Chapter
Almost Split Sequences One of our goals here is t o examine the Green theory in the context of almost split sequences. We are in particular interested how the members of these sequences behave with respect t o the Green correspondence. The notion of an almost split sequence was introduced by Auslander and Reiten (1975). Since then this concept has been extensively used in the general theory of representations of finitedimensional algebras. The last decade witnessed the impact of these techniques on group representations. As an example, we mention a work of Auslander and Carlson (1986) who presented some applications of almost split sequences to the area of modular and integral group representations. As an application of their methods, a new proof of Webb’s theorem was presented. This theorem asserts that the heart of the projective cover of the trivial module is indecomposable except when p = 2 and the Sylow 2subgroup of G is dihedral or 22 x Zz. Special attention is drawn t o providing a detailed proof of the existence of almost split sequences in the context of group algebras over fields. Our presentation, based on an unpublished work of Knorr, does not use any sophisticated technique and avoids the general categorytheory arguments.
1
Preliminary results
Throughout, R denotes an arbitrary ring. We begin by proving the following “five lemma” so named because of the fiveterm exact sequences involved in its formulation. Lemma 1.1.
(The ‘Fve lemma”). Suppose that we are given the fol577
Almost Split Sequences
578
lowing commutative diugram of Rmodule homomorphisms in which the rows are exact :
A  Bf  C  gD  E
h
j
(i) If (Y is surjective and ,8,6are injective, then y is injective. (ii) If ,8,6 are surjective and E is injective, then y is surjective.
(iii) If every vertical arrow but y is an isomorphism, then y is also an isomorphism. Proof. (i) Assume that y(c) = 0 for some c E C. Then 0 = h‘T(c) = Sh(c). Because 6 is injective, c E X e r h = g ( B ) . Writing c = g ( b ) , we have
and so P ( b ) = f’(a’) for some a’ E A‘. Writing a’ = .(a) yields
which forces f(a) = b and c = g(b) = g f ( a ) = 0. (ii) Choose any c’ E C’. Then h’(c’) = S(d) for some d E
D,and
which forces j ( d ) = 0. Hence d = h(c) for some c E C. Now h’(J  ~ ( c )=) h‘(c’)  h ’ 7 ( ~= ) 6(d)  6h(c) = 0 and so c‘  y ( c ) = g‘(b’) for some b’ E B‘. But b’ = P ( b ) for some b E B , so
proving that c’ E y(C). (iii) Apply (i) and (ii).
1 Preliminary results
579
Recall that an Rmodule V if E n d R ( V ) is a local ring. Corollary 1.2.
#
0 is said to be strongly indecomposable
Suppose we have the commutative diagram of Rmodule
homomorphisms :
0  K  v  W  0
f
9
where h ’ is strongly indecomposable and the rows are nonsplit. Then cp and .IC, are isomorphisms. Proof. By Lemma 1.1, it suffices to show that cp is an isomorphism. Replacing Ir‘ by f ( K ) , if necessary, we may assume that f is the inclusion map. Assume by way of contradiction that cp E J ( E n d R ( K ) ) . Then 1  p is a unit of E n d R ( K ) and also
( 1  $)V
Kerg = f(K) = K
But (1  tp)’(l  $)f = l ~so,f is a split injection, a contradiction. With these preliminaries settled, we now introduce and investigate the following import ant notion. An almost split sequence is a nonsplit short exact sequence of Rmodules : O + h ’ +fV % W+O such that the following properties hold : (i) K , W are finitely generated strongly indecomposable Rmodules. (ii) Every Rhomomorphism ,L? : U + W , which is not a split surjection, factors through g , i.e. p = g y for some y E H o ~ R ( U , V ) . Note that W cannot be projective and K cannot be injective, for then the sequence would split. Observe also that if p in (ii) is a split surjection, i.e. PP’ = l w for some ,L?‘ E H o m ~W, ( U ) , then p cannot factor through g .
Almost Split Sequences
580
Indeed, otherwise, l w = g(yP’) for some y E H o r n ~ ( UV, )which contradicts the assumption that the sequence does not split. It should be emphasized that the Rmodule U appearing in (ii) need not be finitely generated. However, if R is an artin algebra (i.e. R is a finitely generated Z(R)module and Z ( R ) is artinian), then condition (ii) can be replaced by (iii) For any finitely generated Rmodule U and for any Rhomomorphism p : U + W which is not a split surjection, p = g y for some y E H o m ~U( ,V ) (see Auslander and Reiten (1975, Theorem 4.2)). We shall avoid the use of this result in our future treatment of group algebras of finite groups over fields by declaring (i) and (iii) as the defining properties of almost split sequences. Of course, in this context “strongly indecomposable” appearing in (i) is the same as “indecomposable” . Turning to uniqueness of almost split sequences, we now prove
Proposition 1.3. Any almost split sequence f 0 + Ii +
v 4 PV + 0
ending in W is unique in the following sense : if
f’ O+K’+V
I
+9’ w + o
is another almost split sequence ending in W , then there exist isomorphisms
x : K’ + K , p : v’
4
v
which render commutative the following diagram :
0K‘V’W0
f’
g’
Proof. We may clearly assume that both f and f’ are inclusion maps. By definition, there is an Rhomomorphism p : V’ + V with g‘ = g p .
1 Preliminary results
581
Let P’ be the restriction of P t o Ii’. Since 0 = g’(I<’) = g(P’(K’)), we see that ,@(I[’) E K . Likewise, there is an Rhomomorphism y : V , V’ with g = g’y and 7 restricts to an Rhomomorphism y’ : K + K’.Letting 1c, = Pr and p = P’y‘, it follows from Corollary 1.2 that 9 is an isomorphism, so 7’is a split injection. But K’ is indecomposable, so y’ is an isomorphism, implying that y is an isomorphism by the “ five lemma”. Setting X = (7’)’ and p = y‘, the result follows. We next demonstrate that the notion of an almost split sequence is selfd u d in the following sense. Proposition 1.4. A nonsplit short exact sequence of Rmodules :
( I i ,W are finitely generated and strongly indecomposable) is an almost split sequence if and only if any Rhomomorphism a : li + K’,which is not a split injection, factors through f, i.e. a = a‘f for some a’ E H o m R ( V ,I[’). Proof. Suppose that the given sequence is almost split. Let
Q
: K ,
I<‘ be any Rhomomorphism which is not a split injection. Consider the following commutative diagram : 0  I <  v  w  0
0

I<‘
f
9
s
Y,Pw0
where P is the pushout, i.e. P = (1;’ $ V ) / A where A = {(a(x), f(x)Ix E I<}, and y, /3 are induced by the canonical injections K‘ K‘ $ V , V + I<’ @ V . The composite f
K‘$
v z, v 9 w
the canonical projection) has kernel Ii’$ li 2 A , and thus induces a surjection S : P + W with kernel y(K’), thereby yielding the above diagram. It is clear that y is injective. (T is
Almost Split Sequences
582
If 6 is split, then so is y, say y’y = 1 h . r for some y‘ E HomR(P,Ii’). Setting a‘ = y’P, we have a‘ E HomR(V,K‘) and a’ f = y’(P f ) = (y’y)a = a Assume that 6 is not split. Then, by definition, 6 = gh for some h E H o r n ~ ( P , v )Hence . we may complete a diagram :
0  K  V  W  0
0  K ‘  P  W  0
f
9
7
6
Applying Corollary 1.2 (with cp = 80, 11 = h p ) , it follows that 8a is an isomorphism. But then a is a split injection, a contradiction. The converse follows by the dual argument and the resuIt is established.
Our next result provides a partial justification of the name “almost split sequence”.
K +f V 5 W 3 0 is an almost Proposition 1.5. Suppose that 0 split sequence. (i) If W’ is a proper submodule of W , then the exact sequence f
0
f
K
+
g’(w’)+ w’
$
0
splits. (ii) If K’ is a nonzero submodule of K , then the exact sequence 0 + h’/K‘
splits.
+
V / f ( K ’ )+ W
+
0
1 Preliminary results
583
Proof. (i) Let W‘ be a proper submodule of W . Then the inclusion map i : W’ 4 W is not surjective. Hence i is not a split surjection. Since 0 + K + V + W + 0 is a n almost split sequence, there is a homomorphism h : W’ + V such that gh = i. The fact that Imh is contained in g’(W’) implies that the exact sequence
0+K
+ g’(W’) +
W’ + 0
splits, as required. (ii) Assume that K’ is a nonzero submodule of K . Then the natural homomorphism n : K + K/K’ is not injective. Hence n is not a split injection. Therefore, by Proposition 1.4, ?r = h f for some homomorphism h : V + K/K’. Because n(K’) = 0, we know that h induces a homomorphism h‘ : V /f(K’) + K/K’ with the property that the composition
K/K’+ v/f (K’) + K/K’ is the identity map. This implies the desired assertion and the result is established. Note that properties (i) and (ii) in the above proposition can be satisfied by nonsplit exact sequences which are not necessarily almost split.
K V 5 W + 0 is an almost split sequence of Rmodules. If U is an Rmodule which has no direct summand isomorphic to W , then the induced sequence Proposition 1.6. Suppose that 0
+
+
0 + IlOmR(U,K ) + HOmR( u,v ) + HOmR( u,
w)
+
0
is exact. Proof. By Proposition 3.4.8 in Vol.1, the above sequence is exact, except possibly a t HomR(U,W). Let ,!? : U + W be any Rhomomorphism. Then ,!? cannot be a split surjection, by hypothesis on U . Hence ,!? = g y for some y E HomR(U,V). This shows that 7 is mapped t o ,!? and the result follows.
Next we prove the following general property of almost split sequences. Proposition 1.7. Suppose that 0 + K + V 5 W + 0 is a n almost split sequence of Rmodules and let E = E n d R ( W ) . Then the induced
Almost Split Sequences
584
sequence O
+
H o ~ R ( W , K ) H o ~ R ( W , V2) E +
+
E/J(E)
+
0
is exact.
Proof. By Proposition 3.4.8 in Vol.1, it suffices to prove that Img, = J ( E )
By definition, for any a E H o m ~ ( W , v )g,(a) , = g o . If ga is an isomorphism, then g is a split surjection, which is impossible. Thus ga is a nonunit of E . Since E is a local ring, we deduce that g a E J ( E ) . Thus Img, C J(E). Conversely, assume that p E J ( E ) . Then ,B : W + W is not a split surjection, since otherwise p is an isomorphism. Hence L,3 = g y for some y E Hom&V,V), i.e. p = g,(7). It follows that J ( E ) 2 Img,, as we wished to show. We close by establishing the duals of the last two results. f V + W + 0 is an alProposition 1.8. Suppose that 0 + K + most split sequence of Rmodules. If U is an Rmodule which has no direct summand isomorphic to K , then the induced sequence
i s exact.
Proof. By Proposition 3.4.5 in Vol.1, the above sequence is exact, except possibly at H o ~ R ( K , U ) Let . @ : K + U be any Rhomomorphism. Then p cannot be a split injection, by hypothesis on U . Hence, by Proposition 1.4, p = P’f for some p’ E H o ~ R ( VU ,) . This shows that p’ is mapped to p and the result follows. f, V + W + 0 is an alProposition 1.9. Suppose that 0 3 K most split sequence of Rmodules and let E = EndR(K). Then the induced sequence 0
+
HomR(W,I ( )
+
HomR(V,K) 5: E
+ E / J ( E ) +
0
2 Projective homomorphisms and the trace map
585
is exact. Proof. By Proposition 3.4.5 in Vol.1 it suffices to verify that Imf*= J ( E ) By definition, for any a E H o m ~ ( V , h ' )f*(a) , = of. If af is a n isomorphism, then f is a split injection, which is impossible. Thus af is a nonunit of E . Since E is a local ring, we deduce that af E J ( E ) . Hence Imf' E J ( E ) . Conversely, suppose that P E J ( E ) . Then P : K K is not a split injection, since otherwise /3 is an isomorphism. Hence, by Proposition 1.4, f*(P') = P. It follows that /3 = P'f for some P' E H o r n ~ ( v , K )i.e. , J ( E ) E I m f * , as required. f
2
Projective homomorphisms and the trace map
This and the next section are based on an unpublished work of Knorr. Our aim is to prove a general result (Theorem 2.6) which will easily imply the existence of almost split sequences for group algebras over fields. Throughout, G denotes a finite group and F an arbitrary field of characteristic p > 0. All FGmodules are assumed to be finitely generated. For any FGmodule U , we write t r = tru : E n d F ( U ) + F for the ordinary trace map. Given FGmodules U and V ,we write H o r n ~ ~ , l (VU ),for the Fsubspace of H o ~ F G ( UV, ) consisting of all projective FGhomomorphisms from U to V . We also put
Horn&'( U ,
v)= H o m F G ( U , V)/HomFG,l(U,v )
Of course, if U = V , then HOmFG,l(U,V) = E n d ~ ~ , l (isv an ) ideal of E n d ~ c ( Vand ) so Hom&(U, V ) = E n d i G ( V )is an Falgebra. We remind the reader that, by Theorem 6.10.15 (with X = (1)) HomFG,l(U,V ) = Tr,G(HomF(U,V ) ) where Trf : Ilom(U,V) + H o m ~ c ( U , vis) a relative trace map. To emphasize the pair U, V of FGmodules, we shall write T(U,V )instead of T r f . Again, assume that U and V are FGmodules. Then we put
I ( U , V ) = {f E HomFG(U,V)ltr(Xf)= 0 for all X E KerT(V,U)}
Almost Split Sequences
586
We are now ready to record the following preliminary result.
Lemma 2.1. For any FGmodules U and V ,
Proof. Put T = T(U,V)and S = T(V,U ) . Given f E HomFG,1(U,V), write f = T ( p ) for some p E H o ~ F ( U , V )Let . X E HomF(V,U) be such that S(X) = 0. Then
WWPN
=
c c
tr [(XgPkJI]
SEG
=
WglAgCL)
SEG
WWCL) = o =
(since S(X) = 0)
Thus HomFG,l(U,V) I(U,V). To prove the opposite inclusion we put
kl = dimFKer T ( U , V ) and k2
Note that
= dimFKer T(V,U )
< X,p >= t r ( X p ) defines a nonsingular bilinear form
<, >: HomF(V,U ) x HomF(U,V ) + F and that by definition
It therefore follows that
n  k1 = dimF(U,V ) dimFKer T (U , V ) = dimFIm T(U,V ) < dimFI(U,V) (by the first paragraph)
2 Projective homomorphisms and the trace map
587
< dimF [ K e r T(V ,U)]' = nka since
<, > is nonsingular.
as required.
By symmetry, equality holds and, in particular,
rn
Given an FGmodule U , we put
Lemma 2.2.
A given FGmodule U is projective if and only if K e r T(U, U ) C UO
Proof. We know that U is projective if and only if EndFG,I(U) = EndFG(U). Hence, by Lemma 2.1, U is projective if and only if I ( U , U ) = E ~ F G ( U )Since . I ( U , U ) is an ideal of EndFG(U), the latter is equivalent to 1.y f I ( U , U ) , i.e. t o 0 = 2r(X ' l u ) = tr(X) for all X E IierT(U, U ) , i.e. to K e r T ( U , U ) 2 UO. In what follows, we write R for the IIeller operator (see Chapter 7 of Vol.1). Thus, for any FGmodule U , R(U) is the kernel of the projective cover P ( U ) U of U . Since F G is a symmetric algebra, R(U) contains no nonzero projective direct summands (see Proposition 7.1.1 in Vol.1). Now let U be an FGmodule and let the exact sequence f
represent a projective cover of U (hence W 2 P ( U ) , V Z R ( U ) and /3 is an essential epimorphism). We also fix X E HOrnF(W,V), p E liOrnF(U,W) with @p = l u and Xcu = l v . We put
T = TrF where
Trf : EndF(W)
+
EndFG(W)
Almost Split Sequences
588
is the relative trace map. This situation will be fixed for the rest of
this section. Since W is a projective module, T is surjective. For any given y E EndFG(W),choose 6 E E n d F ( W ) with T(6)= y. Define the map
{
EndFG(W) a
2 F H
tr(6)
Owing t o Lemma 2.2, this map is welldefined and is clearly Flinear. Next we introduce another Flinear map HOrnFG(U,V)
7

4, F tO(WP)
Using the above map, we now provide the following description of projective homomorphisms pertaining t o our special choice of U and V .
Lemma 2.3. For any FGmodule Ii, we have : (i) HOmFG,l(U, K ) =
(ii) HornFG,l(K,V ) =
Proof. (i) By Lemma 2.1, H o m ~ c , l ( U , K = ) I ( U , I i ) . Assume that p E H o ~ F G , ~ ( U ,and K ) let 7 E H o m ~ ~ ( l i , V )Since . W is projective, T T ? ( ~=) /3 for some f E Hom&V,U). Define r = fay : K + U . Then
T r P ( r ) = T T P ( f ) a y = pay = 0 But p E I(U, K ) , so t r ( r p ) = 0 and thus
K) t ( y p ) = 0 for all y E Conversely, suppose p E H ~ ~ F G ( U , satisfies H o r n ~ ~ ( K , v Let ) . r : K + U with T r p ( r ) = 0. We must show that
2 Projective homomorphisms and the trace map
589
tr(rp) = 0. To this end, observe that pr : K W , so 77 = T r f ( p r ) : W and ~ 7 =7 Trf;(PpT)= T r f ( T >= o +
Therefore y = (Y’T : K
+
+
V is welldefined. By assumption,
as required. (ii) This follows by applying the same argument as in (i).
Lemma 2.4. For any p E EndFG(U), there exist cp‘ E EndFG(W), p” E EndFG(V) which render commutative the following diagram :
ovw
ovw
a
P
u  0
(Y
P
u  0
More0 ver, (i) p is projective if and only if 9’’is projective. (ii) The map End;,(U) + E n d J G ( V )induced by p H p” is an isomorphism of Falgebras. (iii) For any 0 E H o ~ F G ( U , V )t (, @ ) = t(p”0). Proof. By Theorem 7.4.13 in Vol.1, it suffices to prove (iii). Let f E E n d F ( W ) be such that T r f (f ) = p’. Then
t(e9) = = = = as required.
t o ( a e p p )= to(aepp’) tr(aePj)= tr(faep) to(p’aep) = t o ( a p v P ) t(9rIe),
Almost Split Sequences
590
Given a vector space M over F, we put M' = H o r n p ( M , F ) . Note that if M is an (A,B)bimodule for Falgebras A and R , then M' is a ( B , A ) bimodule via
( b e B , ~ ME, f E M * ) (a E A,m E M , f E M * )
( b f > ( m= ) f(mV (fa)(m) = f ( a m >
In what follows, we write f for the image off E H O ~ F GU(, A') in
Lemma 2.5. hold : (i) The map
U , A'),
For any FGmodules K and U , the following properties
[,I : HOm;,(U,
I<) x
EIO~;G(I<,
n(u))
4
F
given by [7,6]= t(6y), 7 E H o ~ F G ( U , K 6) ,E Ho~FG(K,R(U)) is a welldefined nonsingular bilinear form. 4 ( H o m i G ( U ,I<))* given by (ii) The map $ : HomiG(K,R(U))
+(m = [%jl
is an (End;,(U), End$G(lr'))bimoduleisomorphism. (iii) The map 8 : H o ~ + ~ ( U , I + C ) (Hom;,(A',Q(U)))*
given by
e(r)($) = [%SI is an (End&( K ) , End;G(U))bimodule isomorphism. (iV) H o m ; ~ ( u , K )2 EXt;G(K, n2(u))*U S (End;G(Ir'), End$G(U))bimodules. (v) EndiG(U) EztiG(U,R2(U))* as ( E n d i G ( U ) ,EndiG(U))bimodules. In partucdar, if U is a nonprojective indecomposable FG module, then
E 4 G ( U ,O 2 ( U > ) has a simple socle when viewed as a left or right End$G(U)niodu/e. Proof. (i) This is a direct consequence of Lemma 2.3. (ii) It is clear that $ is an Fisomorphism. Put V = R(U). We know that Hom$,(U, K ) is an ( E n d i G ( K ) EndJG(U))bimodule. , Hence U , K)) is an ( E n d i G ( U ) ,EndiG(K))bimodule. Again, H o m i G ( K ,V ) is an ( E n d & ( V ) , End;G(I()  bimodule
2 Projective homomorphisms and the trace map
591
Moreover, by Lemma 2.4, E n d J G ( U ) Z E n d J G ( V )and we can use this isomorphism, denoted by w to define an ( E n d i G ( U ) ,EndjG(li))bimodule structure on H o m J G ( ~V ,) . Now let y E H O m F G ( U , I<), 6 E H O ~ F G ( K , Q E E n d ~ ~ ( l and <) y E E ~ ~ FU G ) . Then (
v),
= 1% Siil = t(b7)
[email protected])(Y) and
[email protected])?(Y)
which shows that
= +(S)(iiY) = [iiY, 61 = t(617Y)
+(67) = +(6)V,
i.e.
+ is a homomorphism
E n d ; ~ (K)modules. Next we note that, by definition, (pd = w(p)6 = yII6. Hence
+ ( p m = +(Cp"J)(Y) = [Y, 7781 = t(cp"6y) while
proving that ?I, is a homomorphism of left End&( U)modules. (iii) Apply the same argument as in (ii). (iv) We first note that, by Proposition 7.4.6 in Vol.1,
EztiG(lr',Q2(u)) = lfomjG(fl(K), nz((v)) It follows that
of right
Almost Split Sequences
592
2 HOmjG(
u,Ir')
(by Theorem 7.4.13 in Vol.1) as required. (v) The desired isomorphism follows by applying (jv) with K = U . Assume that U is nonprojective and indecomposable. Then EndFG(U) is a local ring and J(EndFG(U))is the unique maximal (left or right) ideal of EndFG(U). Note also that E n d j G ( U ) # 0 since is nonprojective. It follows that as a left or right EndjG(U)module, E n d j G ( U ) has a unique maximal submodule. Since EztjG(U,R*(U))is isomorphic t o the dual of E n d j G ( U ) ,we see that the socle of E z t i G ( U , f 1 2 ( U )is ) simple.
u
We now establish the following result which is of independent interest and which will imply existence of almost split sequences. Theorem 2.6. Let U be a nonprojective indecomposable FGmodule. Then there exists a nonprojective FGhomomorphism T : U t R ( U ) such that, for every FGmodule 'h and every nonsplit FGepimorphism q : I< + U , the homomorphism T q : Ir' + R(U)
is projective. Proof. Since U is not projective, E n d J G ( U ) # 0. Rut U is indecomposable, so E = EndJG(U)is a local ring. Choose 0 # f E E* such that f ( J ( E ) ) = 0. Then, by Lemma 2.5(ii) (with K = U ) , there exists T E H o ~ F G ( U , R ( U )such ) that f = $(?). Clearly T is not projective, since otherwise 5 = 0 and f = 0. Now choose any nonsplit FGepimorphism q : A' + U . If we assume that p E HomFG(U,lr'), then qp E EndFG(u) and in fact qp E J(EndFG(U)). Hence 0 = =
f(GP) = ?")(GP) [ G P , 51 = t ( W )
This holds for all p E HomFG(U, I < ) , so by Lemma 2.3(ii) we conclude that T q E HomFG,l(I
3 Existence and introductory properties
3
593
Existence and introductory properties
Auslander and Reiten (1975) proved the existence of almost split sequences in the context of artin algebras, i.e. those algebras A for which Z(A) is artinian and A is a finitely generated Z(A)module. The corresponding proof in this generality is rather involved. However, in the special case of group algebras over fields, one can provide a very transparent proof. Since our interest lies exclusively in group algebras, we shall be content to concentrate on these algebras. We shall also record a number of introductory properties of alomst split sequences. Throughout, G denotes a finite group and F an arbitrary field of characteristic p > 0. As before, all FGmodules are assumed to be finitely generated. The proof of the following result provides a rather explicit contruction of almost split sequences for the group algebra FG. It turns out that the existence of such sequences is closely related to projective homomorphisms. In fact, our construction proceeds as follows. We fix a nonprojective indecomposable FGmodule U and consider the projective cover of Q ( U ) :
0 + R2( U ) 3 P(R( U ) ) 4R( U ) + 0 We then choose any nonprojective FGhomomorphism T : U + n ( U ) such that, for every FGmodule K and every nonsplit FGepimorphism 11 : Ii' + U , the homomorphism rq : Ir' + R(U) is projective (see Theorem 2.6). Finally, we form the pullback E of ( n , r ) :
and demonstrate that
is an almost split seqeunce (here q ( x ) = ( p ( z ) , O ) , + ( Z , Y ) = Y). All the relevant details are contained in the proof of the following result. Theorem 3.1. Let U be a nonprojective indecomposable FGmodule. Then there exists a unique, up to isomorphism of short exact sequences, almost split sequence O+L+E+U+O
Moreover, L
E R2(U).
Almost Split Sequences
594
Proof. The uniqueness is a special case of a more general result, namely Proposition 1.3. Hence it suffices to exhibit an almost split sequence 0 + 0 2 ( U )i E
+
U 0
To this end, consider the following projective covers : 0
4
n(u)5 P ( U ) B u 4
0 + n"u> .% P(R(U)) (hence 0 2 ( U ) is indecomposable). Let and consider the pullback of ( T , T ) :
T
+
0
1R(U)
f
0
: U + O ( U ) be as in Theorem 2.6,
i.e. E = ( ( 5 , ~ E) P ( R ( U ) )CB U ( T ( Z )= T(Y)}, T I ( Z , Y ) = 2, $(Z,Y) = Y. Note that $ is surjective since 7r is. Define cp : 0 2 ( U ) + E by cp(x) = ( p ( z ) , O ) . Then the diagram
commutes and the rows are exact. Moreover, the bottom row is nonsplit, since otherwise there exists u E H o r n ~ ~ ( u , with E ) $0 = 1 ~but ; then ~ ( ~ 1 0=)T $ U = T , i.e. T factors through a projective FGmodule P ( Q ( U ) ) , a contradiction. Finally, if x : K 4 U is not a split surjection, then T X : K R(U) f
3 Existence and introductory properties
595
is projective by Theorem 2.6. Hence there exists y E H o ~ F G ( K , P ( R ( U ) ) with T X = ny. Since E is the pullback, it follows from Proposition 3.7.3 in Vol.1 that there exists S E HOWZFG(K, E ) with $6 = x. This shows that the bottom row of the diagram is an almost split sequence,
Remark 3.2. (a) Assume that R is a complete discrete valuation ring of chamcteristic 0. Then almost split sequences exist in the category of RGlattices. Moreover, given an almost split sequence of RGlattices
O+L+E+U+O we have L 2 R(U), (see Auslander (1978), Roggenkamp and Schmidt (1976), and Roggenkamp (1 977)). Thus the constructions of almost split sequences for FGmodules and RGmodules are very diflerent. (6) It should be pointed out that the above result can be considerably generalized. Namely, assume that R is a complete noetherian local ring and A is an Ralgebra which is a finitely generated Rmodule. If U is a finitely presented Amodule which is indecomposable and nonpmjective, then there is a n almost split sequence ending in U and whose initial term is an artinian module (see Auslander (1986, Theorem 5)). Of course, here in the definition of a n almost sequence, we delete the requirement that both initial and ending terms are finitely generated. Let P be a projective indecomposable FGmodule with P # Soc(P). Then S o c ( P ) is simple and so Soc(P) J ( P ) = J ( F G ) P , since J ( P ) # 0. We put H ( P )= J ( P ) / S o c ( P ) and refer to H ( P ) as the heart of P. We next provide an example of an almost split sequence involving the heart of P.
P
Example 3.3. For every projective indecomposable FGmodule P with Soc(P) there is an almost split sequence :
#
0 4 J ( P )+ P @ H ( P )+ P / S o c ( P )+ 0
Up to isomorphism, this is the only almost split sequence having P as a direct summand of the middle term. Proof. Setting U = P/Soc(P), we have
u 2 Rl(Soc(P))
Almost Split Sequences
596
by virtue of Theorem 15.5.1 in Vol.1. Since P # Soc(P),we see that U is a nonprojective indecomposable FGmodule. Moreover,
R(U) z [email protected]) 2 P / J ( P )2 U / J ( U ) which implies that
Now identify R(U) with P / J ( P ) and iet T : U + P / J ( P ) be as in Theorem 2.6. Let 0 : U P / J ( P ) be induced by e!t natural homomorphism P + P / J ( P ) . Then, by (l),T = 48 for some automorphism II, : P / J ( P ) + P / J ( P ) . Hence 0 satisfies the same property as T and so we may choose = e. Our identification of R(U) with P / J ( P ) leads to the identification of P/J(P)be the natural homomorphism so P(R(U)) with P. Let T : P that we have an exact sequence : f
0 + R2(U) + P
4P / J ( P ) + 0
Then the pullback E of (a,0 ) satisfies :
Thus E is the direct sum of the submodules :
El = {(2,2 and
EZ = ((0,
t
+
SOC(P))la:
E P}
=P
+ Soc(P))JzE K e r T } E H ( P )
Since E is the middle term of an almost split sequence ending in P / S o c ( P ) and begining at R2(U) 2 J ( P ) ,the first assertion is established. The second assertion is a consequence of a more general result to be proved in the next section (see Proposition 4 4 i ) ) . H Corollary 3.4. Let V be a simple nonprojective FGmodule and let P = P ( V ) . Then there is an almost split sequence : 0
f
R(V)
+
H ( P ) €3 P
+ R'(V)
f
0
3 Existence and introductory properties
597
U p to isomorphism, this is the only almost split sequence having P as a diwct summand of the middle term. Proof. By Theorem 15.5.1 in Vol.1, Soc(P) S V . Hence P is a projective indecomposable FGmodule with P # Soc(P). As we have seen in the proof of Example 3.3, P/ Soc(P) 2 n  l ( s o c ( P ) )5?
R'(v)
and so
s22(P/soc(P))Ei Q ( V ) Hence the desired conclusion follows by Example 3.3. Recall that, for any FGmodule V , the factor module V / J ( V ) is said to be the head of V . The next observation, contained in Erdmann (1988) provides circumstances under which the socle and the head of the middle term of a n almost split sequence can be computed in terms of the initial and ending members of that sequence.
Proposition 3.5. FGmodule and let
Assume that U is a nonprojective indecomposable 0 + S22(U) .% v
P f
u
3
0
be an almost split sequence ending in U . (i) If U is not simple, then Soc(V) 2 Soc(U) ed soc(s22(u)) (ii) If n 2 ( U ) is not simple, then
V / J ( V )2 U / J ( U ) ed n2(u)/J(n2(u)) Proof. (i) Since U is not simple, the inclusion map 1c, : S o c ( U ) 3 U is not surjective. Hence 1c, is not a split surjection and so $ = /?A for some X E H o r n ~ c ( S o c ( U ) , V )Since . X and a are injective i t suffices to show that Soc( V ) = A( Soc( U ) ) gj a( Soc( Q2( U ) ) ) It is clear that the submodules in the righthand side have zero intersection. If w E Soc(V), then p ( v ) E Soc(U) and so p(w) = 1c,(p(w))= pXp(v). Hence
w  x ( p ( ~ ) )E
a(n2(u))n SOC(V)
= soc(a(n2(u))) = a(Soc(s22(U))),
Almost Split Sequences
598
as required. (ii) Apply the dual argument. H Another very useful approach to the proof of the existence of almost split sequence is as follows. Given a nonprojective indecomposable FGmodule U , put E = End$,(U), i.e. E is the factor algebra of EndFG(U) with respect to the ideal of all projective endomorphisms of U . Then, by Lemma 2.5(v),
E 4 G ( U ,Q 2 ( W may be viewed as a (left or right) Emodule and as such it has a simple ~ ( Uthe )) socle, say S. Of course, S is also the socle of E x ~ $ ~ ( U , ~ ~when latter is viewed as an EndFG(U)module. Moreover, the exact sequence 0 + Qyu)+
v
t
u +0
is almost split if and only if it generates S. This last statement can be easily verified by elementary arguments. We close by recording the following two results due to Auslander and Reiten (1975).
Theorem 3.6. Let U be a nonprojective indecomposable FGmodule. Then (i) End;,(U) is a division algebrn if and only if E n d J G ( Q 2 ( u ) )is a division algebra. (ii) If E n d j G ( U ) is a division algebra, then any nonsplit sequence 0+
Qyu) v u +
+
+
0
is a n almost split sequence. (iii) If U or Q 2 ( U ) is simple, then any nonsplit sequence 0 + Q"U)
*
v u +0 +
is an almost split sequence.
Proof. (i) It is a consequence of Theorem 7.4.13 in Vol.1 that
End:G(u) 2 EndJG(Q2(U))
as Falgebras
This obviously implies the required assertion. (ii) Assume that E n d i G ( U )is a division algebra. Then Ezt&(U, Q 2 ( U ) )
3 Existence and introductory properties
599
is a onedimensional vector space over End&( U ) since it has a simple socle. Thus any nonsplit extension generates Ext&(U, Rz( U ) ) and so any nonsplit extension is almost split. (iii) Apply (i) and (ii). W We now examine almost split sequences begining with a simple module.
Theorem 3.7. Let U be a nonprojective indecomposable FGmodule and let P 0 0 " U ) 5 + u + 0
v
+
be an almost split sequence. Then the following conditions are equivalent : (i) R2(U) is simple. (ii) If 0 R(U) P(U) U + 0 is a projective cover for U and J = J ( F G ) R ( U ) ,then f
f
f
0
f
R(U)/J
+
P(U)/J
+
U
+
0
is an almost split sequence. (iii) /3 : V + U is an essential epimorphism. Proof. (i) sequence
j (ii)
0
: Since $
P ( U ) + U is an essential epimorphism, the
R(U)/J
+
P(U)/J + u + o
does not split. Since F G is a symmetric algebra, we also know that
by virtue of Proposition 15.5.3(ii) in Vol.1. Since R2(U) is simple, we have
0"U)
Ei
R(U)/J
It then follows from Theorem 3.6(iii) that
0
f
R(U)/J
+
P(U)/J + u
f
0
is an almost split sequence. (iii) : Taking into account that (ii)
0 + R(U)/J
+
P(U)/J
+
u
+
0
Almost Split Sequences
600
and
0
f
nyu) + v
P
+
u
+
0
are both almost split sequences, it follows from Theorem 3.1 that these sequences are isomorphic. The fact that P ( U ) / J + U is an essential epimorphism implies that /3 : V U is an essential epimorphism. (iii) =+ (i) : Suppose that f12(U) is not simple. Then Proposition 1.5(ii) implies that the sequence +
0 + s22(U)/J(FG)s22(U)+ V/J(FG)V
+
U/J(FG)U
+
0
is exact. Because Q 2 ( U ) / J ( F G ) R 2 ( U # ) 0, this shows that ,L? : V not an essential epimorphism.
f
U is
The following result is contained in Gabriel (1980, Corollary 3.4).
Corollary 3.8. Let U be a nonprojectiue indecomposable FGmodule and let P 0 + Q 2 ( U ) v + u 0
z
f
be an almost split sequence. Then U is simple if and only if. is an essential monomorphism. Proof. Apply Theorem 3.7 and Lemma 4.1 (in the next section). H We shall pursue our study of general properties of almost split sequences in the next section.
4
More properties of almost split sequences
Throughout, G denotes a finite group and F an arbitrary field of characteristic p > 0. All FGmodules below are assumed to be finitely generated. Given an FGmodule V, we write V' for the contragredient of V . If U , V are FGmodules and f : U + V is an FGhomomorphism, then the dual homomorphism f * : V' + U* is defined by f*(cr) = cr o f
for all
Lemma 4.1. If 0 + U f V 5 W FGmodules, then so is the dual sequence +
0+
f
(Y
E V*
0 is an almost sequence of
w*< v*I: u* + 0
4 More properties of almost split sequences
601
Proof. It is clear that both W* and U* are indecomposable and that the dual sequence is nonsplit. Let p : Ii + U* be an FGhomomorphism which is not a split surjection. Then p* : U + K* is not a split injection (here we identify U** with U via the canonical isomorphism p : U + U** given by p(u)(X) = X(u) for all u E U , X E U*). Hence, by Proposition 1.4, p* = p’f for some p’ E Hom&V, K * ) . The desired conclusion now follows by taking dual homomorphisms of both sides. H Let U , V be FGmodules. An FGhomomorphism f : U + V is said to be irreducible , provided f is neither a split injection nor a split surjection, and whenever f = f’ o f“ is a factorization of f, then either f’ is a split surjection or f ” is a split injection. Note that if both U and V are indecomposable, then “f is neither a split injection nor a split surjection” is equivalent to “f is not an isomorphism”. The notion of an irreducible homomorphism is closely related to that of almost split sequences. This is illustrated by the following result.
Proposition 4.2. Let V be a nonprojective indecomposable FGmodule and let P 0 + f12(V) 5 + +0
w v
be an almost split sequence. Then for any indecomposable FGmodule U , we have (i) A n FGhomomorphism f : U V is irreducible if and only iff = Py for some split injection y : U + W . (ii) A n FGhomomorphism f : f12(V) 4 U is irreducible if and only if f = 6a for some split surjection 6 : W + U . (iii) There exists an irreducible homomorphism U + V if and only if there exists an irreducible homomorphism f12(V) + U (both conditions are equivalent to the requirement that U is a component of W ) . +
Proof. Observe that (iii) follows from (i) and (ii). We shall establish
(ii), which will imply the dual statement (i) by applying Lemma 4.1. Assume that f : f12(V) + U is irreducible. Since, by hypothesis, f is not a split injection, f = 6a for some 6 E H o r n ~ ~ (Uw) ., Now a is not a split injection, so 6 must be a split surjection by the definition of irreducible homomorphisms. Conversely, let f = Sa for some split surjection 6 : W + U . Since a is not a split injection, f is not an isomorphism. Suppose that f = hg for
Almost Split Sequences
602
some h E H o ~ F G ( K U ,) and g E H o ~ F G ( Q ~ ( K V ) ., Assume that g is not a split injection. We must show that h is a split surjection. To this end, write g = g'cu for some g' E HomFG(W,K). Then
6a = f = hg = hg'a
+
and therefore hg' = 6 f'p for some f' E H o ~ F G ( V , U ) However, . 5 is a split surjection, so l u = 68 for some 8 E HomFG(U,W ) . Therefore
hg'8 = l u
+ f'p8
On the other hand, f ' p 8 is not an automorphism of the indecomposable FGmodule U . Hence hg'8 is an automorphism of U , since EndFc(U) is local. But then h is a split surjection and the result follows. W Next we record the following elementary result concerning irreducible homomorphisms.
Lemma 4.3. Let U , V be FGmodules and let f : U + V be an irreducible homomorphism. Then either f is surjective with K e r f indecomposable, or f is injective with V /f ( U ) indecomposable.
Proof. The factorization of f :
U % U/Kerf k V shows that f is either an injection or surjection. Assume that f is a surjection. To prove that K e r f is indecomposable, suppose by way of contradiction that K e r f = U1 @ U.Lis a nontrivial decomposition. Then f factors as follows :
u 5 U/UI 1:U / ( U l @ U2)
v
Because X is a surjection, it is not a split injection. Hence p is a split surjection, i.e. p has a right inverse p : V + U / U l . Similarly, we obtain a right inverse p' : V + U/U2 for the homomorphism p' : U/U2 + V . But U is the pullback of p and p', so p and p' determine a homomorphism V + U which is a right inverse to f,a contradiction. The dual argument produces the corresponding proof in case f is an injection. W We now apply two preceding results to establish some further properties of almost split sequences.
4 More properties of almost split sequences
603
Proposition 4.4. Let V be a nonprojective indecomposable FGmodule and let 0 n2(v) + W t v 0 $
$
be an almost split sequence. Then (i) If P is a projective indecomposable FGmodule which is a component of W , then V 2 P / S o c ( P ) . (ii) If u is an indecomposable component of then d i m F U # dimFV and dimFU # dimFn2(V ) .
w,
Proof. (i) By Proposition 4.2(iii), there exists an irreducible homomorphism f : P + V . If f is an injection then (since P is projective, hence
injective) f splits, which is a contradiction. Hence, by Lemma 4.3, f is surjective with K e r f indecomposable (in fact, we need only K e r f # 0). Since S o c ( P ) is simple, we have Soc(P) K e r f . Thus f factors as follows :
P
$
P/Soc( P ) + v
Since f is irreducible, we deduce that P/Soc( P ) 2 V . (ii) By Proposition 4.2(iii), there exist irreducible homomorphisms
n2(v)% u, u * v $
+
By Lemma 4.3, cp and are either surjective or injective. Moreover, neither cp nor is a n isomorphism, which implies the required assertion.
+
Our next result (contained in Erdmann (1986)) ties together vertices and irreducible maps.
Proposition 4.5. Let U , V be indecomposable FGmodules with vertices Q and P , respectively. Assvme that there exists an irreducible homomorphism f:U+V Then either Q is Gconjugate to a subgroup of P , or P is Gconjugate to a subgroup of Q .
Proof. We know that there is a factorization :
Almost Split Sequences
604
Because f is irreducible, it follows that either p is a split surjection or p is a split injection. In the first case, V is Qprojective and therefore P is Gconjugate to a subgroup of Q. In the second case, U is a component of ( V Q ) ~Hence . VQ has an indecomposable direct summand whose vertex is Q. Suppose that S is a Psource of V. Considering the Mackey decomposition of (SG),gives that Q C gPg'
n Q for some
gE
G
as desired. We next record the following general property due to Gabriel and Riedtmann (1979) (see also Benson (1991, p.138)).
Let 0 t @(V) + W t V + 0 be an almost split sequence of FGmodules and let H be a subgroup of G. Then Proposition 4.6.
0 + 6?'(V)H
$
WH
+
VH
+
0
is a split exact sequence if and only if I1 does not contain a vertex of V Proof. The sequence splits on restriction to H if and only if for any FHmodule U , the sequence
0 4 HOmFH(U,R2(V)H) i HOmFH(U,WH)
f
HOmFH(U,Vff)
f
0
is exact. Owing to Proposition 18.1.6 in Vol.1, this happens if and only if the sequence
o
+
H O ~ F G ( U ~R, ~ ( v ) )+ H O ~ F G ( U W ~,
) + H O ~ F G ( U ~V),
o
+
is exact. By the defining property of almost split sequences, the above happens if and only if V is not a component of U G for any FHmodule U . So the proposition is true. Some further general properties of almost split sequences are given by the following two theorems contained in Auslander and Reiten (1975). P
Theorem 4.7. Let 0 + A 5 B + C of FGmodules. Then (i) Given any nonsplit exact sequence
+
0 be an almost split sequence
O  + X ~ Y ~ C + O
4 More properties of almost split sequences
605
and a commutative diagram 0  A  B  C  0 a
P
Y
6
0  x  y  c  0
there is a commutative diagram 0  x  Y  c  0
Y
6
such that h f = 1~ and j g = 1 ~In. particular, f and g are split injections. (ii) Given any nonsplit exact sequence
and a commutative diagram
0  A  B  C  0 a there is a commutative diagram
P
Almost Split Sequences
606
0  A  B  C  0 a
P
such that g j = lc and f h = 1 ~ I n. particular, g and f are split surjections. Proof. (i) We first show that f : A + X is a split injection. Assume the contrary. Then there exists a homomorphism X : B + X with f = Xa. But then the sequence
splits since the diagram is a pushout diagram for
a
A
W B
f X This contradiction shows that f : A + X is a split injection. . z be the element Choose a homomorphism h : X + A with h f = 1 ~ Let
O+A+B+C+O in Est&(C,A), and let g be the element
O+XtY+C+O
4 More properties of almost split sequences
607
This implies that there is a commutative diagram
0  A  B  C  0 such that hf = 1~and j g = l g , as required. (ii) This is the dual of (i). We now investigate the special case of almost split sequences
in which the middle term B is projective. P C + 0 be an almost split sequence Theorem 4.8. Let 0 + A 2 B + of FGmodules. Then the following conditions are equivalent : (i) B is projective. P (ii) B , C is a projective cover. (iii) If a homomorphism X : X f C is not a split surjection, then X is projective.
Proof. (i) e (ii) : It is clear that (ii) implies (i). Assume that (i) holds. Since A 2 K e r p is indecomposable and p : B + C is a surjective homomorphism, it follows from Proposition 7.1.2 in Vol.1 that (ii) holds. (ii) (iii) : Assume that (ii) holds. If X : X + C is not a split surjection, then X = P7 for some 7 E H o r n ~ ~ (Bx) ., Since B is projective, it follows that X is a projective homomorphism, which implies (iii). Suppose that (iii) holds. Since p : B + C is not a split surjection, p is projective. Hence there is a commutative diagram
Almost Split Sequences
608
0  A  B  C  0
where Y is projective. Replacing Y by a projective cover of C, if necessary, we may assume that the bottom row does not split. Hence, by Theorem 4.7(i), g is a split injection. Since Y is projective, it follows that B is projective, as required. H
5
Induction of almost split sequences
In what follows, G denotes a finite group and F an arbitrary field of characteristic p > 0. All FGmodules are assumed to be finitely generated. The problem that motivates this section can be explained as follows. Suppose that H is a subgroup of G and V is a nonprojective indecomposable FHmodule. Then the corresponding almost split sequence 0 f R"V)
s w +P v
+
0
yields the short exact sequence
Now assume that U is a nonprojective indecomposable FGmodule such that U is a component of V G and let 0
f
n2(u)
$
I<
f
u +0
be an almost split sequence ending in U . What are the connections between the last two sequences ? It is the aim of this section to provide an answer upon certain further restrictions on U and V. We begin by demonstrating that the Heller operators R", n 2 1, commute with induction and restriction if we ignore projective direct summands.
Lemma 5.1. Let H be a subgroup of G, let V be an FHmodule and let U be an FGmodule. Then, for any given integer n 2 1, there exist a
5 Induction of almost split sequences
609
projective FGmodule X and a projective FHmodule Y such that : (i) O"(V)G E n y v G ) @
(ii)
nn(u)H
x.
On(UH)@ Y .
Proof. (i) The exact sequence 0 + n(v) + P ( V ) + v
+0
yields an exact sequence
Since P(V)G is projective, it follows from Proposition 7.1.2 in Vol.1 that
for some projective FGmodule W . This proves the case n = 1. Assume that n > 1 and that the result is true for n  1. Applying the case n = 1 t o O"l(V) instead of V, we have
for some projective FGmodule X . On the other hand, by hypothesis,
for some projective FGmodule X'. Hence
n(n"'(v)G) E s2"pG) This proves the general case, by comparing (1) and (2). (ii) The exact sequence
0
+ R(U)
+
P(U)+ u
f
0
yields the exact sequence
Since P ( U ) H is projective, it follows from Proposition 7.1.2 in Vol.1 that
Almost Split Sequences
610
for some projective FHmodule W . This proves the case n = 1. The general case now follows by induction as in the proof of (i).
Corollary 5.2. Let H be a subgroup of G, let V be an FHmodule and let U be an FGmodule. Assume that n 2 1 is an integer, W is a nonprojective indecomposable FGmodule and L a nonprojective indecomposable FHmodule. Then (i) The multiplicity of W in V Gis both equal to the multiplicity of Rn(W ) in Rn(V)G and the multiplicity of R"(W) in Rn(VG). (ii) The multiplicity of L in UH is both equal to the multiplicity of R"(L) in R n ( U ) ~and the multiplicity of Rn(L) in R n ( U ~ ) . Proof. (i) We may write V G E mlU1 @ @ m r U r , where UI = W,U2,. .. ,Ur are nonisomorphic indecomposable FGmodules and each m; 2 0 is an integer. Then
Since 171 is nonprojective and indecomposable, it follows from Theorem 7.3.5 in Vol.1 that ml is equal to the multiplicity of the nonprojective indecomposable FGmodule R"(U1) in Rn(VG). Hence the desired conclusion follows by applying Lemma 5.1(i). (ii) This follows from Lemma 5.l(ii) and the above argument. Let R be an arbitrary ring. Given two short exact sequences of Rmodules
Ea : 0 + xi
3 y, 4 2;
+
0
( i = 1,2)
their direct sum El @ E2 is defined to be the short exact sequence
where (91 @ g 2 ) ( ~ 1 , ~ 2=) ( g l ( Y l ) , g 2 ( y 2 ) ) for all y i E Y17 y z E Y2 ( f i @ f 2 is defined in a similar fashoin). Assume that U is a nonprojective indecomposable FGmodule. Then we denote the almost split sequence ending in U by A ( U ) . If H is a subgroup of G, V is a nonprojective indecomposable FHmodule and
A ( V ) : 0 + R2(V) 2
w+ P v
f
0
5 Induction of almost split sequences
611
then the induced short exact sequence A(V)' is defined to be
The following result can be found in Erdmann (1987, Lemma 1.5) and Green (1985, Theorem 7.8); its special case where H 2 N G ( P ) ,P a vertex of V , was proved by Benson and Parker (1984).
Theorem 5.3. Let H be a subgroup of G, let U be a nonprojective indecomposable FGmodule and let V be a nonprojective indecomposable FHmodule. Suppose that V is a component of ( V G )with ~ multiplicity 1 and that U is a component of V G such that V is a component of U H . Then
A(v)2 ~ A ( U ) @ E for some split sequence E Proof. Assume that
is an almost split sequence ending in V . We claim that : (a) The multiplicity of 0 2 ( V )in ( s 2 z ( V ) Gis) ~ equal to 1. (b) s22(V)G2 R2(U)@Xand @ ( V ) is not a component of X H , for some FGmodule X. Indeed, applying Corollary 5.2(ii) (with L = V , U = V G and n = a), we see that the multiplicity of Q 2 ( V in ) R 2 ( V G ) is ~ equal to 1. Since Q 2 ( V )is a nonprojective indecomposable FHmodule and, by Lemma 5.1(i), (s22(V)G)H
= R2(VG)H @
x'
for some projective FHmodule X',it follows that (a) holds. Since U is a component of V G ,it follows from Lemma 5.l(i) that Q 2 ( U ) is a component of R2(V)G,i.e.
s22(V)G2 R2(U) @ X
(for some FGmodule X )
On the other hand, since V is a component of U H , Lemma 5.l(ii) tells us that R2(U)? ~ Q 2 ( V )@ Y (for some FHmodule Y ) Applying (a), it follows that Q 2 ( V is ) not a component of
X H ,proving (b).
Almost Split Sequences
612
Next we introduce the following notation. If A = B @ C for some F G modules A , B and C, then
~ B : B + A and
pg:A+B are the inclusion and projection maps, respectively, so that p D is an FHmodule, then
~ =p1 ~~If.
~ D : D ~  + D
and
xD : D + D~
are the canonical FHhomomorphisms obtained via identification of D with [email protected] We now show that ( c ) Any FGhomomorphism II, : A + V G ,such that nv$ is not a split surjection, factors through 1 8 P : W G V G . Since A ( V ) is an almost split sequence and nv$ : A + V is not a split surjection, there exists an FHhomomorphism q : A + W with nv$ = ,817. Then Xwq : A W Gis an FHhomomorphism of FGmodules such that f
+
proving ( c ) . Now write V G= U @ M and put A = M . Then ( c ) applies t o II,= p ~ . Indeed, if T V ~ is M a split surjection, then V is a component of M H and ( V G ) , = UH @ M H . Since V is a component of U H , it would then follow that the multiplicity of V in ( V G )is~a t least two, a contradiction. It now follows from (c) that M splits off. Similarly, by the dual version of (c) and, by (a) and (b), the homomorphism II,= px : f12(V)G + X factors through 1 @ a,that is X splits off. By the foregoing, there is a commutative diagram with exact rows :
5 Induction of almost split sequences
0 
 W  V  PO
cr
fi2W
613
c
h
P
0

P
P
T
A
P 1
 u 2  u P1 o
a1
Ul
PlJ
PU Y
Let $ : N + U be an FGhomomorphism which is not a split surjection. To complete the proof, we must verify that $ factors through P I . This will
follow from (c), provided we show that (d) nv(pu$) is not a split surjection. Assume by way of contradiction that there exists an FHhomomorphism 7 : V t N such that w ( P U $ ) r l = 1v (3) We shall derive a contradiction to Lemma 8.3.2 (which is obviously valid for fields). Namely, we show that if u = $ T ~ E ( ~ , v r v )then pl~, (e) 0 E J ( E n d F G ( U ) ) ( f ) if 6 = (.VPU).($7), then 6 # J @ d F H ( V ) ) To this end, note that (e) is true since otherwise II, would split. To prove (f), let w E V and let
~U(~7(.)) =
c
t 8 wt
(.t
EV)
t€T
where T is a left transversal for H in G with 1 E T . Owing to (3), the map 6 takes an element w E V to 2,
t
c
t€T{1}
nv%whwt) E v
It follows that 6  1 factorizes through the module
Almost Split Sequences
614
Thus S  1 E J(EndFH(V)),which forces
as desired. H Corollary 5.4. Let H be a normal subgroup of G and let V be a nonprojective indecomposable FH module such that the inertia group of V is H .
If 0
+P
( V )5 w
% v
f
0
is an almost split sequence ending in V , then
is an almost split sequence ending in V G . Proof. By Theorem 10.6.4 in Vo1.3, V Gis indecomposable. Since the inertia group of n 2 ( V )is also H , we see that s22(V)Gis also indecomposable. Hence, by Lemma 5.1(i),
!I2( V G ) 2 s22( V ) G It is clear that V is a component of ( V G )with ~ multiplicity 1. Hence the desired conclusion follows by Theorem 5.3 (with U = V G ) . The requirement that V is a component of ( V c ) with ~ multiplicity 1 contained in Theorem 5.3 is rather restrictive. It certainly holds in the context of the Green correspondence. Our next result shows that it also holds under less restrictive conditions described below. Proposition 5.5. (Green 1985)). Let H be a subgroup of G and let V be an indecomposable FHmodule with vertex P and a Psource S . If H contains the inertia group of S in N G ( P ) , then V is a component of ( V G ) ~ with multiplicity 1. Proof. Let L be the inertia group of S in N G ( P ) ,i.e.
L = ( 9 E N G ( P ) I ~2S S }
5 Induction of almost split sequences
615
Let T be a full set of double coset representatives for ( H , H ) in G. Then, by Mackey decomposition,
Therefore if the proposition is false, then there exists t E T  H such that V is a component of W H ,where W = ( t V ) ~K , = tHt' n H . But S is a component of Vp because S is a source of V (see Theorem 6.2.11). Thus S is a component of ( w H >[email protected] [ ( " ~ > , , r 1 n P I
P
with z running a full set of double coset representatives for ( P , K ) in H . Hence there exists z E H such that S is a component of [ ( z W ) z K z  ~ n p ] p . Because P is the vertex of S , we must have z K z  ' n P = P , i.e. P zKz'. Thus S is a component of ( " W ) p 2 ("")p
On the other hand, V is a component of S H , since S is a source of V . Hence XtV is a component of ("tS)M,where M = xtHt'z'. Thus S is a component of
with z running over a full set of double coset representatives for (ztPt'z', P ) in M . Hence there exists z E M such that S is a component of
But P is the vertex of S , so zztPt'z'z' n P = P , i.e. zztPt'z'z' = P , and S is a component of zztS. This can happen only if S 2 z"tS, which means that z z t E L 5 H . Therefore z t ( z t )  ' z ( z t ) E H . But z E M = ztHt'z' implies ( z t )  ' z ( z t ) E H . Since z E H , it follows that t E H , a contradict ion.
Corollary 5.6. Let H be a subgroup of G and let V be a nonprojective indecomposable FHmodule with vertex P and a Psource S . Suppose that H contains the inertia group of S in N G ( P ) . Choose any indecomposable component U of V G such that V is a component of U H . Then U is a nonpmjective indecomposable FGmodule such that
A( V ) G2 A( U ) @ E for some split sequence E
Almost Split Sequences
616
Proof. If U is projective, then so is UH which is impossible since V is a component of UH and V is nonprojective. Thus U is a nonprojective indecomposable FHmodule. Moreover, by Proposition 5.5, V is a component of ( V G )with ~ multiplicity 1. Hence the desired conclusion follows by virtue of Theorem 5.3.
6
The Green correspondence and almost split sequences
In what follows, we fix a finite group G and an arbitrary field F of characteristic p > 0. We also fix a nontrivial psubgroup P of G and denote by H any subgroup of G with H 2 N G ( P ) . Next we introduce the following sets participating in the Green correspondence : X = {gPgl n Plg E G  H } Y = {gPg' n H l g E G  H } For any nonzero FGmodule U , we write f ( U ) for the Yprojectivefree part of U H . Thus, if U is an indecomposable FGmodule which is Pprojective but not Xprojective, then f ( U ) is the Green correspondent of U with respect to ( G , P , H ) . Our assumption that P # 1 guarantees that U (and hence f ( U ) ) is nonprojective. All the details concerning the Green correspondence can be found in Sec.2 of Chapter 7. Now assume that U is an indecomposable FGmodule which is Pprojective but not Xprojective, and put V = f ( U ) . Our aim is t o relate the middle terms of the almost split sequences A(U) : 0 n 2 ( U )3 M A U 0 f
f
A ( V ) : 0 ,0 2 ( V )3 N k V + 0 ending in U and V , respectively. We remind the reader that the induced short exact sequence A(V)Gis defined to be
Lemma 6.1. With the notation above, we have (i) A(V)GE A(U)@ E for some split sequence E . (ii) N G M @ L for some Xprojective FGmodule L. (iii) If Q1 is a vertex of an indecomposable component of M and Q is a vertex of U , then either Q1 is Gconjugate to a subgroup of Q , or Q is Gconjugate to a subgroup of Q1.
6 The Green correspondence and almost split sequences
617
Proof. (i) By Theorem 7.2.1, V is a component of ( V G )with ~ multiplicity l, U is a component of V G and V is a component of U H . Since both U and V are nonprojective, the required assertion follows by virtue of Theorem 5.3. (ii) Let E : 0 + L1 + L , L2 + 0 be a split exact sequence appearing in (i). Then
By Theorem 7.2.l(iii), L2 is Xprojective. Moreover, by Theorem 7.4.3,
and so, by Theorem 7.2.l(iii), L1 is Xprojective. Since L 2 L1 @ L2, it follows from Corollary 6.10.2 that L is Xprojective. (iii) Let U1 be an indecomposable component of M with vertex Q1. Then, by Proposition 4.2, there exists an irreducible homomorphism U1 + U . Hence the desired conclusion follows by virtue of Proposition 4.5. Assume that cp : U1 + U2 is a homomorphism of FGmodules, and let Q be any subgroup of G. We say that cp is Qprojective if cp factors through some Qprojective FGmodule. Thus, if S = {Q], then cp is Qprojective if and only if cp is Sprojective. Also, by Theorem 6.10.15, cp is Qprojective if and only if Y E q ( H o m F Q ( U 1 , U2)) Next we record the following preliminary observation.
Lemma 6.2. Let U1, U2 be indecomposable FGmodules with vertices Q1, Q2 respectively, and assume that there is an irreducible homomorphism cp : u1 + u2. (i) Suppose that Q1 is Gconjugate to a subgroup of Qz, and let X O be the set of all subgroups D of G such that Q1 is not Gconjugate to a subgroup of D . Then cp is Q1projective but not Xoprojective. (ii) Suppose that Qz is Gconjugate to a subgroup of Q1 and let Yo be the set of all subgroups D of G such that Q2 is not Gconjugate to a subgroup of D . Then cp is Q2projective but not Yoprojective.
Proof. (i) Since Ul is Q1projective, it follows from Lemma 6.10.16 that cp is Q1projective. Assume by way of contradiction that 9 is Xoprojective.
Almost Split Sequences
618
Then, by definition, cp factors through an Xoprojective FGmodule, say W . Since cp is irreducible, either U1 or U2 is a component of W . Since direct summands of Xoprojective modules are Xoprojective (Corollary 6.10.2) it follows that either U1 or Uz is Xoprojective. However, this is impossible since, by the definition of Xo and our assumption on Q1 and Q2, neither U1 nor U, is Xoprojective. This proves (i). (ii) Apply the same argument as in (i). We are now ready to prove our main result, in which f ( U ) denotes the Green correspondent of U with respect to (G, P , H ) , and v x ( W ) is a fixed vertex of W . The following theorem was originally stated for the special case where Q = P and H = N , ( P ) .
Theorem 6.3. (Kawata (1989)). Let P # 1 be a psubgroup of G, let H be any subgroup of G with H 2 N c ( P ) and let V = f ( U ) , where U is an indecomposable FGmodule with vertex Q E P and U is not Xprojective. Consider the almost split sequences : 0 + S22(U) f M
0
f
Q 2 ( V )+ N
+
u
+
+ V +
0 0
Then M and N admit direct decompositions :
M = (&1M;) @ ($$,M,!), N = (&IN;) @I (€35,1Nj) into indecomposable direct summands such that : (i) Q is Gconjugate to a subgroup of v x ( M ; ) , 1 5 i 5 t , vz(M,!) is Gconjugate to a proper subgroup of Q , 1 5 j 5 k, Q is Hconjugate to a subgroup of v x ( N ; ) , 1 i t , v x ( N j ) is Hconjugate to a proper subgroup ofQ, 1SjSr. (ii) Any direct decomposition of NY into indecomposable direct summands contains a unique summand W; with a vertex Q; 2 Q , 1 i t. Moreover, W; 2 M ; , 15 i 5 t . (iii) M; 2 M j if and only if N; 2 N j (1 5 i,j 5 t ) .
< <
< <
Proof. For the sake of clarity, we divide the proof into a number of steps. Step 1. Here we establish a weaker version of (i). By Lemma 6.l(iii), we have the following direct decompositions :
6 The Green correspondence and almost split sequences
619
and
N = ($:='=,Ni) @ ($>=IN:) where each M ; , M j , N; and N i is indecomposable, Q is Gconjugate to a subgroup of v z ( M ; ) , 1 5 i 5 t , v z ( M j ) is Gconjugate to a proper subgroup of Q, 1 5 j 5 k, Q is Hconjugate to a subgroup of v z ( N ; ) ,1 5 i 5 s, and vz(Nj)is Hconjugate to a proper subgroup of Q, 1 5 j 5 T . Thus to prove (i), we are left to verify that s = t . Step 2. Here we show that each NF has an indecomposable direct summand W; with a vertex Q; 2 Q, 1 i s. Indeed, fix i € ( I , . , . ,s } . By Corollary 6.2.5, N y has an indecomposable component W; such that W; and N; have a vertex in common. Since Q is Hconjugate to a subgroup of v z ( N ; ) ,it follows that Wi has a vertex Q ; 1 Q. Step 3. Now we demonstrate that (i) and (ii) will follow provided we prove that any direct decomposition of N p into indecomposable direct summands has at most one summand with a vertex containing Q, 1 5 i 5 t. Indeed, assume that the above assertion is true. Since U is not Xprojective, Q is not Gconjugate to a subgroup of gPg' n P,for all g E G  H . Hence every indecomposable FGmodule with a vertex containing Q cannot be Xprojective. On the other hand, by Lemma 6.1(ii), we have
< <
N G = ($L,N?) 2i M $ L
$
($;=l(Ny)
for some Xprojective FGmodule L . Moreover, by Corollary 6.2.2(i) for each j E (1,. .. ,T } no indecomposable component of (Nj)" has a vertex containing Q. We may now obviously assume that W; 2 M; and so (i) and (ii) follow by applying Steps 1 and 2. Step 4. Let Z be the set of all psubgroups of H of order smaller than lQl. Our aim is to prove that for each N ; ,
Hom$H(Ni, vG) Define Z A H by Z A H = of Z , Z A H = Z . IEence
H O ~ F H ( N ; G ,V
)
(as Fspaces)
{gSg'nH]SE Z,gE G}. Then, by the definition
Hom&(N;G, V ) 2 Hom&(N;G, V G ) (by Proposition 6.10.18(iv)) 2
Hom&&v;,
VG)
(by Proposition 6.10.19( iv))
Almost Split Sequences
620
as required. Step 5. Let 2 be as in Step 4. We show that for each N i ,
H O ~ ; , ( N ; , V ~F ) H O ~ $ , ( N ~ , V ) (as Fspaces) To this end, write
( V G )2~V @ ( e t E ~ L 1 ) ( L t is an FHmodule) Taking into account that
it suffices to show that Horn;&$,
Lt) = 0
for all t E T
Expressed otherwise, we have to show that each a E H o m F H ( N ; , L t ) is 2projective for a fixed t E T . Now V is Qprojective, so by Lemma 7.1.1, Lt is gQglnHprojective for some g E G  H . Setting K = gQg' n H , it follows that a is Kprojective and therefore a = T ~ g ( pfor ) some p E H o r n ~ ~ ( N i , L t ) . Since g # N , we see that K is not Hconjugate to Q. Hence, by Proposition 4.6, the exact sequence 0 + Q 2 ( V ), ~ N K , VK + 0
splits. Thus ( N ~ ) K is a component of ( Q 2 ( V )@ V ) K . Hence ( N ~ ) Kis 2'projective, where 2' is the set of all subgroups of K of order smaller than IQI. Indeed, this follows from the fact that Q 2 ( V ) @ Vis Qprojective and by applying Lemma 6.1.2(i). But then p is 2'projective and, since 2' C 2, it follows from Theorem 6.10.15 that a is Zprojective. So the requird assertion follows. Step 6. Assume that some N? has a decomposition
N;G = w,@ w2 @
w,
where W1 is indecomposable with vertex Q1 2 Q and W2 is indecomposable with vertex Qz 2 Q. We shall prove that this configuration is impossible, which will establish both (i) and (ii) by applying Step 3.
6 The Green correspondence and almost split sequences
621
By Corollary 6.2.2(ii), we may assume that N; is a component of ( W l ) ~ . Now write
( W I ) H2 N ; @ A for some FHmodule A . Because WZis a component of M , there exists an irreducible homomorphism p : W2 + U . By Lemma 6.2(ii), p is not 2projective. We may regard p as an element of H O ~ F G ( N ?V, G ) . Consider the isomorphism T T : H ~ O ~ F H ( NV Y ,) + H o ~ ~ F G ( NV? G , ) given by Proposition 6.10.18. Its inverse H o m F G ( N Y , VG)+ HomFH(N?, is induced by Pg : V G.+ V , C I Et ~ €4 vt in G with 1 E T, vt E V . Now
+ v1,
v)
T is a left transversal for H
and P$ o p is not 2projective by Proposition 6.10.18(ii), since 2 A H = 2. This means that H o m g H ( W 2 , V )# 0. On the other hand, by Steps 4 and 5, we must have
and
This forces Hom$,(Wz, V )= 0, which is a desired contradiction. Step 7 . Here we shall establish (iii), which will complete the proof. If N; 2 N j , then clearly M; Z M j . To prove the converse, assume by way of contradiction that Mi 2 Mj but N; 34 Nj for some i , j . Because N ; is a component of ( M ; ) H and Nj is a component of ( M ~ ) Hit, follows that N ; @ N j is a component of ( M ; ) H ,which in turn is a component of ( N y ) , y . Write (N?)H
2 N; @
Nj @ w
for some FHmodule W . Then, by Steps 4 and 5, we have
H o r n g & V ; , V ) E Horn&(N;,V)
@
Hom&(N,,V)
@ Hom&(W,V)
Almost Split Sequences
622
In particular, we must have HOrn&(Nj,
V )= 0
But this is a contradiction, since there is an irreducible homomorphism V , which is not 2projective by Lemma 6.2.
Nj
3