Second symmetric powers of chain complexes

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by  Anders J. Frankild

SECOND SYMMETRIC POWERS OF CHAIN COMPLEXES
ANDERSJ.FRANKILD,SEANSATHER-WAGSTAFF,ANDAMELIATAYLOR
0
1
0 Abstract. WeinvestigateBuchbaumandEisenbud’sconstructionofthesec-
2 ond symmetric power S2R(X) of a chain complex X of modules over a com-
n mutative ringR. Westate andproveanumber ofresultsfromthefolkloreof
a the subject for which we know of no good direct references. We also provide
J severalexplicitcomputationsandexamples. Weusethisconstructiontoprove
thefollowingversionofaresultofAvramov,Buchweitz,andS¸ega: LetR→S
8
be a module-finite ring homomorphism such that R is noetherian and local,
1
andsuchthat2isaunitinR. LetXbeacomplexoffiniterankfreeS-modules

] such that Xn =0for each n<0. If ∪nAssR(Hn(X⊗SX))⊆Ass(R) andif
C Xp≃Sp foreachp∈Ass(R),thenX ≃S.
A
.
h
t Introduction
a
m
Multilinear constructionslike tensor products andsymmetric powers areimpor-
[ tant tools for studying modules over commutative rings. In recent years, these

3 notions have been extended to the realm of chain complexes of R-modules. (Con-
v sultSection1forbackgroundinformationoncomplexes.) Forinstance,Buchsbaum
8 and Eisenbud’s description [5] of the minimal free resolutions of Gorenstein ideals
1
of grade 3 uses the second symmetric power of a certain free resolution.
0
Inthispaper,weinvestigateBuchsbaumandEisenbud’ssecondsymmetricpower
0
. functor: for a chain complex X of modules over a commutative ring R, we set
9
S2(X) = (X ⊗ X)/(Y +Z) where Y is the graded submodule generated by all
0 R R
8 elements of the form x ⊗ x′ − (−1)|x||x′|x′ ⊗ x and Z is the graded submodule
0 generated by all elements of the form x⊗x where x has odd degree.1
: Our main result is the following version of a result of Avramov, Buchweitz and
v
i S¸ega [2, (2.2)] for complexes. It is motivated by our work in [10] extending the
X results of [2]. Note that S2(X) does not appear in the statement of Theorem A;
R
r however,it is the key tool for the proof, given in 3.7.
a
Theorem A. Let R → S be a module-finite ring homomorphism such that R is
noetherian and local, and such that 2 is a unit in R. Let X be a complex of finite
rank free S-modules such that X =0 for each n<0. If ∪ Ass (H (X⊗ X))⊆
n n R n S
Ass(R) and if X ≃S for each p∈Ass(R), then X ≃S.
p p
Muchofthispaperisdevotedtostatementsandproofsofresultsfromthefolklore
of this subject. Section 2 contains basic properties of S2(X), most of which are
R
motivated by the behavior of tensor products of complexes and the properties of
2000 Mathematics Subject Classification. 13C10,13D25.
Key words and phrases. chaincomplex,symmetricpower,symmetricsquare.
ThisworkwascompletedaftertheuntimelypassingofAndersJ.Frankildon10June2007.
1Note that the definition of S2R(X) given in [5] does not yield the complex described on [5,
p.452]unless2isaunitinR. Thecorrecteddefinitioncanbefound,forinstance,in[4,(3.4.3)].
1
2 A.J.FRANKILD,S.SATHER-WAGSTAFF,ANDA.TAYLOR
symmetricpowersofmodules. This sectionends withanexplicitdescriptionofthe
modules occuring in S2(X); see Theorem 2.9. Section 3 examines the homological
R
properties of S2(X), and includes the proof of Thoerem A. The paper concludes
R
with Section 4, which is devoted to explicit computations.
1. Complexes
Throughout this paper R and S are commutative rings with identity. The term
“module” is short for “unital module”.
Thissectionconsistsofdefinitions,notationandbackgroundinformationforuse
in the remainder of the paper.
Definition 1.1. An R-complex is a sequence of R-module homomorphisms
X =···−∂−nX−+→1 X −∂−nX→X −∂−nX−−→1 ···
n n−1
such that ∂X ∂X =0 for each integer n. A complex X is degreewise-finite if each
n−1 n
X is finitely generated; it is bounded-below if X =0 for n≪0.
n n
The nth homology module of X is H (X):= Ker(∂X)/Im(∂X ). The infimum
n n n+1
of X is inf(X):=inf{i∈Z|H (X)6=0}, and the large support of X is
n
Supp (X)={p∈Spec(R)|X 6≃0}=∪ Supp (H (X)).
R p n R n
Foreachx∈X ,weset|x|:=n. AnR-complexX ishomologically degreewise-finite
n
if H (X) is finitely generated for each n; it is homologically finite if the R-module
n
⊕n∈ZHn(X) is finitely generated.
Foreachintegeri,theithsuspension (orshift)ofX,denotedΣiX,isthecomplex
with (ΣiX) =X and ∂ΣiX =(−1)i∂X . The notation ΣX is short for Σ1X.
n n−i n n−i
Definition 1.2. Let X and Y be R-complexes. A morphism from X to Y is a
sequence of R-module homomorphisms {f : X →Y } such that f ∂X =∂Yf
n n n n−1 n n n
for each n. A morphism of complexes α: X → Y induces homomorphisms on
homology modules H (α): H (X) → H (Y), and α is a quasiisomorphism when
n n n
each H (α) is bijective. Quasiisomorphisms are designated by the symbol “≃”.
n
Definition 1.3. Let X and Y be R-complexes. Two morphisms f,g: X →Y are
homotopic if there exists a sequence of homomorphisms s = {s : X → Y }
n n n+1
suchthat f =g +∂Y s +s ∂X for eachn; here we saythat s is a homotopy
n n n+1 n n−1 n
from f to g. The morphism f is a homotopy equivalence if there is a morphism
h: Y →X such that the compositions fh and hf are homotopic to the respective
identity morphisms id and id , and then f and h are homotopy inverses.
Y X
Definition 1.4. Given two bounded-below complexes P and Q of projective R-
≃
modules, we write P ≃Q when there is a quasiisomorphismP −→Q.
Fact 1.5. The relation ≃ from Definition 1.4 is an equivalence relation; see [3,
(2.8.8.2.2’)] or [8, (6.6.ii)] or [9, (6.21)].
Let P and Q be bounded-below complexes of projective R-modules. Then any
≃
quasiisomorphism P −→ Q is a homotopy equivalence; see [3, (1.8.5.3)] or [8,
(6.4.iii)]. (Conversely,it is straightforwardto show that any homotopyequivalence
between R-complexes is a quasiisomorphism.)
Definition1.6. LetX beahomologicallybounded-belowR-complex. Aprojective
≃
(or free) resolution of X is a quasiisomorphism P −→ X such that each P is
n
SECOND SYMMETRIC POWERS OF CHAIN COMPLEXES 3
≃
projective (or free) and P is bounded-below; the resolution P −→X is degreewise-
finite if P is degreewise-finite. We saythatX hasfinite projective dimension when
≃
it admits a projective resolution P −→X such that P =0 for n≫0.
n
Fact 1.7. LetX beahomologicallybounded-belowR-complex. ThenX hasafree
≃
resolution P −→ X such that P = 0 for all n < inf(X); see [3, (2.11.3.4)] or [8,
n
≃ ≃
(6.6.i)] or [9, (2.6.P)]. (It follows that P 6= 0.) If P −→ X and Q −→ X are
inf(X)
≃
projective resolutions of X, then there is a homotopy equivalence P −→ Q; see [8,
(6.6.ii)] or [9, (6.21)]. If R is noetherian and X is homologically degreewise-finite,
then P may be chosen degreewise-finite; see [3, (2.11.3.3)] or [9, (2.6.L)].
Definition1.8. LetX beanR-complexthatishomologicallybothbounded-below
and degreewise-finite. Assume that R is noetherian and local with maximal ideal
≃
m. A projective resolution P −→ X is minimal if the complex P is minimal, that
is, if Im(∂P)⊆mP for each n.
n n−1
Fact 1.9. Let X be an R-complex that is homologically both bounded-below and
degreewise-finite. Assume that R is noetherian and local with maximal ideal m.
≃
Then X has a minimal free resolution P −→ X such that P = 0 for all n <
n
≃ ≃
inf(X); see [1, Prop. 2] or [3, (2.12.5.2.1)]. Let P −→X and Q−→X be projective
resolutionsof X. If P is minimal, then there is a bounded-below exactcomplex P′
of projective R-modules such that Q∼=P ⊕P′; see [3, (2.12.5.2.3)]. It follows that
X hasfinite projectivedimensionifandonlyif everyminimalprojectiveresolution
of X is bounded. It also follows that, if P and Q are both minimal, then P ∼= Q;
see [3, (2.12.5.2.2)].
Definition 1.10. Let X and Y be R-complexes. The R-complex X ⊗ Y is
R
(X ⊗ Y) = X ⊗ Y
R n p p R n−p
with nth differential ∂X⊗RY given on geLnerators by
n
x⊗y 7→∂X(x)⊗y+(−1)|x|x⊗∂Y (y).
|x| |y|
Fix two more R-complexes X′,Y′ and morphisms f: X → X′ and g: Y → Y′.
Define the tensor product f ⊗ g: X ⊗ Y →X′⊗ Y′ on generators as
R R R
x⊗y 7→f (x)⊗g (y).
|x| |y|
One checks readily that f ⊗ g is a morphism.
R
Fact 1.11. Let P andQ be bounded-below complexes of projectiveR-modules. If
≃
f: X −→ Y is a quasiisomorphism, then so are f ⊗ Q: X ⊗ Q → Y ⊗ Q and
R R R
P ⊗ f: P ⊗ X → P ⊗ Y; see [3, (1.10.4.2.2’)] or [8, (6.10)] or [9, (7.8)]. In
R R R
≃
particular,ifg: P −→Qisaquasiisomorphism,thensoisg⊗g: P⊗ P →Q⊗ Q;
R R
see [8, (6.10)]. This can be used to show the following facts from [9, (7.28)]:
inf(P ⊗ Q)>inf(P)+inf(Q)
R
HR (P ⊗ Q)∼=H (P)⊗ H (Q).
inf(P)+inf(Q) R inf(P) R inf(Q)
Assume that R is noetherian and that P and Q are homologically degreewise-
finite. One can use degreewise-finite projective resolutions of P and Q in order
to show that each R-module H (P ⊗ Q) is finitely generated; see [9, (7.31)]. In
n R
particular,if R is local, Nakayama’sLemma conspireswith the previous display to
produce the equality inf(P ⊗ Q)=inf(P)+inf(Q); see [9, (7.28)].
R
4 A.J.FRANKILD,S.SATHER-WAGSTAFF,ANDA.TAYLOR
The following technical lemma about power series is used in the proofs of Theo-
rem 3.8 and Corollary 3.10.
Lemma 1.12. Let Q(t) = ∞ r ti be a power series with nonnegative integer
i=0 i
coefficients, and assume r > 0. If either Q(t)2+Q(−t2) or Q(t)2−Q(−t2) is a
0
P
non-negative integer, then r =0 for all i>0. Furthermore,
i
(a) Q(t)2+Q(−t2)6=0;
(b) If Q(t)2−Q(−t2)=0, then Q(t)=1;
(c) If Q(t)2+Q(−t2)=2, then Q(t)=1; and
(d) If Q(t)2−Q(−t2)=2, then Q(t)=2.
Proof. We begin by showing that r = 0 for each n > 1, by induction on n. The
n
coefficients of Q(−t2) in odd degree are all 0. Hence, the degree 1 coefficient of
Q(t)2±Q(−t2) is
0=r r +r r =2r r .
1 0 0 1 1 0
Itfollowsthat r =0, since r >0. Inductively,assume thatn>1 andthatr =0
1 0 i
for eachi=1,...,n. Since the degree n+1 coefficientof QRX(−t2) is either ±rn+1
2
(whenn+1iseven)or0(whenn+1isodd),theinductionhypothesisimpliesthat
this coefficient is 0. The degree n+1 coefficient of Q(t)2±Q(−t2) is
0=r r +r r +···+r r +r r =2r r
n+1 0 n 1 1 n 0 n+1 n+1 0
=0
and so r =0.
n+1 | {z }
The previousparagraphshowsthat Q(t)=r , and soQ(t)2±Q(−t2)=r2∓r .
0 0 0
The conclusions in (a)–(d) follow readily, using the assumption r >0. (cid:3)
0
2. Definition and Basic Properties of S2(X)
R
We begin this section with the definition of the second symmetric power of a
complex. It is modeled on the definition for modules.
Definition 2.1. Let X be an R-complex and let αX: X ⊗ X →X ⊗ X be the
R R
morphism described on generators by the formula
x⊗x′ 7→x⊗x′−(−1)|x||x′|x′⊗x.
The weak second symmetric power of X is defined as s2(X) := Coker(αX). The
R
second symmetric power of X is defined as S2(X) := s2(X)/hx⊗x||x| is oddi.
R R
For each i∈Z, let ωX: s2(X) →S2(X) be the natural surjection.
i R i R i
Remark 2.2. Let X be an R-complex. Since s2(X) is defined as a cokernel of a
R
morphism, it is an R-complex. Also, for each n∈Z and x∈X , one has
2n+1
∂X⊗RX(x⊗x)=αX (∂X (x)⊗x).
4n+2 4n+1 2n+1
It follows that S2(X) is an R-complex, and that the sequence {ωX} describes a
R i
morphism ωX: s2(X)→S2(X).
R R
Herearecomputationsforlateruse. Section4 containsmoreinvolvedexamples.
Example 2.3. If M is an R-module, then computing S2(M) and s2(M) as com-
R R
plexes (considering M as a complex concentrated in degree 0) and computing
S2(M)as amodule givethe sameresult. Inparticular,we haveS2(0)=0=s2(0)
R R R
and S2(R)∼=R∼=s2(R).
R R
SECOND SYMMETRIC POWERS OF CHAIN COMPLEXES 5
Example 2.4. For 06=x,y ∈ΣR we haveαΣR(x⊗y)=x⊗y+y⊗x. Hence, the
natural tensor-cancellation isomorphism R⊗ R −∼=→ R yields the vertical isomor-
R
phisms in the following commutative diagram:
(ΣR)⊗ (ΣR) αΣR // (ΣR)⊗ (ΣR) pΣR //s2(ΣR)
R R R
∼= β ∼= β ∼= β
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
Σ2R (2) //Σ2R // Σ2R/(2)
It follows that s2(ΣR)∼=Σ2R/(2).
R
By definition, the kernel of the natural map ωX: s2(X) → S2(X) is generated
R R
by 1⊗1∈s2(ΣR) . Since we have β 1⊗1 =1, it follows that S2(ΣR)=0.
R 2 2 R
More generally, we have s2(Σ2n+1R) ∼= Σ4n+2R/(2) and S2(Σ2n+1R) = 0 for
R (cid:0) (cid:1) R
each integer n. In particular, if 2R6=0, then
s2(Σ2n+1R)∼=Σ4n+2R/(2)6≃Σ4n+2R∼=Σ4n+2s2(R).
R R
Contrastthiswiththebehaviorofs2(Σ2nX)andS2(Σ2nX)documentedin(2.5.2).
R R
Thefollowingpropertiesarestraightforwardtoverifyandwillbeusedfrequently
in the sequel.
Properties 2.5. Let X be an R-complex.
2.5.1. If 2 is a unit in R, then the natural morphism ωX: s2(X) → S2(X) is an
R R
isomorphism, and the morphism 1αX is idempotent.
2
2.5.2. For each integer n, there is a commutative diagram
(Σ2nX)⊗ (Σ2nX)αΣ2nX//(Σ2nX)⊗ (Σ2nX)
R R
∼= β ∼= β
(cid:15)(cid:15) (cid:15)(cid:15)
Σ4n(X ⊗ X) Σ4nαX // Σ4n(X ⊗ X)
R R
withβ(x⊗y)=x⊗y. Theresultingisomorphismofcokernelsyieldsanisomorphism
β: s2(Σ2nX)−∼=→Σ4ns2(X)
R R
given by β(x⊗y) = x⊗y. In particular, the equality β(x⊗x) = x⊗x implies
that β induces an isomorphism
S2(Σ2nX)∼=Σ4nS2(X).
R R
2.5.3. There is an exact sequence
0→Ker(αX)−j−X→X ⊗ X −α−X→X ⊗ X −p−X→s2(X)→0
R R R
where jX and pX are the natural injection and surjection, respectively.
2.5.4. A morphism of complexes f: X →Y yields a commutative diagram
αX //
X ⊗ X X⊗ X
R R
f⊗Rf f⊗Rf
(cid:15)(cid:15) (cid:15)(cid:15)
αY //
Y ⊗ Y Y ⊗ Y.
R R
6 A.J.FRANKILD,S.SATHER-WAGSTAFF,ANDA.TAYLOR
Hence, this induces a well-defined morphism on cokernels s2(f): s2(X) → s2(Y),
R R R
given by s2(f)(x⊗y) = f(x)⊗f(y). The equality s2(f)(x⊗x) = f(x)⊗f(x)
R R
shows that s2(f) induces a well-defined morphism S2(f): S2(X) → S2(Y) given
R R R R
by S2(f)(x⊗y) = f(x)⊗f(y). From the definition, one sees that the operators
R
s2(−)andS2(−)arefunctorial,butExample4.7showsthattheyarenotadditive,
R R
as one might expect.
The nexttworesultsshowthatthe functors s2(−)andS2(−)interactwellwith
R R
basic constructions.
Proposition 2.6. Let X be an R-complex.
(a) If ϕ: R → S is a ring homomorphism, then there are isomorphisms of S-
complexes s2(S⊗ X)∼=S⊗ s2(X) and S2(S⊗ X)∼=S⊗ S2(X).
S R R R S R R R
(b) If p ⊂ R is a prime ideal, then there are isomorphisms of R -complexes
p
s2Rp(Xp)∼=s2R(X)p and S2Rp(Xp)∼=S2R(X)p.
Proof. (a) The vertical isomorphisms in the following commutative diagram are
given by β((s⊗x)⊗(t⊗y))=(st)⊗(x⊗y):
αS⊗RX //
(S⊗ X)⊗ (S⊗ X) (S⊗ X)⊗ (S⊗ X)
R S R R S R
∼= β ∼= β
(cid:15)(cid:15) (cid:15)(cid:15)
S⊗ (X ⊗ X) S⊗RαX //S⊗ (X ⊗ X).
R R R R
Thisdiagramyieldsthefirstisomorphisminthenextsequence. Thesecondisomor-
phism is due to the right-exactnessof S⊗ −, and the equalities are by definition.
R
s2(S⊗ X)=Coker(αS⊗RX)∼=Coker(S⊗ αX)
S R R
∼=S⊗ Coker(αX)=S⊗ s2(X).
R R R
By definition, the induced isomorphismβ: s2(S⊗ X)−∼=→S⊗ s2(X) is given by
S R R R
β (s⊗x)⊗(t⊗y) =(st)⊗(x⊗y).
(cid:16)Let Y ⊆ s2(S ⊗(cid:17) X) be the S-submodule generated by elements of the form
S R
u⊗u such that u ∈ S ⊗ X has odd degree. That is, Y = Ker(ωS⊗X) where
R
ωS⊗X: s2(S ⊗ X)→S2(S⊗ X) is the natural surjection. It is straightforward
S R S R
to show that Y is generated over S by all elements of the form (1⊗x)⊗(1⊗x).
Let Z ⊂ s2(X) be the R-submodule generated by elements of the form x⊗x
R
with x∈X of odd degree. That is, we have an exact sequence of R-morphisms
0→Z →s2(X)−ω−X→S2(X)→0.
R R
Tensoring with S yields the next exact sequence of S-morphisms
S⊗ Z →S⊗ s2(X)−S−⊗−R−ω−X→S⊗ S2(X)→0
R R R R R
and it follows that Ker(S ⊗ ωX) is generated over S by all elements of the form
R
1⊗(x⊗x) with x∈X of odd degree. Thus, the equality β (1⊗x)⊗(1⊗x) =
1⊗(x⊗x) shows that β induces an S-isomorphismS2(S⊗(cid:16)X)∼=S⊗ S2(X(cid:17)).
S R R R
(b) This follows from part (a) using the ring homomorphism R→R . (cid:3)
p
SECOND SYMMETRIC POWERS OF CHAIN COMPLEXES 7
Proposition 2.7. If X and Y are R-complexes, then there are isomorphisms
(2.7.1) s2(X ⊕Y)∼=s2(X)⊕(X ⊗ Y)⊕s2(Y)
R R R R
(2.7.2) S2(X ⊕Y)∼=S2(X)⊕(X ⊗ Y)⊕S2(Y).
R R R R
Proof. (2.7.1)Tensor-distributionyields the horizontalisomorphismsinthe follow-
ing commutative diagram
∼= //
(X ⊕Y)⊗ (X ⊕Y) (X ⊗ X)⊕(X ⊗ Y)⊕(Y ⊗ X)⊕(Y ⊗ Y)
R R R R R
αX 0 0 0
αX⊕Y  00 id−Xθ⊗XRYY id−Yθ⊗YRXX 00 
 0 0 0 αY 
(cid:15)(cid:15)  (cid:15)(cid:15)
∼= //
(X ⊕Y)⊗ (X ⊕Y) (X ⊗ X)⊕(X ⊗ Y)⊕(Y ⊗ X)⊕(Y ⊗ Y)
R R R R R
where θ : U⊗ V →V ⊗ U is the tensor-commutativity isomorphismgiven by
UV R R
u⊗v 7→(−1)|u||v|v⊗u. This diagramyields the first isomorphismin the following
sequence while the first equality is by definition
s2(X ⊕Y)=Coker(αX⊕Y)
R
αX 0 0 0
∼=Coker 0 idX⊗RY −θYX 0
0 −θXY idY⊗RX 0 !
0 0 0 αY
∼=Coker(αX)⊕Coker idX⊗RY −θYX ⊕Coker(αY)
−θXY idY⊗RX
∼=s2(X)⊕(X ⊗ Y)⊕(cid:16)s2(Y). (cid:17)
R R R
Thesecondisomorphismisbyelementarylinearalgebra. Forthethirdisomorphism,
using the definition of s2(−), we only need to prove Coker(β)∼=X ⊗ Y where
R R
β = idX⊗RY −θYX : (X ⊗ Y)⊕(Y ⊗ X)→(X ⊗ Y)⊕(Y ⊗ X).
−θXY idY⊗RX R R R R
(cid:16) (cid:17)
We set
γ =(id θ ): (X ⊗ Y)⊕(Y ⊗ X)→X ⊗ Y
X⊗RY YX R R R
which is a surjective morphism such that Im(β) ⊆ Ker(γ). Thus, there is a well-
defined surjective morphism γ: Coker(β)→X⊗ Y given by
R
x⊗y 7→x⊗y+(−1)|x′||y′|x′⊗y′.
y′⊗x′
(cid:16) (cid:17)
It remains to show that γ is injective. To this end, define δ: X ⊗ Y →Coker(β)
R
by the formula x⊗y 7→ x⊗y . It is straightforward to show that δ is a well-
0
defined morphism and that δγ =id . It follows that γ is injective, hence an
(cid:0) (cid:1) Coker(β)
isomorphism, as desired.
(2.7.2) The isomorphism β: s2(X ⊕Y) −∼=→ s2(X)⊕(X ⊗ Y)⊕s2(Y) from
R R R R
part (2.7.1) is given by the formula
β (x,y)⊗(x′,y′) = x⊗x′,x⊗y′+(−1)|x′||y|x′⊗y,y⊗y′ .
(cid:16) (cid:17) (cid:16) (cid:17)
8 A.J.FRANKILD,S.SATHER-WAGSTAFF,ANDA.TAYLOR
Thus, for an element (x,y)∈X ⊕Y of odd order |x|=|(x,y)|=|y|, we have
β (x,y)⊗(x,y) = x⊗x,x⊗y+(−1)|x||y|x⊗y,y⊗y
(cid:16) (cid:17)=((cid:16)x⊗x,x⊗y−x⊗y,y⊗y) (cid:17)
=(x⊗x,0,y⊗y).
It follows that
S2(X ⊕Y)
R
s2(X)⊕(X ⊗ Y)⊕s2(Y)
∼= R R R
h(x⊗x,0,y⊗y)|x∈X and y ∈Y have odd degreei
s2(X) (X ⊗ Y) s2(Y)
∼= R ⊕ R ⊕ R
hx⊗x|x∈X odd degreei 0 hy⊗y|y ∈Y odd degreei
∼=S2(X)⊕(X ⊗ Y)⊕S2(Y).
R R R
as desired. (cid:3)
Example 2.4shows why we must assumethat 2 is a unit inR in the next result.
Proposition 2.8. Assume that 2 is a unit in R, and let X be an R-complex.
(a) The following exact sequences are split exact
0→Ker(αX)−j−X→X ⊗ X −q−X→Im(αX)→0
R
0→Im(αX)−i−X→X ⊗ X −p−X→S2(X)→0
R R
where iX andjX arethe naturalinclusions, pX is thenaturalsurjection, and
qX is induced by αX. The splitting on the right of the first sequence is given
by 1iX, and the splitting on the left of the second sequence is given by 1qX.
2 2
In particular, there are isomorphisms
Im(αX)⊕Ker(αX)∼=X⊗ X ∼=Im(αX)⊕S2(X).
R R
(b) If X is a bounded-below complex of projective R-modules, then so are the
complexes Im(αX), Ker(αX) and S2(X).
R
Proof. (a) The given exact sequences come from Properties (2.5.1) and (2.5.3).
The fact that 1αX is idempotent tells us that iX is a split injection with splitting
2
givenby 1qX and qX is a split surjection with splitting givenby 1iX. The desired
2 2
isomorphisms follow immediately from the splitting of the sequences.
(b) With the isomorphisms from part (a), the fact that X ⊗ X is a bounded-
R
below complex of projective R-modules implies that Im(αX), Ker(αX) and S2(X)
R
are also bounded-below complexes of projective R-modules. (cid:3)
The nexttwo results explicitly describe the modules in s2(X) andS2(X). Note
R R
that the difference between parts (a)–(b) and part (c) shows that the behavior
documented in Example 2.4 is, in a sense, the norm, not the exception.
Theorem2.9. LetX beacomplexofR-modules. Fixanintegernandseth=n/2
and V = (X ⊗X ).
m<h m n−m
(a) If n is odd, then s2(X) ∼=V.
L R n
(b) If n≡0 (mod 4), then s2(X) ∼=V S2(X ).
R n R h
(c) Assume that n≡2 (mod 4).
L
SECOND SYMMETRIC POWERS OF CHAIN COMPLEXES 9
(c1) There is an isomorphism
X ⊗ X
s2(X) ∼=V h R h
R n hx⊗x′+x′⊗x|x,x′ ∈X i
h
M
and there is a surjection τ: s2(X) →V ⊕∧2(X ) with Ker(τ) gener-
R n h
ated by {x⊗x∈s2(X) |x∈X }.
R n h
(c2) If X is projective, then s2(X) ∼= V ∧2(X ) K for some R-
h R n h
module K that is a homomorphic image of X /2X .
h h
(c3) If X is projective and 2 is a unit in R, tLhen s2(X)L∼=V ∧2(X ).
h R n h
Proof. (a)Assumethatnisodd. Letγ: (X⊗X) →V ⊕V begivenoLngenerators
n
by the formula
(x⊗x′,0) if |x|<h
γ(x⊗x′)=
((0,x′⊗x) if |x|>h.
Since n is odd, this is a well-definedisomorphism. Let g: V ⊕V →V ⊕V be given
by g(v,v′)=(v−v′,v′−v). This yields a commutative diagram
(X ⊗ X) αXn //(X ⊗ X)
R n R n
(2.9.1)
∼= γ ∼= γ
(cid:15)(cid:15) (cid:15)(cid:15)
g //
V ⊕V V ⊕V.
Note that the commutativity depends on the fact that n is odd, because it implies
that |x||x′| is even for each x⊗x′ ∈(X ⊗ X) .
R n
Themapf: V ⊕V →V givenbyf(v,v′)=v+v′ isasurjectivehomomorphism
withKer(f)=h(v,0)−(0,v)|v ∈Vi=Im(g). This explainsthe lastisomorphism
in the next sequence
s2(X) =Coker(αX)∼=Coker(g)∼=V.
R n n
The other isomorphism follows from diagram (2.9.1).
(b)–(c) When n is even, we have a similar commutative diagram
(X ⊗ X) αXn // (X ⊗ X)
R n R n
(2.9.2)
∼= γ′ ∼= γ′
(cid:15)(cid:15) (cid:15)(cid:15)
g′ //
V ⊕V ⊕(X ⊗X ) V ⊕V ⊕(X ⊗X ).
h h h h
where γ′ and g′ are given by
(x⊗x′,0,0) if |x|<h
γ′(x⊗x′)= (0,x′⊗x,0) if |x|>h

(0,0,x⊗x′) if |x|=h.
g′(v,v′,x⊗x′)=(v−v′,v′−v,x⊗x′−(−1)h2x′⊗x)
=(v−v′,v′−v,x⊗x′−(−1)hx′⊗x).
In other words, we have g′ =g⊕α where α: X ⊗ X →X ⊗ X is given by
h R h h R h
α(x⊗x′):=x⊗x′−(−1)hx′⊗x.
e e
e
10 A.J.FRANKILD,S.SATHER-WAGSTAFF,ANDA.TAYLOR
The following sequence of isomorphisms follows directly
s2(X) =Coker(αX)∼=Coker(g′)∼=Coker(g)⊕Coker(α)∼=V ⊕Coker(α).
R n n
If n≡0 (mod 4), then h is even, so we have
e e
X ⊗ X
Coker(α)∼= h R h ∼=S2(X ).
hx⊗x′−x′⊗x|x,x′ ∈X i R h
h
For the remainder of ethe proof, we assume that n ≡ 2 (mod 4), that is, that h is
odd. In this case, we have
X ⊗ X
(2.9.3) Coker(α)∼= h R h .
hx⊗x′+x′⊗x|x,x′ ∈X i
h
It is straightforwardto show that
e
hx⊗x′+x′⊗x|x,x′ ∈X i⊆hx⊗x|x∈X i.
h h
Hence, there is an epimorphism
X ⊗ X
τ : Coker(α)→ h R h ∼=∧2(X )
1 h
hx⊗x|x∈X i
h
such that
e
(2.9.4) Ker(τ )=hx⊗x∈Coker(α)|x∈X i∼=hx⊗x∈s2(X) |x∈X i.
1 h R n h
The conclusions of part (c1) follow from setting τ =id ⊕τ .
V 1
e
For the rest of the proof, we assume that X is projective. It follows that
h
∧2(X ) is also projective, hence the surjection τ splits. Setting K = Ker(τ ), we
h 1 1
have s2(X) ∼= V ∧2(X )⊕K. Using (2.9.3) and (2.9.4) we see that the map
R n h
π: X → Ker(τ ) given by x 7→ x⊗x is surjective with 2X ⊆ Ker(π). It follows
h 1 h
L
that K is a homomorphic image of X /2X , which establishes part (c2). Finally,
h h
part (c3) follows directly from (c2): if 2 is a unit in R, then X /2X =0. (cid:3)
h h
Theorem 2.10. Let X be a complex of R-modules. Fix an integer n and set
h=n/2 and V = (X ⊗X ).
m<h m n−m
(a) If n is odd, then S2(X) ∼=V.
L R n
(b) If n≡0 (mod 4), then S2(X) ∼=V S2(X ).
R n R h
(c) If n≡2 (mod 4), then S2(X) ∼=V ∧2(X ).
R n L h
Proof. Set Y = x⊗x∈s2(X) x∈X oddLdegree ⊆s2(X).
R R
(a)–(b) If n is odd or n ≡ 0 (mod 4), then Y = 0; hence S2(X) ∼= s2(X) ,
(cid:10) (cid:12) n (cid:11) R n R n
and the desired conclusions follo(cid:12)w from Theorem 2.9(a)–(b).
(c) Assume that n≡2 (mod 4). The surjection τ: s2(X) →V ⊕∧2(X ) from
R n h
Theorem 2.9(c1) has Ker(τ) generated by {x⊗x ∈ s2(X) | x ∈ X }; that is
R n h
Ker(τ)=Y , so we have
n
V ∧2(X )∼=s2(X) /Y ∼=S2(X)
h R n n R n
as desired. M (cid:3)
We state the next result for S2(X) only, because Theorem 2.9 shows that it is
R
only reasonable to consider such formulas for s2(X) when 2 is a unit; in this case
R
the formulas are the same because of the isomorphism s2(X)∼=S2(X).
R R