Essential cohomology for elementary abelian p-groups

Journal of Pure and Applied Algebra 213 (2009) 2238–2243 Contents lists available at ScienceDirect Journal of Pure and Applied Algebra journal homepage: www.elsevier.com/locate/jpaa Essential cohomology for elementary abelian p-groups F. Altunbulak Aksu a , D.J. Green b,∗ a Department of Mathematics, Bilkent University, Bilkent, 06800, Ankara, Turkey b Mathematical Institute, Friedrich-Schiller-Universität Jena, 07737 Jena, Germany article info Article history: Received 25 September 2008 Received in revised form 19 February 2009 Available online 17 May 2009 Communicated by M. Broué abstract For an odd prime p the cohomology ring of an elementary abelian p-group is polynomial tensor exterior. We show that the ideal of essential classes is the Steenrod closure of the class generating the top exterior power. As a module over the polynomial algebra, the essential ideal is free on the set of Mùi invariants. © 2009 Elsevier B.V. All rights reserved. MSC: Primary: 20J06 13A50 55S10 1. Introduction Let G be a finite group and k a field whose characteristic p divides the order of G. A cohomology class x ∈ H n (G, k) is called essential if its restriction ResH (x) is zero for every proper subgroup H of G. The essential classes form an ideal, called the essential ideal and denoted by Ess(G). It is standard that restriction to a Sylow p-subgroup of G is a split injection (see for example Theorem XII,10.1 of [1]), and so the essential ideal can only be non-zero if G is a p-group. Many p-groups have non-zero essential ideal, for instance the quaternion group of order eight. The essential ideal plays an important role and has therefore been the subject of many studies: two such being Carlson’s work on the depth of a cohomology ring [2], and the cohomological characterization due to Adem and Karagueuzian of those p-groups whose order p elements are all central [3]. The nature of the essential ideal depends crucially on whether or not the p-group G is elementary abelian. If G is not elementary abelian, then a celebrated result of Quillen (Theorem 7.1 of [4]) implies that Ess(G) is a nilpotent ideal. By contrast, the essential ideal of an elementary abelian p-group contains non-nilpotent classes. Work to date on the essential ideal has concentrated on the non-elementary abelian case. In this paper we give a complete treatment of the outstanding elementary abelian case. As we shall recall in the next section, the case p = 2 is straightforward and well known. So we shall concentrate on the case of an odd prime p. So let p be an odd prime and V a rank n elementary abelian p-group. We may equally well view V as an n-dimensional Fp -vector space. Recall that the cohomology ring has the form H ∗ (V , Fp ) ∼ = S (V ∗ ) ⊗Fp Λ(V ∗ ), (1) ∗ 1 2 where the exterior copy of the dual space V is H (V , Fp ), and the polynomial copy lies in H (V , Fp ): specifically, the polynomial copy is the image of the exterior copy under the Bockstein boundary map β . Our first result is as follows: Theorem 1.1. Let p be an odd prime and V a rank n elementary abelian p-group. Then the essential ideal Ess(V ) is the Steenrod closure of Λn (V ∗ ). That is, Ess(V ) is the smallest ideal in H ∗ (V , Fp ) which contains the one-dimensional space Λn (V ∗ ) ⊆ H n (V , Fp ) and is closed under the action of the Steenrod algebra. ∗ Corresponding author. E-mail addresses: [email protected] (F. Altunbulak Aksu), [email protected] (D.J. Green). 0022-4049/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.jpaa.2009.04.016 F. Altunbulak Aksu, D.J. Green / Journal of Pure and Applied Algebra 213 (2009) 2238–2243 2239 Our second result concerns the structure of Ess(V ) as a module over the polynomial subalgebra S (V ∗ ) of H ∗ (V , Fp ). It was conjectured by Carlson (Question 5.4 in [5]) – and earlier in a less precise form by Mùi [6] – that the essential ideal of an arbitrary p-group is free and finitely generated as a module over a certain polynomial subalgebra of the cohomology ring. In [7], the second author demonstrated finite generation, and for most p-groups of a given order was able to prove freeness as well: specifically the method works provided the group is not a direct product in which one factor is elementary abelian of rank at least two. Our second result states that Carlson’s conjecture holds for elementary abelian p-groups too, and gives explicit free generators. Theorem 1.2. Let p be an odd prime and V a rank n elementary abelian p-group. Then as a module over the polynomial part S (V ∗ ) of the cohomology ring H ∗ (V , Fp ), the essential ideal Ess(V ) is free on the set of Mùi invariants, as defined in Definition 3.3. Structure of the paper. In Section 2 we briefly cover the well-known case p = 2. We introduce the Mùi invariants in Section 3. After proving Theorem 1.2 in Section 4 we consider the action of the Steenrod algebra on the Mùi invariants in order to prove Theorem 1.1 in Section 5. 2. Elementary abelian p-groups and the case p = 2 The cohomology group H 1 (G, Fp ) may be identified with the set of group homomorphisms Hom(G, Fp ). This set is an Fp -vector space, and – assuming that G is a p-group – the maximal subgroups of G are in bijective correspondence with the one-dimensional subspaces: the maximal subgroup corresponding to α : G → Fp being ker(α). Of course, the cohomology class α ∈ H 1 (G, Fp ) has zero restriction to the maximal subgroup ker(α). Note that in order to determine Ess(G) it suffices to consider restrictions to maximal subgroups. Definition. Denote by Ln the polynomial Ln (X1 , . . . , Xn ) = det X1 p X1 .. . pn−1 X1 X2 p X2 .. . pn−1 X2 ··· ··· .. . ··· Xn Xnp .. . ∈ Fp [X1 , . . . , Xn ]. n −1 Xnp There is a well-known alternative description of Ln . Lemma 2.1. Ln is the product of all monic linear forms in X1 , . . . , Xn . So for an n-dimensional Fp -vector space V we may define Ln (V ) ∈ S (V ∗ ) up to a non-zero scalar multiple by Ln (V ) = Y (2) x. [x]∈PV ∗ Proof. First part: Here we call a linear form monic if the first non-zero coefficient is one. The right hand side divides the p p2 pn−1 left. Both sides have the same total degree. And the coefficient of X1 X2 X3 · · · Xn follows.  is +1 in both cases. The second part Let V be an elementary abelian 2-group. Then H ∗ (V , F2 ) ∼ = S (V ∗ ), where the dual space V ∗ is identified with H 1 (V , F2 ). 1 Pick x1 , . . . , xn to be a basis for H (V , F2 ). The following is well-known: Lemma 2.2. For an elementary abelian 2-group V , the essential ideal is the principal ideal in H ∗ (V , F2 ) generated by Ln (x1 , . . . , xn ). Moreover, Ess(V ) is the free S (V ∗ )-module on Ln (V ), and the Steenrod closure of this one generator. Proof. Ln (V ) is essential, because every non-zero linear form is a factor and every maximal subgroup is the kernel of a nonzero linear form. Now suppose that y is essential, and let x ∈ V ∗ be a non-zero linear form. Let U ⊆ V ∗ be a complement of the subspace spanned by x. So y = y′ x + y′′ with y′ ∈ S (V ∗ ) and y′′ ∈ S (U ). Hence ResH (y′′ ) = 0 for H = ker(x), as y is essential and ResH (x) = 0. But the map ResH : V ∗ → H ∗ satisfies ker(ResH ) ∩ U = 0, and so ResH is injective on S (U ). Hence y′′ = 0, and x divides y. By unique factorization in S (V ∗ ) it follows that Ln (V ) divides y. So Ess(V ) is the principal ideal generated by Ln (V ), and the free module on this one generator. Finally, the definition of the essential ideal means that it is closed under the action of the Steenrod algebra.  We finish off this section by recalling the action of the Steenrod algebra on the cohomology of an elementary abelian p-group in the case of an odd prime. So let p be an odd prime and V an elementary abelian p-group. Recall that the mod-pcohomology ring is the free graded commutative algebra H ∗ (V , F p ) ∼ = Fp [x1 , . . . , xn ] ⊗Fp Λ(a1 , . . . , an ), 2240 F. Altunbulak Aksu, D.J. Green / Journal of Pure and Applied Algebra 213 (2009) 2238–2243 where ai ∈ H 1 (V , Fp ), xi ∈ H 2 (V , Fp ), and n is the rank of V . That is, a1 , . . . , an is a basis of the exterior copy of V ∗ , and x1 , . . . , xn is a basis of the polynomial copy. The product a1 a2 · · · an ∈ H n (V , Fp ) is a basis of the top exterior power Λn (V ∗ ). p The Steenrod algebra A acts on the cohomology ring, making it an unstable A-algebra with β(ai ) = xi and P 1 (xi ) = xi . Observe that Ln (x1 , . . . , xn ) is essential, for the same reason as in the case p = 2. 3. The Mùi invariants Let k be a finite field and V a finite dimensional k-vector space. Consider the natural action of GL(V ) on V ∗ . The Dickson invariants generate the invariants for the induced action of GL(V ) on the polynomial algebra S (V ∗ ). But there is also an induced action on the polynomial tensor exterior algebra S (V ∗ ) ⊗k Λ(V ∗ ), and the Mùi invariants are SL(V )-invariants of this action: see Mùi’s original paper [8] as well as Crabb’s modern treatment [9]. We shall need several properties of the Mùi invariants. For the convenience of the reader, we rederive these from scratch: but see Mùi’s papers [8,10] and Sum’s work [11]. Notation. Often we shall work with the direct sum decomposition H ∗ (V , Fp ) = n M Nr (V ) , r =0 where n is the rank of V and we set Nr (V ) = S (V ∗ ) ⊗Fp Λr (V ∗ ) . Observe that restriction to each subgroup respects this decomposition. This means that the essential ideal is well-behaved with respect to this decomposition: Ess(V ) = n M Nr (V ) ∩ Ess(V ) . (3) r =0 Definition. Recall that Ln (x1 , . . . , xn ) is the determinant of the n × n-matrix  x1 x2  . C =  .. .. . pn−1 pn−1 x1 x2 xn ··· .. . ··· n−1 xnp ps−1  ..  . , where Cs,i = xi for 1 ≤ s ≤ n. For each such s, define E (s) to be the matrix obtained from C by deleting row s and then
prefixing a1 a2 · · · an as new first row: so det E (s) = n X (−1)i+1 γs,i ai , i=1 where γs,i is the determinant of the minor of C obtained by removing row s and column i. Now define the Mùi invariant Mn,s ∈ H ∗ (V , Fp ) by Mn,s = det E (s). Note that our indexing differs from Mùi’s: our Mn,s is his Mn,s−1 . a1 x1 Example. So M4,3 = p a2 x2 p a3 x3 p a4 x4 p x1 x2 x3 x4 p3 p3 x2 p3 x3 p3 x4 x1 and γ2,3 = x1 x2 p2 x1 p3 p2 x2 p3 x1 x2 x4 p2 x4 . p3 x4 Lemma 3.1. Mn,s ∈ N1 (V ) ∩ Ess(V ). Proof. By construction Mn,s ∈ N1 (V ). Restricting to a maximal subgroup of V involves killing a non-zero linear form on V ∗ : That is, one imposes a linear dependence on the ai and consequently the same linear dependency on the xi . So one obtains a linear dependency between the columns of E (s), meaning that restriction kills Mn,s = det E (s).  Lemma 3.2. Ess(V )2 = Ln (V ) · Ess(V ). F. Altunbulak Aksu, D.J. Green / Journal of Pure and Applied Algebra 213 (2009) 2238–2243 2241 Proof. As Ln (V ) is essential, the left hand side contains the right. Now let H be a maximal subgroup of V . Then H = ker(a) for some non-zero a ∈ H 1 (V , Fp ). Let x = β(a) ∈ H 2 . Observe that the kernel of restriction to H is generated by a, x. Suppose that f , g both lie in this kernel: then we may write f = f ′ a + f ′′ x, g = g ′ a + g ′′ x, and so fg = (f ′′ g ′ ± f ′ g ′′ )ax + f ′′ g ′′ x2 , that is fg = xh for h = (f ′′ g ′ ± f ′ g ′′ )a + f ′′ g ′′ x ∈ ker ResH . ∗ Since H ∗ (V , Fp ) is a free module over the unique factorization   ring S (V ), this means that fg = Ln (V ) · y for some y ∈ H ∗ (V , Fp ). So h = y ∈ Ess(V ). Ln (V ) x · y. As ResH (h) = 0 and ResH Ln (V ) x is a non-zero divisor, we deduce that ResH (y) = 0. So  Definition 3.3. Let S = {s1 , . . . , sr } ⊆ {1, . . . , n} be a subset with s1 < s2 < · · · < sr . In view of Lemmas 3.1 and 3.2 we may define the Mùi invariant Mn,S ∈ Nr (V ) ∩ Ess(V ) by Mn,S = 1 Ln (V )r −1 Mn,s1 Mn,s2 · · · Mn,sr . Note in particular that Mn,∅ = Ln (V ). Remark. Observe that Mn,S Mn,T =  ±Ln (V )Mn,S ∪T 0 if S ∩ T = ∅; otherwise. (4) 4. Joint annihilators In this section we study the joint annihilators of the Mn,S with |S | = r as a means to prove Theorem 1.2. Lemma 4.1. The joint annihilator of Mn,1 , . . . , Mn,n is Nn (V ). Proof. The element a1 . . . an is a basis for Λn (V ) and is clearly annihilated by each Mn,s . Conversely, suppose that y 6= 0 is annihilated by every Mn,s . As Mn,s Nr (V ) ⊆ Nr +1 (V ) we may assume without loss of generality that y ∈ Nr (V ) for some r. If r ≤ n − 1, then for some i we have 0 6= ai y ∈ Nr +1 (V ). So as ai y also lies in the joint annihilator, it will suffice by iteration to eliminate the case y ∈ Nn−1 (V ). Suppose therefore that 0 6= y ∈ Nn−1 (V ) lies in the joint annihilator. Denote by K the field of fractions of S (V ∗ ), and let W = K ⊗k Λn−1 (V ∗ ). Each Mn,s induces a linear form φs : W → K given by φs (w)a1 · · · an = Mn,s w . By assumption, y 6= 0 lies in the kernel of every φs . A basis for W consists of the elements a1 · · · abr · · · an for 1 ≤ r ≤ n, where the hat denotes omission. Now, Mn,s · a1 · · · abr · · · an = (−1)r +1 γs,r ar · a1 · · · abr · · · an , and so φs (a1 · · · abr · · · an ) = γs,r . Now consider the matrix Γ ∈ Mn (K ) given by Γs,r = γs,r . If one transposes and then multiplies the ith row by (−1)i and the jth column by (−1)j , then one obtains the adjugate matrix of C . As the determinant of C is Ln (V ) and in particular non-zero, it follows that det Γ 6= 0. So by construction of Γ , the φs form a basis of W ∗ . So their common kernel is zero, contradicting our assumption on y.  Corollary 4.2. The joint annihilator of {Mn,S : |S | = r } is L s≥n−r +1 Ns (V ). Proof. By induction on r, Lemma 4.1 being the case r = 1. As Mn,S ∈ N|S | (V L ) and Nr (V )Ns (V ) ⊆ Nr +s (V ), the annihilator is at least as large as claimed. Now suppose that y ∈ H ∗ (V , Fp ) does not lie in s≥n−r +1 Ns (V ). We may therefore write y= n X ys s=0 with ys ∈ Ns (V ), and we know that s0 ≤ n − r for s0 = min{s | ys 6= 0}. As ys0 6= 0 and ys0 6∈ NnL (V ), Lemma 4.1 tells us that ys0 Mn,t 6= 0 for some 1 ≤ t ≤ n. As ys0 Mn,t ∈ Ns0 +1 (V ), we conclude that yMn,t lies outside s≥n−r +2 Ns (V ). So the inductive hypothesis means that there is some T with |T | = r − 1 and yMn,t Mn,T 6= 0. So yMn,S 6= 0 for S = T ∪ {t } and |S | = r: Note that t ∈ T is impossible.  Corollary 4.3. Every Mn,S is non-zero. For S = n = {1, . . . , n} we have Mn,n is a non-zero scalar multiple of a1 a2 · · · an . 2242 F. Altunbulak Aksu, D.J. Green / Journal of Pure and Applied Algebra 213 (2009) 2238–2243 Proof. Observe that Mn,n is a scalar multiple of a1 · · · an for degree reasons. The case r = n of Corollary 4.2 says that 1 ∈ N0 (V ) does not annihilate Mn,n and therefore Mn,n 6= 0. But from Eq. (4) we see that every Mn,S divides Ln (V )Mn,n 6= 0.  Proof of Theorem 1.2. In view of Eq. (3) it suffices to show that for each r the Mùi invariants Mn,S with |S | = r are a basis of the S (V ∗ )-module Nr (V ) ∩ Ess(V ). We observed in Definition 3.3 that these Mn,S lie in this module. So suppose that y ∈ Nr (V ) ∩ Ess(V ). We should like there to be fS ∈ S (V ∗ ) such that y= X (5) fS Mn,S . |S |=r Note that for T = n − S we have Mn,S Mn,T = ±Ln (V )Mn,n by Eq. (4). Define εS ∈ {+1, −1} by Mn,S Mn,T = εS Ln (V )Mn,n . So Eq. (5) implies that we should define fS by fS Mn,n = 1 Ln (V ) εS yMn,T , since T ∩ S ′ 6= ∅ and therefore Mn,S ′ Mn,T = 0 for all S ′ 6= S with |S | = r. Note that this definition of fS makes sense, as yMn,T lies in both Nr (V )Nn−r (V ) = Nn (V ) and Ln (V ) Ess(V ), the latter inclusion coming from Lemma 3.2. With this definition of fS we have y− X fS Mn,S |S |=r ! Mn,T = 0 P P fS Mn,S by Corollary 4.2. for every |T | = n − r. As y − |S |=r fS Mn,S lies in Nr (V ), this means that y = |S |=rP Finally we show linear independence. Suppose that gS ∈ S (V ∗ ) are such that |S |=r gS Mn,S = 0. Pick one S and set T = n − S. Multiplying by Mn,T , we deduce that gS = 0.  5. The action of the Steenrod algebra To prepare for the proof of Theorem 1.1 we shall study the operation of the Steenrod algebra on the Mùi invariants. Lemma 5.1. β(Mn,s ) =  Ln (V ) 0 s=1 otherwise β(Ln (V )) = 0. (6) For 0 ≤ s ≤ n − 2 we have: ps P (Mn,r ) =  Mn,r −1 0 r =s+2 otherwise s P p (Ln (V )) = 0. (7) Proof. One sees Eq. (6) by inspecting the determinants in the definition of Mn,s and Ln (V ). The proof of Eq. (7) is also based ps on an inspection of these determinants. Recall that P m (ai ) = 0 for every m > 0, and that P m (xi ) is zero too except for s ps ps+1 P p (xi ) = xi . We may use the Cartan formula P m (xy) = X P a (x)P b (y) a+b=m ps to distribute P over the rows of the determinant. As ps cannot be expressed as a sum of distinct smaller powers of p, we s only have to consider summands where all of P p is applied to one row and the other rows are unchanged. This will result ps+1 in two rows being equal unless it is the row consisting of the xi that is missing. Lemma 5.2. Let S = {s1 , . . . , sr } with 1 ≤ s1 < s2 < · · · < sr ≤ n. 1. Suppose that 1 6∈ S. Then Mn,S = β(Mn,S ∪{1} ). 2. Ln (V )r −1 P m (Mn,S ) = P m (Mn,s1 · · · Mn,sr ) for each m < pn−1 . 3. For 2 ≤ u ≤ n set X = {s ∈ S | s ≤ u} and Y = {s ∈ S | s > u}. Then Ln (V )P p u−2 u−2 (Mn,S ) = P p n− 1 4. For 1 ≤ r ≤ n and 0 < m < p 5. For 2 ≤ u ≤ n one has P pu−2 (Mn,X ) · Mn,Y . one has P m (Mn,{1,...,r } ) = 0. (Mn,{1,...,u−2,u} ) = Mn,{1,...,u−1} .  2243 F. Altunbulak Aksu, D.J. Green / Journal of Pure and Applied Algebra 213 (2009) 2238–2243 Proof. Recall that Ln (V )r Mn,S = Ln (V )Mn,s1 · · · Mn,sr . (8) The first two parts follow by applying Eqs. (6) and (7). s Recall that by the Adem relations each P m may be expressed in terms of the P p with ps ≤ m. So the third part follows from the second, since we deduce from Eq. (7) that P m (Mn,s ) = 0 if 0 < m ≤ pu−2 and s > u. Fourth part: By induction on r. Follows for r = 1 from the Adem relations and Eq. (7). Inductive step: Enough to consider s P p for 0 ≤ s ≤ n − 2. By the inductive hypothesis and a similar argument to the third part, deduce that s s Ln (V )P p (Mn,{1,...,r } ) = Mn,{1,...,r −1} P p (Mn,r ). But this is zero by Eq. (7), since Mn,{1,...,r −1} Mn,r −1 = 0. Fifth part: Using the fourth part and an argument similar to the third, deduce that Ln (V )P p u−2 (Mn,{1,...,u−2,u} ) = Mn,{1,...,u−2} P p but this is Ln (V )Mn,{1,...,u−1} . u−2 (Mn,u ) = Mn,{1,...,u−2} Mn,u−1 :  Proof of Theorem 1.1. We shall show that for every Mn,S there is an element θ of the Steenrod algebra with Mn,S = θ(Mn,n ). We do this by decreasing induction on r = |S |. It is trivially true for r = n, so assume now that r < n. Amongst the S with |S | = r we shall proceed by induction over u, the smallest element of n − S. So S = {1, . . . , u − 1} ∪ Y with s > u for every s ∈ Y . Part 1 of Lemma 5.2 covers the case u = 1, so assume that u ≥ 2. Set T = {1, . . . , u − 2, u}. We complete the induction by u −2 showing that Mn,S = P p Ln (V )P p But P p u−2 u−2 (Mn,T ∪Y ). Part 3 of Lemma 5.2 gives us (Mn,T ∪Y ) = P p u−2 (Mn,T )Mn,Y . u −2 (Mn,T ) = Mn,{1,...,u−1} , by Part 5 of that lemma. So P p (Mn,T ∪Y ) = Mn,S , as claimed.  Remark. Theorem 1.2 shows that the S (V ∗ )-module generated by the Mùi invariants Mn,S is the essential ideal and therefore closed under the action of the Steenrod algebra. One may however see more directly that this S (V ∗ )-module is Steenrod closed. This is observed for example in [11]. In view of Lemma 5.2 and Eqs. (6) and (7) it only remains to show that Pp n−1 n −1 (Mn,s ) lies in our S (V ∗ )-module. Now P p (Mn,n ) = 0 by the unstable condition, so suppose s < n. Recall that Mn,s is pn−1 a determinant, the last row of the matrix having entries xi pn pr . So applying P p n−1 pn replaces these entries by xi . But it is well known that xi is an S (V ∗ )-linear combination of the xi for r ≤ n − 1, and that the coefficients are independent of i: this is the ‘‘fundamental equation’’ in the sense of [12], and the coefficients are the Dickson invariants cn,r in S (V ∗ ). Applying S (V ∗ )linearity of the determinant in the bottom row of the matrix, one deduces that P p of the Mn,r . n−1 (Mn,s ) is an S (V ∗ )-linear combination Acknowledgements The first author was supported by a Ph.D. research scholarship from the Scientific and Technical Research Council of Turkey (TÜBİTAK-BAYG). The first author wishes to express her gratitude to her research supervisor Prof. Ergün Yalçın for advice and guidance. Both authors thank him for suggesting this problem. 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