CHAPTER 3. Hopf Algebras, Algebraic, Formal, and Quantum Groups

Transcription

1 CHAPTER 3 Hopf Algebras Algebraic Formal and Quantum Groups 83

2 84 3. HOPF ALGEBRAS ALGEBRAIC FORMAL AND QUANTUM GROUPS 3. Dual Objects At the end of the rst section in Corollary we saw that the dual of an H module can be constructed. We did not show the corresponding result for comodules. In fact such a construction for comodules needs some niteness conditions. With this restriction the notion of a dual object can be introduced in an arbitrary monoidal category. Denition Let èc; è be a monoidal category M 2Cbe an object. An object M 2Ctogether with a morphism ev : M M! I is called a left dual for M if there exists a morphism db : I! M M in C such that èm èm 1db db 1! M M M 1ev! M ev 1 M M! Mè =1 M! M è=1 M : A monoidal category is called left rigid if each object M 2Chas a left dual. Symmetricallywe dene: an object M 2Ctogether with a morphism ev : M M! I is called a right dual for M if there exists a morphism db : I! M M in C such that èm 1db! M ev 1 M M! Mè =1 M è db 1 M! M M M! 1ev Mè=1 M: A monoidal category is called right rigid if each object M 2Chas a left dual. The morphisms ev and db are called the evaluation respectively the dual basis. Remark If èm ; evè is a left dual for M then obviously èm;evè is a right dual for M and conversely. One uses the same morphism db : I! M M. Lemma Let èm ; evè be a left dual for M. Then there is a natural isomorphism Mor C è M;è = MorC è; M è; i. e. the functor M : C!Cis left adjoint to the functor M : C!C. Proof. We give the unit and the counit of the pair of adjoint functors. We dene èaè :=1 A db : A! A M M and èbè :=1 B ev : B M M! B. These are obviously natural transformations. We have commutative diagrams and èa M èb M FèAè= 1Adb 1M GèBè= 1B1M db A M M M B M M M thus the Lemma has been proved by Corollary A FèAè= 1A1M ev GèBè= 1Bev 1M The converse holds as well. If M is left adjoint to M A Mè =1 AM B M è=1 BM then the unit gives a morphism db := èiè :I! M M and the counit gives a morphism ev := èiè :M M! I satisfying the required properties. Thus we have

3 3. DUAL OBJECTS 85 Corollary If M : C!Cis left adjoint to M : C!Cthen M is a left dual for M. Corollary èm ; evè is a left dual for M if and only if there is a natural isomorphism Mor C èm ; è = MorC è;m è; i. e. the functor M : C!Cis left adjoint to the functor M : C!C. The natural isomorphism if given by and èf : M N! P è 7! èè1 M fèèdb 1 N è:n! M M N! M P è èg : N! M P è 7! èèev 1 P èè1 M gè :M N! M M P! P è: Proof. We have a natural isomorphism Mor C èm ; è = MorC è;m è; i èm;evè is a right dual for M èas a symmetric statement to Lemma 3.3.3è i èm ; evè is a left dual for M. Corollary If M has a left dual then this is unique up to isomorphism. Proof. Let èm ; evè and èm! ; ev! è be left duals for M. Then the functors M and M! are isomorphic by Lemma A.9.7. In particular we have M = I M = I M! = M!. If we consider the construction of the isomorphism then we get in particular that èev! 1èè1 dbè : M!! M! M M! M is the given isomorphism. Problem Let èm ; evè be a left dual for M. Then there is a unique morphism db : I! M M satisfying the conditions of Denition Denition Let èm ; ev M è and èn ; ev N è be left duals for M resp. N. For each morphism f : M! N we dene the transposed morphism èf : N! M è:=èn 1db M! N 1f 1 M M! N N M ev N 1! M è: With this denition we get that the left dual is a contravariant functor since we have Lemma Let èm ; ev M è èn ; ev N è and èp ; ev P è be left duals for M N and P respectively. 1. We have è1 M è =1 M. 2. If f : M! N and g : N! P are given then ègfè = f g holds. Proof. 1. è1 M è = èev 1èè1 1 1èè1 dbè = 1 M.

4 86 3. HOPF ALGEBRAS ALGEBRAIC FORMAL AND QUANTUM GROUPS 2. The following diagram commutes M N f dbn 1 dbn 1 N N M 11f N N N g11 P N N 1evN N 1evN P g Hence we have gf =è1 ev N èèg 1 fèèdb N 1è. Thus the following diagram commutes P 1db P N N 1g1 P P N Z ZZ 1db P M M 1gf1 P P M M ev 1 1db 1 1f 1ev 1 1ev 1db P N N M M 1g1f 1 XXXX Xz P P N N M ev 1 N N M 1g1 1db P P N M M 9 1 1f 1 ev 1 1db = N M M Z ZZ ev 1 Z ZZZ~ N Problem In the category of Ngraded vector spaces determine all objects M that have a left dual. 2. In the category of chain complexes KComp determine all objects M that have a left dual. 3. In the category of cochain complexes KCocomp determine all objects M that have a left dual. 4. Let èm ; evè be a left dual for M. Show that db : I! M M is uniquely determined by M M and ev. èuniqueness of the dual basis.è 5. Let èm ; evè be a left dual for M. Show that ev : M M! I is uniquely determined by M M and db.

5 3. DUAL OBJECTS 87 Corollary Let M; N have the left duals èm ; ev M è and èn ; ev N è and let f : M! N be a morphism in C. Then the following diagram commutes I dbm M M dbn f 1 N N N M : 1f Proof. The following diagram commutes f M db 1 db 1 N H HHHj 1 N N M 11f N N N 1ev N This implies èf 1 M èdb M = èè1 N ev N èè1 N 1 N fèèdb N 1 M è 1 M èdb M = è1 N ev N 1 M èè1 N 1 N f 1 M èèdb N 1 M 1 M èdb M =è1 N ev N 1 M èè1 N 1 N f 1 M èè1 N 1 N db M èdb N =è1 N èev N 1 M èè1 N f 1 M èè1 N db M èè db N =è1 N f èdb N. Corollary Let M; N have the left duals èm ; ev M è and èn ; ev N è and let f : M! N be a morphism in C. Then the following diagram commutes N M f 1 M M 1f evm N N I: evn Proof. This statement follows immediately from the symmetry of the denition of a left dual. Example Let M 2 R M R be an RRbimodule. Then Hom R èm:; R:è is an RRbimodule by èrfsèèxè = rfèsxè. Furthermore we have the morphism ev : Hom R èm:; R:è R M! R dened by evèf R mè=fèmè. èdual Basis Lemma:è A module M 2M R is called nitely generated and projective if there are elements m 1 ;::: ;m n 2 M und m 1 ;::: ;m n 2 Hom R èm:; R:è such that 8m 2 M : nx i=1 m i m i èmè =m: Actually this is a consequence of the dual basis lemma. But this denition is equivalent to the usual denition.

6 88 3. HOPF ALGEBRAS ALGEBRAIC FORMAL AND QUANTUM GROUPS Let M 2 R M R. M as a right Rmodule is nitely generated and projective im has a left dual. The left dual is isomorphic to Hom R èm:; R:è. If M R is nitely P generated projective then we use db : R! M R Hom R èm:; R:è n with dbè1è = m i=1 i R m i. In fact we have è1 R evèèdb R 1èèmè = è1 R P evèè m i R m i P R mè= m i m i èmè =m. Wehave furthermore èev R 1èè1 R dbèèfèèmè = èev P n R 1èè f i=1 R m i P P R m i èèmè = fèm i èm i èmè =fè m i m i èmèè = fèmè for all m 2 M hence èev R 1èè1 R dbèèfè =f. Conversely if M has a left dual M then ev : M R M! R denes a homo : M! Hom R èm:; R:è in R M R by èm èèmè = evèm R mè. We dene Pmorphism n m i=1 i m i := dbè1è 2 M M then m =è1evèèdb 1èèmè =è1 P evèè m i m i P mè = m i èm i èèmè so that m 1 ;::: ;m n 2 M and èm 1 è;::: ;èm n è 2 Hom R èm:; R:è form a dual basis for M i.e. M is nitely generated and projective asanrmodule. Thus M and Hom R èm:; R:è are isomorphic by the map. Analogously Hom R è:m; :Rè is a right dual for M i M is nitely generated and projective as a left Rmodule. Problem Find an example of an object M in a monoidal category C that has a left dual but no right dual. Denition Given objects M; N in C. An object ëm; Në is called a left inner Hom of M and N if there is a natural isomorphism Mor C è M; Nè = Mor C è; ëm; Nëè i. e. if it represents the functor Mor C è M; Nè. If there is an isomorphism Mor C èp M; Nè = MorC èp; ëm; Nëè natural in the three variable M; N; P then the category C is called monoidal and left closed. If there is an isomorphism Mor C èm P; Nè = MorC èp; ëm; Nëè natural in the three variable M; N; P then the category C is called monoidal and right closed. If M has a left dual M in C then there are inner Homs ëm; ë dened by ëm; Në :=N M. In particular left rigid monoidal categories are left closed. Example The category of nite dimensional vector spaces is a monoidal category where each object has a èleft and rightè dual. Hence it is èleft and rightè closed and rigid. 2. Let Ban be the category of ècomplexè Banach spaces where the morphisms satisfy k fèmè kk m k i. e. the maps are bounded by 1 or contracting. Ban is a monoidal category by the complete tensor product M bn. InBan exists an inner Hom functor ëm; Në that consists of the set of bounded linear maps from M to N made into a Banach space by an appropriate topology. Thus Ban is a monoidal closed category. 3. Let H be a Hopf algebra. The category HMod of left Hmodules is a monoidal category èsee Example è. Then Hom K èm; Nè is an object in HMod by the multiplication èhfèèmè := X h è1è fèmsèh è2è è

7 3. DUAL OBJECTS 89 as in Proposition Hom K èm; Nè is an inner Hom functor in the monoidal category HMod. The isomorphism : Hom K èp; Hom K èm; Nèè = HomK èp M; Nè can be restricted to an isomorphism Hom H èp; Hom K èm; Nèè = HomH èp M; Nè; Pbecause èfèèhèp mèè P P = èfèè h è1è p P h è2è mè = fèh è1è pèèh è2è mè = èhè1è èfèpèèèèh è2è mè= h è1è èfèpèèsèh P è2è èh è3è mèè = hèfèpèèmèè P = hèèfè èpmèè and conversely èhèfèpèèèèmè = h è1è èfèpèèsèh è2è èmèè = h è1è èèfè èp P Sèh è2è èmèè = èfèèh è1è p h è2è Sèh è3è èmè =èfèèhp mè =fèhpèèmè. Thus HMod is left closed. If M 2 HMod is a nite dimensional vector space then the dual vector space M := Hom K èm; Kè again is an Hmodule by èhfèèmè :=fèsèhèmè: Furthermore M is a left dual for M with the morphisms db : K 3 1 7! X i m i m i 2 M M and ev : M M 3 f m 7! fèmè 2 K where m i and m i are a dual basis of the vector space M. Clearly we have è1evèèdb 1è=1 M and èev 1èè1dbè = 1 M since M is a nite dimensional vector space. We have to show that db and ev are Hmodule homomorphisms. We have P P èh dbè1èèèmè =èhè m i P m i èèèmè =è h è1è m i h è2è m i èèmè = P P èhè1è m i èèèh è2è m i èèmèè = èh è1è P m i èèm i èsèh è2è èmèè = hè1è Sèh è2è èm = "èhèm = "èhèè m i m i èèmè ="èhè dbè1èèmè = dbè"èhè1èèmè = dbèh1èèmè; hence h dbè1è = dbèh1è. Furthermore we have h evèf mè =hfèmè = P h è1è fèsèh è2è èh è3è mè= P èh è1è fèèh è2è mè= P evèhè1è f h è2è mè = evèhèf mèè: 4. Let H be a Hopf algebra. Then the category of left Hcomodules èsee Example è is a monoidal category. Let M 2 HComod be a nite dimensional vector space. Let m i be a basis for M and let the comultiplication of the comodule be èm i è= P h ij m j. Then we have èh ik è= P h ij h jk. M := Hom K èm; Kè becomes a left Hcomodule èm j è:= P Sèh ij èm i. One veries that M is a left dual for M. Lemma Let M 2Cbe an object with left dual èm ; evè. Then 1. M M is an algebra with multiplication r := 1 M ev 1 M : M M M M! M M

8 90 3. HOPF ALGEBRAS ALGEBRAIC FORMAL AND QUANTUM GROUPS and unit u := db : I! M M ; 2. M M is a coalgebra with comultiplication and counit :=1 M db 1 M : M M! M M M M " := ev : M M! I: Proof. 1. The associativity isgiven by èr1èr =è1 M ev 1 M 1 M 1 M èè1 M ev 1 M è=1 M ev ev 1 M =è1 M 1 M 1 M ev 1 M èè1 M ev 1 M è=è1rèr. The axiom for the left unit is rèu1è = è1 M ev 1 M èèdb 1 M 1 M è=1 M 1 M. 2. is dual to the statement for algebras. Lemma Let A be an algebra inc and left M 2Cbe a left rigid object with left dual èm ; evè. There is a bijection between the set of morphisms f : A M! M making M a left Amodule and the set of algebra morphisms e f : A! M M. 2. Let C be acoalgebra inc and left M 2Cbe a left rigid object with left dual èm ; evè. There is a bijection between the set of morphisms f : M! M C making M a right Ccomodule and the set of coalgebra morphisms e f : M M! C. Proof. 1. By Lemma the object M M is an algebra. Given f : A M! M such that M becomes an Amodule. By Lemma we associate e f := èf 1èè1 dbè : A! A M M! M M. The compatibility of e f with the multiplication is given by the commutative diagram A H ef e f M M M M 11db A A M M 1 e ef 11 The unit axiom is given by 1f 1 A M M 1db 11 r r11 f 1 A M M M M u I A f 11 db 1db 1 XXXXXXXXz 11ev 1 1ev 1 M M H HHHHj u1 A M M A M M f 1 M M 11 A M M f 1 M M 1 f 1 M M ef

9 3. DUAL OBJECTS 91 Conversely let g : A! M M be an algebra homomorphism and consider eg := è1 evèèg 1è : A M! M M M! M. Then M becomes a left Amodule since A A M 1eg A M XX X 1g1 X gg1 XXXXX Xz A M M M 11ev g111 g1 : r1 M M M M M 111ev M M M M H HHHH u1 db 1 XXXXXXX 1 X XXX Hj Xz A M M M M g1 eg 1ev 11 M M M 1ev XXXXXXXXXXX Xz 1ev M A M g1 A AA 1ev A AA eg A AU M commute. 2. èm ; evè is a left dual for M in the category C if and only if èm ; dbè is the right dual for M in the dual category C op. So if we dualize the result of part 1. we have to change sides hence 2.