INJECTIVE RESOLUTIONS OF UNSTABLE MODULES Errokh Mohamed Jmoui Chariya P. Peterson July 9, 1993 abstract We characterize unstable modules over the Steenrod algebra in terms of their in* *jective resolutions. A Module of type n is an unstable module admitting an injective resolution whos* *e first n + 1 terms are reduced injective. The class of modules of type 2 contains the Dickson alge* *bras and various rings of invariants. 1 Introduction Let A denote the (mod 2) Steenrod algebra. The category of unstable modules U is abelian and has enough injectives. The injective unstable modules are divided i* *nto two classes: The reduced injective modules and the nilpotent injective modules.* * A reduced injective module is a direct sum of direct factors of polynomial algebr* *as. Injective modules which are not of this type are nilpotent. For any positive in* *teger n, an unstable module M is said to be of type n if M admits an injective resolu* *tion such that the first n + 1 terms are reduced injective. Modules of type n can be characterized by the vanishing of various derived functors in U. Let M be an un* *stable module, then M is of type n if and only if ExtiU(N; M) = 0 for 0 i n and for all nilpotent module N. An unstable module M is said to be nilpotent if the top squaring operation acts nilpotently on M. More precisely, for every element x there exist a positive in* *teger r such that (Sq0)rx = 0, where the operation Sqi : M ! M is the linear map Sqi(x) = Sq|x|-ix. M is said to be reduced if M has no nontrivial nilpotent uns* *table submodule, or equivalently, for any element x 2 M, Sq0x = 0 if and only if x = * *0. If furthermore, Sq1x = 0 if and only if x = 0 or x = Sq0y for some y 2 M, then M is said to be nilclosed. Modules of type 0 and 1 are precisely the reduced modules* * and the nilclosed modules respectively. In section 3, we give a complete characteri* *zation of modules of type n 2 in terms of Steenrod operations. 1 Theorem 1.1 An unstable module is of type 2 if and only if M is nilclosed and Ker(Sq2) is precisely the vector subspace spanned by (Sq20)M and Sq1M. Examples of modules of type 2 are given in section 4. Among them are the co- homology rings H*(BO(n); Z=2) and H*(BSO(2n - 1); Z=2) and various rings of invariants including the Dickson algebra. In our future work we intend to apply* * our classification of unstable modules to the computation of the MacLane cohomology in U=N il. We would like to thank S. Zarati for his interest and numerous correspondence on this problem, J. Lannes for the prove of Lemma 3.4 and the idea of using cohomo* *logy of finite groups in section 4, and also L. Schwartz for his suggestion and his * *example of an unstable module of type 1. The second author wish to thank, Nick Kuhn, Stewart Priddy and L. Smith for their valuable suggestions. The example of fami* *lies of unstable modules of type 1 is due to Nick Kuhn. She also would like to thank* * the Sonderforschungsbereichs Geometrie und Analysis 170, George Augus Universit"at and the Department of Mathematics at Northwestern University for their hospital* *ity. 2 Unstable modules of type n Definition 2.1 Let M be an unstable module. An element x of M is said to be nilpotent if (Sq0)r(x) = 0 for some positive integer r, where (Sq0)r = Sq0 . .S* *.q0, r times and where, for i 0, Sqi denotes the linear map Sqi : M ! M; Sqi(x) = Sq|x|-ix. M is said to be nilpotent if every element of M is nilpotent. M is sa* *id to be reduced if HomU(N; M) = 0 for every nilpotent module N and M is said to be nilclosed if ExtiU(N; M) = 0 for i = 0; 1 and for every nilpotent module N. In [5], lannes and Zarati showed that an unstable module M is reduced if and on* *ly if Sq0 : M ! M is an injective linear map. They also showed that M is nilclosed if* * and only if M is reduced and for any element x 6= 0, Sq1x = 0 if and only if x = Sq* *0y for some element y of M. In [3], Lannes and Schwartz proved that a reduced U-inject* *ive module is a direct sum of indecomposable direct factors of the mod 2 cohomology* * of an elementary abelian 2-group V , which will be denoted by H*(BV ) or sometimes P (V ) or F2[x1; : :;:xn] to emphasize the fact that it is a polynomial ring on* * V with a choice of basis x1; : :;:xn for V . 2 Definition 2.2 For any integer 0 n 1, an unstable A-module M is said to be of type n if there exists a U-injective resolution of M; 0 ! M ! I0 ! I1 ! . .!.Ii ! . .;. such that Ii is reduced U-injective for i n. Let Un denote the full subcategor* *y of U whose objects are of type n. Examples of modules of type infinity. (1) It is clear that any reduced U-injective is of type 1. (2) An example suggested by Lionel Schwartz is the following. The module L(2) h* *as a set of basis consisting of polynomials of the form uivj+ujvi for i 6= j and i* *+j 4. Let W denote an ideal in F2[x; y; z; t] generated by the monomial xyzt. W is cl* *early a direct factor of F2[x; y; z; t] hence it is reduced injective. We will const* *ruct a non-split surjective A-linear map OE : W ! L(2), the kernel of which is then, * *an unstable module of type 1 which is not injective. Let j : W ! F2[u; v] be a map defined by: j(f(x; y; z; t)) = f(u; u; u; v) + f(u; v; v; v). j is clearly A-l* *inear. The image of j is the unstable submodule M F2[u; v] spanned by elements of the form uivj+ ujvi = j(xjyi-j-1ztj) assuming i > j. j does not split, since the only el* *ement of degree 4 in W is xyzt, which cannot be in L(2), e.g. (Sq4 + Sq3Sq1)(xyzt) 6=* * 0 while (Sq4+Sq3Sq1)(u3v +uv3) = 0, where j(xyzt) = u3v +uv3. Furthermore, L(2) is a direct summand as unstable submodule of M. We therefore have a nonsplit surjective A-linear map OE : W ! L(2). This example serves as a counter example for the conjecture that any surjective A-linear map H*(V ) ! H*(W ) splits. (3) Recall that the reduced indecomposable injective modules are indexed by 2- regular partitions. Let I(A) the an indecomposable reduced injective module cor* *re- sponding to a 2-regular partition A, which is not an injective envelope of F (n* *) for any n. This happens when A is not a diagram of node n, denoted by (n). Let J(A) be the injective envelope of the projective cover of I(A). We have a A-linear * *epi- morphism J(A) i I(A) by the injectivity of I(A). J(A) is the sum of injectives * *of the form J(n). Since A 6= (n), there is no non-trivial monomorphism I(A) ! J(A) and hence the epimorphism does not split. Let M(A) be the kernel of this epi- morphism. For any partitions A1; . .;.As which are not of the form (n) for any * *n, M(A1) . . .M(As) is of type infinity. Using the results of [2], Nick Kuhn can show that such a module has injective dimension precisely s. 3 Proposition 2.3 For any unstable module M, the following statements are equiv- alent: (1) M is of type n, n 0. (2) ExtiU(N; M) = 0 for any N 2 N il and 0 i n. Proof: First assume M is of type n with minimal injective resolution I*. By d* *ef- inition, Ext iU(N; M) = Hi(Hom U(N; I*)). However, Hom U (N; Ii) = 0 for each 0 i n, since Ii is reduced. Conversely, for any nilpotent module N and any cosyzygy Cj of M, Ext iU(N; M) = Ext i-j-1U(N; Cj) = 0 for 0 j i n. Con- dition for nilclose property implies immediately that Ii is reduced for 0 i n. Corollary 2.4 Let 0 ! M0 ! M ! M00! 0 be a short exact sequence of unstable modules then: (1) If M0 and M00are of type n, then M is of type n. (2) If M is of type n then M0 is of type n if and only if M00is of type n - 1. Corollary 2.5 Let M 2 Um and N 2 Un, then M N 2 Uk where k = min (n; m). Proof: For any reduced injective L, L M is of type n. This follows easily from [5] that the tensor product of reduced injectives is again reduced injective an* *d the fact that tensor is an exact functor. Assume n m. We will prove by induction on n that M N is of type n. Since U1 is closed under tensor product, it is tr* *ue for n = 1. Assume true for n = i - 1. Let M be of type i, consider the short ex* *act sequence 0 ! M ! L ! M0 ! 0, with L reduced injective, which exists because M is nilclosed. By corollary 2.4 M0 is of type i - 1. Tensoring with N we have 0 ! M N ! L N ! M0 N ! 0. The second and the third terms are of type n and i - 1 respectively. Again by corollary 2.4 M N is of type i. An A-linear map f : M1 ! M2, Mi 2 U, i = 1; 2, is an F-isomorphism if kerf 2 N * *il and cokerf 2 N il. As a consequence of proposition 2.3 we have: 4 Corollary 2.6 Let f : M1 ! M2 be an F-isomorphism. Then the following proper- ties are equivalent. (1) M is of type n. (2) The induced maps fi : Ext iU(M2; M) ! Ext iU(M1; M) is bijective for i < * *n, and injective for i = n. Proof: Assume M 2 Un. Let N1, N2 and C be the kernel, cokernel and image of f respectively. The long exact sequence obtained from applying Hom U( ; M) to t* *he short exact sequence 0 ! N1 ! A ! C ! 0 is . .E.xtnU(C; M) ! Ext nU(A; M) ! Ext nU(N1; M) ! . . . N1 being nilpotent implies Ext iU(A; M) ~= Ext iU(C; M) for 0 i n. Similarly, using the short exact sequence 0 ! C ! B ! N2 ! 0 we have Ext iU(A; M) ~= Ext iU(C; M) ~= Ext iU(B; M) for 0 i < n, and Ext nU(B; M) ,! Ext nU(C; M) ~= Ext nU(A; M). Conversely, if N is nilpotent, the map f : 0 ! N is an F-isomorph* *ism. Hence Ext iU(N; M) = 0 for 0 i n. So be proposition 2.3 M is of type n. To give other characterization of Un, we need some notations and definitions. * *Let be the double functor which associates to M 2 U the unstable A-module M defined as follows: ( n_ 2; if n (mod 2) i i_ (M)n = M and Sq x = Sq 2x; x 2 M; 0 othersise, where a_2= 0 if a 6 1 (mod 2). We denote by : U ! U the suspension functor, by e, (resp. e ) the right adjoint of (resp. ) and by Rie (resp. Rie) the deri* *ved functor of e (resp. e). There is a natural exact sequence ([8]): 0 ! F (m + 1) ! F (m + 1) ! F (m) ! 0; (1) Proposition 2.7 Let M be an unstable A-module. Then for any n 1, the follow- ing statements are equivalent. (a) M 2 Un. (b) e(M) ~=M and Rie(M) = 0 for 1 i < n. (c) Rie(M) = 0 for every 0 i n. Proof: By definition, RieM and RieM in degree m are Ext iU(F (m); M) and Ext iU(F (m); M) respectively. Applying Hom U( ; M) to (1) we get: 0 ! eMm ! Mm+1 ! eMm+1 ! Ext 1U(F (m); M) ! 0; 5 and that Ext iU(F (m); M) ~=Ext i+1U(F (m); M). Hence (b) and (c) are equivalen* *t. The map in (1) is an F-isomorphism, hence by corollary (2.6.2), (a) ) (b). Now assume M satisfies (b). e is left exact and hence RieL = 0 for any injective un* *stable module L. Let K0 ! K1 ! K2 ! . .b.e a minimal injective resolution of M. Since M is nilclosed, K0 is reduced. Let Ci be the ith cozyzygy. Apply e to the sh* *ort exact sequences 0 ! Ci ! Ki ! Ci+1 ! 0, putting M = C0. We get, e C1 ~=C1 and RieC1 ~= Ri+1e M for i 1. Hence, by induction, Ci satisfies condition (b) with n replaced by n - i. Moreover, this implies that Ci, and hence Ki, is redu* *ced for i n. Remark: By definition, ExtiU(F (j); M) ~= Rie(M)j for all i; j 0. Hence M 2 Un if and only if for all j 0, Ext iU(F (j); M) = 0 for 0 < i < n and Hom iU(F (j); M) = Mj. For each k > 0, let F(k)* be the minimal projective res* *o- lution of F (k). Each F(k)a is a finite sum of F (i) for some i's, and each dif* *ferential is a matrix whose entries are elements of A. We can then, write Hom U(F(k)*; M) as a cocomplex of vector spaces, consisting of Mi, with the same differential, * *and R*e M is the cohomology of this cocomplex. Hence, in theory, Proposition 2.7 (b) provides a way to completely characterize Un in terms of Steenrod operations. T* *hat is, if one knows the first n terms of F(k)* for all k. However, this proved to * *be quite a difficult task. Nevertheless, if one knows F(k)* even for a single k, one may* * obtain a useful criterion for an unstable module to be of type n. Even for k = 1, whic* *h we have. Proposition 2.8 If M 2 Un, then M is Sq1 acyclic in degree n+1. In particula* *r, reduced injective modules are square 1 acyclic in each degree. Sq1 Sq1 Sq1 Proof: . . .! F (i) ! F (i - 1) ! . . .! F (1) ! Z=2 ! 0 is the minimal projective resolution. Applying Hom U ( ; M) to this resolution one gets a co* *m- Sq1 2 Sq1 plex of vector spaces, M* = M1 ! M ! . .:.Therefore, if M 2 Un then ExtiU(Z=2; M) = H*(M* ) = 0 for 0 i n by proposition (2.3.2). Examples: Let V be an F2 vectorspace of rank 2 and let H*(BV ) ~= F2[u; v] denote its F2 cohomology ring. The full ring of invariants is known to be, D(2)* * =: H*(BV )GL(V )~=F2[w2; w3], where w2 = u2+ uv + v2 and w3 = uv(u + v). D(2) fails to be Sq1-acyclic in degree 4, since w22is not in the image of Sq1 while Sq1w22* *= 0. Hence, D(2) cannot be in U3. Lather we will show that D(2) is precisely of type* * 2. 6 3 Steenrod operations on Modules of type n The type n property of unstable modules depends intimately on the Steenrod oper- ations. In this section we will show that, at least for small n, n = 0 ; 1 and* * 2, the type n property is encoded in the kernel of the linear map Sqi for 0 i n. Rec* *all that, a module M 2 U0 if and only if it is reduced. The following proposition * *is immediate. Proposition 3.1 An unstable module M is of type 0 if and only if for each x 2* * M, Sq0(x) = 0 if and only if x = 0. As shown in [3], every reduced injective module is a direct sum of direct facto* *rs of H*(BV ), hence by Cartan formula, any element in a reduced injective module is annihilated by Sq0 if and only if it is zero. So, U0 is in some sense a "Sq0 approximation" of U1 . For n = 1, it is clear that an unstable module M 2 U1 if* * and only if M is nilclosed. We have the following characterization of modules of ty* *pe 1, whose proof will be given in the sequel. Proposition 3.2 Compare with [1], [5] and [11 ]. An unstable module M is of t* *ype 1 if and only if M is reduced and for each x 2 M, Sq1(x) = 0 if and only if x =* * 0 or x = Sq0(y) for some y 2 M. As in the case of U0, using Cartan formula one can easily deduce that any eleme* *nt of a reduced injective module satisfies the condition in proposition (3.2). So, U1* * may be thought of as a "Sq1 approximation" of U1 . We wil now show that there is a similar characterization for U2. Definition 3.3 An A-linear map of unstable modules f : M ! N is said to be Sqn- closed if for any x 2 M, f(x) = Sqn(y) for some y 2 N if and only if x = Sqn(z) for some z 2 M. In particular, for any module M of type 1, the injective hull M ,! K is Sq0-clo* *sed. Before stating the main theorem of this section we need the following two lemmas suggested by Lannes. 7 Lemma 3.4 Let Hd = F2[u1; : :;:ud] denote the mod 2 cohomology of an element* *ary abelian 2-group of rank d. Then: _@__ @ Sqn = Sqn-1 ____ @ui @ui for 1 i d and n 0. Proof: Let x = uff11. .u.ffddbe a monomial in H, @Sqnx__ @ Xn ff ffd ffi = ____ Sqj(u1 1. .^.ui.u.d.)Sqn-jui @ui @uij=0 Xn ff ! @ = Sqj(uff11. .^.ui.u.f.fdd) i ____u2ffi-n+ji j=0 n - j @ui Xn ff ! = Sqj(uff11. .^.ui.u.f.fdd) i (n - j)u2ffi-n+j-1i j=0 n - j Xn ff - 1 ! = Sqj(uff11. .^.ui.u.f.fdd) i ffiu2ffi-n+j-1i j=0 ffi- (n - j) Xn = Sqj(uff11. .^.ui.u.f.fdd)Sqn-1-jffiuffi-1i j=0 @ ff ffd = Sqn-1(____)u1 1. .u.d; @ui where (uff11. .^.ui.u.f.fdd) denotes the monomial uff11. .u.ffi-1i-1uffi+1i+1.u* *.f.fdd. More gen- erally, we have the following formula, _____@k_____ @k Sql= Sql-k____________ @ui1. .@.uik @ui1. .@.uik The following formula for any monomial x 2 Hd is also useful: Xd @x ! 2 Sq1x = ui ____ : (2) i=1 @ui Lemma 3.5 Let K be a reduced injective module. For i = 0; 1; 2 consider the * *linear maps Sqi : K ! K. Then (i) Ker(Sq0) = 0; (ii) Ker(Sq1) = ; (iii) Ker(Sq2) = : 8 Proof: It suffices to prove the special case K = Hd. (i) follows from the Car* *tan relation that Sq0(x) = x2 hence Sq0(x) = 0 if and only if x = 0. (ii), If Sq1(x* *) = 0 then, from (2) _@x_@u= 0 for each i, hence x must be a square. (iii) First, ass* *ume x i is of even degree. Sq2x = 0 implies Sq1Sq2x = Sq1x = 0 hence by (ii) x = Sq0y f* *or some y. However, Sq2x = Sq2Sq0y = Sq0Sq1y = 0. Hence Sq1y = 0 and we have y = Sq0z for some z. So x = (Sq0)2z. On the other hand, if x is of odd degree. Consider the following equations: _@__ @x Sq2x = Sq1____= 0: (3) @ui @ui Apply (i) to the second equality we get _@x_@u= z2ifor some zi 2 Hd. Hence x = P d i 2 j=1ui(zi) : Compare this last result with the left hand side of (2), we see t* *hat (iii) is true if the vector "z= (z1; : :;:zd) is a gradient, i.e. zi = _@y_@ufor each* * 1 i d i and some fixed polynomial y. In mod 2 calculus, "zis a gradient implies both "r* *x "z and "r . z = 0. Conversely, if "r x "z= 0 and @zi_@u= 0 for 1 i d then "zis a i gradient. so we need to show ( @zj_ @zi_ for i 6= j, = @uj (4) @ui 0 otherwise. To show this, consider Xd X2 Sq2x = Sq2-juiSqj(zi)2 i=1j=0 Xd = u2i(Sq1zi)2 = 0: i=1 P d Hence i=1uiSq1zi = 0: Denote this sum by S. Apply formula (2) to S we get; Xd @z !2 S = uiuj __i_ j;i=1 @uj !2 !2 X @zi X @zi = ui____ + uiuj ____ = 0 (5) i @ui i