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On the arithmetic of integral representations of finite groups

Dmitry Malinin

Department of Mathematics UWI

Spa, Belgium, June 23, 2017

Dmitry Malinin On the arithmetic of integral representations

Absolutely irreducible orders

R – a Dedekind domain with quotient field K, H – R-order, L – absolutely irreducible H-representation module (i.e., a finitely generated H-module that is torsion free as an R-module and such that K ⊗R L is an absolutely irreducible K ⊗R H-module). The most interesting case H = RG, G - a finite group and charK ∤ |G| Jordan-Zassenhaus Theorem: Every isomorphism class of KH-representation modules splits in a finite number of isomorphism classes of RH-representation modules if the ideal class group of R is finite. What happens if cl(R) is infinite?

Dmitry Malinin On the arithmetic of integral representations

p-groups G

• Theorem. Let K denote either Qp(ζp∞) the extension of Qp
• btained by adjoining all roots ζpi, i = 1, 2, 3, ... of p-primary
• rders of 1. Let us fix the degree t of matrix representations.

Let G be any finite nonabelian p-group, and let OK be the ring

• f integers of K. Then there is an infinite number of integral

pairwise inequivalent absolutely irreducible representations of finite groups G in GLn(OK) . The constructed representations are contained in the kernel of reduction It(modP) modulo some prime divisor P of p.

Dmitry Malinin On the arithmetic of integral representations

One combinatorial construction for p-groups

Let K be a finite extension of either the rational p-adic field Qp,

• r the field Q of rationals, and let OK be its ring of integers.

Consider a group G0 generated by two elements a and b of

• rder t = pm, at=bt=1 such that the commutator c = [a, b] is

contained in the center of G0, and ct=1. Let ζ be a primitive root of 1 of degree t.

Dmitry Malinin On the arithmetic of integral representations

A combinatorial construction

C =

• n≥i≥j≥1

(−1)i−j n − j i − j

• eij,

C1 =

• n≥i≥j≥1

n − j i − j

• eij.

Let X = diag(1, x, x2, . . . , xt−1), then C1XC =

• n≥i≥j≥1

n − j i − j

• xj−1(1 − x)i−jeij,

and if we take x = 1, this will imply C−1 = C1.

Dmitry Malinin On the arithmetic of integral representations

A combinatorial construction

C =

• n≥i≥j≥1

(−1)i−j n − j i − j

• eij,

C1 =

• n≥i≥j≥1

n − j i − j

• eij.

Let X = diag(1, x, x2, . . . , xt−1), then C1XC =

• n≥i≥j≥1

n − j i − j

• xj−1(1 − x)i−jeij,

and if we take x = 1, this will imply C−1 = C1.

Dmitry Malinin On the arithmetic of integral representations

The construction for 2-generated p-groups

The following representation of G0 is faithful and absolutely irreducible. A = ∆(a) =        1 . . . . . . . . . ... . . . . . . 1 1 . . .        , B = ∆(b) = diag(1, ζ, . . . , ζt−1)

Dmitry Malinin On the arithmetic of integral representations

The construction for 2-generated p-groups

Let X = diag(1, x, x2, . . . , xt−1), then C−1XC =

• n≥i≥j≥1

n − j i − j

• xj−1(1 − x)i−jeij,

If we take x = ζ, we will obtain: ∆′(b) = C−1∆(b)C = C−1BC =

• n≥i≥j≥1

n − j i − j

• ζj−1(1−ζ)i−jeij,

∆′(a) = C−1AC =         1 − t 1 . . . − t

2

• 1

1 . . . . . . . . . ... ... . . . . . . − t

t−1

• . . .

1 1 . . . 1         ,

Dmitry Malinin On the arithmetic of integral representations

The construction for 2-generated p-groups

Let X = diag(1, x, x2, . . . , xt−1), then C−1XC =

• n≥i≥j≥1

n − j i − j

• xj−1(1 − x)i−jeij,

If we take x = ζ, we will obtain: ∆′(b) = C−1∆(b)C = C−1BC =

• n≥i≥j≥1

n − j i − j

• ζj−1(1−ζ)i−jeij,

∆′(a) = C−1AC =         1 − t 1 . . . − t

2

• 1

1 . . . . . . . . . ... ... . . . . . . − t

t−1

• . . .

1 1 . . . 1         ,

Dmitry Malinin On the arithmetic of integral representations

The construction for 2-generated p-groups

Consider a finite extension Lh = K(ζ) of K, its maximal order OLh, a prime divisor P of p and its prime element πh, this prime element may be chosen as ζpm − 1 in a cyclotomic field K = Q(ζpm) or K = Qp(ζpm), where ζpm is a primitive p-root of 1. Let D = diag(1, πh, π2

h, . . . , πt−1 h

), then ∆h(a) = D−1

h ∆′(a)Dh =

        1 − t πh . . . − t

2

• π−1

h

1 πh . . . . . . . . . ... ... . . . . . . − t

t−1

• π2−t

h

. . . 1 πh . . . 1         , ∆h(b) = D−1

h ∆′(b)Dh =

• n≥i≥j≥1

n − j i − j

• ζj−1(1 − ζ)i−jπj−i

h eij.

Dmitry Malinin On the arithmetic of integral representations

The construction for 2-generated p-groups

Consider a finite extension Lh = K(ζ) of K, its maximal order OLh, a prime divisor P of p and its prime element πh, this prime element may be chosen as ζpm − 1 in a cyclotomic field K = Q(ζpm) or K = Qp(ζpm), where ζpm is a primitive p-root of 1. Let D = diag(1, πh, π2

h, . . . , πt−1 h

), then ∆h(a) = D−1

h ∆′(a)Dh =

        1 − t πh . . . − t

2

• π−1

h

1 πh . . . . . . . . . ... ... . . . . . . − t

t−1

• π2−t

h

. . . 1 πh . . . 1         , ∆h(b) = D−1

h ∆′(b)Dh =

• n≥i≥j≥1

n − j i − j

• ζj−1(1 − ζ)i−jπj−i

h eij.

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

The phenomenon of global irreducibility was first distinguished by J. G. Thompson in the course of constructing the sporadic finite simple group F3. An essential ingredient of his construction was an even unimodular lattice Λ248 of rank 248 with Aut(Λ248) = Z2 × F3. Thompson observed that (Λ248, F3) is an example of pairs (G, Λ), G is a finite group and Λ is a torsion free ZG-module of finite rank, which satisfy the following condition: Λ/pΛ is an irreducible FpG-module, for every prime p. Many further results and examples are due to Pham Huu Tiep. We are interested in generalizations of the concept of global irreducibility for arithmetic rings introduced by F . Van Oystaeyen and A.E. Zalesskii.

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

Definition 1. We say that M(n, R) is a Schur ring if M(n, R) = GR, the R-span of G, for some finite group G ⊂ GL(n, R). Definition 2. Let F be an algebraic number field and let R be its ring of integers. A subgroup G ⊂ GL(n, F) is called globally irreducible if for each non-archimedean valuation v of F a reduction of G modulo v is absolutely irreducible. Problem 1. Let R be an arithmetic ring and M(n, R) the matrix ring over R. For which n there exists a finite group G ⊂ GL(n, R) such that the R-span of G is just M(n, R)?

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

Problem 2. Let G be a finite irreducible linear group over C, and let R be the ring spanned by the traces of the elements of

• G. Under what conditions GR ∼

= M(n, R)? Problem 3. Determine globally irreducible finite subgroups of GL(n, C). The case of prime n is of particular importance for Problem 1.

• Lemma. Let G ⊂ GL(k, R) and GR = M(k, R). Then there

exists Hm ⊂ GL(km, R) such that HmR = M(km, R).

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

F . Van Oystaeyen, A.E. Zalesskii, Finite groups over arithmetic rings and globally irreducible representations, J. Algebra, vol. 215, p. 418-436 Theorem 1. M(n, Z) is a Schur ring if and only if n is a multiple

• f 8 or n = 1. If n is a multiple of 8 then M(n, R) is a Schur ring

for every R. Observe that it is sufficient to examine maximal finite subgroups

• f GL(n, R) in order to check whether M(n, R) is a Schur ring.

Theorem 2. Let G ⊂ GL(n, R). Then GR = M(n, R) if and

• nly if G is globally irreducible.

One may expect that the existence of a globally irreducible subgroup G ⊂ GL(n, K) should imply that M(n, R) is a Schur

• ring. However, G is not always conjugate to a subgroup of

GL(n, R). Examples: Cliff, Ritter, Weiss, Feit, Serre.

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

One may expect that the existence of a globally irreducible subgroup G ⊂ GL(n, K) should imply that M(n, R) is a Schur ring, for this it would be sufficient to prove that G is conjugate to a subgroup of GL(n, R). But the examples given by Cliff, Ritter, Weiss, Feit and Serre show that the latter is not true. Theorem 3. Let p > 2 be an odd prime. Let Kπ denote the torsion subgroup of K ∗, the multiplicative group of K. Then M(p, R) is a Schur ring if and only if one of the following holds: (i) K contains an odd root of 1; (ii) there exists a field extension L of K such that [L : K] = p and Lπ/Kπ contains a cyclic group of order st where s, t are distinct primes; (iii) q = 2p + 1 is a prime and K contains √−q.

Dmitry Malinin On the arithmetic of integral representations

Globally irreducible representations over arithmetic rings

Theorem 4. Let p be an odd prime and G a primitive finite irreducible subgroups of GL(p, C). Then G is globally irreducible exactly in the following cases: (1) G ∼ = 3+Alt(6), 3-fold cover of the alternating group; (2) G ∼ = PSL(2, q), p = (q − 1)/2 with q prime or (3) G ∼ = PSp(2l, 3), p = (3l − 1)/2 and l > 1 is an odd prime. The trace fields in the cases (1), (2) and (3) are Q( √ −3), Q(√−q) and Q( √ 3) respectively.

Dmitry Malinin On the arithmetic of integral representations

Some related problems

D.K. Faddeev,1965,1995– Generalized integral representations Problem A. (W. Feit, J.-P . Serre). Given a linear representation ρ : G → GLn(K) of finite group G over a number field K/Q, is it conjugate to a representation ρ : G → GLn(OK) over OK? Global irreducibility, Schur rings – F . Van Oystaeyen and A.E.

• Zalesskii. R-span of a group G ⊂ M(n, R) ⇔ Brauer reduction
• f G(modp) is abs. irreducible for each prime ideal p of R.

Problem B. Describe the possible n and arithmetic rings R such that there is a globally irreducible G ⊂ M(n, R). What happens for n = 2? Theorem (Feit, Cliff, Ritter, Weiss). If G = Q8, K = Q( √ −35) and ρ : G → GL2(K), Problem 1 has a negative answer. It is possible to describe the fields K where G = Q8 is realizable in GL2(OK), but Serre asked a question, whether it is possible to give the realization explicitly in the terms of lattices. Explicit constructions are recently given by D. M. and F . Van Oystaeyen

Dmitry Malinin On the arithmetic of integral representations

Some related problems

D.K. Faddeev,1965,1995– Generalized integral representations Problem A. (W. Feit, J.-P . Serre). Given a linear representation ρ : G → GLn(K) of finite group G over a number field K/Q, is it conjugate to a representation ρ : G → GLn(OK) over OK? Global irreducibility, Schur rings – F . Van Oystaeyen and A.E.

• Zalesskii. R-span of a group G ⊂ M(n, R) ⇔ Brauer reduction
• f G(modp) is abs. irreducible for each prime ideal p of R.

Problem B. Describe the possible n and arithmetic rings R such that there is a globally irreducible G ⊂ M(n, R). What happens for n = 2? Theorem (Feit, Cliff, Ritter, Weiss). If G = Q8, K = Q( √ −35) and ρ : G → GL2(K), Problem 1 has a negative answer. It is possible to describe the fields K where G = Q8 is realizable in GL2(OK), but Serre asked a question, whether it is possible to give the realization explicitly in the terms of lattices. Explicit constructions are recently given by D. M. and F . Van Oystaeyen

Dmitry Malinin On the arithmetic of integral representations

Some related problems

D.K. Faddeev,1965,1995– Generalized integral representations Problem A. (W. Feit, J.-P . Serre). Given a linear representation ρ : G → GLn(K) of finite group G over a number field K/Q, is it conjugate to a representation ρ : G → GLn(OK) over OK? Global irreducibility, Schur rings – F . Van Oystaeyen and A.E.

• Zalesskii. R-span of a group G ⊂ M(n, R) ⇔ Brauer reduction
• f G(modp) is abs. irreducible for each prime ideal p of R.

Problem B. Describe the possible n and arithmetic rings R such that there is a globally irreducible G ⊂ M(n, R). What happens for n = 2? Theorem (Feit, Cliff, Ritter, Weiss). If G = Q8, K = Q( √ −35) and ρ : G → GL2(K), Problem 1 has a negative answer. It is possible to describe the fields K where G = Q8 is realizable in GL2(OK), but Serre asked a question, whether it is possible to give the realization explicitly in the terms of lattices. Explicit constructions are recently given by D. M. and F . Van Oystaeyen

Dmitry Malinin On the arithmetic of integral representations

Quaternions and 2-dimensional representations.

Theorem A. Let G = Q8, K = Q( √ −d), and d > 0. Then 1) G is realizable over K, ρ : G → GL2(K), if and only if d = a2 + b2 + c2 for some integers a, b, c. 2) G is realizable over OK, ρ : G → GL2(OK), if and only if d = a2 + b2 for some integers a, b or d = a2 + 2b2 for some integers a, b. Theorem B. Let G ⊂ GL2(C) be a globally irreducible group. Then one of the following holds: (i) G is an extention of quaternion group by S3 × C2 with χ(G) = Q( √ −1); (ii) G ≡ GL2(3) with χ(G) = Q( √ −2); (iii) G ≡ E48, the binary octaedral group, with χ(G) = Q( √ 2); (iv) G ≡ SL2(5) with χ(G) = Q( √ 5); (v) G has an abelian normal subgroup A such that |A/Z(A)| is not a prime power.

Dmitry Malinin On the arithmetic of integral representations

G = E48 and G = SL(2, 5)

• Proposition. The binary octahedral group G = E48 and the

binary icosahedral group G = SL(2, 5) are realizable in GL(2, OK) over the rings of integers OK of K = Q( √ 2, √ −1) and K = Q( √ 5, √ −1) respectively, i.e. any representation ρ : G → GL(2, K) of the group G over the field K/Q is conjugate in GL(2, K) to a representation ρ : G → GL(n, OK) over OK.

Dmitry Malinin On the arithmetic of integral representations

• Theorem. 1) Let G = Q4m, and let H = Q8. Then there is a

quadratic subfield K1 ⊂ K and an OK1H-module I which is an ideal in an extended field L1 = K1(i), such that: G = Q4m is realizable over OK if and only if H is realizable over OK1, and all Hilbert symbols −d,NL1/Q(I)

p

• = 1 for all p|d.

2) If H is not realizable over OK1, the minimal realization field containing K1 such that H is realizable over its ring of integers is E = Q(√d1, √d2), d = d1d2, di = ± 1, di = ± d. 3) The explicit computation of the above ideal I in the extended field L1 = K1(i) depends on a representation of the integer d = a2 + b2 + c2 as a sum of squares of integers. In particular, in the following 2 cases NL1/K1I is a principal ideal in OK1 : (a) b = c; then d = 2a2 + b2 (a and b are coprime) or equivalently, d has no prime factors p ≡ 5(mod8) and p ≡ 7(mod8). (b) c = 0; then d = a2 + b2 (a and b are coprime) or equivalently, d has no prime factors p ≡ 3(mod4).

Dmitry Malinin On the arithmetic of integral representations

Genera

F - an algebraic number field and K/F a finite abelian extension its Galois group - Gal(K/F). K ′ - the Hilbert class field of K, CK be the ideal class group of K. By class field theory the fields between K and K ′ are in one-one correspondence with the subgroups of CK. The genus field L of K/F is the maximal abelian extension of F contained in K. The subgroup of CK corresponding to L is the principal genus and the quotient of CK modulo the principal genus is the group of

• genera. H. Hasse: for F = Q in the principal genus can be

determined by arithmetic characters.

Dmitry Malinin On the arithmetic of integral representations

Genera

We construct OK1H-module I and OKG-module M in such a way that the action of G on M agrees with the action of H on I, and M can be obtained from I by extending the ground ring adjoining ζ2m. In K1/Q can be selected as generators of the genus group, qi represent cosets of the ideal class group CK1 modulo C2

K1, and the number of different genera is 2t−1. If

NL1/K1(I) is not in the principal genus, it is contained in some non-principal genus qi1qit . . . qitC2

• K1. The minimal extension of

K1 where NL1/K1(I) becomes a square is a quadratic extension K1(√qi1qi2 . . . qit) = K1(√d1) for d1|d, and since the ideal qi1qi2 . . . qit is not principal, it cannot be a unit ideal or ( √ −d). The Steinitz class of I in Q(√d1, √d2) becomes trivial, and H is realizable over the ring of integers of Q(√d1, √d2) for d = d1d2.

Dmitry Malinin On the arithmetic of integral representations

Classification of primitive representations of Galois Groups

Classification of absolutely irreducible primitive representations

• f the absolute Galois groups of local fields.
• Theorem. G – a finite group, H – its normal p-subgroup, G/H

supersolvable, ρ : G → GLn(K) – faithful primitive. Then: n = pd The center Z = Z(H) is cyclic of order pz, and for c ∈ Z of order p there are elements u1, v1, . . . , ud, vd which together with Z generate H and satisfy the generating re- lations: [ui, uj] = [vi, vj] = 1, [ui, vj] = cδi,j, (i, j = 1, . . . , d), and the generators from different pairs commute. There are 2 possibilities: 1) up = vp = 1 for p = 2 2) up = vp = c (quaternion type), or up = vp = 1 (dihedral type) for p = 2. H/Z is p-elementary abelian of order p2d H has (p − 1)pz−1 inequivalent faithful irr. representations

Dmitry Malinin On the arithmetic of integral representations

Classification of primitive representations of Galois Groups

Classification of absolutely irreducible primitive representations

• f the absolute Galois groups of local fields.
• Theorem. G – a finite group, H – its normal p-subgroup, G/H

supersolvable, ρ : G → GLn(K) – faithful primitive. Then: n = pd The center Z = Z(H) is cyclic of order pz, and for c ∈ Z of order p there are elements u1, v1, . . . , ud, vd which together with Z generate H and satisfy the generating re- lations: [ui, uj] = [vi, vj] = 1, [ui, vj] = cδi,j, (i, j = 1, . . . , d), and the generators from different pairs commute. There are 2 possibilities: 1) up = vp = 1 for p = 2 2) up = vp = c (quaternion type), or up = vp = 1 (dihedral type) for p = 2. H/Z is p-elementary abelian of order p2d H has (p − 1)pz−1 inequivalent faithful irr. representations

Dmitry Malinin On the arithmetic of integral representations

Embedding problem in Galois theory

Let 1 → B → G

ϕ

→ F → 1 be an exact sequence of p-groups, K/k be a Galois extension

• f a local field with the Galois group F, and p be the

characteristic of the residue field Qp of k. The embedding problem consists in constructing an extension L of K (or Galois algebra over K) having the Galois group G over k, such that the automorphisms g ∈ G, being restricted on K, coincide with ϕ(g).

• Example. Let G = Q8, B = Z/4Z, k = Q, K = Q(

√ −d) if and

• nly if d = a2 + b2 + c2 for some integers a, b, c. All the

solutions are fields if and only if d = a2 + b2 for some integers a, b.

Dmitry Malinin On the arithmetic of integral representations

Embedding problem in Galois theory

The associated abelian problem is a similar problem for the sequence 1 → B/B′ → G/B′ → F → 1 where B′ is the commutator subgroup of B; the solution of the abelian embedding problem is known. Let p > 2, and let F be the Demushkin group of k, that is, the Galois group of the maximal p-extension of k. The number d(F) of generators of the group F is equal to [k : Qp] + 2. Let d(F) be the number of generators of F. If d(F) ≥ d(F) + 3 then the embedding problem is equivalent to the associated abelian problem. In the proof they used the generalized Hilbert symbol and

• rthogonality of elements of k∗/k∗p for an option of a basis

k∗/k∗p and abelian radical extensions of k and for the fulfillment of the Faddeev-Hasse compatibility conditions.

Dmitry Malinin On the arithmetic of integral representations