Journal of Mathematical Sciences, Vol. 116, No. 1, 2003

´ SERIES FOR SOME ALGEBRAS OF INVARIANTS OF THE HILBERT–POINCARE CYCLIC GROUPS N. L. Gordeev

UDC 512.743; 512.547

Let ρ be a linear representation of a finite group over a field of characteristic 0. Further, let R ρ be the corresponding algebra of invariants, and let P ρ(t) be its Hilbert–Poincar´ e series. Then the series P ρ(t) represents a rational function Ψ(t)/Θ(t). If R ρ is a complete intersection, then Ψ(t) is a product of cyclotomic polynomials. Here we prove the inverse statement for the case where ρ is an “almost regular” (in particular, regular) representation of a cyclic group. This yields an answer to a question of R. Stanley in this very special case. Bibliography: 3 titles.

1. Introduction Let G be a finite group, and let ρ : G → GL(V ) be a linear representation of it in a finite-dimensional vector space V over an algebraically closed field K of characteristic zero. Then the group ρ(G) acts naturally on the symmetric algebra S = S(V ). Let Rρ = Sρ(G) be the corresponding algebra of invariants: Rρ = {s ∈ S | ρ(σ)s = s for every σ ∈ G}. P∞ Then, Rρ is a graded algebra = n=0 Rρ,n , where Rρ,n is the homogeneous component of degree n. The Hilbert–Poincar´e series of Rρ (see [2, 3]), Pρ (t) =

∞ X

(dim Rρ,n)tn ,

n=0

determines a rational function Ψ(t)/Θ(t), and in the sequel we identify Pρ (t) and the function Ψ(t)/Θ(t). We can take different representations of the Q function Pρ (t) in the form Ψ(t)/Θ(t). In particular, we can take Ψ and Θ such that Ψ(t) ∈ Z[t] and Θ(t) = j (1 − tmj ) (see [2, 3]). In the cases where Rρ is a complete intersection, Q the numerator Ψ(t) of Pρ (t) can also be written in the form i (1 − tei ). Note that this statement is true for every graded complete intersection but not necessarily for an algebra of invariants. In [2], R. Stanley formulated the question as to under what additional conditions the converse statement holds, i.e., if the numerator of a Q Q Cohen–Macaulay graded algebra is of the form i (1 − tei )/ j (1 − tmj ) for some {ei }, {mj } ⊂ N , then this algebra is a complete intersection. He pointed out two special cases where it is true and gave an example when it is false. It seems that it would be reasonable to look at a naturally defined class of graded algebras, say, algebras of invariants, from the point of view of Stanley’s question. However, this is a difficult problem even for the case where G is a cyclic group. Here, we give an answer in the affirmative to Stanley’s question in the cases where G is a cyclic group and ρ is close to the regular representation. Let ρ : G = hσi → GL(V ) be a representation of a cyclic group of order n. Further, let d | n and let εd be a fixed primitive dth root of unity. For g ∈ G we put md (g) = the multiplicity of the eigenvalue εd of the operator ρ(g). By md we denote max{md (g) | g ∈ G}. Definition. We say that a representation ρ is almost regular if (1) mn (σ) ≥ 1; (2) ρ is Aut G-invariant (i.e., ρ(f(σ)) = ρ(σ) for every f ∈ Aut G); (3) for every divisor d of n, the equality md (g) = md holds if and only if the order of g is equal to d. Example 1. Let ρ be the regular representation. Then ρ is almost regular. Indeed, conditions (1) and (2) obviously hold. Further, let g = σ` , where ` | n. Then the restriction of ρ to the group hgi is the sum of ` copies Published in Zapiski Nauchnykh Seminarov POMI, Vol. 272, 2000, pp. 144–160. Original article submitted May 4, 2000. c 2003 Plenum Publishing Corporation 1072-3374/03/1161-2961 $25.00

2961

of the regular representation of hgi. Hence, md (g) = 0 if d is not a divisor of n/` and md (g) = ` if d | (n/`). Thus, the maximum of the function md (g) is equal to n/d and it is attained only on elements of order d. Example 2. Let ρ be a representation satisfying conditions (1) and (2). Further, V =

X

Vd ,

d|n

where Vd is a vector space generated by eigenvectors of ρ(σ) with eigenvalues that are primitive dth roots of unity. Suppose the following condition holds: (30 ) dimVd1 < dimVd2 for every divisor d1 6= d2 of n such that 1 6= d1 | d2. We show that ρ is almost regular. Let n = dk` for some d, k, and `, where d 6= 1, k 6= 1; let g = σ` . Put Dd,` = { a | n | ε`a is a primitive dth root of unity} (here, εa is a primitive ath root of unity). Since ρ is Aut G-invariant, we obtain X

md (g) =

(dim Va )/ϕ(d).

a∈Dd,` 0 If a ∈ Dd,` , then ε`k ak is a primitive dth root of unity. Using (3 ), we obtain

X

md (gk ) ≥

(dimVak )/ϕ(d) > md (g).

a∈Dd,`

Thus, the function md (g) attains its maximum only on elements of order d. Example 3. Let ρ be a representation satisfying conditions (1) and (2), and let V = Vn + V1 , i.e., V is the sum of V1 = V ρ(G) and the vector space Vn generated by eigenvectors of ρ(σ) with eigenvalues that are primitive nth roots of unity. Then md (g) 6= 0 if and only if the element g has order d. Thus, ρ is almost regular. The main result of this paper is the following theorem. Theorem 1. Let ρ be an almost regular representation of a cyclic group of order n. Let Pρ (t) = Ψ(t)/Θ(t) be the corresponding Hilbert–Poincar´e series. Assume that Ψ(t) is a product of cyclotomic polynomials. Then either n = 2, dim V /V ρ(G) = 1, and Rρ is a polynomial algebra or n = 2, 3, 4, 6, dim V /V ρ(G) = 2, and Rρ is a hypersurface. Remark 1. If the numerator of Pρ (t) (in some representation) is a product of cyclotomic polynomials, then multiplying Ψ(t) and Θ(t) for some appropriate cyclotomic polynomials we can obtain the form Pρ (t) =

Y Y (1 − tei )/ (1 − tmj ) i

j

for some ei and mj . Thus, we obtain an answer in the affirmative to Stanley’s question in this very particular case. In the case where n = p is a prime number, we have a stronger assertion. Theorem 2. Let ρ : G = hσi → GL(V ) be a faithful representation of a cyclic group of prime order p. Assume that ρ(f(σ)) = ρ(σ) for some f ∈ Aut G, f 6= 1, and the numerator Ψ(t) of the Hilbert–Poincar´e series Pρ (t) is a product of cyclotomic polynomials. Then f(σ) = σ−1 , dim V /V ρ(G) = 2, and the algebra Rρ is a hypersurface. We prove both results, using the approach developed in [1] for the classification of finite groups whose algebra of invariants is a complete intersection. This method is based on arithmetical properties of algebraic numbers Ψ(ε), where ε is a root of Θ. 2962

2. Arithmetical characteristics of cyclotomic polynomials 2.1. Let p be a prime number, and let α be an algebraic integer (i.e., α is a root of a polynomial xn + a1 xn−1 + · · · + an with ai ∈ Z). Further, let (α) and (p) be the principal divisors of the field Q(α) corresponding to α. Put vp (α) = max{r ∈ Q+ | (p)r | (α)}. Assume that Q(α)/Q is a normal extension. Then (p) = P1e P2e . . . P se,

(α) = P1m1 P2m2 . . . P sms D

for some prime divisors P1 , . . . , P s of Q(α) and a divisor D of Q(α) such that (Pi, D) = 1 for every i (here, e, mi ∈ Z, e > 0, and mi ≥ 0). In this case, vp (α) = mi0 /e, where mi0 = min{mi }. Note that if L/Q is a normal extension that contains the normal extension Q(α)/Q, then vp (α) = vp,L (α), where vp,L (α) = max{r ∈ Q+ | (p)rL | (α)L} and (p)L and (α)L are the corresponding principal divisors of L. Thus, if Q(α) ⊂ L and L/Q, Q(α)/Q are normal, we can substitute vp,L (α) for vp (α). Below we consider quantities vp (Ψm (εn )), where Ψm is the mcyclotomic polynomial and εn is a primitive nth root of unity (for some m and n). We can calculate such quantities, considering Ψm (εn ) as an element of L = Q(εn , εm ). Q 2.2. Let n = pr n1 6= m = ps m1 , where (n1 , p) = (m1 , p) = 1. Using the identity n = 1≤k≤n−1 (1 − εkn ), it is easy to obtain the following relations:  0 if n1 6= m1 ,     1 if n1 = m1 , s > r, vp (Ψm (εn )) = (1) s  ϕ(p )   if n1 = m1 , s < r,  ϕ(pr ) where ϕ is the Euler function (see [1], Lemma 5). 2.3. Assume that Ψ(t) =

Y

Ψmi (t),

(2)

mi ∈M

where Ψmi (t) is the mi -cyclotomic polynomial and M = {m1 , . . . , m k } is a finite sequence of natural numbers (possibly, mi = mj for i 6= j). Let p be a prime number, and let d = pa d1 , where (p, d1 ) = 1 (possibly, a = 0). Put Mp,d = {mi ∈ M | mi = ps d1 , s ≥ 0},

Y

Ψp,d =

Ψmi .

(3)

mi ∈Mp,d

The following statement is a variation of Lemmas 6 and 7 from [1]. Proposition 1. Suppose Ψ(εn ) 6= 0 for every n = pb d1 , 0 ≤ b ≤ a + a0 , where a0 ∈ Z, a0 ≥ 0. Then for every such n, vp (Ψ(εn )) = vp (Ψ(εd )) and deg Ψp,d ≥ vp (Ψ(εd )) · ϕ(d)

ϕ(pa+a0 +1 ) . ϕ(pa )

Moreover, if e = vp (Ψ(εd )) and P is a prime divisor of p in Q(εd ), then 0 6≡ Ψ(εn )/pe ≡ Ψ(εd )/pe (mod P ). Remark 2. We do not use the latter congruence below. However, this congruence could be useful if we consider Hilbert–Poincar´e series for different types of linear groups (see [1]). 2963

Proof. Let mi ∈ Mp,d . Then mi = ps d1 for some s ≥ 0. Since Ψ(εn ) 6= 0 for every n = pb d1 , b ≤ a + a0 , we have Ψmi (εn ) 6= 0 for every n = pb d1 , b ≤ a + a0 , and, therefore, s > a + a0 . From (1) we obtain vp (Ψmi (εn )) = vp (Ψmi (εd )) = 1

(4)

for every n = pb d1 , b ≤ a + a0 . If mi ∈ / Mp,d , then again from (1) we have vp (Ψmi (εn )) = vp (Ψmi (εd )) = 0.

(5)

The relation from the proposition now follows from (2), (4), and (5). From (1), (3), and (4) we obtain X

deg Ψp,d =

ϕ(mi ) ≥ |Mp,d | min{ϕ(mi )| mi ∈ Mp,d }

mi ∈Mp,d

= vp (Ψ(εd )) · min{ϕ(mi )| mi ∈ Mp,d } ≥ vp (Ψ(εd ))ϕ(pa+a0 +1 d1 ) = vp (Ψ(εd ))ϕ(d)

ϕ(pa+a0 +1 ) . ϕ(pa )

Now we prove the congruence. If mi = ps d1 , s > a, then Ψmi (εn ) = Ψmi (εd )) = Ψmi (εd1 ). Relation (4) implies that Ψmi (εn )/p = Ψmi (εd )/p is an algebraic integer of the field Q(εd ) and 0 6≡ Ψmi (εn )/p ≡ Ψmi (εd )/p (mod P )

(6)

if mi ∈ Mp,d . Further, εpb ≡ 1 (mod P ) and, therefore, Ψmi (εn ) ≡ Ψmi (εd ) for every mi . Moreover, (5) implies

(mod P )

Ψmi (εn ) 6≡ 0 (mod P )

(7)

(8)

for mi ∈ / Mp,d . From (4)–(8) we have 

Y







Y



Ψ(εn )  Ψmi (εn )   = Ψmi (εn ) pe p mi ∈Mp,d mi ∈M / p,d    Y  Ψm (εd )  Y Ψ(εd ) i  ≡ Ψmi (εd ) = 6≡ 0 (mod P ). p pe mi ∈Mp,d

mi ∈ / Mp,d

3. The Hilbert–Poincar´ e series of an algebra of invariants Let ρ : G → GL(V ) be a linear representation of a finite group G over a field K (below we assume that K is an algebraically closed field of characteristic zero). The linear group ρ(G) acts naturally on the symmetric algebra S(V ). Let Rρ = S(V )ρ(G) be the corresponding algebra of invariants. By Maschke’s theorem, the Hilbert–Poincar´e series Pρ (t) of Rρ can be expressed by the formula   1 X 1  Pρ (t) = |G| det(1 − ρ(g)t) g∈G

(see [2, 3]). Thus, Pρ (t) = 2964

Ψ(t) , Θ(t)

where Θ(t) = LCM {Θg (t) = det(1 − ρ(g)t) | g ∈ G} and Ψ=

1 X Θ(t)Θ−1 g (t). |G| g∈G

On the other hand, Rρ is a Cohen–Macaulay algebra (see [2, 3]), and therefore e Ψ(t) , ej j=1 (1 − t )

Pρ (t) = Qm where

e Ψ(t) = a0 tn + a1 tn−1 + · · · + an

for some n, a0 , . . . , a n ∈ N ∪ {0}, m = dim V , and e1 , . . . , e m are the degrees of a homogeneous Rρ-sequence e (see [2, 3]). Moreover, the algebra Rρ is Gorenstein if and only if the polynomial Ψ(t) is symmetric, i.e., ai = an−i for every i (see [2]). Obviously, every cyclotomic polynomial is symmetric. Further, if f = g and h, where f, g, h ∈ F [t] for some field F , and if f, h are symmetric polynomials, then g is also symmetric. Thus, if the numerator of Pρ (t) in some representation Pρ (t) = Ψ0 (t)/Θ0 t) is a product of cyclotomic polynomials, then the algebra Rρ is Gorenstein. Further, if the group ρ(G) does not contain a reflection, then the algebra Rρ is Gorenstein if and only if ρ(G) ≤ SL(V ) (see [2, 3]). Thus, we have the following fact. Lemma 1. Let the numerator of Pρ(t) (in some representation) be a product of cyclotomic polynomials. If the group ρ(G) has no reflections, then ρ(G) ≤ SL(V ). 4. Cyclic groups of primary orders We obtain Theorem 2 from a more general assertion. Theorem 3. Let ρ : G = hσi → GL(V ) be a faithful linear representation of a cyclic group of primary order ps , p 6= 2, and let Pρ (t) = Ψ(t)/Θ(t) be the corresponding Hilbert–Poincar´e series. Assume that (a) ρ(f(σ)) = ρ(σ) for an automorphism f ∈ Aut G such that the subgroup of automorphisms hfi acts without fixed points on G \ {1} (i.e., f a (σb ) 6= σb if f a 6= 1 and σb 6= 1); (b) Ψ(t) is a product of cyclotomic polynomials; (c) (Ψ(t), Θ(t)) = 1 and Θ(εpe ) = 0 for every 1 ≤ e ≤ s. Then dim V /V ρ(G) = 2, the matrix ρ(σ) is similar to a matrix diag (εps , ε−1 ps , 1, 1, . . . , 1), and the algebra R ρ ps ps is a hypersurface generated by the monomials x1 , x2 , x1 x2 , x3 , . . . , x m for some basis x1 , x2 , . . . , x m of V . Proof. We have 1 Pρ (t) = |G|

X g∈G

1 Θg (t)



e where Θ(t) = LCM {Θg | g 6= 1} and h(t) =



X 1 1 + (1 − t)m Θg (t) g6=1   h(t) 1 1 + , = s e p (1 − t)m Θ(t) 1 = s p

X Θ(t) e . Θg (t)

 (9)

(10)

g6=1

Since |hσi| = ps , we have

Θg (t) = (1 − ε1g t)(1 − ε2g t) . . . (1 − εmg t),

where {εig } are ps th roots of unity (not necessarily primitive). The Galois group Gal (Q(εps /Q) acts on the set e e {Θg | g 6= 1} by permutations. Thus, Θ(t) ∈ Z[t], and from (10) we have h(t) ∈ Z[t]. Since all the roots of Θ(t) are ps th roots of unity, we have 1 e Θ(t) = Ψaps0 Ψaps−1 . . . Ψ aps−1 (1 − t)as . (11) 2965

Let

e d(t) = (h(t), Θ(t)),

From (11) we obtain

h(t) = d(t)h0 (t),

e e 0 (t). Θ(t) = d(t)Θ

e 0 (t) = Ψbp0s Ψb1s−1 . . . Ψ bps−1 (1 − t)bs . Θ p

(12)

e 0 (t), h0 (t)) = 1, and condition (c) of the theorem imply Now (9), (12), (Θ b0 , b1 , . . . , b s−1 ≥ 1, bs < m. Moreover,

e 0 (t), Θ(t) = (1 − t)m−bs Θ  1  Ψ(t) = s Ψbp0s · · · Ψ bps−1 + (1 − t)m−bs h0 (t) . p

(13)

(14)

Let us show that b0 = b1 = · · · = bs−1 = 1.

(15)

Assume that bi > 1 for some i. Then from (13) and (14) we have ` = vp (Ψ(1)) = b0 + · · · + bs−1 − s > 0.

(16)

Applying Proposition 1 with d = ps and d1 = 1, from (16) we obtain deg Ψ(t) ≥ deg Ψp,ps ≥ ` · (p − 1)ps .

(17)

On the other hand, (9) and (14) imply deg Ψ(t) ≤ b0 · (p − 1)ps−1 + b1 (p − 1)ps−2 + . . . + bs−1 (p − 1) = (p − 1)[b0 ps−1 + b1 ps−2 + . . . + bs−1 ].

(18)

If we fix the number b = b0 + . . . + bs−1 , then sum (18) is maximal if b 1 = 1, . . . , b s−1 = 1 and b0 = b − (s − 1) = (` + s) − (s − 1) = ` + 1. Thus,   ps − 1 s−1 s−2 s−1 . (19) deg Ψ(t) ≤ (p − 1)[(` + 1)p +p + . . . + 1] = (p − 1) `p + p−1 From (17) and (19) we obtain `ps−1 +

ps − 1 ≥ `ps p−1

(20)

for an integer ` > 0. Thus, ps − 1 ≥ `ps−1 (p − 1)2 . If p > 2, then (p − 1)2 > p and we have a contradiction: ps − 1 > `ps . Thus, (20) and, therefore, (16) are false. Hence we have (15). From (14) and (15) it follows that 1 Ψ(t) = s p



 s 1 − tp m−bs h0 (t) . + (1 − t) 1−t

(21)

All the coefficients of the polynomial h0 (t) are divisible by c = |hfi|. Indeed, condition (a) of the theorem implies that the set G \ {1} decomposes into the union ∪α∈AOα of hfi-orbits with |Oα | = |hfi| = c (here A is a set of representatives of orbits). Moreover, Θg1 = Θg2 for every g1 , g2 ∈ Oα . Thus, from (7) we obtain h(t) = c

X Θ(t) e Θg (t)

! ,

α∈A

where gα ∈ Oα . Hence, all the coefficients of h are divisible by c. Since h0 = h/d and d is a product of cyclotomic polynomials, we obtain our assertion. 2966

Thus, for every n the divisor (h0 (εn )) of the field Q(εn ) is divisible by the divisor (c). Let q be a prime number that divides c, and let n = qpa , 1 ≤ a ≤ s. Note that (c, p) = 1 and therefore (q, p) = 1 (because c
| |G \ {1}|; this follows from condition (a)). Assume that q 6= 2. Since (c) | (h0 (εm )) and s vq (1 − εpn )/(1 − εn ) = 1/q − 1, formula (21) implies that Ψ(εn ) 6= 0 and vq (Ψ(εn )) = 1/q − 1.

(22)

Comparing (22) with (1), we obtain Ψpa | Ψ. This is a contradiction with (14). Thus, q = 2. For every n = 2 · pa, 1 ≤ a ≤ s, we have 2 | (Ψ(εn )) (possibly, Ψ(εn ) = 0). Thus, for every 1 ≤ a ≤ s we have Ψ2ea pa | Ψ for some ea ≥ 1 (this follows from (1)). Since deg Ψ ≤ ps − 1, we obtain ea = 1 for every 1 ≤ a ≤ s and, therefore, Ψ = Ψ2ps · Ψ2ps−1 · · · Ψ 2p . Now, from (9), (12), (14), (15), and (21), we arrive at s

Pρ (t) =

Ψ2ps Ψ2ps−1 · · · Ψ2p 1 − t2p = . s 2 m p Ψps Ψps−1 · · · Ψp (1 − t) (1 − t ) (1 − t2 )(1 − t)m−2

(23)

Consider the representation ρe : G = hσi → GL(V ) defined by the formula ρe(σ) = diag (εps , ε−1 ps , 1, . . . , 1). One can easily check that the algebra of invariants Rρ˜ is a hypersurface and its Hilbert–Poincar´e series has the form (23), i.e., P ρ˜(t) = Pρ (t). We show now that the representation ρ is equivalent to ρe (for some choice of εps ). We have dim V ρ(G) = Pρ0 (0) = Pρ˜0 (0) = m − 2. Since char K = 0, we have V = V1 ⊕ V ρ(G) , where V1 is a 2-dimensional ρ(G)-module. Let ρ1 : G = hσi → GL(V1 ) be the restriction of ρ. Then Pρ (t) =

1 Pρ (t). (1 − t)m−2 1

Condition (a) implies that the group ρ(G) has no reflections. Thus, the group ρ1 (G) has no reflections as well. Since the numerator of Pρ1 (t) is a product of cyclotomic polynomials, we may apply Lemma 1. Therefore, ρ1 (G) ⊂ SL(V1 ) and ρ1 (σ) ∼ diag (εps , ε−1 ps ). Now we show how Theorem 3 implies Theorem 2. Since f 6= 1 in the assumption of Theorem 2, we have p 6= 2. In the case s = 1, conditions (a) and (b) of Theorem 3 follow from the assumption of Theorem 2. Consider condition (c) of Theorem 3. If s = 1, from (9) we obtain Pρ (t) =

f(t) (1 − t)m Ψkp

for some k ≥ 0. We may assume that (f, Ψp ) = 1), deg f ≤ deg Ψkp . If k = 0, then Pρ (t) = 1/(1 − t)m . But this means that dim V ρ(G) = Pρ0 (0) = m. Thus, k > 0 and we have condition (c). This proves Theorem 2. 5. Proof of Theorem 1 Let ρ : G = hσi → GL(V ) be an almost regular representation, and let m = dim V . Further, let g ∈ G and let ερ(g)1 , . . . , ε ρ(g)m be the eigenvalues of the operator ρ(g). Then det(1 − ρ(g)t) =

m Y (1 − ερ(g)i t). i=1

Put Θg (t) =

m Y (1 − ερ(g)i t),

Θ(t) = LCM {Θg (t)}.

i=1

2967

From the Mashke formula, we have Pρ (t) =

1 X Θ(t)Θ−1 g (t) . n Θ(t)

(24)

g∈G

Further, from the definition of an almost regular representation we have Θ(t) = (1 − t)m

Y

d Ψm d .

(25)

d|n,d6=1

(Here Ψd is the d-cyclotomic polynomial.) Let d | n. Since ρ is Aut G-invariant, all polynomials Θ g (t) for elements of the same order d have the form Y m d Ψk k,d (1 − t)m1,d . (26) Θd (t) = Ψm d k|d,k6=1,d

Note that md in (26) is the same as in part (3) of the definition of an almost regular representation. Also, this definition implies that (27) mk,d < mk for every k | d, k 6= 1. Put

e d (t) = Θ(t)Θd (t)−1 . Θ

Then from (24) we conclude that Pρ (t) =

e d (t) Ψ(t) 1 X ϕ(d)Θ = . n Θ(t) Θ(t)

(28)

d|n

From (25), (26), (27), and (28), we obtain (Ψ(t), Θ(t)) = 1 and 0 6= Ψ(εd ) = =

1 ϕ(d)(1 − εd )m−m1,d n

1 e d (εd ) ϕ(d)Θ n

Y

Ψmk −mk,d (εd )

Y

` Ψm ` (εd ).

(29)

`/ |d

k|d,k6=1

Lemma 2. Let p2 | n for some prime p. Then p = 2, n = 4, the matrix ρ(σ) is similar to diag (i, −i, 1, . . . , 1), and Rρ is a hypersurface. Proof. Let n = pa `, where a > 1 and (`, n) = 1. Further, let g = σ` . From part (3) of the definition of an almost regular representation, we have mpa (g) = mpa . Obviously, b

mpa−b = mpa−b (gp ) = pb mpa for every b < a. Put s0 = mpa , s1 = mpa−1 , · · · , sa−1 = mp . Thus, si = pi s0 .

(30)

From (1), (29), and (30), we have   vp (Ψ(1)) = vp Ψsp0a (1)Ψsp1a−1 (1) . . . Ψspa−1 (1) − vp (n) = s0 + · · · + sa−1 − a = s0 (1 + p + · · · + pa−1 ) − a = s0

pa − 1 − a. p−1

From (1), (26), and (29), we have −e1 −e2 vp (Ψ(εpa )) = vp (Ψsp1a−1 (εpa )Ψsp2a−2 (εpa ) · · · Ψspa−1−ea−1 (εpa ))

2968

(31)

+vp ((1 − εpa )m−e0 ) + vp (ϕ(pa )) − vp (n),

(32)

where e1 = mpa−1 ,pa ,

e2 = mpa−2 ,pa , . . . , e a−1 = mp,pa ,

e0 = m1,pa .

From (30) and (32) we obtain vp (Ψ(εpa )) =

s1 − e1 s2 − e2 sa−1 − ea−1 m − e0 + + ··· + −1 2 a−1 p p p (p − 1)pa−1 a−1 X ei m − e0 + − 1. = s0 (a − 1) − i p (p − 1)pa−1 i=1

(33)

From (26) we arrive at m = deg Θpa = mpa ϕ(pa ) +

a−1 X

mpa−i ,pa ϕ(pa−i ) + m1,pa

i=1

= s0 ϕ(pa ) + e1 ϕ(pa−1 ) + e2 ϕ(pa−2 ) + . . . ea−1 ϕ(p) + e0 = s0 (p − 1)pa−1 + (p − 1)(e1 pa−2 + e2 pa−3 + · · · ea−1 ) + e0 . Hence, m − e0 = s0 (p − 1)pa−1 + (p − 1)(e1 pa−2 + · · · + ea−1 ).

(34)

From (33) and (34), we have vp (Ψ(εpa )) = s0 a − 1.

(35)

Proposition 1 and formulas (31) and (35) imply s0 and, therefore,

pa − 1 − a = s0 a − 1 p−1

pa − 1 a−1 = a+ . p−1 s0

(36)

We have (pa − 1)/(p − 1) ≥ 2a − 1. On the other hand, a + (a − 1)/s0 ≤ 2a − 1. Thus, (36) implies that a = 2 and therefore p + 1 = 2 + 1/s0 . Then, p = 2 and s0 = 1. Now n = 4 · ` with (2, `) = 1. Recall that s0 = mpa = m4 is the multiplicity of the eigenvalue ε4 of the operator ρ(σ` ). Since s0 = 1, we have ` = 1. Further, V = V4 + V2 + V1 , where Vd is the vector space generated by eigenvectors with eigenvalues that are primitive dth roots of unity. From part (3) of the definition of an almost regular representation, we obtain dim V2 < dim V4 . But dim V4 = 2, because s0 = 1 and ρ is Aut G-invariant. Since dim V2 6= 1 (because ρ(G) ≤ SL(V ); see Lemma 1), we have dim V2 = 0. Thus, V = V4 + V1 . If x1 and x2 are eigenvectors corresponding to i and −i and x3 , . . . , x m are linearly independent ρ(G)-invariants, then in the basis x1 , x2 , x3 , . . . , x m the matrix ρ(σ) has the form diag (i, −i, 1, . . . , 1) and the algebra of invariants R ρ is generated by the monomials x41 , x1 x2 , x42 , x3 , . . . , x m . Now we assume that n = p1 p2 · · · p k . Lemma 3. Let k > 1. Then n = 6, the matrix ρ(σ) is similar to diag (ε6 , ε−1 6 , 1, . . . , 1), and R ρ is a hypersurface. Proof. If k > 1, then n > 2. Then the operator ρ(σ) has ϕ(n) > 1 distinct eigenvalues, which are primitive nth roots of unity. Thus, ρ(σe ) is not a reflection for every e. By Lemma 1, ρ(G) ≤ SL(V ). Therefore, all coefficients at tm in the polynomials Θg (t) are equal to (−1)m . Now, from (24) and (28) we obtain deg Ψ(t) = deg Θ(t) − m.

(37) 2969

From (25) and (37) we conclude that

X

deg Ψ(t) =

ϕ(d)md .

(38)

d|n,d6=1

Now, let 1 6= d | n and let p | d for some prime p. Put d1 = d/p. From (1), (25), and (26), we have d vp (Ψ(εd1 )) ≥ vp (Ψm d (εd1 )) − vp (n) = md − 1.

(39)

Let Ψp,d (t) be the product of polynomials of the form Ψd1 ps (t), s > 0, which divides the polynomial Ψ. Then s > 1 and by Proposition 1 and formula (39) we derive deg Ψp,d ≥ (md − 1)pϕ(d).

(40)

(Here, a = 1 and a0 = 0 in the assumption of Proposition 1.) Note that if d1 6= d2 are two divisors of n and if p1 | d1 and p2 | d2 are primes (possibly, p1 = p2 ), then (Ψp1 ,d1 , Ψp2 ,d2 ) = 1 (recall that s > 1). Thus, if for every divisor 1 6= d | n we fix any prime pd | d, from (40) we obtain deg Ψ ≥

X

(md − 1)pd ϕ(d).

(41)

d|n,d6=1

If md > 1, then (md − 1)pd ≥ md ;

(42)

(md − 1)pd > md if pd 6= 2.

(43)

moreover, Thus, if md > 1 for every d | n, then (41)–(43) contradict (38). Note that md > 1 for every d | n, d 6= n, n/2, because md ≥ ϕ(n/d)mn . Therefore, we may assume that mn = 1. Suppose p | n for some prime p > 2 and n 6= 2p. Then, mp > 1 and (mp − 1)pϕ(p) = (mp − 1)2ϕ(p) + (mp − 1)(p − 2)ϕ(p) ≥ mp ϕ(p) + (ϕ(n/p) − 1)(p − 2)ϕ(p) = mp ϕ(p) + (1 − ϕ(p)/ϕ(n))(p − 2)ϕ(n) ≥ mp ϕ(p) +

(44)

(p − 2) ϕ(n) 2

(ϕ(p)/ϕ(n) ≤ 12 , because n 6= 2p). If we compare (41)–(44) with (38), we obtain p ≤ 3 in the cases of odd n or mn/2 > 1 and p ≤ 5 in the case mn/2 = 1. Thus, we have only the possibility that n = 2 · 3 · 5 and mn/2 = 1. Now we substitute in (41) the last sum of (44) for p = 3 and 5 instead of the corresponding members. The inequality obtained contradicts (38). Hence n = 2p for some prime p. Recall that we may assume mn = 1. From (38) and (41), we obtain ϕ(2p) + mp ϕ(p) + m2 ϕ(2) = (mp + 1)ϕ(p) + m2 ≥ (mp − 1)pϕ(p) + (m2 − 1)2ϕ(2) = (mp − 1)pϕ(p) + 2(m2 − 1). Hence mp = 1 or p≤

mp + 1 . mp − 1

The latter inequality is possible only if p = 3 and mp = 2. Assume mp = 1. Then deg Ψ = 2ϕ(p) + m2 . Let q | p − 1, and let `q = vq (Ψ(ε2p )). From (1) and (29), we obtain `q ≥ 1 for every q | p − 1. By Proposition 1, deg Ψq,2p ≥ `q ϕ(2pq). 2970

Moreover, deg Ψ2,1 ≥ 2(m2 − 1). Since the cyclotomic polynomials Ψq,2p are prime to each other and Ψ2,1 , we have the inequality X `q ϕ(2pq) + 2(m2 − 1) deg Ψ = 2ϕ(p) + m2 ≥ q | p−1

(recall that in this sum q represents prime numbers). Note that m2 ≥ ϕ(p). Therefore, this inequality holds only if m2 = 2 and there exists only one prime q that divides p − 1; moreover, q = p − 1. Hence, m2 = 2 and p = 3. Now we have n = 6 and ρ is an Aut G-invariant representation with m6 = m3 = 1 and m2 = 2. Obviously, ρ(σ) ∼ diag (ε6 , ε−1 6 , 1, . . . , 1), where ε6 is a primitive 6th root of unity. In this case, the algebra Rρ is a hypersurface generated by the monomials x 61 , x1 x2 , x62 , x3 , . . . , x m. Assume now that n = 6 and m3 = 2. Then, deg Ψ = ϕ(6) + 2ϕ(3) + m2 ϕ(2) = 6 + m2 . By Proposition 1, deg Ψ3,1 ≥ 3ϕ(6) = 6, deg Ψ2,1 ≥ 2(m2 − 1), deg Ψ2,6 ≥ 2 · ϕ(6) = 4. Hence, 6 + m2 ≥ 6 + 4 + 2(m2 − 1), a contradiction. Consider the case k = 1, i.e., n = p is a prime number. If p ≥ 3, the result follows from Theorem 2. Let p = 2. If dim V /V ρ(G) = 1, then ρ(σ) is a reflection and Rρ is a polynomial algebra. If dim V /V ρ(G) = 2, then Rρ is a hypersurface generated by the monomials x21 , x1 x2 , x22 , x3 , . . . , x m . Let m2 = dim V /V ρ(G) > 2. Then deg Ψ = m2 . On the other hand, by Proposition 1 deg Ψ2,1 ≥ 2(m2 − 1). Thus, m2 ≥ 2(m2 − 1). This is a contradiction. Now Theorem 1 is proved. The paper was supported by the SFB 343 “Diskrete Strukturen in der Mathematik” and the INTAS-93-2618Ext. REFERENCES 1. N. Gordeev, “Finite linear groups whose algebra of invariants is a complete intersection,” Izv. Akad. Nauk USSR, Ser. Mat., 50, 343–392 (1986). 2. R. P. Stanley, “Hilbert functions of graded algebras,” Adv. Math., 28, 57–83 (1978). 3. R. P. Stanley, “Invariants of finite groups and their application to combinatorics,” Bull. Amer. Math. Soc., 1, 475–511 (1979).

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