aa r X i v : . [ m a t h . A T ] J a n IS D SYMMETRIC MONOIDAL?
N. P. STRICKLAND
Brooke Shipley has considered [1] a certain functor D : Sp Σ (Ch + ) → Ch, and proved that it is stronglymonoidal: we have a natural isomorphism φ A,B : D ( A ) ⊗ D ( B ) → D ( A ⊗ B ), which is compatible with theassociativity and unit isomorphisms in Sp Σ (Ch + ) and Ch. The proof of one result in [1] implicitly used theclaim that D is symmetric monoidal, so we have a commutative diagram D ( A ) ⊗ D ( B ) φ A,B / / τ D ( A ) ,D ( B ) (cid:15) (cid:15) D ( A ⊗ B ) D ( τ A,B ) (cid:15) (cid:15) D ( B ) ⊗ D ( A ) φ B,A / / D ( B ⊗ A ) . However, in a later note Brooke suggests that D is not in fact symmetric monoidal, so the relevant theoremneeds to be weakened and given a more elaborate proof.I think that D is in fact symmetric monoidal; this note is an attempt to verify this.We first give the relevant definitions. Definition 1.
Ch is the category of chain complexes of abelian groups, with differential decreasing degreesby one. Ch + is the subcategory of chain complexes C ∗ with C k = 0 for k <
0. These are symmetric monoidalcategories, with the usual rules: ( A ⊗ B ) n = M n = i + j A i ⊗ B j d ( a ⊗ b ) = d ( a ) ⊗ b + ( − | a | a ⊗ d ( b ) τ ( a ⊗ b ) = ( − | a || b | b ⊗ a. We write Z [ n ] for the chain complex generated additively by a single element e n ∈ Z [ n ] n with d ( e n ) = 0.For n ≥ n act on Z [ n ] and Z [ − n ] by σ ∗ ( e n ) = sgn( σ ) e n and σ ∗ ( e − n ) = sgn( σ ) e − n . Definition 2.
We let J be the following symmetric monoidal category. The objects are natural numbers,where n is identified with { , , . . . , n − } in the usual way. The morphisms are injective functions. Themonoidal product is given on objects by n (cid:3) m = n + m . For morphisms f : p → q and g : m → n we define f (cid:3) g : p + m → q + n by ( f (cid:3) g )( i ) = ( f ( i ) if 0 ≤ i < pg ( i − p ) + q if p ≤ i < p + m. The associator maps ( m (cid:3) n ) (cid:3) p → m (cid:3) ( n (cid:3) p ) are identities, and the twist maps τ m,n : m (cid:3) n → n (cid:3) m aregiven by τ m,n ( i ) = ( i + n if 0 ≤ i < mi − m if m ≤ i < m + n. This is a permutation of signature ( − nm .We write F for the subcategory with the same objects as J , where the morphisms are bijections. Definition 3.
We write SS(Ch + ) for the category of functors A : F → Ch + . We make this a symmetricmonoidal category by the usual procedure, due to Day: given A, B ∈ SS(Ch + ) we have a functor F × F A × B −−−→ Ch + × Ch + ⊗ −→ Ch + , nd we take the left Kan extension along the functor (cid:3) : F × F → F to get a functor A ⊗ B : F → Ch + .This can be unwrapped as follows: Given a ∈ A ni and b ∈ B mj and a bijection θ : n (cid:3) m → p , we have anelement θ ∗ ( a ⊗ b ) ∈ ( A ⊗ B ) p,i + j , and A ⊗ B is generated by elements of this form, subject to the rule( θ ◦ ( α (cid:3) β )) ∗ ( a ⊗ b ) = θ ∗ ( α ∗ ( a ) ⊗ β ∗ ( b )) . The twist map τ A,B : A ⊗ B → B ⊗ A is given by τ A,B ( θ ∗ ( a ⊗ b )) = ( θ ◦ τ m,n ) ∗ ( τ A n ,B m ( a ⊗ b )) = ( − ij ( θ ◦ τ m,n ) ∗ ( b ⊗ a ) . The F -action is just γ ∗ ( θ ∗ ( a ⊗ b )) = ( γ ◦ θ ) ∗ ( a ⊗ b ). Remark 4.
If we tried to define the twist map by τ A,B ( θ ∗ ( a ⊗ b )) = θ ∗ ( τ A n ,B m ( a ⊗ b )) = ( − ij θ ∗ ( b ⊗ a ) . then this would not be well-defined (because of the relation ( θ ◦ ( α (cid:3) β )) ∗ ( a ⊗ b ) = θ ∗ ( α ∗ ( a ) ⊗ β ∗ ( b ))) andvarious other things would go wrong. It is conceivable that this is behind the idea that D is not symmetricmonoidal. Remark 5.
If we have an isomorphism θ : n (cid:3) m → p then we must have p = n (cid:3) m = n + m , and in thatcontext there is an obvious choice of θ , namely the identity. However, this is an artificial consequence of thefact that we are working with a skeletal category, and it is usually best to avoid thinking in those terms.One exception, however, is that the above observation allows us to make sense of sgn( θ ), which is needed forour next definition. Definition 6.
We can regard Z [ ∗ ] as an object of SS(Ch + ) as follows. All morphisms in F are automorphismsand thus permutations; for α ∈ F ( n, n ) = Σ n we define α ∗ : Z [ n ] → Z [ n ] by α ∗ ( e n ) = sgn( α ) e n as before.We define µ : Z [ ∗ ] ⊗ Z [ ∗ ] → Z [ ∗ ] by µ ( θ ∗ ( e n ⊗ e m )) = sgn( θ ) e n + m . Lemma 7.
The map µ makes Z [ ∗ ] a commutative, associative and unital monoid in SS(Ch + ) .Proof. The only point that needs checking is commutativity. We have τ ( θ ∗ ( e n ⊗ e m )) = ( − nm ( θτ m,n ) ∗ ( e m ⊗ e n ) , so µτ ( θ ∗ ( e n ⊗ e m )) = ( − nm sgn( θτ m,n ) e n + m = sgn( θ ) e n + m = µ ( θ ∗ ( e n ⊗ e m )) . Thus µτ = µ , as required. (cid:3) Definition 8.
An object A ∈ Sp Σ (Ch + ) is a module in SS(Ch + ) over Z [ ∗ ]. To unpack this a little, we havea sequence of chain complexes A n ∈ Ch + for n ≥
0, with an action of Σ n on A n and maps σ : Z [ k ] ⊗ A n → A k + n satisfying ( α (cid:3) β ) ∗ σ ( e k ⊗ a ) = sgn( α ) σ ( e k ⊗ β ∗ ( a ))for all α ∈ Σ k and β ∈ Σ m , as well as the unit condition σ ( e ⊗ a ) = a and the associativity condition σ ( e i ⊗ σ ( e j ⊗ a )) = σ ( e i + j ⊗ a ) . Given
A, B ∈ Sp Σ (Ch + ) we define A ⊗ B to be their tensor product over Z [ ∗ ], which is the coequaliser of a pairof maps A ⊗ Z [ ∗ ] ⊗ B → A ⊗ B . This makes Sp Σ (Ch + ) into a symmetric monoidal category. The elements of( A ⊗ B ) nk can again be written in the form θ ∗ ( a ⊗ b ), but there are some extra relations. To be more explicit,consider a permutation φ ∈ Σ n + m + p (which we can regard as an isomorphism ( n + m ) (cid:3) p → n + m + p or asan isomorphism n (cid:3) ( m + p ) → n + m + p ) and elements a ∈ A ni and b ∈ B pk . We then have φ ∗ ( a ⊗ σ ( e m ⊗ b )) = ( − im ( φ ◦ ( τ m,n (cid:3) p )) ∗ ( σ ( e m ⊗ a ) ⊗ b ) . Definition 9.
Given A ∈ Sp Σ (Ch + ), we define a functor D ( A ) : J →
Ch as follows. On objects, we justput D ( A ) n = Z [ − n ] ⊗ A n . Now consider a morphism α : n → m in J (so n ≤ m ). Let ρ : n → m be givenby ρ ( i ) = ( m − n ) + i . One can choose α ′ ∈ Σ m with α ′ ρ = α . We would like to define α ∗ ( e − n ⊗ a ) = sgn( α ′ ) . ( e − m ⊗ α ′∗ σ ( e m − n ⊗ a )) . o see that this is well-defined, let α ′′ be another permutation such that α ′′ ρ = α . Then α ′ = α ′′ ◦ ( δ (cid:3)
1) forsome δ ∈ Σ m − n , so sgn( α ′ ) = sgn( α ′′ ) sgn( δ ) and( δ (cid:3) ∗ σ ( e m − n ⊗ a ) = sgn( δ ) σ ( e m − n ⊗ a ) . It follows that sgn( α ′ ) . ( e − m ⊗ α ′∗ σ ( e m − n ⊗ a ))= sgn( α ′′ ) sgn( δ ) . ( e − m ⊗ α ′′∗ ( δ (cid:3) ∗ σ ( e m − n ⊗ a ))= sgn( α ′′ ) . ( e − m ⊗ α ′′∗ σ ( e m − n ⊗ a ))as required. One can check directly that this construction gives a chain map.We next check that we have a functor. Suppose we have injective maps n α −→ m β −→ p . We can fit these ina diagram as follows: n α / / ρ (cid:31) (cid:31) ❅❅❅❅❅❅❅❅ m β / / ρ (cid:31) (cid:31) ❅❅❅❅❅❅❅❅ pm α ′ O O ρ (cid:31) (cid:31) ❅❅❅❅❅❅❅❅ p β ′ O O p p − m (cid:3) α ′ O O Here α ′ and β ′ are bijections. Put γ = βα and γ ′ = β ′ ◦ (1 (cid:3) α ′ ). These are related as required for thedefinition of γ ∗ , so γ ∗ ( e − n ⊗ a ) = sgn( γ ′ ) . ( e − p ⊗ γ ′∗ σ ( e p − n ⊗ a )) . Here sgn( γ ′ ) = sgn( β ′ ) sgn( α ′ ). Also, we can write σ ( e p − n ⊗ a ) as σ ( e p − m ⊗ σ ( e m − n ⊗ a )), and applying(1 (cid:3) α ′ ) ∗ to this gives σ ( e p − m ⊗ α ′∗ σ ( e m − n ⊗ a )). Thus γ ∗ ( e − n ⊗ a ) = sgn( β ′ ) e − p ⊗ β ′∗ σ (sgn( α ′ ) e p − m ⊗ α ′∗ σ ( e m − n ⊗ a )) . We see from the definitions that this is the same as β ∗ α ∗ ( e − n ⊗ a ).Note that for β ∈ J ( n, n ) = Σ n we have β ∗ ( e − n ⊗ a ) = sgn( β ) .e − n ⊗ β ∗ ( a ), whereas ρ ∗ ( e − n ⊗ a ) =( − ( m − n ) n e − m ⊗ σ ( e m − n ⊗ a ).We now define D ( A ) to be the colimit of the functor D ( A ) : J →
Ch. More explicitly, an element a ∈ A n,k gives an element e − n ⊗ a ∈ D ( A ) n,k − n and thus an element ξ ( a ) ∈ D ( A ) k − n . The complex D ( A ) is generatedby elements of this form, subject to the rules ξ ( β ∗ ( a )) = sgn( β ) ξ ( a ) (for β ∈ Σ n ) and ξ ( σ ( e k ⊗ a )) = ξ ( a ).As d ( e − n ⊗ a ) = ( − n e − n ⊗ da , we have dξ ( a ) = ( − n ξ ( da ). Definition 10.
We define φ A,A ′ : D ( A ) ⊗ D ( A ′ ) → D ( A ⊗ A ′ ) as follows. Given a ∈ A n,i and a ′ ∈ A ′ n ′ ,i ′ , wehave an element ι ∗ ( a ⊗ a ′ ) ∈ ( A ⊗ A ′ ) n + n ′ ,i + i ′ , and we define φ A,A ′ ( ξ ( a ) ⊗ ξ ( a ′ )) = ( − n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ a ′ )) . To see that this is well-defined, we must check various identities. Firstly, given α ∈ Σ n and α ′ ∈ Σ n ′ wemust have ( − n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ a ′ )) = ( − n ′ n + n ′ i sgn( α ) sgn( α ′ ) ξ ( ι ∗ ( α ∗ ( a ) ⊗ α ′∗ ( a ′ ))) . Then, we must show that for r, r ′ ≥ − n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ a ′ )) = ( − n ′ ( n + r )+ n ′ ( i + r ) ξ ( ι ∗ ( σ ( e r ⊗ a ) ⊗ a ′ ))( − n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ a ′ )) = ( − ( n ′ + r ′ ) n +( n ′ + r ′ ) i ξ ( ι ∗ ( a ⊗ σ ( e r ′ ⊗ a ′ ))) . For the first of our three identities, we note that ι ∗ ( α ∗ ( a ) ⊗ α ′∗ ( a ′ )) = ( α (cid:3) α ′ ) ∗ ι ∗ ( a ⊗ a ′ ) ξ ( ι ∗ ( α ∗ ( a ) ⊗ α ′∗ ( a ′ ))) = ξ (( α (cid:3) α ′ ) ∗ ι ∗ ( a ⊗ a ′ )) = sgn( α (cid:3) α ′ ) ξ ( ι ∗ ( a ⊗ a ′ ))= sgn( α ) sgn( α ′ ) ξ ( ι ( a ⊗ a ′ )) . or the second, we note that ( − n ′ ( n + r )+ n ′ ( i + r ) = ( − n ′ n + n ′ i and that ι ∗ ( σ ( e r ⊗ a ) ⊗ a ′ ) = σ ( e r ⊗ ι ∗ ( a ⊗ a ′ )),so ξ ( ι ∗ ( σ ( e r ⊗ a ) ⊗ a ′ )) = ξ ( ι ∗ ( a ⊗ a ′ )). The third identity can be deduced from the second using the identity ι ∗ ( a ⊗ σ ( e r ′ ⊗ a ′ )) = ( − ir ′ ( τ r ′ ,n (cid:3) n ′ ) ∗ ( σ ( e r ′ ⊗ a ) ⊗ a ′ )which holds in ( A ⊗ A ′ ) r ′ + n + n ′ ,r ′ + i + i ′ .We next claim that φ A,B is a chain map. Indeed, we have dφ A,A ′ ( ξ ( a ) ⊗ ξ ( a ′ )) = ( − n ′ n + n ′ i dξ ( ι ∗ ( a ⊗ a ′ ))= ( − n ′ n + n ′ i + n ′ + n ξ ( d ( ι ∗ ( a ⊗ a ′ ))= ( − n ′ n + n ′ i + n ′ + n ξ ( ι ∗ ( d ( a ) ⊗ a ′ )) + ( − n ′ n + n ′ i + n ′ + n + i ξ ( ι ∗ ( a ⊗ d ( a ′ ))) φ A,A ′ ( d ( ξ ( a ) ⊗ ξ ( a ′ ))) = φ A,A ′ ( d ( ξ ( a )) ⊗ ξ ( a ′ )) + ( − n + i φ A,A ′ ( ξ ( a ) ⊗ d ( ξ ( a ′ )))= ( − n φ A,A ′ ( ξ ( d ( a )) ⊗ ξ ( a ′ )) + ( − n ′ + n + i φ A,A ′ ( ξ ( a ) ⊗ ξ ( d ( a ′ )))= ( − n + n ′ n + n ′ ( i − ξ ( ι ∗ ( d ( a ) ⊗ a ′ )) + ( − n ′ + n + i + n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ d ( a ′ ))) . These are easily seen to be the same.
Proposition 11. D is symmetric monoidal.Proof. We must show that the following diagram commutes: D ( A ) ⊗ D ( B ) φ A,B / / τ D ( A ) ,D ( B ) (cid:15) (cid:15) D ( A ⊗ B ) D ( τ A,B ) (cid:15) (cid:15) D ( B ) ⊗ D ( A ) φ B,A / / D ( B ⊗ A ) . Consider a ∈ A n,i and b ∈ B m,j as before. We have τ D ( A ) ,D ( B ) ( ξ ( a ) ⊗ ξ ( b )) = ( − mn + mi + nj + ij ξ ( b ) ⊗ ξ ( a )and so φ B,A τ D ( A ) ,D ( B ) ( ξ ( a ) ⊗ ξ ( b )) = ( − mn + mi + nj + ij φ A,B ( ξ ( b ) ⊗ ξ ( a ))= ( − mn + mi + nj + ij ( − nm + nj ξ ( ι ∗ ( b ⊗ a ))= ( − mi + ij ξ ( ι ∗ ( b ⊗ a )) . On the other hand, we have φ A,B ( ξ ( a ) ⊗ ξ ( b )) = ( − mn + mi ξ ( ι ∗ ( a ⊗ b )), and τ A,B ( ι ∗ ( a ⊗ b )) = ( − ij τ m,n ∗ ( b ⊗ a ), so D ( τ )( φ A,B ( ξ ( a ) ⊗ ξ ( b ))) = ( − mn + mi D ( τ A,B )( ξ ( ι ∗ ( a ⊗ b )))= ( − mn + mi ξ ( τ A,B ( ι ∗ ( a ⊗ b )))= ( − mn + mi + ij ξ ( τ m,n ∗ ( b ⊗ a ))= ( − mn + mi + ij sgn( τ m,n ) ξ ( ι ∗ ( b ⊗ a ))= ( − mi + ij ξ ( ι ∗ ( b ⊗ a )) . It follows that the diagram commutes, as required. (cid:3)
Definition 12.
Given C ∈ Ch we define τ ≥ C ∈ Ch + by( τ ≥ C ) n = n < d : C → C − ) if n = 0 C n if n > . It is well-known that this gives a functor, right adjoint to the inclusion of Ch + in Ch. efinition 13. We define R : Ch → Sp Σ (Ch + ) as follows. Given C ∈ Ch and n ≥ RC ) n = τ ≥ ( Z [ n ] ⊗ C ). We let Σ n act on this via the action on Z [ n ], so α ∗ ( e n ⊗ c ) = sgn( α ) .e n ⊗ c . The map µ : Z [ n ] ⊗ Z [ m ] → Z [ n + m ] induces a map σ : Z [ n ] ⊗ ( RC ) m → ( RC ) n + m , given by σ ( e n ⊗ ( e m ⊗ c )) = e n + m ⊗ c .This defines an object RC ∈ Sp Σ (Ch + ), and there is an obvious way to make it a functor. Lemma 14.
There is a natural map ψ = ψ C,C ′ : RC ⊗ RC ′ → R ( C ⊗ C ′ ) given by ψ ( α ∗ (( e p ⊗ c ) ⊗ ( e p ′ ⊗ c ′ ))) = ( − p ′ k sgn( α ) e p + p ′ ⊗ c ⊗ c ′ for all c ∈ C p,k and c ′ ∈ C ′ p ′ ,k ′ . Moreover, this makes R a lax symmetric monoidal functor.Proof. First, we have a map e ψ p,p ′ = ( − pp ′ . ( Z [ p ] ⊗ C ⊗ Z [ p ′ ] ⊗ C ′ ⊗ τ ⊗ −−−−→ Z [ p ] ⊗ Z [ p ′ ] ⊗ C ⊗ C ′ µ ⊗ ⊗ −−−−→ Z [ p + p ′ ] ⊗ C ⊗ C ′ ) . If we restrict the domain to the subcomplex ( RC ) p ⊗ ( RC ′ ) p ′ (which lies in Ch + ) then the map must factoruniquely through τ ≥ ( Z [ p + p ′ ] ⊗ C ⊗ C ′ ) = R ( C ⊗ C ′ ) p + p ′ . We thus get a map e ψ p,p ′ : ( RC ) p ⊗ ( RC ′ ) p ′ → R ( C ⊗ C ′ ) p + p ′ . This is easily seen to be equivariant for Σ p × Σ p ′ , or in other words, natural for ( p, p ′ ) ∈ F . Our descriptionof ( RC ) ⊗ ( RC ′ ) as a Kan extension shows that e ψ induces a map ψ : ( RC ) ⊗ ( RC ′ ) → R ( C ⊗ C ′ ), with theformula ψ ( α ∗ (( e p ⊗ c ) ⊗ ( e p ′ ⊗ c ′ ))) = ( − pp ′ + p ′ ( p + k ) α ∗ ( e p + p ′ ⊗ c ⊗ c ′ ) = ( − p ′ k sgn( α ) e p + p ′ ⊗ c ⊗ c ′ . We must show that this factors through ( RC ) ⊗ ( RC ′ ), or in other words, that the following diagram com-mutes: RC ⊗ Z [ ∗ ] ⊗ RC ′ τ ⊗ / / ⊗ σ (cid:15) (cid:15) Z [ ∗ ] ⊗ RC ⊗ RC ′ σ ⊗ (cid:15) (cid:15) RC ⊗ RC ′ ψ / / R ( C ⊗ C ′ ) RC ⊗ RC ′ ψ o o The complex ( RC ⊗ Z [ ∗ ] ⊗ RC ′ ) n is spanned by elements x = α ∗ (( e p ⊗ c ) ⊗ e m ⊗ ( e p ′ ⊗ c ′ ))with c ∈ C k and c ′ ∈ C ′ k ′ and α : p (cid:3) m (cid:3) p ′ → n in F . For such x we have( τ ⊗ x ) = ( − m ( p + k ) ( α ◦ ( τ m,p (cid:3) ∗ ( e m ⊗ ( e p ⊗ c ) ⊗ ( e p ′ ⊗ c ′ )) . Applying σ ⊗ τ m,p ∗ e p + m = ( − mp e p + m ) gives( − mp + mk ( α ◦ ( τ m,p (cid:3) ∗ (( e p + m ⊗ c ) ⊗ ( e p ′ ⊗ c ′ )) = ( − mk α ∗ (( e p + m ⊗ c ) ⊗ ( e p ′ ⊗ c ′ )) . Applying ψ now gives ( − mk + p ′ k sgn( α ) e p + m + p ′ ⊗ c ⊗ c ′ . Going the other way around the diagram, we first get(1 ⊗ σ )( x ) = α ∗ (( e p ⊗ c ) ⊗ ( e m + p ′ ⊗ c ′ )) , and applying ψ to this gives ( − ( m + p ′ ) k sgn( α ) e p + m + p ′ ⊗ c ⊗ c ′ , which is the same as before. We thus have a natural map ψ = ψ C,C ′ : RC ⊗ RC ′ → R ( C ⊗ C ′ )as claimed. We leave it to the reader to check the relevant associativity property, showing that this makes R a monoidal functor. We will, however, check the commutativity property, which says that the following iagram commutes: RC ⊗ RC ′ ψ / / τ (cid:15) (cid:15) R ( C ⊗ C ′ ) R ( τ ) (cid:15) (cid:15) RC ′ ⊗ RC ψ / / R ( C ′ ⊗ C ) . The group ( RC ⊗ RC ′ ) n is spanned by elements x = α ∗ (( e p ⊗ c ) ⊗ ( e p ′ ⊗ c ′ )) with c ∈ C k and c ′ ∈ C ′ k ′ say.For such x we have τ ( x ) = ( − ( p + k )( p ′ + k ′ ) ( α ◦ τ p ′ ,p ) ∗ (( e p ′ ⊗ c ′ ) ⊗ ( e p ⊗ c )) . Applying ψ to this gives e p + p ′ ⊗ c ′ ⊗ c multiplied by the sign( − pp ′ + pk ′ + kp ′ + kk ′ ( − pk ′ sgn( α ) sgn( τ p ′ ,p ) = ( − kp ′ + kk ′ sgn( α ) . Going the other way around, we have ψ ( x ) = ( − kp ′ sgn( α ) e p + p ′ ⊗ c ⊗ c ′ , and R ( τ ) exchanges c and c ′ witha sign of ( − kk ′ , giving an overall sign of kp ′ + kk ′ as before. (cid:3) Lemma 15.
There is a natural map η : A → RD ( A ) given by η ( a ) = e n ⊗ ξ ( a ) for a ∈ A n,k .Proof. As | ξ ( a ) | = k − n and | e n | = n , the specified rule certainly gives a natural map of graded abeliangroups. We also have dη ( a ) = ( − n e n ⊗ d ( ξ ( a )) = ( − n e n ⊗ ξ ( d ( a )) = η ( d ( a )) , so η : A n → ( RD ( A )) n is a chain map. For α ∈ Σ n we have α ∗ ( η ( a )) = α ∗ ( e n ⊗ ξ ( a )) = sgn( α ) e n ⊗ ξ ( a ) ξ ( α ∗ ( a )) = sgn( α ) ξ ( a ) η ( α ∗ ( a )) = e n ⊗ ξ ( α ∗ ( a )) = sgn( α ) e n ⊗ ξ ( a );so η is equivariant. We also have σ ( e p ⊗ a ) ∈ A p + n,p + k with ξ ( σ ( e p , a )) = ξ ( a ), so η ( σ ( e p ⊗ a )) = e p + n ⊗ ξ ( σ ( e p ⊗ a ))= e p + n ⊗ ξ ( a ) = σ ( e p ⊗ ( e n ⊗ ξ ( a )))= σ ( e p ⊗ η ( a )) . Thus, η is a morphism in Sp Σ (Ch + ). (cid:3) Lemma 16.
There is a natural map ǫ : DR ( C ) → C given by ǫ ( ξ ( e n ⊗ c )) = c .Proof. We must first check that the definition is compatible with the relations ξ ( α ∗ ( a )) = sgn( α ) ξ ( a ) and ξ ( σ ( e m ⊗ a )) = ξ ( a ). This is easy, because when a = e n ⊗ c we have α ∗ ( a ) = sgn( α ) a and σ ( e m ⊗ a ) = e n + m ⊗ c .We thus have a well-defined map of graded abelian groups, and ǫ ( d ( ξ ( e n ⊗ c ))) = ǫ (( − n ξ ( d ( e n ⊗ c ))) = ( − n ǫ ( ξ (( − n e n ⊗ dc )) = ǫ ( ξ ( e n ⊗ dc )) = dc = dǫ ( ξ ( e n ⊗ c )) , so ǫ is a chain map. (cid:3) Proposition 17.
The following diagrams commute, showing that D is left adjoint to R : RC η RC / / $ $ ■■■■■■■■■ RDRC Rǫ C (cid:15) (cid:15) DA Dη A / / $ $ ■■■■■■■■■ DRDA ǫ DA (cid:15) (cid:15) RC DA
Proof.
This is immediate from the formulae. (cid:3) efinition 18. We let χ : D ( A ⊗ A ′ ) → DA ⊗ DA ′ be conjugate to ψ : RC ⊗ RC ′ → R ( C ⊗ C ′ ). Moreexplicitly, χ is given by the composite D ( A ⊗ A ′ ) D ( η ⊗ η ) −−−−−→ D ( RDA ⊗ RDA ′ ) D ( ψ ) −−−→ DR ( DA ⊗ DA ′ ) ǫ −→ DA ⊗ DA ′ . From the definitions, we see that when a ∈ A n,i and a ′ ∈ A n ′ ,i ′ we have χ ( ξ ( α ∗ ( a ⊗ a ′ ))) = ( − n ′ i + n ′ n sgn( α ) ξ ( a ) ⊗ ξ ( a ′ ) . Remark 19.
As (
R, ψ ) is lax symmetric monoidal, it is formal to check that (
D, χ ) is lax symmetriccomonoidal.
Lemma 20. χ : D ( A ⊗ A ′ ) → DA ⊗ DA ′ is inverse to φ : DA ⊗ DA ′ → D ( A ⊗ A ′ ) .Proof. Recall the formula φ ( ξ ( a ) ⊗ ξ ( a ′ )) = ( − n ′ n + n ′ i ξ ( ι ∗ ( a ⊗ a ′ )) . It follows immediately that χφ = 1, and if we recall that ξ ( α ∗ ( a ⊗ a ′ )) = sgn( α ) ξ ( ι ∗ ( a ⊗ a ′ )), it also followsthat φχ = 1. (cid:3) Remark 21.
As (
D, χ ) is symmetric comonoidal and χ is an isomorphism, it follows that ( D, ψ ) = (
D, χ − )is symmetric monoidal, giving an alternative proof of Proposition 11. References [1] Brooke Shipley, H Z -algebra spectra are differential graded algebras , Amer. J. Math. (2007), no. 2, 351–379. MR2306038(2007), no. 2, 351–379. MR2306038