 2 S; ms of degree k respectively to ai and l respectively to a+i g, then ; (W ) = W is also the linear span of monomials that are irreducible under ; (S ). ; (S ) is clearly adapted to 1 an : Indeed, if it were not the case, we could write 1 = an 1 , and we would have an2 1 6> an 1 an , which is impossible, unless 1 = an2 , and if we iterate 15
we get 1 = an l . Now, by formula (2), Nn 1 = 0, thus l = 0, 1 = 1, and therefore N~1 = 1 , which contradicts (P0 ). Thus an 1 > 1 an . But lm([N~1 ; an ]) = an 1 . Therefore an 1 must be reducible. Four cases are to be distinguished : (i) 1 = a+i 1 , i < n (ii) 1 = ai1, i < n (iii) 1 = a+n 1 (iv) 1 = an1 Case (i) : 1 = a+i 1 is irreducible (for it is in W ), and an 1 = an a+i 1 is reducible. Since the presentation is quadratic con uent, we nd that an a+i must be reducible. This is only possible if the family (1; ?1)a , or (1; ?1)b is in R. Case (ii) : Doing as above, we see that an ai must be reducible. So R contains one of the set of type (1,1). Case (iii) : Writing [N~1 ; a1 ] we see that lm([N~1 ; a1 ] = a1 a+n 1 , which must be reducible. Thus we are in the (1,-1) case. Case (iv) : Here 1 has to be studied. a) If 1 = 1 ai then for i 6= j : [N~1 ; a+j ] = 1 an 1 ai a+j + 22 a+j + : : : ? 1 a+j an 1 ai ? 2 a+j 2 ? : : : +[u; a+j ] 2 I We see that the leading monomial is an 1 ai a+j , but now an 1 is irreducible, and as a consequence, a set of type (1,-1) must be in R. b) If 1 = 1 a+i , with i 6= j we nd : lm([N~1 ; a+j ]) = an 1 a+i a+j , this time it is the case (1,1). QED.
4 Demonstration of Theorem 1 Since the demonstration is a bit long, we only give a sample of it, refering the interested reader to [Bes]. If I is such that (P0 ); : : : ; (P4 ) holds, it has a standard presentation. We have to study every such presentation that is not yet ruled out by lemma 10. In most cases, it is possible to show that (P0 ) or (P3 ) cannot hold. Nonetheless, in some cases it is necessary to call upon a reduction system to calculate in a basis of irreducible monomials. 16
It is usually easier to deal with relations in B rather than with generators of the ideal I . We will do so all the time. Most of the time, the relations of type (0; 0) depend on a non-zero constant named . It is then possible to suppose that = 1 thanks to the homomorphism 1 ;1 , for instance. We will sometimes do so. In cases containing (1; ?1)a , and if r 6= 0, we set q = ? sr . If s 6= 0, we set q0 = ? rs . From now on we suppose n 2, we will look at the case n = 1 afterwards. (1; 1)c [ (1; ?1)b [ (2; 0) : I is generated by the relations : 8 > ai aj = a+i a+j = 0 < +2 2 =0 a = a i > : a a+ = ai+ a = 0 i j i j together with relations of type (0; 0). If relations of type A2 , C , B2 or D with or 6= 0 are present, it is easily seen by multiplying on the left or on the right these relations by ai that ai = 0, so that (P0 ) is not satis ed.
A :
8 a a+ + P a+ a = 1 (l ) 1 < 1 1 i i : : : : a a+ + P a+ a = 1 (l ) n n n i i Let's multiply (l1 ) on the right by an , we get an = 0, thus B = 0. This also rules out the case A1 [ B1 [ F = A3 [ B1 . 3
E :
8 a a+ + a+ a + P a+ a = 1 (l ) 1 < 1 1 1 1 1 2 i6=1 i i : : : : a a+ + a+ a + P a+ a = 1 (l ) n 1 n n 2 n n i6=n i i We do as above. F: X X ai a+i + a+i ai = 1 We rst multiply the relation F by ai on the left, then on the right, and we get : ai a+i ai = ai a+i ai = ai If 6= , we have B = 0. 2
17
We thus can suppose that = = 1. We have ai a+i ai = ai , 8i, i.e., going back to Ln , ai a+i ai ? ai 2 I . But, if there exists a quadratic con uent reduction system for I , it is of type (a) or (b) (see proposition 2). But in both cases, there exists i such that ai a+i and a+i ai are irreducible, so ai a+i ai ? ai must be irreducible, and this is a contradiction. M Cl(1; 1), and has a cubic remark : This algebra is isomorphic to in
1
con uent presentation. (1; 1)b [ (1; ?1)b [ (2; 0) :
8 a a ?a a =0 > i j j i > > < a+i a+j ? a+j a+i = 0 +2 2 > > > : ai+ = ai+ = 0 ai aj = a i aj = 0 Making use of either the relations (1; ?1)b or (2; 0), we see that A2 or B2 with 6= 0 leads to B = 0. C: X + ai ai = 1 X ) a12 a+1 + a1aia+i = a1 i>1 X ) aia1 a+i = a1 i>1
)0=a )B=0
1
All other presentations containing relations of type C are thus also ruled out (in particular E20 = E1 [ C ). F: X + X + ai ai + ai ai = 1 1in 1in X X ) a1ai a+i + a1 a+i ai = a1 i>1 1in X ) ai a1a+i + a1 a+1a1 = a1 i>1
18
) a a a = a + 1 1 1
1
Multiplying by a1 to the right we would obtain in the same way : ()
a1 a+1 a1 = a1
thus = . Furthermore, multiplying () by ai , i 6= 1 we get :
a1 ai = a1 a+1 a1 ai = a1 a+1 aia1 = 0 But, the presentation being standard, fai aj ? aj ai ; a+i a+j ? a+j a+i gi6=j is a basis of I2(1;1) . Now a1 ai 2= Vectfai aj ? aj ai ; a+i a+j ? a+j a+i g. Contradiction. We can do the same for F , and thus for A3 , B3 and E2 . (1; 1)a [ (1; ?1)b [ (2; 0) : This case is similar to the preceding one. (1; 1)c [ (1; ?1)a [ (2; 0) : 8 + + > < ai aj = ai a2j = 0 8i 6= j + 2 0 > : raaai + =+ asai + a= = 0 8i 6= j i j
j i
A : Let's multiply (l ) to the right by a : 3
1
1
a1 a+1 a1 = a1 () Now a1 a+1 = ai a+i ) a1 = ai a+i a1 , 8i. Then if s 6= 0, we have a1 = ? rs ai a1 a+i = 0, for i 6= 1, and B = 0. We can thus suppose that s = 0. Now ai a+j = 0, and multiplying (l1 ) to the left by a1 , we nd a1 = a1 a+1 a1 , consequently we have = 1, by ().
We are then in the case (b) of the theorem. It is easily seen that the reduction system f(ai aj ; 0); (a+i a+j ; 0); (ai 2 ; 0); (a+i 2 ; 0); (ai a+j ; 0); (ai a+i ; 1? P a+k ak )g is con uent and adapted to the deglex-ordering coming from a+1 < : : : < a+n < a1 < : : : < an , which we will denote by > < a1 a+1a + 1 1 a1 +1a 21 = a1 (ii) 1 1 1 1 1 1 + > (iii) 2 a1 a1 a1 = a1 > : 2 a+1 a1 2 = a1 (iv) Thanks to the 3 last formulas we get ( 2 ? 1 ? 1)a1 = 0. 1 in the case E 0 ) B = 0. So, if 2 ? 1 ? 1 6= 0, (i.e. 1 6= n ? 2 n Moreover, (iii) and (iv) ) 2 6= 0, or else B = 0. 23
The only remaining case is 2 = 1 + 1 6= 0. We de ne : B ! n M K [xi; yi ]=hxi yi ? 1= 2 i by (ai ) = xi, (a+i ) = yi . is well de ned i=1 P P since (ai a+i + 1 a+i ai + 2 j 6=i a+j aj ) = (1+ 1 )xi yi + 2 j 6=i xj yj = 1. Thus (P3 ) cannot hold. E1 [ C [ D : 8 a a+ + : : : + a a+ = (l ) 1 < +1 1 n n + a ( l 2) 1 a1 + : : : + an an = : a a+ + a+ a = (l ) i i
;i
i i
3
We must have = by C [ D. Now, if we multiply (l3;i ) by aj with j 6= i, we nd aj = 0, therefore = 0, or B = 0. But ( + ) = n = 0 and must be non-zero by lemma 3, then + = 0. It can thus be supposed that = 1 and = ? = 1. We can then use the same as in the case E2 . (P3 ) does not hold. F: X X ai a+i ? a+i ai = 1 ) ai2 a+i ? aia+i ai = ai And that would be impossible if (P4 ) was satis ed, for there must always exist i s.t. ai a+i and a+i ai are irreducible.
(1; 1)a ou b [ (1; ?1)b : Such a presentation is never standard. Indeed, we always have relations of the form : X X i ai a+i + i a+i ai + = 0 in
in
1
1
with 6= 0. Therefore : X
a1
in
1
i aia+i + 1 a1 a+1 a1 + a1 = 0
) a a a a + a a a a + a a = 0 ) a (a a )a a (a a )a + a a = 0 ) a a = 0 )a a =0 + 2 1 2 2 2
2 2
+ 1 2
+ 1 1 1 1 2
2
1 1
1 2
This is impossible.
1 2
24
+ 1 2
1 2
2
1 2
(1; 1)c [ (1; ?1)a :
a a = a+ a+ = 0 i 6= j i j i j raia+j + sa+j ai = 0 i 6= j This case is very easy, and we only state the results.
C , C [ D, A [ D : Multiplying the relation C or D by ai, one can 1
prove that the presentation is not standard. A3 or A1 [ B1 [ F : It can be proven that ai 2 = 0, thus the presentation is not standard. E2 or E20 : It can be shown that :
rs( 2 ? 1 ? 1)a1 = 0 Therefore we have rs( 2 ? 1 ? 1) = 0. If 2 ? 1 ? 1 = 0 the homomorphism of (1; 1)c [ (1; ?1)b [ E2 can be used to exclude this case. If rs = 0 and 2 ? 1 ? 1 6= 0, the presentation is shown to be non-standard. F : We have, 8i :
ai 2a+i ai + ai a+i ai2 = ai 2 From which we conclude that the presentation is not standard. (1; 1)b [ (1; ?1)a : 8 > < +ai a+j = a+j ai+ a a =a a > : ra ai + j+ sa+ja i= 0 i 6= j i j j i A2 : ra2 = ra2 a1 a+1 = ra1 a2 a+1 ) ?sa1a+1 a2 = ra2 ) (r + s)a2 = 0 Thus r + s 6= 0 ) B = 0. If r + s = 0 an homomorphism : B ! C , with C = K [x1 ; : : : ; xn ; y1 ; : : : ; yn ]=hx1 y1 ? 1; : : : ; xn yn ? 1i, is de ned by setting (ai ) = xi , (a+i ) = yi . Therefore (P3 ) cannot hold, by lemma 4. We do the same for A2 [ B1 and A2 [ B2 . 25
A , A [ D and A [ D : In these 3 cases we have : 3
1
2
+ ai a+i = P aj a+j ai ai + j aj aj = 1 with = 0 for A2 [ D and A1 [ D, and = 1 for A3 . If we set x = a1 a+1 = : : : = an a+n , we easily show that 8i, raix + sxai = 0, and 8i 6= 1, ra1 a+i ai + sa+i ai a1 = 0. Then : +
(ra1a+1 a1 + sa+1 a1 2 ) = (r + s)a1 ) (ra12 a+1a1 + sa1a+1a12 ) = (r + s)a1 2 ) (ra1 x + sxa1)a1 = (r + s)a12 ) 0 = (r + s)a1 2 Therefore r + s = 0. Let's show that x := ai a+i and yi := a+i ai are central elements of B . For x it is trivial : we calculate the commutator of x with ak or a+k by writing x = ai a+i with i 6= k. Now yi clearly commutes with aj and a+j for j 6= i. As for j = i, we just need to write :
yi = 1 ? 1 x ?
X k6=i
a+k ak
()
Now we show that B 0 Z (B ). Let m 2 B 0 be a monomial, and write m as m = b1 : : : b2l , with bi = ai or a+i . There must exists i < j such that bj = b+i (:= J (bi )) and 8k, i < k < j , bk 6= bi and bk 6= b+i . Then bi commutes with every bk s.t. i < k < j , and we can write : m = b1 : : : bi?1 bi bi+1 : : : bj?1 bj bj+1 : : : b2l = b1 : : : bi?1 bi+1 : : : bj ?1(bi bj )bj +1 : : : b2l Now bi bj = bi b+i = x or yi, thus bi b+i 2 Z (B ). Consequently m = b1 : : : bi?1 bi+1 : : : bj?1 bj+1 : : : b2l (bi b+i ). Let m = m0(bi b+i ), with bi b+i 2 Z (B ) and m0 2 B 0 . By easy induction, we nd m 2 Z (B ). Then B 0 Z (B ) (the other inclusion is always true). As a consequence, (P3 ) cannot hold (unless B = 0). A1 [ B1 [ F : As above, we must have r + s = 0. If + 6= 0, lemma 4 can be used, with the help of : B ! K [x; y]=hxy ? 1=( + )i, de ned by (ai ) = x, (a+i ) = y. If + = 0 we have (with = 1):
a1a+1 ? a+1 a1 = 1 26
)a a a ?a a =a ) a ai ai ? a a = a ; i 6= 1 ) ai ai a ? a a = a )a a ?a a =0=a )B=0 C : We do as in case (1; 1)c [ (1; ?1)b [ C . + 1 1 1
1
+
+ 2 1 1
+
1
1
+ 2 1 1
1
+ 2 1 1
+ 2 1 1
1
+ 2 1 1
1
C) C [ D : aa aa ++ :: :: :: ++ aanaan == ((D ) n n ra (C ) + s(C )a ) ra a + sa a a = (r + s)a ra (D) + s(D)a ) ra a a + sa a = (r + s)a 1
1
+ 1 1 + 1 1
+
+
1
1
2 + 1
+ 1 1 1
1
+ 1 1 1
+ 2 1 1
1
1
From this two relations we get : 2 + ra1 a1 a1 + sa1 a+1 a1 2 = (r + s)a1 2 ra1 2 a+1 a1 + sa1 a+1a1 2 = (r + s)a1 2
() ()
) (r + s)( ? )a = 0 This shows that (r + s)( ? ) = 0. { = : We can use as in case C . { =6 ; r + s = 0 then () and () ) a a = a a a = a a . 1
2
1
Thus :
2 + 1
+ 1 1 1
a1 : : : an = aP1 : : : an P aia+i = P ai 2 a+i a1 : : : ai?1 ai+1 : : : an = a+i ai (a1 : : : an ) = a1 : : : an Therefore a1 : : : an = 0, and, by induction : B = 0. E2 or E20 : On one hand ra1(l1 ) + s(l1 )a1 gives (r + s)a1 = s 1 a+1 a1 2 + ra1 2 a+1 + (s + r 1 )a1 a+1 a1 on the other hand ra1 (l2 ) + s(l2 )a1 gives : s 2 a+1 a1 2 + r 2 a1 a+1 a1 = (r + s)a1 From (ii) we get : 2 = 0 ) r + s = 0 27
+ 2 1 1
(i) (ii)
Furthermore : 2 (i) ? 1 (ii) , (S ) : 2 a1 (ra1 a+1 + sa+1 a1 ) = ( 2 ? 1 )(r + s)a1 2 (ra1 a+1 + sa+1a1 )a1 = (r + s)a1 (ii) Now : + + 2 1 a1 )a1 = ( 2 ? 1 )(r + s)a1 (S ) ) 2 a1 ( raa1 a(1ra+asa + + 2 2 1 1 1 + sa1 a1 )a1 = (r + s)a1
) ( ? ? 1)(r + s)a = 0 Then we must have (r + s)( ? ? 1) = 0. { ? ?1=0 : = 6 0, we de ne : Ln ! C , with C = K [x; y]=hxy ? 2 i, 2
1
1
2
2
2
1
1
1
2
such that (a1 ) = x, (a+1 ) = y, and (ai ) = (a+i ) = 0, 8i > 1. We then see that is well de ned non-zero morphism to C . 2 = 0 (which entails r + s = 0); 1 = ?1. We then have the presentation :
fai aj ? aj ai; ai aj ? aj ai ; ai aj ? aj ai; ai ai ? ai ai ? 1g + +
+ +
+
+
+
+
We recognize the Weyl algebra An , which is a well-known solution to our problem, with Ni = a+i ai + i :1, i 2 C. { 1 + 1 ? 2 6= 0, r + s = 0 + 0 (S ) , 2 a[a1 [a; a1 ;+a]a1 ] = 2 1 1 = 0 1 where [:; :] denotes the commutator. 1 + 1 + (n ? 1) 2 6= 0 (which always holds in the case E20 ) : We can then de ne a homomorphism from B to C = K [x1 ; : : : ; xn ]=hx2i ? 1 + + 1(n ? 1) i by setting : (ai ) = 1 2 (a+i ) = xi. 1 + 1 + (n ? 1) 2 = 0 X X X (l1 )+: : :+(ln ) , ai a+i + 1 a+i ai +(n?1) 2 a+i ai = n X X , a+i ai ? aia+i = n 28
,
X
[ai ; a+i ] = ?n But 2 6= 0, (or else 1 + 1 ? 2 = 0), then from (S ) we get : X ?na1 : : : an = [ai ; a+i ]a1 : : : an X ?na1 : : : an = ([ai ; a+i ]ai a1 : : : ai?1ai+1 : : : an) = 0 i
) a : : : an = 0 1
And we can iterate to get B = 0. E1 [ C [ D : From C [ D, the only remaining case to study is when = , that is to say n = ( + ). Moreover (by the case E2 ), 2 = 0, we must have r + s = 0. Then we see that we can use as in the case E2 , setting (ai ) = (a+i ) = xi , with x2i = n .
F : We have :
P + P + (r + s)aj = raP P ai a+i + P a+i ai)aj j ( ai ai + ai ai ) + s(P = i6=j ai (raj a+i + sa+i aj ) + i6=j (raj a+i + sa+i aj )ai +(r + s )aj a+j aj + r aj 2 a+j + s a+j aj 2 = (r + s )aj a+j aj + r aj 2 a+j + s a+j aj 2
Now, may the reduction be of the form (a) or (b), there exists j s.t. the monomials in the last expression are irreducible. We then have r = s = 0 and r + s = 0, and that entails that either r and s or and must be zero, which is impossible. (1; 1)a [ (1; ?1)a : This case is quite similar to the preceding one, and we leave it to the reader. In the case E2 we nd the pseudo-boson solution. (1; 1)c [ (2; 0) : ( ai aj = a+i a+j = 0 ai 2 = a+i 2 = 0 A3 : Let's look at the di erent reduction systems we can have :
29
{ (a) : On one hand we have aj aj = 0, and on the other hand : aj aj = aj (1 ? P Pin ai ai ) = aj ? Pi6 j aj ai ai ? (aj aj )aP j ai ai ? aj + in ai aiaj = aj ? i6 j ajP = (1 ? )aj ? i6 j aj ai ai 2 +
+
2 +
=
=
+
+
+
=
2
+
+
and this last expression is irreducible. Since it is non-zero, this case is ruled out. { (b) : We can use the same argument if we choose j = j0 . A1 [ B1 [ F : From (l3 )ai and ai(l3 ) we get = . We set to 1, and we de ne : Ln ! L1 = K ha; a+ i by (ai ) = a, (a+i ) = a+ . This induces a non-zero morphism from B to the Cli ord algebra Cl(1; 1). Indeed, (ai a+i ? aj a+j ) = 0 = (a+i ai ? a+j aj ), (ai aj ) = a2 = 0, (a+i a+j ) = a+ 2 = 0, (ai a+i + a+i ai) = aa+ + a+ a = 1. We conclude by lemma 7. E2 : (l1 )a2 a+2 , a1 a+1a2a+2 = a2a+2 Moreover, a1 a+1 (l2 ) , a1 a+1 a2 a+2 = a1 a+1 . Thus ai a+i = aj a+j , which is not possible. E1 [ C [ D : From C and D we have = = 0. Then n = + = 0, which is impossible. F : We have : P a a+ a ? a = 0 i i i j j Pi aj a+i ai ? aj = 0 But at least one one of these two expressions is irreducible. Absurde. Remark : A con uent cubic reduction system could be found in this case. Furthermore, in the case n = 2 and 6= 1, (P3 ) is shown to be satis ed (see [Bes]). (1; 1)b [ (2; 0) : C : B = 0 (see case (1; 1)b [ (1; ?1)a [ (2; 0)). E2 : Distinguons les cas suivant les di erents systemes de reductions :
30
{ (a) : On one hand we have ai ai = 0, and on the other hand : X ai ai = ai ? ai ai ai ? ai aj aj 2 +
2 +
1
+
+
2
j 6=i X = ai ? 1 ( ? 1 a+i ai ? 2 a+j aj )ai j 6=i X
X
?
2
ai a+j aj
j 6=i X
= (1 ? 1 )ai + 1 2 aj aj ai + 1 2 a+j ai aj J I X + +
?
2
j 6=i
ai aj aj
where I is the set of indices j s.t. aj < ai and J the set of all j 's s.t. aj > ai . The last expression being irreducible, we come to a contradiction. { (b) : We do the same with aia+i 2 . { (c) : (n = 2) We have a+j aj 2 = 0, and also a+j aj 2 = 1 (aj ? 2 1 a+i aiaj ? aia+i aj ). The term a+i ai aj may be reduced to a+i aj ai , but the whole expression cannot reduce to 0. { (d) :
a+i ai2 = 0
= 1 (ai ? 2 a+j aj ai ? ai a+i ai ) 1 1 = (ai ? 2 a+j aj ai ? 1 (ai ? 2 ai a+j aj ? ai 2 a+i )) 1 1 1 1 + ((1 ? )a ? a a a + 2 a a+ a ) =
1 i 2 j j i 1 i j j Con uence implies that 1 = 1, 2 = 0. In this case, I is generated by the relations a1 a+1 + a+1 a1 = , a2 a+2 + a+2 a2 = , a1 2 = a2 2 = a+1 2 = a+2 2 = 0, ai aj ? aj ai = a+i a+j ? a+j a+i = 0, 8i 6= j . We can then send B onto Cl(1; 1) by ai 7! a, a+i 7! a+. 1
We then use lemma 7.
E0 : { (a) : Same method as in (a) of case E . 2
2
31
{ (b) : We write : 0 = ai0 2 a+i0 = ai0 ?
X I
ai0 ai a+i ?
X J
aiai0 a+i
with I = fijai0 < ai g and J = fijai0 > ai g. We thus have a non-zero irreducible expression. { (c) : Same as above. F: { (a) : It is harmless to suppose that an is bigger than any other ai . Then i0 = n, and we have : X X 0 = an 2 a+n = an ? ai an a+i ? an a+i ai i aj : 2
2
1
2
2
?aj aiaj
+
.
aiaj a+j
&
ai
(P4 ) cannot hold. A2 [ D :
ai0
.
ai0 a+i0 ai0
&
(P4 ) cannot hold.
ai0 ? Pi6=i0 ai0 a+i ai
A [ D : We de ne : B ! Cl(n; 0) by (ai ) = (ai ) = p1n xi, we +
1
then use lemma 6. A3 : { (a) : Suppose that ai is the biggest of all ak 's, and choose j 6= i :
?aj aiaj
+
.
ai aj a+j
&
ai ? ai a+i ai ? X #
X k6=i
aia+k ak
X (1 ? )ai + 2 a+k ak ai ? ai a+k ak k6=i
This contradicts (P4 ). { (b) : If n 3, let i; j 6= i0 be s.t. ai < aj . Then aj aia+i = ?aiaj a+i , which is irreducible. And aj aia+i = aj ai0 a+i0 = ?ai0 aj a+i0 , which is also irreducible. This case is then ruled out. If n = 2, one can easily show that i0 6= j0 leads to a contradiction. We can then 43
suppose that i0 = j0 = 2. If we de ne < by a2 < a1 < a+1 < a+2 , we see that S = f(a1 a2 ; ?a2 a1 ); (a+2 a+1 ; ?a+1 a+2 ); (a1 a+1 ; a2 a+2 ); (a+2 a2 ; ? a2 a+2 ? a+1 a1 )g is con uent and adapted to
