THE DONALD–FLANIGAN PROBLEM

FOR FINITE REFLECTION GROUPS

Murray Gerstenhaber, Anthony Giaquinto, and Mary E. Schaps

July 25, 2000

Abstract. The Donald–Flanigan problem for a finite group H and coefficient ring k asks for a deformation of the group algebra kH to a separable algebra. It is solved here for dihedral groups and Weyl groups of types B_{n} and D_{n} (whose rational group algebras are computed), leaving but six finite reflection groups with solutions unknown. We determine the structure of a wreath product of a group with a sum of central separable algebras and show that if there is a solution for H over k which is a sum of central separable algebras and if S_{n} is the symmetric group then i) the problem is solvable also for the wreath product H ≀ S_{n} = H × ⋯ × H (n times) ⋊ S_{n} and ii) given a morphism from a finite abelian or dihedral group G to S_{n} it is solvable also for H ≀ G. The theorems suggested by the Donald–Flanigan conjecture and subsequently proven follow, we also show, from a geometric conjecture which although weaker for groups applies to a broader class of algebras than group algebras.

1. Introduction: The Donald–Flanigan problem. The Donald–Flanigan conjecture was one of the most intriguing of those suggested by algebraic deformation theory because it sought to relate the behavior of the group algebra of a finite group in characteristic p with that of its complex group algebra. If G is a finite group and k a commutative unital ring in which the order of G is invertible, then Maschke’s theorem asserts that the group algebra kG is separable; in particular, ℂG is a direct sum of matrix algebras. By contrast, if k is a field of characteristic p dividing #G then kG always has a non-trivial radical. Donald and Flanigan conjectured that for this ‘modular’ case the group algebra kG can always be deformed to a separable algebra, [DF]. In the examples they exhibited they observed that the separable algebra they constructed was a direct sum of copies of the coefficient field and thus resembled the complex group algebra. Schaps suggested that one should look for a deformation of kG with matrix blocks in natural bijection with those of ℂG, corresponding blocks having the same dimensions. This concept is made precise in the definition of a global solution.

The Donald–Flanigan conjecture fails for the 8-element quaternion group {±1, ±i, ±j, ±k}, [GGi], but the problem remains of determining when kG can be deformed to a separable algebra of the “right form” (one resembling the complex group algebra). The result is known for several classes of groups and algebras. (In the algebra case, the problem is again to deform it to a separable algebra.) These

Typeset by AMS-TEX

Key words and phrases. deformations, Donald–Flanigan problem, wreath products, Weyl groups, Coxeter groups, finite reflection groups, finite representation type.

include abelian groups and symmetric groups S_{n} [GSps] (where in both cases there is a global solution connecting the behavior at all primes), blocks with dihedral defect group [ES], and blocks with abelian normal defect group [MS]. Paradoxically, certain implications of the Donald–Flanigan conjecture, cf. [GG], were subsequently verified, in particular that in the modular case kG always has a non-inner derivation, [FJL]. The paradox might be resolved by the fact that this also follows (§5) from a more geometric conjecture which, although weaker for groups, is more general.

Finite Coxeter groups are identical with finite reflection groups; we may use the terms interchangeably. There are four infinite classes of irreducible ones (i.e., which are not direct products), namely the Weyl groups of type A_{n}(= S_{n+1}), B_{n}, D_{n} and the dihedral groups. (For a general reference and excellent exposition, cf. [H].) In addition, there are four exceptional Weyl groups, F_{4}, E_{6}, E_{7}, E_{8} and two noncrystallographic groups, H_{3} of order 120 (the symmetry group of the icosahedron) and H_{4} of order 14,400 (the symmetry group of a regular 120-sided solid in ℝ^{4}). One can omit G_{2} because it is identical with 12-element dihedral group D_{6}. Here we give another solution for the Donald–Flanigan problem for the dihedral groups (extending certain work of Erdmann and Erdmann–Schaps, cf. [ES]), and using our earlier solution for S_{n} we solve it for the groups B_{n} and D_{n}. Thus there are only six finite Coxeter groups for which the Donald–Flanigan problem remains open.

An essential step in the solution is the determination of the structure of a wreath product of a direct sum of central separable algebras with a a finite group. Using this we show that if H is a group and k a ring for which the Donald–Flanigan problem has a solution which is a sum of central separable algebras then i) the problem is solvable for the wreath product H ≀ S_{n} = H × ⋯ × H (n times) ⋊ S_{n} and ii) given a morphism from a finite abelian group or dihedral group G to S_{n} it is solvable also for H ≀ G. While not important for us here, the condition on the solution for H will later be relaxed somewhat, since it is enough that kH be deformed to an algebra A which becomes a direct sum of central separable algebras after suitable extension of the coefficient ring. For this it is sufficient that the center of kH be faithfully projective and of constant rank over the new coefficient ring, an extension of k resulting from the deformation process.

Determining the structure of H ≀ S_{n} (and more generally, of H ≀ G where G operates by permutation of the factors of H^{×n}) parallels the classical problem of finding the irreducible representations of a semidirect product N ⋊ G in which the normal subgroup is abelian. The basic work (fundamental to particle physics) is due to Wigner, who computed the irreducible representations of the Poincaré group (semidirect product of the Lorentz group and ℝ^{4}), cf. [St], §§3.8, 3.9. Wreath products are special cases of semidirect products but here the coefficient rings are generally not fields and H need not be commutative. Nevertheless, it may be instructive to compare our treatment with that of Sternberg, [St], particularly our example of §3, where H is commutative, with that in [St] (pp. 139–142) of the eight element dihedral group C_{4} ⋊ C_{2}, where C_{n} is the n-element cyclic group. The Weyl group D_{n} is only a semi-direct product and not a wreath product, but the Donald–Flanigan problem for it can be reduced to the wreath product case. In the course of solving the Donald–Flanigan problem for B_{n} and D_{n} we must, in effect, compute their rational group algebras, which are given here explicitly.

Many related problems remain. In particular, suppose that a finite group G acts as automorphisms (possibly trivially) of a finite group H, so that we can form the semidirect product H ⋊ G. Then G also operates as automorphisms of the group algebra kH for any coefficient ring k. An equivariant deformation of kH is one in which the operations of G continue to be automorphisms of the deformed algebra. When can an equivariant solution to the Donald–Flanigan problem for H be extended to one of the semidirect product? More generally, suppose that a finite group G acts on a separable k-algebra A and let A#kG, the smash product, be the algebra with underlying k-module A ⊗ kG and multiplication given by (a ⊗ σ)(b ⊗ τ) = aσ(b)στ; a, b ∈ A, σ, τ ∈ G. (This is a special case of the usual smash product in Hopf theory, cf. [M], Ch. 7, for note that kG is a Hopf algebra with Δh = h ⊗ h, h ∈ H.) For example, C_{2} × ⋯ × C_{2}( n times) = C_{2}^{×n} is acted upon in a natural way not only by S_{n} but by S_{n+1}; the semidirect product is the Weyl group D_{n}. At present we do not even know if under this operation 𝔽_{2}C_{2}^{×n} has an equivariant deformation to a separable algebra. It seems unlikely. On the other hand, an equivariant deformation may be more than one needs. It would be sufficient that the action of the group deform simultaneously with the structure of the algebra in such a way that the group continues to act as automorphisms of the deformed algebra. This is what happens in our solution to the Donald–Flanigan problem for D_{n} but the behavior at the prime p = 2 is different from that at other primes.

As preliminaries we reexamine what it means to deform an algebra to a separable algebra and the concept of a global deformation introduced in [GSps].

2. Tempered successive deformations. Recall that a one-parameter family of deformations of an algebra A over a ring k is a k[[t]]-algebra A_{t} whose underlying module is A[[t]] and which reduces modulo t to the original algebra A. It follows that multiplicaton in A_{t} can be written in the form a∗b = ab+tF_{1}(a, b)+t^{2}F_{2}(a, b)+…, where the F_{i} are k-bilinear maps from A to A which tacitly are extended to be k[[t]]-bilinear. Assume that A is free and of finite rank as a k-module. Then as a k[[t]]-module A_{t} is also free on the original generators. One can similarly define multiparameter deformations A_{t,u,...,v} of A in which the “∗” multiplication is given by power series in several variables. But no matter how many parameters are used, if A is not separable then no such deformation can be separable, for A remains a quotient of the deformed algebra and a quotient of a separable algebra is always again separable.

For k a field, saying that “A has been deformed to a separable algebra” tacitly means that there is some multiplicatively closed subset S of k[[t, u, …, v]] such that S^{−1}A_{t,u,...,v} is separable over the new coefficient ring R = S^{−1}k[[t, u, …, v]]. Here R is free over k but when k is only a (commutative, unital) ring this seems too strong a condition; it should be sufficient (as in all our examples) that R be projectively faithful over k although, of course, not of finite rank. This insures, in particular, that k remains a subring of R and that a prime of k is invertible in R if and only if it is already so in k. Again, if k is a field and there is only one parameter t then it is sufficient to invert t; the coefficient ring R then becomes the Laurent series field k((t)). Using the separability idempotent, one can see that when k is arbitrary and there are several parameters it is still necessary to invert only a single element of k[[t, u, …, v]]. For any single element of S^{−1}A_{t,u,...,v} is already contained in the algebra obtained by a single such inversion and the separability idempotent only involves a finite number of elements. But even in the case of a single parameter, it may not be simply t that one wants to invert but some polynomial in t (whose constant term is not a unit), cf. [GSps].

Suppose now that we try to deform an algebra A to a separable algebra in two stages: first reducing the inseparability by deforming to some A_{t} and forming B = S^{−1}A_{t} where S is generated by some single element of k[[t]], and then by deforming B to some B_{u} and inverting an element of the new cofficient ring. Can the same be done by first performing some two-parameter deformation A_{t,u} of A and then inverting an element of k[[t, u]]? We need this in order to justify deforming an algebra to a separable one in successive steps. To understand the reason for caution, suppose that A is a finite-dimensional algebra over a field k, that we deform it, and then specialize the deformation parameter. If dim_{k} A = n and we have chosen a basis a_{1}, …, a_{n} then the algebra can be described by its multiplication constants relative to this basis: a_{i}a_{j} = ∑ c_{ijl}a_{l}. The constants c_{ijl} may be viewed as determining a k-point in the variety alg_{n} of structure constants of n-dimensional associative algebras. The deformed algebra has structure constants c_{ijl}(t) lying in k[[t]] which reduce at t = 0 to the original constants. These in general will be transcendental over k and define a subvariety in the same component of alg_{n} as the original. The result of specializing and then performing another deformation can, however yield a point in a different component, since the specialization may lie on the intersection of two components. But intuitively, if one deforms and then, starting from a generic point of the original deformation deforms again, one must remain on the same component.

So suppose that A is free as a k-module with basis a_{1}, …, a_{n}, that we have deformed it to A_{t} and have formed B = S^{−1}A_{t} where S is generated by a single element f(t) ∈ k[[t]]. Let B now be deformed to B_{u}. Set S^{−1}k[[t]] = R. We may suppose that we have a basis b_{1}, …, b_{n} of B with structure constants given by b_{λ}b_{μ} = ∑ γ_{λμν}b_{ν} where the γ_{λμν} initially lie in R but after deformation are elements of R[[u]]. The problem is that the coefficients of these power series in u may contain negative powers of f(t) and the negative powers may be unbounded. The hypothesis we make about the deformation B_{u} is that there is some fixed N such that all f(t)^{N}γ_{λμν}(u) lie in R[[t, u]]. Now we can write each of the b_{λ} as a linear combination of the a_{i} with coefficients in R. Again, these coeficients will generally involve negative powers of f(t). As there are only finitely many coefficients, the hypothesis insures that we can write out the multiplication in B_{u} in the form a_{i} ∗ a_{j} = ∑ c_{ijl}(t, u)a_{l} where there is some fixed N′ such that all f(t)^{N′}c(t, u) ∈ k[[t, u]]. But all of the c(t, 0) lie in k[[t]]. Therefore, replacing u by f(t)^{N′}v we have an ordinary two-parameter deformation A_{t,u} of A over k[[t, u]]. A second deformation with the foregoing boundedness properties will be called tempered (relative to the first). If, finally, inverting an element of R[[u]] with bounded powers of f(t) in the denominators of the coefficients makes B_{u} separable, then there is an element F (t, u) ∈ k[[t, u]] inverting which makes A_{t,u} separable. To return to a one-parameter family of deformations we can now replace u by some element of k[[t]] such that F (t, u) ≠ 0.

3. An example. As a simple illustration, let C_{2} = {1, a, a^{2} = 1} be the two element group and G = C_{2} ≀ C_{2} = (C_{2} × C_{2}) ⋊ C_{2}. The right C_{2} operates by interchange of the two left factors. Denoting its elements by {1, σ}, we will write the 8 elements of the group as {(1, 1), (a, 1), (1, a), (a, a), (1, 1)σ, (a, 1)σ, (1, a)σ, (a, a)σ} where, for example, (a, 1)σ · (1, a)σ = (a, 1)(a, 1)σ^{2} = (1, 1), the unit element of G. Now letting k = 𝔽_{2}, we wish to deform the group algebra 𝔽_{2}G to a separable algebra. We can deform 𝔽_{2}C_{2} to a separable algebra by setting a^{2} = ta+1+t; then (1+a)^{2} = t(1+a) so after inverting t we have orthogonal idempotents e = (1+a)/t and f = 1 + e. Doing the same to both factors in C_{2} × C_{2} (and noting that if G_{1}, G_{2} are finite groups then k(G_{1} × G_{2}) = kG_{1} ⊗ kG_{2}) we can perform a first deformation of 𝔽_{2}G to an algebra generated by the four orthogonal idempotents e⊗e, f ⊗e, e⊗f, f ⊗f and a “switch” element σ with σ^{2} = 1 such that σ(x⊗y)σ = (y ⊗ x) for all x ⊗ y ∈ 𝔽_{2}C_{2} ⊗ 𝔽_{2}C_{2}. The resulting algebra is still in a natural sense a wreath product. Let A be the algebra to which 𝔽_{2}C_{2} has been deformed. Its coefficients are now in 𝔽_{2}(t). We still have C_{2} operating on A ⊗ A by interchange of the tensor factors. Denote the algebra resulting from this first deformation by A ≀ C_{2}. Further, A is separable over 𝔽_{2}(t) (part of the original inseparability has been removed) and is a direct sum of two subalgebras, A = Ae ⊕ Af, each of which is trivially central separable. While A ⊗ A has four central primitive idempotents, A ≀ C_{2} is a direct sum of 3 subalgebras corresonding to the orbits of these idempotents under C_{2}, namely {e ⊗ e}, {f ⊗ f}, and {e ⊗ f, f ⊗ e}. The isotropy group of the last orbit is reduced to the identity element of C_{2} and the orbit gives rise to a four dimensional summand of A ≀ C_{2} spanned over 𝔽_{2}(t) by {e⊗f, f ⊗e, (e⊗f)σ, (f ⊗e)σ}. It is easy to check that this is isomorphic to the 2×2 matrix algebra M_{2}(𝔽_{2}(t)) and hence is central separable over 𝔽_{2}(t). The isotropy groups of the other orbits are non-trivial, being in fact all of C_{2}. The subalgebra corresponding to {e ⊗ e} is spanned by {e ⊗ e, (e ⊗ e)σ} and should be viewed as isomorphic to M_{1}(𝔽_{2}(t))⊗𝔽_{2}(t)C_{2}, where C_{2} here is the isotropy group of the orbit, and similarly for the orbit of {f ⊗ f}. This will be generalized in §§8, 9. We can now perform a second deformation, treating the direct summands individually, so that the whole algebra now becomes separable over the new coefficient ring. The non-trivial matrix summand is already central separable so it will be left unchanged except for the necessary extension of the coefficient ring. In this second deformation we can not simply set σ^{2} = uσ + 1 + u (since there is to be no change in the matrix part); in the first summand we set [(e ⊗ e)σ]^{2} = u(e ⊗ e)σ + (1 + u)(e ⊗ e) and similarly with f in the second.

It is easy to see that this second deformation is tempered relative to the first. In fact, 𝔽_{2}(t)C_{2} is actually defined over 𝔽_{2}; it is obtained from 𝔽_{2}C_{2} just by extension of coefficients. So the only way that t enters into the second deformation is in the choice of the basis for the algebra obtained by the first deformation. Since there are only a finite number of basis elements, the condition of being tempered will automatically be satisfied whenever they all have coefficients (relative to the original basis) in some S^{−1}k[t] where S is generated by a single f(t) ∈ k[t]. Nevertheless, it may be useful in this example to write explicitly the final result as a two parameter deformation over 𝔽_{2}[[t, u]] (in fact, over k[t, u]) followed by inversion of tu. Since now σ ∗ σ is no longer the unit element of the algebra, we compute it explicitly. We can write σ = [(e + f) ⊗ (e + f)]σ = [e ⊗ f + f ⊗ e]σ ⊕ (e ⊗ e)σ ⊕ (f ⊗ f)σ relative to the decomposition after the first deformation. The second deformation has respected this decomposition, so we can square each direct summand separately. The square of the first is e⊗f +f ⊗e, that of the second is (1+u)(e⊗e)+u(e⊗e)σ, and similarly for the third with f replacing e. Thus σ ∗ σ = (e + f) ⊗ (e + f) + u(e ⊗ e + f ⊗ f)(1 + σ). Writing out σ ∗ σ in terms of the original basis elements the first summand remains unchanged (being just the unit element of both the original and the deformed algebra) but the second summand must be written as t^{−1}u(a ⊗ 1 + 1 ⊗ a + t · 1 ⊗ 1)(1 + σ). So we need here to replace u by tv, which in fact will work for all the other products. With this we have a true two-parameter deformation of the original algebra (the parameters now being t and v) which becomes separable after inversion of t and of u = tv, or simply after inversion of tv. To get a one-parameter deformation to a separable algebra we could now set u = t^{2}. (Later we will actually use the deformation given by a^{2} = (q − q^{−1})a + 1 with q = 1 + t; the results are the same.)

While we must repeatedly use successive deformations, in the cases we consider it will be evident, as it is here, that the second deformation is tempered, so we may simply omit the discussion of that fact.

4. Global solutions and split global solutions. In the modular case we would like to deform kG to a separable algebra which is a direct sum of matrix algebras whose summands are in one-one correspondence with the matrix blocks of the complex group algebra ℂG, with corresponding blocks having the same dimension. More precisely, let k now be a field. Write kG = A, denote the deformed algebra by A_{t}, and supppose that it becomes separable when coefficients are extended to k((t)). It need not be a sum of matrix algebras but will become one when coefficients are extended to the algebraic closure of k((t)), i.e., when one forms A_{t} ⊗_{k[[t]]} \overline{k((t))}. But in general there need be no relation between the matrix blocks of this algebra and those of ℂG. For example, G may be abelian but kG may have non-commutative separable deformations. Suppose, however, that we have a global solution to the Donald–Flanigan problem for G in the sense of [GSps]: a deformation A_{t} of the integral group ring ℤG, together with a multiplicatively closed subset S of ℤ[[t]] which does not meet the ideal generated by any rational prime dividing #G, such that S^{−1}A_{t} is separable over S^{−1}ℤ[[t]]. If such a global solution exists then one may assume that S consists of the powers of a single element s. More important, one can reduce S^{−1}A_{t} modulo any rational prime p not dividing #G. The image s̄ of s must be of the form t^{m}ϵ, where ϵ is a unit of 𝔽_{p}[[t]]. Since a quotient of a separable algebra is separable, the result will be a deformation of 𝔽_{p}G which becomes separable when coefficients are extended to 𝔽_{p}((t)). Denote the resulting separable algebra by A_{t}(p). Now if S^{−1}A_{t} is already split, i.e., a direct sum of matrix algebras, then not only must the same be true of each A_{t}(p) but the correspondence between the matrix blocks of S^{−1}A_{t} and those of A_{t}(p) is simply that those of the latter are the quotients of those of the former. Corresponding blocks then certainly also have the same dimensions. Since one can embed ℤ[[t]] in ℂ it also becomes clear that the blocks are the same as those of ℂG. If S does contain some primes p_{1}, p_{2}, … dividing #G then we say that we have a global solution away from p_{1}, p_{2}, . . .. An arbitrary deformation of kG to a separable algebra will be called a weak solution to the Donald–Flanigan problem. Different weak solutions are sometimes possible. For example 𝔽_{2}(C_{2} × C_{2}) is deformable both to a direct sum of four copies of the new coefficient ring and to a 2 × 2 matrix ring over the coefficient ring. A principal result of [GSps] is that there does exist a split global solution to the Donald–Flanigan problem for the symmetric groups S_{n}. In general, however, a global solution if it exists will not be split. For letting t → 0, the existence of a split global solution would imply that all irreducible representations of G are rational, something true of S_{n} but not in general. This problem would disappear if we had a positive answer to the following question. If A is a separable algebra over a domain R, is there always a finite integral extension R̂ of R over which A is split? For if ℤG is deformed to a separable algebra over some R = S^{−1}Z[[t, u, …, v]], where S is a multiplicatively closed set not intersecting the ideal generated by any rational prime dividing #G, then the same will still be true if the coefficients are extended to R̂; if this splits the separable deformation then we can proceed as before. Known results do not get us quite this far; see §10. Note that for our purposes one can assume that R has characteristic zero and even that all rational primes not dividing #G are already invertible in R, but it does not seem that these assumptions should be necessary. The original definition of a global solution is also too restrictive in that it requires that we start with the integral group ring ℤG of the group G. It is useful to broaden the definition by allowing replacement of ℤ by some finite integral extension O. This need not be the full ring of integers of some number field but we may generally assume that it is. When G and a single p|#G are given then a global solution away from all other primes dividing #G will be called a local solution at p. If either it is already split or the conjecture above holds then this gives a canonical correspondence between the matrix blocks of ℂG and those of the global solution. Suppose now that we have local solutions at several primes dividing #G. Then their blocks must be in natural correspondence with each other, corresponding blocks having the same dimensions, since for each prime they are in correspondence with those of ℂG. The requirement that the local or global solution be split can be eased. All one needs is that it be a direct sum of Azumaya, i.e., central separable algebras. Such an algebra, if it has constant rank at each prime ideal of its coefficient ring is a just a twisted form of a matrix algebra, cf, e.g., [KO]. We return to this in §10.

The global solution (with O just ℤ itself) for the Donald–Flanigan problem for S_{n} given in [GSps] is essentially its Hecke algebra, H_{n}(q). Setting ℤ_{q} = ℤ[q, q^{−1}], this is the free module over ℤ_{q} with basis elements T_{w} indexed by the elements w ∈ S_{n} and multiplication given as follows: The length ℓ(w) is the number of factors in a shortest expression of w as a product of generators s_{i} := (i, i+1), i = 1, …, n−1 of S_{n}. Now set (i) T_{s}T_{w} = T_{sw} if s = (i, i + 1) for some i and w ∈ S_{n} is an element with ℓ(sw) > ℓ(w), and (ii) T_{s}^{2} = (q − q^{−1})T_{s} + 1. This implies that T_{s}T_{w} = (q −q^{−1})T_{w} +T_{sw} when ℓ(sw) < ℓ(w). (The same definition extends to any Coxeter group, and in particular to any finite reflection group where instead of the transpositions s_{i} = (i, i+1) one takes the basic generators s.) Writing 1+t for q one sees that H_{n}(1 + t) is a deformation of ℤS_{n}. Set i_{q} := (1 − q^{i})/(1 − q) and similarly i_{q^{2}} := (1−q^{2i})/(1−q^{2}); set n_{q^{2}}! := n_{q^{2}}(n−1)_{q^{2}} … 2_{q^{2}} and ℤ_{q,n} := ℤ[q, q^{−1}, 1/n_{q^{2}}!]. A main result of [GSps] was that over ℤ_{q,n} the Hecke algebra H_{n}(q) becomes a direct sum of matrix algebras. It follows that H_{n}(1 + t) together with the multiplicatively closed subset of ℤ[[t]] generated by n_{q^{2}}! = n_{(1+t)^{2}}! is a global solution to the Donald– Flanigan problem for the symmetric group. Setting q = 1 or equivalently t = 0 yields the special case that all the irreducible complex representations of the symmetric group are actually defined over ℚ. This is the reason that no extension of ℤ was needed for G = S_{n}. In general i_{q^{2}}j_{q^{2}} ≠ (ij)_{q^{2}}, so adjoining the inverse of m_{q^{2}} generally does not bring with it the adjunction of i_{q^{2}} for any factor i of m. In particular, the multiplicatively closed set S above is not generated by n!_{q^{2}} (≠ n_{q^{2}}!). However, if m is even, say m = 2r, then (2r)_{q^{2}} = (1 + q^{r})r_{q^{2}}, so inverting (2r)_{q^{2}} also inverts r_{q^{2}}. To make the Hecke algebra of S_{n} separable we therefore essentially adjoined to its original ring of definition ℤ_{q} the inverses of all i_{q^{2}} for i = 2, …, n. There is no proof that all of these adjunctions are necessary, but we suspect that they are. The numbers 2, …, n are are also the “degrees” of S_{n}, i.e., degrees of its basic invariant polynomials. This suggests the following refinement of a conjecture in [GSps]. Let W be a finite Coxeter group and H_{W} (q) be its Hecke algebra with coefficient ring ℤ_{q} = ℤ[q, q^{−1}]. If S is the multiplicatively closed subset generated by all i_{q^{2}} where i runs through the degrees of W then S^{−1}H_{W} (q) should be separable. We do not know if the deformations of ℤB_{n} and ℤD_{n} constructed here are in fact their Hecke algebras, but they do have the following property. Setting n_{q^{2}}! = n_{q^{2}}(n − 1)_{q^{2}} ⋯ 3_{q^{2}}2_{q^{2}} and ℤ_{q,n} = ℤ_{q}[1/n_{q^{2}}!] the deformed algebras become direct sums of matrix algebras over ℤ_{q,n}. Setting q = 1 recaptures the result that all the complex irreducible representations of B_{n} and D_{n} are rational. Since this is not the case for the dihedral groups (whose degrees are 2, m), even if the conjecture is true then the resulting solution is not split. On the other hand, since the product of the degrees is equal to the order of W, it is consistent with the fact that the group algebra kW is separable over any ring k in which #W is invertible. If this conjecture is true then in principle one could prove it by exhibiting the separability idempotent, but as noted in [GSps] there is so far no good formula for that even in the known case where W = S_{n}. It may be the case more generally that rings of invariant polynomials associated with a finite group have some connection with the solvability of the Donald–Flanigan problem for that group. The conjecture does not assert that the Hecke algebra is the only global solution to the Donald–Flanigan problem for a finite reflection group (although it may in some sense be the best). For the Weyl groups of type B_{n} the degrees are 2, 4, …, 2n, but we shall see that one can obtain a global solution as soon as one has one for S_{n}, and that requires inverting only all i_{q^{2}} for i = 1, …, n. However, as remarked above, if a ring contains the inverses of all (2i)_{q^{2}} for i = 1, …, n then it already contains the inverses of all i_{q^{2}}.

Infinite Coxeter groups also have Hecke algebras which can be viewed as deformations of their integral group rings. If coefficients are extended to the direct limit ℤ_{q,∞} of the ℤ_{q,n} then these algebras should possess some properties similar to separability. While ℤ_{q,∞} is not a field, all the “quantum integers” n_{q^{2}} have become invertible. While almost a quantum version of the rationals, it is still possible to reduce ℤ_{q,∞} modulo any rational prime. Although reduction modulo t does give the rationals, it is certainly not a deformation of the rationals

5. Geometric rigidity and a generalized global problem. For this and the next section the coefficient ring k will be a field. We review some basic facts about jump deformations and approximate automorphisms.

Recall that a jump deformation A_{t} of a k-algebra A is one which is non-trivial and remains constant for generic t ≠ 0. More precisely, if u is a second variable and coefficients are extended to k((t))[[u]] then there is an isomorphism A_{t} ≅ A_{(1+u)t} which reduces to the identity when u → 0. That is, the trivial deformation of the k((t))-algebra A_{t} is equivalent to the deformation A_{(1+u)t}. (For the essential properties of jump deformations, cf. [G2,3; GSck, §7].) A jump deformation simultaneously “breaks some symmetry” of the algebra and “destroys its own infinitesimal”. The meaning of the latter statement is this. Suppose that we have a deformation A_{t} of A. An n-cocycle ζ ∈ Z^{n}(A, A) can be lifted to an n-cocycle of A_{t} if there are cochains ζ_{i} ∈ C^{n}(A, A) such that ζ_{t} = ζ + tζ_{1} + t^{2}ζ_{2} + ⋯ ∈ Z^{n}(A_{t}, A_{t}). If ζ + tζ_{2} + ⋯ + ζ_{m} is a cocycle modulo t^{m+1} then there is an obstruction cocycle η ∈ Z^{n+1}(A, A) and we must have η = δζ_{m+1} for some ζ_{m+1} ∈ C^{n}(A, A) in order for the construction to continue. Any cocycle which is liftable to a coboundary is called a jump cocycle. This includes, in particular all coboundaries in A but there may be non-trivial ones. One has H^{n}(A_{t}, A_{t}) = (liftable n-cocycles)/(jump n-cocycles). The infinitesimal of a jump deformation is always a jump cocycle, so dim H^{2}(A_{t}, A_{t}) < dim H^{2}(A, A). It follows that A can only undergo finitely many jump deformations.

For n ≥ 3 every jump cocycle in Z^{n}(A, A) is the obstruction at some stage to lifting some ζ ∈ Z^{n−1}(A, A). In characteristic zero this is true also for n = 2. In this case, therefore, if F is the infinitesimal of a jump deformation then F is itself the obstruction to lifting some derivation ϕ ∈ Z^{1}(A, A) to a derivation of A_{t}. Therefore e^{tϕ} is a formal one-parameter family of automorphisms of A which can not be lifted to A_{t}, so this symmetry is broken. In characteristic p > 0 it is possible that after a jump deformation every ϕ ∈ Z^{1}(A, A) remains liftable, but there is still a symmetry that is broken. For example, in A = 𝔽_{2}[x]/x^{4} a derivation is uniquely determined by its value on x, which can be arbitrary. This is still the case if we deform A to A_{t} = 𝔽_{2}[x]/(x^{4} + tx^{2}) (a jump deformation), so the dimension of the space of derivations has not changed. We have, however, lost some “approximate automorphisms” in the following sense. When we have a derivation ϕ of an algebra A of characteristic p > 0 it may not be always be possible (as it was in characteristic zero) to construct a full one-parameter family of automorphisms of the form Φ_{t} = ϕ+tϕ_{1}+t^{2}ϕ_{2}+. . .. The largest m for which we can construct an automorphism of A[t]/t^{m} of the form Φ_{t} = ϕ + tϕ_{1} + ⋯ + t^{m−1}ϕ_{m−1} with a fixed ϕ always has m = p^{r} for some s. We call this m the order of the approximate automorphism Φ_{t} and also the order of ϕ. A jump deformation of A reduces the order of some derivation. In the example above, the order of the derivation ϕ of 𝔽_{2}[x]/x^{4} sending x to 1 was initially 4 but was reduced to 2 after the jump deformation. With a jump deformation the order of some derivation is always reduced; it can not be lifted to an approximate autmorphism of the same order, and the infinitesimal of the deformation is the obstruction to continuing the now truncated approximate automorphism to higher order. In this sense, a jump deformation always breaks some symmetry. If H^{1}(A, A) = 0 there can be no jump deformations.

Remark. Following [G2], denote by Aut_{m−1} A the group of automorphisms of A[t]/t^{m} of the form id_{A} +tϕ_{1} + … t^{m−1}ϕ_{m−1} where the ϕ_{i} are 1-cochains of A, i.e., linear maps A → A (which are tacitly extended to be k[t]-linear). There is a canonical monomorphism Aut_{m−1} A → Aut_{mn−1} defined by replacing t by t^{n} and considering the resulting polynomial as one of degree mn − 1. The direct limit \underrightarrow{lim} Aut_{m} A contains a subgroup P consisting of the images of all elements in all Aut_{m} A which have a prolongation to an element of Aut_{m+1} A. This subgroup is normal by the basic theorem on the additivity of obstructions [G2, Theorem 1], and the quotient Aut_{∗} A is abelian. It consists of classes of obstructed approximate automorphisms, since all full one-parameter families of automorphisms beginning with the identity lie in P. This group has a natural filtration, as does the space RH^{2}(A, A) of “restricted” elements of H^{2}(A, A) (=those which are the obstructions to approximate automorphisms) and the associated graded groups are isomorphic. The group of obstructed approximate automorphisms is thus in a natural way finite dimensional. In characteristic zero the Euler-Poincaré characteristic of an algebra A, if it has one, is just ∑(−1)^{n} dim H^{n}(A, A). If it does, then so does any deformation of A and it is invariant under deformation. In characteristic p > 0 for this to be true we shall probably have to replace dim H^{1}(A, A) with the dimension of the space of approximate automorphisms.

The original weak Donald–Flanigan conjecture implied, by way of the following theorem ([GGr]) that if p|#G then there exists an element g ∈ G whose centralizer C_{G}(g) contains a normal subgroup of index p.

Theorem 1. Let A be a k-algebra which is not itself rigid but which can be deformed to a rigid algebra. Then H^{1}(A, A) ≠ 0, i.e., A has a non-inner derivation.

Proof. The infinitesimal of any deformation of A to a rigid algebra must be a jump 2-cocycle, since by definition no further deformation of a rigid algebra is possible. Since a jump 2-cocycle exists, H^{1}(A, A) ≠ 0. □

Any deformation of a non-separable algebra A to a separable one is necessarily a jump deformation since separable algebras have trivial cohomology and therefore are rigid. It follows that H^{1}(A, A) ≠ 0. However, for A = 𝔽_{p}G we have (cf. [B, Theorem 2.11.12]) an isomorphism of additive groups

H^{n}(𝔽_{p}G, 𝔽_{p}G) ≅ ⨁ H^{n}(C_{G}(g), 𝔽_{p})

where the operation on 𝔽_{p} is trivial and the sum is over a set of representatives g of conjugacy classes in G. So the Donald–Flanigan conjecture implied that for some g ∈ G one has H^{1}(C_{G}(g), 𝔽_{p}) ≠ 0. But a derivation into a trivial module is just a morphism, giving what was asserted. Guided by this Fleischmann, Janiszczak and Lempken proved a stronger result [FJL]: If p|#G then there exists a g ∈ G whose order is divisible by p and whose centralizer C = C_{G}(g) has the property that its commutator subgroup C′ does not contain the p-part of g. Their proof reduces to the case where G is simple and uses the classification theorem for finite simple groups. (Publication of [FJL] preceded that of [GGr] because of the greater lead time for the latter.)

To resolve the paradox that the Donald–Flanigan conjecture fails while its corollary in [GGr] holds we propose a conjecture which for finite groups is weaker than the Donald–Flanagan conjecture but applies more generally and still implies the statement in [GGr]. Recall first the various concepts of rigidity for an algebra A, cf. [GSck]. The first, usually called simply rigidity but more precisely analytic rigidity says that every formal deformation of A is equivalent to the trivial deformation. This will certainly hold if H^{2}(A, A) = 0, often called absolute rigidity. For the second, suppose for the moment that A has dimension n < ∞ over some algebraically closed field k and let V = alg_{n}(k) ⊂ k^{n^{3}} be the variety of structure constants of n-dimensional k-algebras. Now GL(n, k) operates on V and hence on each of its components, with orbits corresponding to isomorphism classes of ndimensional k-algebras. One calls A geometrically rigid if the corresponding orbit is a Zariski open set in the component V_{A} of V containing it. In the terminology of Wigner, every algebra in the (necessarily unique) component of A is a contraction of A. Equivalently, every algebra which can be deformed to A is a contraction of A; this formulation no longer requires finite dimensionality. Analytic rigidity implies geometric rigidity and in characteristic zero they are equivalent [G3, GSck §9], but not in characteristic p > 0. In that case there can exist non-trivial formal deformations of A with multiplication a ∗ b = ab + tF_{1}(a, b) + t^{2}F_{2}(a, ab) + … in which F_{1} is not a coboundary but where the deformation becomes trivial when t is replaced by t^{m} for some m; this m is then necessarily a power of p, cf. [G3, GSck]. Such restricted deformations can occur only when A has some non-inner derivations ϕ, for the cocycle F must be the obstruction at some stage to constructing a formal automorphism of the form Φ = id_{A} +tϕ + t^{2}ϕ_{2} + … whose infinitesimal is ϕ. The obstruction to a derivation is always the infinitesimal of a restricted deformation. (See also Skryabin, [Sk]).

Suppose that an algebra A is defined and is a free module of finite rank over an integral extension O of ℤ (so the structure constants with respect to a suitable choice of basis lie in O), and suppose that A becomes rigid over ℂ. In place of Donald– Flanigan we conjecture now that for any prime p, A ⊗_{O} 𝔽_{p} can be deformed to a geometrically rigid algebra. The conjecture relates the structure of the component V_{A} of A over ℂ to the structure of the variety of structure constants in characteristic p at any point representing the reduction of A mod p. If one has a finite group G then the hypothesis of having a free module is certainly satisfied (with O just ℤ itself) for the integral group algebra ℤG. The conclusion is weaker than that of the Donald–Flanigan conjecture, for if k is a field of characteristic p dividing #G then it says only that kG can be deformed to a geometrically rigid algebra, not a separable one (all of whose cohomology in positive dimensions vanishes). This, however, is enough to imply that 𝔽_{p}G has a non-inner derivation. For if the geometrically rigid algebra has a restricted deformation then we are done; otherwise it is analytically rigid, the deformation is just as before a jump deformation whose infinitesimal is the obstruction to some derivation, so in any case the derivation exists. One is then led again to the group-theoretic statement in [GGr].

For group algebras over a field, the conjecture of the next section would imply that they can actually be deformed to geometrically rigid algebras, but to state it we must introduce a broader concept of equivalence of deformations ∗ and ∗′ of a k-algebra A. We will call them effecively equivalent if there are integers m and n such that replacing t by t^{m} in the first and by t^{n} in the second, the resulting deformations become equivalent in the original sense of [G1]. For finite dimensional algebras A over an algebraicially closed field k this can happen in a non-trivial way only in characteristic p > 0, where its significance is the following. Choosing a basis for A, one has new multiplication constants given by ∗ and by ∗′ which in both cases are now power series in t reducing to the original multiplicaton constants when t = 0. Each set of constants may now be viewed as a generic point of some subvariety of k^{n^{3}}; the deformations are effectively equivalent precisely when these subvarieties coincide. A deformation which is effectively equivalent to a trivial deformation is one previously called restricted but a better name may be effectively trivial. An effective jump deformation will be one which is effectively equivalent to a (non-trivial) jump deformation.

6. Jump algebras. There is an important class of algebras A where (using known deep results) it is easy to show that H^{1}(A, A) ≠ 0. These are those algebras of finite representation type or representation finite algebras which are not already rigid. (Clearly one must exclude, e.g., matrix algebras.) The reason is that the only deformations which these admit are effectively equivalent to jump deformations (and hence, in characteristic zero, they admit only true jump deformations). In this section the coefficient ring k is still a field (although some statements will obviously hold more generally), algebras will be finite dimensional, and a deformation of a k-algebra A will be viewed as an algebra over the Laurent series field k((t)). A class of algebras is called open if every deformation of an algebra in the class is again in the class. For algebras of a fixed dimension n, an open class may be represented as a Zariski open subset of the variety of structure constants of n-dimensional algebras. A theorem of Gabriel asserts that the class of representation finite algebras is open. (For an overview of the theory, cf [GR].)

From the preceding section, in any sequence of deformations of a finite dimensional algebra A over a field, only a finite number can be jump deformations or effective jump deformations. Because a representation finite algebra always has a multiplicative basis (i.e., one where the product of two basis elements is a third or zero) there can only be finitely many in any dimension. It follows that any deformation of an algebra of finite representation type which effectively changes its structure, i.e., which is not effectively trivial, must be an effective jump deformation. Therefore, if a representation finite algebra A is not rigid then H^{1}(A, A) ≠ 0, else its group of obstructed approximate automorphism would already be reduced to the identity.

In view of this, call an algebra an effective jump algebra if it admits only deformations effectively equivalent to jump deformations. Any deformation of an effective jump algebra is again an effective jump algebra since any finite sequence of deformations of a finite dimensional algebra can be gathered into one multiparameter family. It follows that the class of effective jump algebras is open and contains the class of representation finite algebras. Clearly an effective jump algebra can be deformed by a finite number of effectively jump deformations to an effectively rigid algebra. If it is not already rigid then the preceding argument shows that it must have a non-inner derivation. We now ask, Is every group algebra an effective jump algebra? By Higman’s theorem a group algebra is representation finite if and only if its p-Sylow subgroups are cyclic. (For a proof, cf. [P, p.194]) Most group algebras therefore are not representation finite but it is conceivable that all are effective jump algebras. If so, while they can not all be deformed to separable algebras, they all could at least be deformed to effectively rigid ones. Note that while finite representation type is open, it does not imply rigidity. For example, 𝔽_{2}C_{2} has finite representation type but is not rigid. It would be useful to know in particular if the group algebra over 𝔽_{2} of the quaternion group, for which the Donald–Flanigan conjecture fails, is an effective jump algebra.

7. Finite abelian groups and dihedral groups. In this section we show that there is a global solution to the Donald–Flanigan problem for any finite abelian group Γ, provided that one is allowed to begin with a suitable integral extension of ℤ. (Donald and Flanigan exhibited only a weak solution.) Since Γ is a product of groups of prime power order, it is sufficient to consider the case of C_{r} where r = p^{m}, a prime power. Its integral group algebra is A := ℤ[x]/(x^{r} − 1), which can be deformed (over ℤ[t]) to A_{t} := ℤ[x, t]/(x^{r} − tx − 1). This is not yet separable since we can still reduce modulo the ideal generated by p and t to get an algebra with a non-trivial radical. Now a finitely generated algebra A over a commutative ring R is separable if and only if A/𝔪A is separable over R/𝔪 for every maximal ideal 𝔪 of R. Let d be the discriminant of x^{r} − tx − 1 and set R = ℤ[x, t, 1/d]. For A now take A_{t} with coefficients extended to R, i.e., A_{t} ⊗_{ℤ[x,t]} R. Equivalently, letting S be the multiplicatively closed subset consisting of the powers of d, we have A = S^{−1}A_{t}. Reduction of this algebra modulo a maximal ideal of R produces an algebra over some finite field F of the form F [x]/(x^{r} −τx−1), where τ is a non-zero element of F and where x^{r} − τx − 1 has distinct roots because its discriminant is invertible. It is therefore separable. Hence so is A = S^{−1}A_{t}, which is therefore a global solution to the Donald–Flanigan problem for C_{r}. A problem with this solution is that the algebra is generally not split (which here means not isomorphic to a direct sum of copies of R), so there is no natural correspondence between its blocks and those of the complex group algebra. We can modify the foregoing procedure to obtain a split solution. The results may look similar but the solutions will be inequivalent. As an example, let Γ be the cyclic group C_{3} of order 3, so defining A as above, the (non-split) global deformation of its integral group algebra is given by by setting x^{3}−tx−1 = 0. The discriminant of x^{3}−tx−1 is d = −27+4t^{3} and the coefficient ring R will be ℤ[t, d^{−1}]. Recall that separability of an algebra A over a ring R is equivalent to the existence of a separability idempotent, i.e., an element e = ∑ x_{i} ⊗ y_{i} ∈ A ⊗_{R} A such that for all a ∈ A one has ae = ∑ ax_{i} ⊗ y_{i} = ∑ x_{i} ⊗ y_{i}a = ea and such that ∑ x_{i}y_{i} = 1. (There are many equivalent definitions of the separability of an algebra A over a ring R. The existence of a separability idempotent is often the most convenient but we shall need the homological one later.) When, as here, A is a free module over its coefficient ring R, calculating the separability idempotent (which in general need not be unique) is equivalent to solving a system of simultaneous linear equations. If R is a domain and the system is consistent then there is a solution in the quotient field of R, but it is clear from the process that one need invert only a single element of R. In the present case, A ⊗_{R} A is a free module over R spanned by all x^{i} ⊗ x^{j}, i, j = 0, 1, 2 and we seek a linear combination e = ∑ c_{ij}x^{i} ⊗ x^{j} with ∑ c_{ij}x^{i+j} = 1 and with xe = ∑ c_{ij}x^{i+1} ⊗ x^{j} = ∑ c_{ij}x^{i} ⊗ x^{j+1} = ex, the latter condition being sufficient, since x generates A, to insure that ae = ea for all a ∈ A. Replacing x^{3} by tx + 1 and x^{4} by tx^{2} + x gives the equations for the c_{ij}. Direct calculation then yields

e = d^{−1}[(−9 + 4t^{3})1 ⊗ 1 + 2t^{2}x ⊗ x + 6t(x^{2} ⊗ x^{2} + 1 ⊗ x + x ⊗ 1)
− 4t^{2}(1 ⊗ x^{2} + x^{2} ⊗ 1) − 9(x ⊗ x^{2} + x^{2} ⊗ x)].

Note that e is well-defined modulo any prime, and in particular, modulo 3. If the coefficient ring had been one in which one could divide by 3, then at t = 0 the separability idempotent e would reduce to \frac{1}{3}(1 ⊗ 1 + x ⊗ x^{2} + x^{2} ⊗ x), i.e., to the classical separability idempotent \frac{1}{#G} ∑ g ⊗ g^{−1} where here #G = 3 and g runs through the elements of the cyclic group C_{3} = {1, x, x^{2}}. Since the rational group algebra ℚC_{3} is not split the present algebra also could not be split, else extending coefficients to include ℚ we could let t → 0, giving a contradiction.

To remedy this, return to the group C_{r} with r a power of a prime p, let Φ be the cyclotomic polynomial for the primitive rth roots of unity and take O = ℤ[y]/Φ(y). Denoting the image of y by η, instead of x^{r} − tx − 1, take now the polynomial f(x, t) = Π_{i=0}^{r−1}(x−η^{i}(1+t)). Since f(x, 0) = x^{r} −1, O[x, t]/f(x, t) is a deformation of the group algebra OC_{r}. We may now proceed exactly as before, localizing at the multiplicatively closed set S generated by the discriminant. The result is a global solution which is split and whose blocks, therefore, are now in natural one-to-one correspondence with those of the complex group algebra once one fixes a morphism of O into ℂ. This will always be tacitly understood.

To handle the dihedral groups we need one additional refinement: a split global solution which moreover is equivariant with respect to the automorphism sending every element of the cyclic group to its inverse. For this we need that f(x, t) be symmetric in the sense that f(x, t) = (−x)^{r}f(x^{−1}, t), where r = p^{m}, as before. Introduce a parameter q which will be set equal to 1 + t and extend O to O[q^{−1}]. Set

f(x, t) = \{ \begin{matrix}Π_{i=−(r−1)/2}^{(r−1)/2}(x − η^{i}q^{i}) p odd, \\ (x − q)(x − q^{−1}) & Π_{i=1}^{(r/2)−1}(x − q^{2i}η^{i})(x − q^{−2i}η^{−i}) p = 2. \end{matrix}

Together we have the following.

Theorem 2. For every finite abelian group and every finite dihedral group there is a finite integral extension k of ℤ over which there is a split global solution to the Donald–Flanigan problem. □

Now consider again the example of the cyclic group of order 3. The coefficient ring with which one starts is now O = ℤ[y]/(1 + y + y^{2}) := ℤ[ω], where ω is the image of y. If we set s = ω^{2}(q^{−1} − q^{3}) then f(x, t) now has the form x^{3} − sx^{2} + sx − 1. Viewing s as a new parameter, this deformation is not equivalent to the preceding since at the prime 3 no change of variable can remove the quadratic term from f. Thus we have inequivalent deformations leading to distinct separable commutative algebras. The discriminant now is Δ = (s + 1)(s − 3)^{3} and the separability idempotent is

e = Δ^{−1}[(s − 3)(s^{3} − 3s^{2} + s + 3)(1 ⊗ 1) + 2s(s − 2)(s + 1)(s − 3)(x ⊗ x)+
2s(s − 3)(x^{2} ⊗ x^{2}) − s(s − 2)(s + 1)(s − 3)(1 ⊗ x + x ⊗ 1)+

s(s − 1)(s − 3)(1 ⊗ x^{2} + x^{2} ⊗ 1) − (2s − 3)(s + 1)(s − 3)(x ⊗ x^{2} + x^{2} ⊗ x)].

This is again well-defined modulo 3 and reduces to \frac{1}{3}(1 ⊗ 1 + x ⊗ x^{2} + x^{2} ⊗ x) at s = 0. Recalling that q = 1 + t, this is equivalent to setting t = 0. (While O is not separable over ℤ since modulo 3 it contains a central nilpotent element the deformed algebra is separable over O, which is all we want.) This new solution is split when s is replaced by ω^{2}(q^{−1} − q^{3}). Although it is inequivalent to the preceding one, this does not imply that over 𝔽_{3} the algebra defined by x^{3} = 1 lies on the intersection of two components of the variety of algebras of dimension 3. For over an algebraically closed field there is only one separable algebra of dimension 3, namely, the sum of three copies of the coefficient field, so the deformations lie on the same component. But only for the second (split) one is there a fixed correspondence between its summands and those of ℂC_{3}.

We will show, in particular, that i) if the Donald–Flanigan problem has a solution (in the weak sense, i.e., at a fixed prime) for some group K then it also has a solution for the wreath products K ≀Γ, K ≀S_{n} of K with any abelian or symmetric group and ii) if there is a split global solution for K, i.e., one which is a sum of matrix algebras, then the same is true for these wreath products. This will imply, in particular, that the problem has a global solution for the the Weyl groups of type B_{n} and by minor modifications for the dihedral groups and groups of type D_{n}. For the groups B_{n} and D_{n} it will not be necessary to adjoin any roots of unity to ℤ, so we recover the fact that all their irreducible complex representations also are rational. If instead of a split global solution for K we have only a central separable one, then as mentioned in the introduction we shall need some conditions on the center, Z of the deformed algebra. Suppose that the new coefficient ring is R. To insure that Z can be split by a separable, faithfully projective extension of R we shall have to assume, that Z is faithfully projective and of constant rank over R.

8. Some smash and wreath products. We will need to know the structure of a smash product A#kG where G is a finite group and A a k-algebra of the form k^{n} = k ⊕ ⋯ ⊕ k (n times). The only way that a group G can act as kalgebra automorphisms of such an algebra A is by permutation of the summands, in effect, by permutation of the indices 1, …, n. Denoting the unit element of the ith summand of A by e(i), the smash product is spanned by the elements e(i)σ, σ ∈ G. These multiply by the rule σe(i) = e(σi)σ, so e(i)σe(j)τ = δ_{i,σj}e(i)στ. It follows that A#kG is a direct sum of subalgebras corresponding to the orbits of G. Fix some i ∈ {1, …, , n}, denote the orbit of i by Gi, the order of this orbit by m_{i}, and the isotropy group of i by G_{i}. If {σ̄, τ̄, … } are representatives of the cosets G/G_{i} then the various e(σ̄i)σ̄^{−1}(τ̄) span a subalgebra of A#kG isomorphic to M_{m_{i}}(k). For writing e(σ̄i)(τ̄)^{−1} = E_{σ̄,τ̄}, these multiply just like matrix units, i.e., E_{μ̄,ν̄}E_{σ̄,τ̄} = δ_{ν̄,σ̄}E_{μ̄,τ̄}.

Theorem 3. If A = k^{n} and G is a finite group operating by permutation of the summands then

A#kG ≅ ⨁_{i} M_{m_{i}}(k) ⊗ kG_{i}

where the sum is over representatives i of the distinct orbits of G in {1, …, n}, m_{i} is the order of the orbit Gi of i, and G_{i} is the isotropy group of i.

Proof. Since A#G is the direct sum of subalgebras corresponding to the orbits of G, what we must show is that the subalgebra corresponding to the orbit of i is isomorphic to M_{m_{i}} ⊗ kG_{i}. Letting σ̄, τ̄, … be as before representatives of the cosets G/G_{i}, the E_{σ̄,τ̄} = e(σ̄i)σ̄^{−1}τ̄ span an algebra isomorphic to M_{m_{i}}(k). This contains, in particular, the permutation matrices, so if we have any permutation π of {1, …, n} there is a unique matrix T_{π} ∈ M_{m_{i}}(k) giving the same permutation. Thus the operator T_{π}^{−1}π acts as the identity on {1, …, n} and hence commutes with the operation of all elements of M_{m_{i}}(k). Therefore T_{π}^{−1}πT_{ρ}^{−1}ρ = T_{ρ}^{−1}T_{π}^{−1}πρ = T_{πρ}^{−1}πρ. The map π → T_{π}^{−1}π is therefore a group morphism. Since the elements E_{σ̄,τ̄}ρ with ρ ∈ G_{i} form a k-basis for A, this proves the assertion. □

Note here that the elements of the isotropy group do not themselves commute with the elements of M_{m_{i}}. The second tensor factor in the statement of theorem is not the group algebra of the isotropy group G_{i} itself but the group algebra of the isomorphic group of all T_{π}^{−1}π. An analogous argument is used in the next two theorems. One consequence of the above theorem is that if we have an algebra of the form A = k^{n}#kG such that the Donald–Flanigan problem is solvable for all the isotropy groups G_{i} then it is solvable for all of A.

In addition to the foregoing we need to know the structure of A≀G when A is a direct sum of central separable algebras over some coefficient ring k. In this section we compute this when A is itself central separable and find that A≀G is canonically isomorphic to A^{⊗n} ⊗ kG. This makes it possible to extend any deformation of kG to one of A ≀ G. Again, while kG is contained in a natural way in both A ≀ G and A ⊗ G, the isomorphism, while canonical, does not carry one embedding to the other. The result is extended to the general case in the next section.

Denote the group of invertible elements of A by A^{∗}, its group of automorphisms by Aut A, and the center of A by Z. There is a group morphism A^{∗} → Aut A sending a ∈ A^{∗} to the inner automorphism in_{a} defined by in_{a} x = axa^{−1}; the kernel of this morphism is Z^{∗}. For a central simple algebra over a field the Skolem– Noether theorem asserts that the morphism is onto, but this need not be so for an arbitrary central separable algebra. Action of a group G on A is the same as a group morphism f : G → Aut A. The morphism f “factors through” the canonical morphism A^{∗} → Aut A if there is a morphism G → A^{∗} whose composite with the canonical morphism A^{∗} → Aut A is f. This is stronger than asserting that the image of every individual σ ∈ G is inner (which would always be the case for a central simple algebra). For if σ ∈ G then the choice of an element a_{σ} ∈ A conjugation by which has the same effect as σ is determined only up to an element of Z^{∗}. It may not be possible to make the choices in a consistent way to produce a morphism G → A^{∗}, as the following trivial example shows. Let A = M_{2}(k) and

set a = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , b = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ). Then in_{a} and in_{b} commute and both have

square equal to the identity, so they define an operation of C_{2} × C_{2} on M_{2}(k). But ab = −ba and multiplying a, b, or both by an invertible central element can not change this. Therefore (if the characteristic is not 2) the morphism of C_{2} × C_{2} into Aut A does not factor through A^{∗}. The problem is homological. For choosing an a_{σ} for every σ defines an element of H^{2}(G, Z^{∗}). This class does not depend on the choices and may not be trivial. (If k contains a square root of -1, which we may denote by i, then in_{ia} = in_{a}, in_{ib} = in_{b}, and ia, ib generate a group isomorphic to the quaternion group.)

Lemma A. Supppose that a group G operates as a group of automorphisms on a k-algebra A and that the associated morphism G → Aut A factors through A^{∗}. Then A#kG ≅ A ⊗ kG.

Proof. To avoid confusion, denote the image of x ∈ A under the operation of σ ∈ G by x^{σ}. By hypothesis, for each σ ∈ G there exists an invertible element a_{σ} such that a_{σ}xa_{σ}^{−1} = x^{σ} and a_{σ}a_{τ} = a_{στ} all σ, τ ∈ G. It follows that the elements a_{σ}^{−1}σ ∈ A#kG commute with all elements of A and form a multiplicative subgroup of G′ of A#kG isomorphic to G. But then A#kG = A ⊗ kG′ ≅ A ⊗ kG. For if {a_{i}} is a k-basis for A then every element of A#kG can obviously be written uniquely as a linear combination with coefficients in k of the elements a_{i} ⊗ a_{σ}^{−1}σ. □

Suppose now that A is a central separable k-algebra. Every a ∈ A then has a reduced trace Trd(a) ∈ k; when A = M_{n}(k) this is just the usual trace of a matrix (cf. [KO, pp. 90, 110]). (The “ordinary” trace of a comes from the characteristic polynomial of left multiplication by a in the algebra. This can vanish even for a primitive idempotent. Consider e.g. 2 × 2 matrices in characteristic 2. Here the trace of left multiplication by E_{11} is zero but the reduced trace is 1.) Now a central separable k-algebra A always contains k as a k-module direct summand so the map a ↦ Trd a may be viewed as k-module endomorphism of A. Since for central separable A there is an isomorphism A⊗A^{op} → End_{k} A sending x⊗y to the endomorphism carrying a ∈ A to xay, it follows that there exists a unique element T = ∑ x_{i} ⊗ y_{i} ∈ A ⊗ A such that Trd a = ∑ x_{i}ay_{i}. Viewing this “switch” element T as an element of A ⊗ A, it has the properties i) T^{2} = 1 ⊗ 1, the unit element of A ⊗ A and ii) T (a ⊗ b)T = b ⊗ a for all a, b ∈ A. (This result, communicated by A. Frölich to Knus and Ojanguren is credited by them [KO, p.112] to O. Goldman who with M. Auslander developed the present concept of separable algebra, [AG].) For A = M_{n}(k) one has T = ∑_{i,j} e_{ij} ⊗ e_{ji}. It follows from the preceding Lemma that if C_{2} operates on A ⊗ A by interchange of the tensor factors, then A ≀ C_{2} = A^{⊗2}#kC_{2} is canonically isomorphic to A^{⊗2} ⊗ kC_{2}. The isomorphism is canonical because there is a canonical choice of an element in A ⊗ A conjugation by which interchanges the tensor factors, namely T. Note, incidentally, that T is symmetric since T = T^{3} = T (T)T, i.e., T with its tensor factors interchanged.

Theorem 4. Let A be a central separable k-algebra and let S_{n} operate on A^{⊗n} by permutation of the tensor factors. Then there is a canonical isomorphism A ≀ S_{n} ≅ A^{⊗n} ⊗ kS_{n}.

Proof. The group S_{n} is generated by elements T_{12}, T_{23}, …, T_{n−1,n} corresponding to the transpositions (12), (23), …, (n − 1, n) subject only to the conditions that T_{i,i+1} and T_{j,j+1} commute for |i − j| > 1, that T_{i,i+1}^{2} = 1, and the braid relation T_{i,i+1}T_{i+1,i+2}T_{i,i+1} = T_{i+1,i+2}T_{i,i+1}T_{i+1,i+2}. This is the “Artin presentation” of S_{n} in the terminology of Serre (but it is probably due to Serre himself). Now let T_{i,i+1} denote the switch element T operating in places i and i + 1 of A^{⊗n}. It will be sufficient to show that these satisfy the relations of the Artin presentation. Since T^{2} = 1 the first relations are clearly satisfied. For the braid relations it is only necessary to consider the case n = 3 where we must prove that T_{12}T_{23}T_{12} = T_{23}T_{12}T_{23}. If T = ∑ x_{i} ⊗ y_{i} then the left side is ∑(x_{i} ⊗ y_{i} ⊗ 1)(1 ⊗ x_{j} ⊗ y_{j})(x_{k} ⊗ y_{k} ⊗ 1) = ∑ x_{i}x_{k} ⊗ y_{i}x_{j}y_{k} ⊗ y_{j}. However, ∑ x_{i}x_{k} ⊗ y_{i}x_{j}y_{k} = T (1 ⊗ x_{j})T = x_{j} ⊗ 1, so this is just ∑ x_{j} ⊗ 1 ⊗ y_{j} (which would naturally be denoted T_{13}). One checks similarly that the right side gives the same thing. □

It was convenient in the preceding to deal with S_{n} but it is clear that nothing would change if instead of S_{n} itself we had a group G which operated by permutation of the tensor factors, i.e., through a morphism G → S_{n}. This gives the following.

Corollary. If A is central separable over k and a group G operates on A^{⊗n} through a morphism G → S_{n} then A ≀ G is canonically isomorphic to A^{⊗n} ⊗ kG. □

It follows that any deformation of kH induces one of A ≀ H.

9. Wreath product with a sum of central separable algebras. Suppose now that the k-algebra A is a direct sum of central separable algebras, A = A_{1} ⊕ ⋯ ⊕ A_{r} with respective unit elements e_{1}, …, e_{r} and let Z denote the commutative algebra which these generate. Extending the result of the previous section, we wish to determine the structure of A ≀ S_{n} and more generally of A ≀ G where G is a group operating on A^{⊗n} by permutation of the tensor factors. For any multiindex I = (i_{1}, …, i_{n}) set A(I) = A_{i_{1}} ⊗ A_{i_{2}} ⊗ ⋯ ⊗ A_{i_{n}} and denote its unit element e_{i_{1}} ⊗ e_{i_{2}} ⊗ ⋯ ⊗ e_{i_{n}} by e(I). Then A^{⊗n} is the direct sum of the A(I), which are all central separable, one has e(I)e(J) = δ_{IJ}e(I), and the distinct A(I) are mutually orthogonal. The wreath product A ≀ G is spanned by elements of the form a(I) ⊗ σ, a(I) ∈ A(I), σ ∈ G, which we write simply as a(I)σ. One has e(I)σe(J)τ = e(I)e(σJ)στ, where σ(j_{1}, …, j_{n}) = (j_{σ^{−1}1}, …, j_{σ^{−1}n}). We write (a_{1}, …, a_{n}) for a_{1} ⊗ ⋯ ⊗ a_{n} ∈ A^{⊗n} where each a_{i} is in A, and if α = (a_{1}, …, a_{n}) then σα = α^{σ}σ, where (a_{1}, …, a_{n})^{σ} = (a_{σ^{−1}1}, …, a_{σ^{−1}n}).

Set B = A≀G, choose a multiindex I, and let B(I) be the subalgebra generated by A(I) and the elements of G. It is spanned by all α^{σ}τ. One has B(I) = B(σI) for all σ ∈ G, so these subalgebras are really indexed by the orbits of G in the set of multiindices, and B is their direct sum. Determining the structure of B is thus the same as determining that of the various B(I). If there is a global solution to the Donald–Flanigan problem for each of the “orbital” subalgebras B(I) then the same will be true for A ≀ kG, and similarly if there is only a weak solution for each of the subalgebras then the same will be hold for A ≀ kG. Fixing the multiindex I, let G_{I} be the isotropy group of I. Now choose fixed representatives σ̄, τ̄, … for the cosets σG_{I}, τG_{I}, . . .. The elements e(σ̄I)σ̄τ̄^{−1} then multiply precisely like the matrix units E_{σ̄,τ̄} (whose rows and columns are indexed by the chosen coset representatives), namely (e(ρ̄I)ρ̄μ̄^{−1})(e(σ̄I)σ̄τ̄^{−1}) = δ_{μ̄σ̄}e(ρ̄I)ρ̄τ̄^{−1}.

The choice of coset representatives {μ̄, σ̄, … } fixes particular isomorphisms A(I) → A(σ̄I) and hence also isomorphisms A(σ̄I) → A(μ̄I) for all σ̄ and μ̄. We can write the elements of A(σ̄I) in the form aσ̄ with a ∈ A(I). Since (aμ̄μ̄σ̄^{−1})(bσ̄σ̄^{−1}τ̄) = (ab)μ̄μ̄τ̄^{−1} the elements of the form aμ̄μ̄σ̄^{−1} span a subalgebra M(I) of A ≀ G. Let the index of G_{I} in G be m (the cardinality of the set of representatives). Then the subalgebra M(I) is isomorphic to M_{m}(A(I)) under the map aσ̄σ̄τ̄^{−1} ↦ aE_{σ̄,τ̄}. We will show that B(I) ≅ M_{m}(A(I)) ⊗ kG_{I}, but as in the previous section, the tensor factor kG_{I} here is not the usual subalgebra kG_{I} of A≀G. With the foregoing notations and writing α^{π} for π(α), α ∈ A(I), π ∈ G we have the following.

Lemma B. For every π ∈ G_{I} there is a canonical choice of T_{π} ∈ A(I) such that T_{π}αT_{π}^{−1} = α^{π}.

Proof. Suppose that A(I) = A(i_{1}, …, i_{n}) = A_{i_{1}} ⊗ ⋯ ⊗ A_{i_{n}}. For convenience, reorder the tensor factors, putting the indices in increasing order, so that the tensor product has the form A_{1}^{⊗n_{1}} ⊗ ⋯ ⊗ A_{r}^{⊗n_{r}} where n_{1} + ⋯ + n_{r} = n. Since πI = I, after the reordering π permutes the first n_{1} factors amongst themselves, the next n_{2} factors amongst themselves, etc.. From the preceding section we know that there is a canonical T_{π}^{(1)} ∈ A_{1}^{⊗n_{1}} such that for any α_{1} ∈ A_{1}^{⊗n_{1}} we have T_{π}^{(1)}α_{1}(T_{π}^{(1)})^{−1} = (α_{1})^{π}, and similarly for the other A_{i}^{⊗n_{i}}. Then T_{π}^{(1)} ⊗⋯⊗T_{π}^{(r)} with factors returned to their original order is the desired T_{π}. □

Now set υ(π) = ∑ μ̄T_{π}^{−1}πμ̄^{−1}, where the sum runs over all coset representatives μ̄ of G_{I}. Since μ̄T_{π}^{−1}πμ̄^{−1} = (T_{π}^{−1})μ̄μ̄πμ̄^{−1} ∈ A(μ̄I)μ̄πμ̄^{−1} the terms for different μ̄ are mututally orthogonal, from which it is easy to see that υ(π)υ(π′) = υ(ππ′) for π, π′ ∈ G_{I}. Since υ(1) = ∑ μ̄e(I)μ̄^{−1} = ∑ e(μ̄I) is just the identity element of B(I) = ⊕A(μ̄I), the υ(π) form a multiplicative subgroup of B(I) isomorphic to G_{I}. Moreover, if α ∈ A(I) then υ(π) also commutes with all aσ̄σ̄τ̄^{−1}.

Lemma C. There is a canonical isomorphism B(I) ≅ M(I) ⊗ kG_{I} given by aσ̄σ̄τ̄^{−1}υ(π) ↦ aE_{σ̄,τ̄} ⊗ π.

Proof. Observe that aσ̄σ̄τ̄ is orthogonal to all the summands of υ(π) = ∑ μ̄T_{π}^{−1}πμ̄^{−1} except that with μ̄ = τ̄, so aσ̄σ̄τ̄^{−1}υ(π) = (aT_{π}^{−1})σ̄σ̄πτ̄^{−1} = (aT_{π}^{−1})σ̄σ̄(τ̄π^{−1})^{−1}. As a runs through A(I), (aT_{π}^{−1})σ̄ runs through A(σ̄I). As π runs through G_{I} so does π^{−1} and τ̄π^{−1} runs through the coset of τ̄. Consequently, the τ̄π^{−1} exhaust G and therefore so do their inverses, and hence, for any fixed σ̄ so do the σ̄πτ̄^{−1}. So for fixed σ̄, the set of all aσ̄σ̄τ̄^{−1}υ(π) with varying τ̄, π is precisely B(I), from which the result follows. □

We therefore have the following.

Theorem 5. Let A = A_{1} ⊕ ⋯ ⊕ A_{r} be a sum of central separable k-algebras A_{i}, and suppose that a group G operates on A^{⊗n} by permutation of the tensor factors. For every multiindex I = (i_{1}, …, i_{n}) set A(I) = A_{i_{1}} ⊗ ⋯ ⊗ A_{i_{n}}, let G_{I} be the isotropy group of I, and let {Ī, J̄, … } be a set of representatives of the orbits of G in the set of multiindices. Set m_{I} = (G : G_{I}), the index of G_{I} in G. Then A ≀ G is canonically isomorphic to ⨁ M_{mĪ}(AĪ) ⊗ kGĪ where the sum runs over the orbits of G in the set of multiindices. □

There is an overlap between this theorem and Theorem 3, namely, the case where each A_{i} is just k itself and one has a wreath product, but Theorem 3 allows more general smash products. As an immediate corollary of Theorem 5 one gets explicitly the structure of the complex group algebra (and hence the irreducible complex representations) of a wreath product of groups H ≀ G. More generally, if k is a splitting ring for both H and all the G_{I}, i.e., one such that both kH and the kG_{I} are sums of central separable algebras, then the theorem implies that k is also a splitting ring for H ≀ G and it gives the structure of k(H ≀ G). Consider, for example, B_{n} = C_{2} ≀ S_{n}. Here ℚ is a splitting field for C_{2}. Writing C_{2} = {1, a} and setting e = (1 + a)/2, f = (1 − a)/2, we have ℚC_{2} = ℚe ⊕ ℚf, a direct sum of two copies of ℚ. The idempotents e(I) are here just tensor products of length n of e’s and f’s. If e occurs m times and f occurs n − m times then the order of its orbit is \binom{n}{m} and the corresponding isotropy group is isomorphic to S_{m} × S_{n−m}. Since ℚ is also a splitting field for the symmetric group, it is a splitting field for B_{n}, so the irreducible complex representations of B_{n} are all rational. The theorem then gives

ℚB_{n} = ⨁_{m=0}^{n} M\binom{n}{m}(ℚ) ⊗ ℚS_{m} ⊗ ℚS_{n−m}.

This includes the classic result that the representations of B_{n} are indexed by all ordered pairs consisting of a representation of S_{m} and a representation of S_{n−m}. For if λ, μ are partitions of m and n−m, respectively, and if S_{m}(λ) is the block (=matrix algebra summand) of ℚS_{m} corresponding to λ and S_{n−m}(μ) the block of ℚS_{n−m} corresponding to μ, then the blocks of ℚB_{n} are of the form M\binom{n}{m}⊗S_{m}(λ)⊗S_{n−m}(μ). The representations B_{n} are thus indexed by the pairs (λ, μ). Reversing their order gives a different representation. By contrast (cf. §11), the representations of D_{n} may be viewed as indexed by unordered pairs of representations of S_{m} and S_{n−m} (or by pairs in which m ≤ n − m).

10. First applications. The most immediate application of the results of the preceding section is to the case where we have a group H for which there is a split global solution to the Donald–Flanigan problem and we seek one for a group of the form H ≀ G. So suppose that k is some subring of algebraic integers in ℂ and that kH has been deformed into a direct sum A = M_{i_{1}}(R) ⊕ ⋯ ⊕ M_{i_{r}}(R) of matrix algebras. Here the new coefficient ring R is the localization of a power series ring k[[t, u, …, v]] at a single element f. Since R is a domain, the inclusion of k into ℂ can be extended to a monomorphism of R into ℂ. If we have been careful with R to require, e.g., that it is projectively faithful over k, then we can still reduce modulo every prime p of k (see below), so there is a natural correspondence between the blocks of ℂH and those we get by reducing A modulo p. By assumption there is given some morphism G → S_{n}, and through this G operates on H^{⊗n} by permutation of the tensor factors. Keeping the notations, the results of the preceding section immediately yield the following.

Theorem 6. If there is a split global solution for RH and if in H ≀ G for each isotropy group G_{I} there is a split global solution to the Donald–Flanigan problem for RG_{I} then there is one for H ≀ G over k. (If there is only a weak solution for RG_{I} then there is a weak solution for H ≀ G and similarly for local solutions.) □

There are several important cases in which the conditions on G will be fulfilled. First, if G is abelian then so is every isotropy group; we have already constructed split global solutions for finite abelian groups. Second, suppose that G is all of S_{n}. The isotropy group of an index of the form (i_{1}^{n_{1}}, …, i_{r}^{n_{r}}), where i^{m} stands for i repeated m times and n_{1} ⋯ + n_{r} = n is S_{n_{1}} × ⋯ × S_{n_{r}}. Since we have split global solutions for the symmetric groups we also have a split global solutions for these.

Theorem 7. There is a split global solution to the Donald–Flanigan problem for any wreath product H ≀ G where H is finite abelian and G is either finite abelian or G = S_{n}. □

The Weyl groups of type B_{n} are a special case. Since both for C_{2} and S_{n} we can start with the coefficient ring ℤ we do not have to adjoin any roots of unity.

Theorem 8. There is a split global solution to the Donald–Flanigan problem for the Weyl groups of type B_{n} starting with ℤ. For the coefficient ring after deformation one can take R = ℤ_{q,n} since there is a split global deformation of S_{n} defined over this R. It follows that all complex irreducible representations of the Weyl groups of type B_{n} are rational. □

To handle dihedral groups in this context, consider S_{2} acting on an abelian group K by sending every element of the latter to its inverse. We can form the semidirect product H ⋊S_{2}; for H the cyclic group C_{n} this is just the dihedral group D_{n}. Let A be a split equivariant solution to the Donald–Flanigan problem for H. (Here, if e is the exponent of H then we must work over ℤ̄ = ℤ[y]/Φ(y), where Φ is the cyclotomic polynomial for the primitive e- th roots of unity.) This A is then just a direct sum of copies of whatever coefficient ring k has been introduced, and S_{2} acts as automorphisms of the sum. However, since the automorphism group of each summand is reduced to the identity, S_{2} must either leave an individual summand fixed or exchange them in pairs. The equivariant deformation has thus produced a direct sum of copies of kS_{2} plus copies of k ⊕k on which S_{2} operates by interchange of the summands. As a special case of what we have seen before, the latter yields 2 × 2 matrix algebras over k, while kS_{2} itself has a global deformation to a sum of two copies of k (extended by the deformation parameter). Therefore we have

Theorem 9. There is a split global solution to the Donald–Flanigan problem for any semidirect product H ⋊ S_{2} where H is a finite abelian group. □

Notice that the argument implies the familiar fact that all complex irreducible representations of a dihedral group D_{m} have dimension either one or two and and are defined over ℚ(η) where η is a primitive m-th root of unity. Since every subgroup of a dihedral group is again either dihedral or abelian we have the following.

Theorem 10. There is a split global solution to the Donald–Flanigan problem for any group of the form H ≀ D_{m} with H abelian. □

Suppose now that H is a group for which the Donald–Flanigan problem is solvable over the coefficient ring k but where the solution is a non-split separable algebra A over the new coefficient ring R. Now A is Azumaya (central separable) over its center Z, which in turn is separable over R (cf., e.g., [DI], p. 55). We will be able to use the preceding results if we can “split” Z by a suitable extension T of the coefficient ring R, i.e., reduce it to a direct sum of copies of the new coefficient ring. For this we can apply the following basic result: For a commutative R-algebra S the following properties are equivalent: 1) S is separable, faithfully projective (i.e., faithful and projective) and of constant rank over R, 2) there is a faithfully projective and separable R-algebra T splitting S, i.e., such that S ⊗_{R} T ≅ T ⊕ T ⊕ ⋯ ⊕ T (n times) = T^{n} and 3) there is a faithfully flat and separable Ralgebra T such that S ⊗_{R} T ≅ T^{n} (cf. [KO], Théorème 4.7, p. 88). Projectivity and faithfulness of T will insure, in particular, that i) if p is a rational prime then the sequence 0 → pA ⊗ T → A ⊗ T → (A/pA) ⊗ T → 0 is exact and ii) the map of A into A ⊗ T given by a ↦ a ⊗ 1 is an inclusion, and similarly for A/pA in place of A. Also, if we have an inclusion R ↪ ℂ then we still have an inclusion T = R ⊗ T ↪ ℂ ⊗ T (where ℂ is regarded as an R-module by the first inclusion.

If k is a direct sum or product of subrings k_{i} then the Donald–Flanigan problem for a group G will be solvable if it is solvable for each k_{i}G. To avoid the complications of infinite sums or products, we may restrict attention to the case where k has only finitely many idempotents. This holds, in particular, if k is noetherian, which is always the case here. With this, we can reduce to the case where the only non-zero idempotent in k is its identity. A projective module over k must then have constant rank. If k has only finitely many idempotents, so k = k_{1} ⊕ ⋯ ⊕ k_{r}, where each summand has no idempotent but the identity, then any k-algebra A is similarly a sum of k_{i}-algebras A_{i}. If each of these can be deformed to a separable k_{i}-algebra then A can be deformed to a separable k-algebra.

Deformation can not enlarge the center of an algebra but it can diminish it. We will say that a deformation is centrality preserving if elements which were central before deformation continue to be central. (It need not be center preserving because in general the center is deformed.) This must be the case, in particular, for a solution to the Donald–Flanigan problem for kG which resembles the complex group algebra. For any finite group G and coefficient ring k, the center of kG is a free module over k; a typical basis element is just the sum of the elements of some conjugacy class. The center of a centrality preserving deformation of a group algebra thus continues to be a free module over the new coefficient ring R, so condition 1) above will hold without the assumption that k is noetherian. (Of course it holds if the original group is commutative and the deformed group algebra remains so, but there are important examples of deformations where the commutativity is lost.)

11. Weyl groups of type D_{n}. To describe the Weyl groups of type D_{n}, write the elements of B_{n} in the form (c_{1}, …, c_{n})σ with c_{i} ∈ C_{2}, σ ∈ S_{n}; then D_{n} is the subgroup consisting of all elements in which an even number of the c_{i} are equal to the unit element of C_{2}. Equivalently, the product of the c_{i} is equal to the unit element. So viewed, D_{n} is an example of a class of semidirect products which are subgroups of wreath products arising as follows. Let H be any abelian group (not necessarily finite) and consider the morphism μ : H^{n} = H ×⋯×H (n times ) → H sending (c_{1}, …, c_{n}) to the product c_{1}c_{2} … c_{n}. If K is any subgroup of H then μ^{−1}K is stable under the operation of S_{n} on the factors of H^{n} so one can form the semidirect product (μ^{−1}K) ⋊ S_{n}. For n = 2, H cyclic and K reduced to the unit element one gets the dihedral groups. For H = C_{2} and K the unit subgroup one gets the groups D_{n}. Writing simply C_{2}^{n} for C_{2}^{×n}, here μ^{−1}K ≅ C_{2}^{n−1} so the Weyl group D_{n} is a semidirect product C_{2}^{n−1} ⋊ S_{n}. (One has D_{2} = S_{2}, D_{3} = S_{4}.)

It will be convenient to deal with D_{n+1} = C_{2}^{n} ⋊ S_{n+1} rather than D_{n}. The deformation of C_{2}^{n} we employed in constructing a solution for B_{n} was equivariant under the operation of S_{n}, but in D_{n+1} this group is acted upon by S_{n+1} and the deformation ceases to be equivariant. The original operations of S_{n+1} consequently cease to be automorphisms, but the operations can also be so deformed that S_{n+1} continues to operate as automorphisms of the deformed algebra. This would normally be a difficult homological problem but fortunately here there is a natural solution (although it exhibits a rather unnatural phenomenon at the prime p = 2). To exhibit it, however, we first examine the rational group algebra of D_{n+1}.

To describe the operation of S_{n+1} on C^{n}, view S_{n+1} as permutations of the set {1, …, n + 1}, let τ_{i}, i = 1, …, n be the transposition (i, n + 1) and let τ_{n+1} be the identity. Then every element of S_{n+1} can be written uniquely as a product στ_{i} for some i, where σ is a permutation of 1, …, n. Now let C_{2} = {1, a} and view C_{2}^{n} again as the subgroup of C_{2}^{n+1} consisting of all (c_{1}, …, c_{n+1}) where c_{i} is either 1 or a and the number of entries equal to a is even. If c_{1}, …, c_{n} are known then so is c_{n+1}, which we may therefore omit from the notation. With this, we have τ_{i}(c_{1}, …, c_{n}) = (c′_{1}, …, c′_{n}) where c′_{j} = c_{j} for j ≠ i and c′_{i} = 1 if the number of c_{i} equal to a is even and c′_{i} = a if it is odd. Now ℚC_{2} ≅ ℚ ⊕ ℚ where the idempotents of the summands on the right side may be taken to be e = (1 + a)/2 and f = (1 − a)/2, respectively. Using our previous notation, the 2^{n} primitive idempotents of ℚC_{2}^{n} may therefore be written in the form E = e(I), where I is an ordered n-tuple of indices whose values can only be 1 or 2; the jth entry in e(I) is e or f according as the jth entry in I is 1 or 2. Since S_{n+1} operates as automorphisms of ℚC_{2}^{n} it must permute these idempotents e(I). The effect of a permutation σ of {1, …, n} is clearly just to permute the entries of E. We must compute τ_{i}E, i = 1, …, n. For simplicity, suppose that i = n; the argument will be the same for the other values of i. To avoid confusion, write u instead of 1 for the unit of C_{2}, and consider the non-commutative polynomial ring ℚ{u, a} formally generated by these two symbols. If we consider e and f also as abstract symbols, this is the same ring as ℚ{e, f} under the identifications e = (u+a)/2, f = (u−a)/2. Now consider the effect of τ_{n} on elements of the form ξe, ξf, where ξ is a monomial in a and u. If the degree of ξ in a is even then τ_{n}(ξa) = ξa, τ(ξu) = ξu, while if the degree of ξ in a is odd then τ_{n}(ξa) = ξu, τ_{n}(ξu) = ξa. It follows that whatever the degree of ξ in a we have τ_{n}(ξe) = ξe while τ_{n}(ξf) = ±ξf according as the degree of ξ in a is even or odd.

It follows from the foregoing that if ξ is an arbitrary homogeneous polynomial of total degree n−1 in a and u then τ_{n}(ξe) = ξe. In particular, if E′ is a monomial of degree n − 1 in e and f then τ_{n}(E′e) = E′e. In τ_{n}(E′f), however, if E′ is expanded as a polynomial in a and u, then the signs of all the terms of odd degree in a are reversed. This is the same thing as replacing every factor e in E′ by f and every factor f by e. Since the analogous result clearly holds for all i = 1, …, n, we have the following description of the action of S_{n+1} on ℚC_{2}^{n} when the latter is viewed as the set of homogeneous polynomials of total degree n in ℚ{e, f}. (This is, of course, essentially the same thing as the tensor algebra of ℚC_{2}, and the polynomials of degree n are just (ℚC_{2})^{⊗n}.)

Theorem 11. Let E ∈ (ℚC_{2})^{n} be a monomial of degree n in e and f. If the ith factor in E is e, then τ_{i}E = E. If the ith factor in E is f then the ith factor of τ_{i}E remains f but in all other factors e and f are interchanged. □

In this theorem, the fact that the ground field is ℚ is of no consequence; it was used only to show that we had constructed a representation of S_{n+1} as operators on the nth cartesian power of a two-element set. In purely combinatorial terms it can be rephrased as follows.

Theorem 12. There is a permutation representation of S_{n+1} on the set of sequences of length n of 0’s and 1’s in which the subgroup S_{n} of S_{n+1} fixing n + 1 operates by permutation of the entries, and the transpositions (i, n + 1) are represented by operators τ_{i} as follows: If the ith entry of a sequence 𝔰 is 0 then τ_{i}𝔰 = 𝔰.

If the ith entry of 𝔰 is 1 then the ith entry of τ_{i}𝔰 remains 1 but all other entries are replaced by their complements (i.e., 0’s are replaced by 1’s and 1’s by 0’s). □

It is clear that the representation is faithful except in the trivial case where n = 1. The group consisting of all permutations of the entries in the sequences together with all operations which replace a single entry by its complement is the full group of permutations of all 2^{n} of the sequences of 0’s and 1’s.

We are now free to use any coefficient ring k. Referring back to the Hecke algebra construction (§4) we can take k = ℤ_{q,2} = ℤ[q, q^{−1}, 1/2_{q^{2}}] where 2_{q^{2}} = 1/(1 + q^{2}) and eventually q = 1 + t. Then kC_{2} ≅ k ⊕ k where the idempotents generating the summands are, respectively, e = (1 + qa)/2_{q^{2}}, f = (q^{2} − qa)/2_{q^{2}}. Note that when q = 1 these are just our previous e and f, and that as 2_{q^{2}} = 1 + q^{2} = 2 + 2t + t^{2} we can reduce modulo any rational prime, in particular p = 2, and the expressions remain meaningful. When q = 1 or equivalently, t = 0, the operation of S_{n+1} is just the original operation on ℚC_{2}^{n}, but the deformation of ℤC_{2} to its Hecke algebra and its induced deformation of ℤC_{2}^{n} is not equivariant with respect to the original operation. It is so with respect to the subgroup S_{n} which just permutes the factors, which in effect is just the subgroup leaving n + 1 fixed, but not with respect to the τ_{i}. Nevertheless, the theorem effectively defines an operation of all S_{n+1}, but that operation is a deformation of the original one.

It is important now to exhibit the deformation explicitly in order to show what happens at p = 2. To do so, we must write it in terms of the basis {1, a} for kC_{2}. Here e and f are represented by the column vectors of the matrix

X = \frac{1}{2_{q^{2}}} ( \begin{matrix}1 & q^{2} \\ q & −q \end{matrix} ). Putting 1 before a gives a lexicographic order to the 2^{n}

basis elements of k(C_{2}^{n}) = (kC_{2})^{⊗n} consisting of the tensor products of length n of 1’s and a’s. The tensor products of length n of the e’s and f’s arranged in lexicographic order are then represented by the column vectors of the 2^{n} × 2^{n} matrix X^{⊗n}. To illustrate what happens it will be sufficient to consider the case n = 2, so we are tacitly dealing with D_{3}. Write

Y = X^{⊗2} = \frac{1}{(2_{q^{2}})^{2}} ( \begin{matrix}1 & q^{2} & q^{2} & q^{4} \\ q & −q & q^{3} & −q^{3} \\ q & q^{3} & −q & −q^{3} \\ q^{2} & −q^{2} & −q^{2} & q^{2} \end{matrix} ); Y^{−1} = ( \begin{matrix}1 & q & q & q^{2} \\ 1 & −q^{−1} & q & −1 \\ 1 & q & −q^{−1} & −1 \\ 1 & −q^{−1} & −q^{−1} & q^{−2} \end{matrix} ) .

If an element σ ∈ S_{3} is represented in terms of the basis {e⊗e, e⊗f, f ⊗e, f ⊗f} by a matrix M, then its representation in terms of of the basis {1⊗1, 1⊗a, a⊗1, a⊗a} is then given by Y MY^{−1}. For σ = (1, 2) the effect on this basis is to interchange its second and third elements, so the matrix M is the permutation matrix P_{23} =

( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ). One can check immediately that this commutes with Y, so the

operation of (1, 2) is undeformed. More generally, it is easy to see in the case of D_{n+1} that the operation of S_{n}, which operates by the usual permutation of the tensor factors, is undeformed. The transposition (n, n + 1), in our illustration (2, 3), behaves quite differently. In the basis generated by the tensor products of the idempotents e, f, it fixes e ⊗ e and f ⊗ e and interchanges e ⊗ f and f ⊗ e, so it

is represented by P_{24} = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{matrix} ). Its matrix in terms of the basis generated

by the elements 1, a of C_{2} is easily calculated to be

Y P_{24}Y^{−1} = \frac{1}{(2_{q^{2}})^{2}} ( \begin{matrix}(2_{q^{2}})^{2} & 0 & q^{5} − q & 1 − q^{4} \\ 0 & (2_{q^{2}})^{2} & 1 − q^{4} & q^{3} − q^{−1} \\ 0 & 0 & 1 − q^{4} & 2(q^{3} + q) \\ 0 & 0 & 2(q^{3} + q) & 1 − q^{4} \end{matrix} )

Letting q → 1, or equivalently, writing q = 1+t and letting t → 0, this becomes the permutation matrix P_{34}, which one can check is indeed the representation of (2, 3) in the basis generated by 1 and a. This matrix remains formally unchanged after reduction modulo any prime, in particular p = 2. However, reduction modulo 2 and reduction modulo t do not commute here. Reducing first modulo 2 and then modulo

t yields the matrix N = ( \begin{matrix}1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ). Letting q → 1 first recovers the original

representation of (1, 2) and (2, 3) on the 1, a basis by P_{23} and P_{34}; reducing first

modulo 2 gives, respectively, P_{23} and the above N. Setting W = ( \begin{matrix}1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{matrix} ),

one has W P_{23}W^{−1} = P_{23}, W NW^{−1} = P_{34}, so W conjugates the pair P_{23}, N into P_{23}, P_{24}. Therefore, P_{23} and N give, a representation of S_{3} on kC_{2} equivalent to that originally given by P_{23} and P_{24}. Thus, if we had started in characteristic 2 we would still have produced the necessary deformation of the action of S_{3} consistent with the deformation of kC_{2} but its behavior at different primes patches together in an unexpected way. Nevertheless we do have a global deformation; it will behave correctly at every prime.

Having now deformed the integral group ring ℤD_{n} to a smash product A = (k ⊕ k)^{⊗n}#kS_{n+1}, Theorem 3 is applicable. In the next stage of the solution we must examine the orbits and the isotropy groups of the action of S_{n+1}; the decomposition into orbits will give a decomposition of the algebra. If B(I) is the direct sum of the rank one algebras A(J) for J in the orbit of I, then we will again have A = ⊕B(I)#k(S_{n+1})_{I} where the sum ranges over the orbits of S_{n+1} in the set of indices I. Since the A(I) are all just isomorphic to k they have no automorphisms, so if the orbit of I has order m then B(I) ≅ M_{m}(k)#k(S_{n+1})_{I}. For the Weyl groups B_{n} = C_{2} ≀S_{n} the isotropy groups were of the form S_{m} ×S_{n−m}. Since each factor had a split global solution starting with ℤ, so did their product. Here the same is almost true, but for D_{n+1} with n odd there is a special case, the “middle” isotropy group. This is the one case where we have to use Theorem 5.

To examine the orbits and isotropy groups for D_{n+1} consider, for simplicity, the case where E = e(I) is a tensor product of m factors e followed by n−m factors f; write it as e^{m}f^{n−m}. Every e(J) with the same number of e’s and f’s as this is in the orbit of E. Generally this is not the whole orbit since τ_{n}E will contain e now n − m − 1 times and f now m + 1 times. The exceptional or middle case is that where n = 2r + 1 is odd and m = r. One must be careful in counting orbits, since if we let m run from 0 to n then every orbit will be counted twice except the middle one (when n is odd). To count each only once, restrict m by requiring that m ≤ n − m or 2m ≤ n. The middle case must also be handled separately.

In the non-exceptional case, the order of the orbit of e(I) is therefore \binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} and the order of the isotropy group (S_{n+1})_{I} is (m + 1)!(n − m)!. In fact, in the non-exceptional case, we have (S_{n+1})_{I} ≅ S_{m+1} × S_{n−m}. Here the second factor arises because it is evident that permuting the last n − m factors of e(I) will leave it unchanged. For the first m factors, since they are all equal to e, we can not only permute them, but also apply τ_{i} for i = 1, …, m and we have just proven that the group generated by all these operations in S_{m+1}. (It may appear that the case m = n is also exceptional since f does not occur and the orbit of e^{m} is reduced to e^{m} itself, but the formulas remain correct if we interpret S_{0}, whose order is 0!, as the identity.) In the exceptional case where n = 2r + 1 and e(I) = e^{r}f^{r+1}, the order of the orbit is only \binom{2r+1}{r} and the order of the isotropy group is therefore (2r + 2)!/\binom{2r+1}{r} = 2((r + 1)!)^{2}. This isotropy group contains the subgroup isomorphic to S_{r+1} × S_{r+1}, where, as before, the first factor is generated by the permutations of the first r factors together with the τ_{i} for i = 1, …, r and the second factor consists of the permutations of the last r + 1 factors. Being of index 2, it is normal.

Lemma D. In the middle case, where e(I) = e^{r}f^{r+1} one has (S_{2r+2})_{I} ≅ S_{r+1} ≀C_{2}.

Proof. Set σ = (1, r + 1)(2, r + 2) ⋯ (r, 2r) and ρ = στ_{2r+1}. Then ρe(I) = e(I) so ρ is in the isotropy group of I. Clearly ρ^{2} = 1. What one must show is that conjugation by ρ interchanges the factors in the product S_{r+1} × S_{r+1}. It is trivial that ρ interchanges the permutations of places 1, …, r with the permutations of places r + 1, …, 2r. Finally, we claim that ρτ_{i}ρ = (r + i, 2r + 1) for i = 1, …, r. One must show that the operations are the same on every e(J). There are four cases, according as the ith entry of e(J) is e or f and likewise for the (2r + 1)st entry; each is readily checked. □

With this the Donald–Flanigan problem for D_{n} is solved. For with a split global solution for S_{r+1}, the problem for the middle isotropy group is reduced to our previous one of a wreath product of a sum of central separable algebras. The largest isotropy group is S_{n+1}, so we have the following.

Theorem 13. There is a split global solution starting with ℤ to the Donald– Flanigan problem for the Weyl groups D_{n+1}; all their irreducible complex representations are rational. The coefficient ring after deformation is ℤ_{q,n+1}. □

We can also compute the rational group ring of D_{n+1} in the following way, separating the odd and even cases, and just writing n = 2r in the latter. For the middle case we need to compute ℚ(S_{r} ≀ C_{2}). From Theorem 5, for each block S_{r}(λ) of ℚS_{r} there is a pair of summands each isomorphic to S_{r}(λ), and for every pair of distinct blocks S_{r}(λ), S_{r}(μ) of ℚS_{r} there is a unique summand isomorphic to M_{2}(S_{r}(λ)⊗S_{r}(μ)). The order of λ and μ is immaterial; there is only one summand for the pair. Write λ < μ to indicate that λ precedes μ in the lexicographic order of partitions of n.

Lemma E. The rational group ring for the middle component of ℚD_{2r} is given by

ℚ(S_{r} ≀ C_{2}) = ⨁_{λ}\{S_{r}(λ) ⊕ S_{r}(λ)\} ⨁_{\begin{matrix}λ,μ \\ λ<μ \end{matrix}}\{M_{2}(ℚ) ⊗ S_{r}(λ) ⊗ S_{r}(μ)\} □

This gives the following expression for the rational group ring.

Theorem 14. The rational group ring of the Weyl groups of type D_{n} is given by

ℚD_{2r+1} = ⨁_{m=0}^{r} \{M\binom{2r+1}{m}(ℚ) ⊗ ℚS_{2r+1−m} ⊗ ℚS_{m}\}

ℚD_{2r} = ⨁_{m=0}^{r−1}\{M\binom{2r}{m}(ℚ) ⊗ ℚS_{2r−m} ⊗ ℚS_{m}\} ⨁\{M\binom{2r−1}{r}(ℚ) ⊗ ℚ(S_{r} ≀ C_{2})\},

where ℚ(S_{r} ≀ S_{r}) is given by Lemma E. □

From this one can see that the representations of D_{n} may be indexed by unordered pairs of representations of S_{m} and S_{n−m}. If n is odd then the order of the pair will be fixed by the requirement that 2m > n while if n = 2r is even then, as we have just seen, it is true both for m ≠ r and for m = r.

As mentioned at the beginning, now only six finite reflection groups remain for which a solution to the Donald–Flanigan problem is unknown. If the problem is solvable for all of them, one would like to know whether in all cases the Hecke algebra effectively gives a solution. While the Donald–Flanigan problem has been solved here for large classes of groups, the major question remains of what characterizes a group for which it is solvable. That, so far, is a mystery.

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Dept. of Mathematics, University of Pennsylvania, Philadelphia, PA 19104-6395 E-mail address: mgersten@math.upenn.edu

Dept. of Mathematics and Computer Science, Loyola University of Chicago, Chicago, IL 60626–5311

E-mail address: tonyg@math.luc.edu

Dept. of Mathematics and Computer Science, Bar Ilan University, Ramat-Gan 52900, Israel

E-mail address: mschaps@bimacs.cs.biu.ac.il