Toolbox. 2. The categories of R-modules R-Mod, k-vector spaces k-vec or k-mod, groups

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1 CHAPTER 8 Toolbox 1. Categories Denition Let C consist of 1. a class Ob C whose elements A;B;C;:::2 Ob C are called objects, 2. a family fmor C èa; BèjA; B 2 Ob Cg of mutually disjoint sets whose elements f;g;:::2 Mor C èa; Bè are called morphisms, and 3. a family fmor C èa; Bè Mor C èb;cè 3 èf;gè 7! gf 2 Mor C èa; CèjA; B; C 2 Ob Cg of maps called compositions. C is called a category if the following axioms hold for C 1. Associative Law: 8A; B; C; D 2 Ob C;f 2 Mor C èa; Bè;g2 Mor C èb;cè;h2 Mor C èc; Dè : hègf è=èhgèf; 2. Identity Law: 8A 2 Ob C91 A 2 Mor C èa; Aè 8B;C 2 Ob C; 8f 2 Mor C èa; Bè; 8g 2 Mor C èc; Aè : 1 A g = g and f1 A = f: Examples The category of sets Set. 2. The categories of Rmodules RMod, kvector spaces kvec or kmod, groups Gr, abelian groups Ab, monoids Mon, commutative monoids cmon, rings Ri, elds Fld, topological spaces Top. Since modules are highly important for all what follows, we recall the denition and some basic properties. Denition and Remark Let R be a ring èalways associative with unitè. A left Rmodule R M is an èadditively writtenè abelian group M together with an operation R M 3 èr;mè 7! rm 2 M such that 1. èrsèm = rèsmè, 2. èr + sèm = rm + sm, 3. rèm + m 0 è=rm + rm 0, 4. 1m = m for all r;s2 R, m; m 0 2 M. Each abelian group is a Zmodule in a unique way. 1

2 2 8. TOOLBOX A homomorphism of left Rmodules f : R M,! R N is a group homomorphism such that fèrmè =rfèmè. Analogously we dene right Rmodules M R and their homomorphisms. We denote by Hom R è:m; :Nè the set of homomorphisms of left Rmodules R M and RN. Similarly Hom R èm:;n:è denotes the set of homomorphisms of right Rmodules M R and N R. Both sets are abelian groups by èf + gèèmè :=fèmè+gèmè. For arbitrary categories we adopt many of the customary notations. Notation f 2 Mor C èa; Bè will be written as f : A,! B or A f,! B. A is called the domain, B the range of f. The composition of two morphisms f : A,! B and g : B,! C is written as gf : A,! C or as g f : A,! C. Denition and Remark A morphism f : A,! B is called an isomorphism if there exists a morphism g : B,! A in C such that fg =1 B and gf =1 A. The morphism g is uniquely determined by f since g 0 = g 0 fg = g. We write f,1 := g. An object A is said to be isomorphic to an object B if there exists an isomorphism f : A,! B. If f is an isomorphism the so is f,1. If f : A,! B and g : B,! C are isomorphisms in C then so is gf : A,! C. Wehave: èf,1 è,1 = f and ègfè,1 = f,1 g,1. The relation of being isomorphic between objects is an equivalence relation. Example In the categories Set, RMod, kvec, Gr, Ab, Mon, cmon, Ri, Fld the isomorphisms are exactly those morphisms which are bijective as set maps. In Top the set M = fa; bg with T 1 = f;; fag; fbg; fa; bgg and with T 2 = f;;mg denes two dierent topological spaces. The map f =id:èm; T 1 è,! èm; T 2 èis bijective and continuous. The inverse map, however, is not continuous, hence f is no isomorphism èhomeomorphismè. Many well known concepts can be dened for arbitrary categories. We are going to apply some of them. Here are two examples. Denition A morphism f : A,! B is called a monomorphism if 8C 2 Ob C; 8g;h 2 Mor C èc; Aè : fg = fh =è g = h èf is left cancellableè: 2. A morphism f : A,! B is called an epimorphism if 8C 2 Ob C; 8g;h 2 Mor C èb;cè: gf = hf =è g = h èf is right cancellableè: Denition Given A; B 2C. An object A B in C together with morphisms p A : A B,! A and p B : A B,! B is called a ècategoricalè product of A and B if for every object T 2Cand every pair of morphisms f : T,! A and

3 1. CATEGORIES 3 g : T,! B there exists a unique morphism èf;gè:t,! AB such that the diagram,,, f, A ç p A T èf;gè pb B commutes. An object E 2Cis called a nal object if for every object T 2Cthere exists a unique morphism e : T,! E èi.e. Mor C èt;eè consists of exactly one elementè. A category C which has a product for any two objects A and B and which has a nal object is called a category with nite products. Remark If the product èab;p A ;p B èoftwo objects A and B in C exists then it is unique up to isomorphism. If the nal object E in C exists then it is unique up to isomorphism. Problem Let C be a category with nite products. Give a denition of a product of a family A 1 ;::: ;A n èn ç 0è. Show that products of such families exist in C. Denition and Remark Let C be a category. Then C op with the following data Ob C op := Ob C, Mor C opèa; Bè :=Mor C èb;aè, and f op g := g f denes a new category, the dual category to C. Remark Any notion expressed in categorical terms èwith objects, morphisms, and their compositionè has a dual notion, i.e. the given notion in the dual category. Monomorphisms f in the dual category C op are epimorphisms in the original category C and conversely. A nal objects I in the dual category C op is an initial object in the original category C. Denition The coproduct of two objects in the category C is dened to be a product of the objects in the dual category C op. Remark Equivalent to the preceding denition is the following denition. Given A; B 2C. An object A q B in C together with morphisms j A : A,! A q B and j B : B,! A q B,! B is a ècategoricalè coproduct of A and B if for every object T 2Cand every pair of morphisms f : A,! T and g : B,! T there exists a unique morphism ëf;gë:aqb,! T such that the diagram ja ç j B A A q B ëf;gë, T

4 4 8. TOOLBOX commutes. The category C is said to have nite coproducts if C op is a category with nite products. In particular coproducts are unique up to isomorphism.

5 2. FUNCTORS 5 2. Functors Denition Let C and D be categories. Let F consist of 1. a map Ob C3A 7! FèAè 2 Ob D; 2. a family of maps ff A;B : Mor C èa; Bè 3 f 7! F A;B èfè 2 Mor D èfèaè; FèBèèjA; B 2Cg ëor ff A;B : Mor C èa; Bè 3 f 7! F A;B èfè 2 Mor D èfèbè; FèAèèjA; B 2Cgë F is called a covariant ëcontravariant ë functor if 1. F A;A è1 A è=1 FèAè for all A 2 Ob C, 2. F A;C ègfè =F B;C ègèf A;B èfè for all A; B; C 2 Ob C: ëf A;C ègfè =F A;B èfèf B;C ègè for all A; B; C 2 Ob Cë. Notation: We write A 2C instead of A 2 Ob C f 2C instead of f 2 Mor C èa; Bè Fèfè instead of F A;B èfè. Examples Id : Set,! Set 2. Forget : RMod,! Set 3. Forget : Ri,! Ab 4. Forget : Ab,! Gr 5. P : Set,! Set; PèMè :=power set of M. PèfèèXè :=f,1 èxè for f : M,! N;X ç N is a contravariant functor. 6. Q : Set,! Set; QèMè :=power set of M. QèfèèXè :=fèxè for f : M,! N;X ç M is a covariant functor. Lemma Let X 2C. Then Ob C3A 7! Mor C èx; Aè 2 Ob Set Mor C èa; Bè 3 f 7! Mor C èx; fè 2 Mor Set èmor C èx; Aè; Mor C èx; Bèè; with Mor C èx; fè : Mor C èx; Aè 3 g 7! fg 2 Mor C èx; Bè or Mor C èx; fèègè = fg is a covariant functor Mor C èx; è. 2. Let X 2 C. Then Ob C3A 7! Mor C èa; Xè 2 Ob Set Mor C èa; Bè 3 f 7! Mor C èf;xè 2 Mor Set èmor C èb;xè; Mor C èa; Xèè with Mor C èf;xè : Mor C èb;xè 3 g 7! gf 2 Mor C èa; Xè or Mor C èf;xèègè =gf is a contravariant functor Mor C è;xè. Proof. 1. Mor C èx; 1 A èègè =1 A g = g = idègè; Mor C èx; fèmor C èx; gèèhè = fgh = Mor C èx; fgèèhè. 2. analogously.

6 6 8. TOOLBOX Remark The preceding lemma shows that Mor C è; è is a functor in both arguments. A functor in two arguments is called a bifunctor. We can regards the bifunctor Mor C è; è as a covariant functor Mor C è; è : C op C,! Set: The use of the dual category removes the fact that the bifunctor Mor C è; è is contravariant in the rst variable. Obviously the composition of two functors is again a functor and this composition is associative. Furthermore for each category C there is an identity functor Id C. Functors of the form Mor C èx; è resp. Mor C è;xè are called representable functors ècovariant resp. contravariantè and X is called the representing object èsee also section 8.8è.

7 3. NATURAL TRANSFORMATIONS 7 3. Natural Transformations Denition Let F : C,! D and G : C,! D be two functors. A natural transformation or a functorial morphism ' : F,!Gis a family of morphisms f'èaè :FèAè,!GèAèjA 2Cgsuch that the diagram FèAè Fèfè FèBè 'èaè 'èbè GèAè GèBè Gèfè commutes for all f : A,! B in C, i.e. Gèfè'èAè ='èbèfèfè. Lemma Given covariant functors F = Id Set : Set,! Set and G = Mor Set èmor Set è,;aè;aè:set,! Set for a set A. Then ' : F,!G with is a natural transformation. 'èbè :B 3 b 7! èmor Set èb;aè 3 f 7! fèbè 2 Aè 2GèBè Proof. Given g : B,! C. Then the following diagram commutes B g C 'èbè 'ècè Mor Set èmor Set èb;aè;aè Mor Set èmor Set èc; Aè;Aè Mor Set èmor Set èg;aè;aè since 'ècèfègèèbèèf è='ècègèbèèf è=f gèbè ='èbèèbèèf gè =ë'èbèèbèmor Set èg;aèëèfè = ëmor Set èmor Set èg;aè;aè'èaèèbèëèfè: Lemma Let f : A,! B be a morphism in C. Then Mor C èf;è : Mor C èb;è,! Mor C èa; è given by Mor C èf;cè : Mor C èb;cè 3 g 7! gf 2 Mor C èa; Cè is a natural transformation of covariant functors. Let f : A,! B be a morphism in C. Then Mor C è;fè : Mor C è;aè,! Mor C è;bè given by Mor C èc; fè : Mor C èc; Aè 3 g 7! fg 2 Mor C èc; Bè is a natural transformation of contravariant functors.

8 8 8. TOOLBOX Proof. Let h : C,! C 0 be a morphism in C. Then the diagrams Mor C èb;cè Mor C èf;cè Mor C èa; Cè Mor C èb;hè Mor C èa;hè and Mor C èb;c 0 è Mor C èc 0 ;Aè Mor C èf;c 0 è Mor C èc 0 ;f è Mor C èa; C 0 è Mor C èc 0 ;Bè commute. Mor C èh;aè Mor C èc; Aè Mor C èc;f è Mor C èh;bè Mor C èc; Bè Remark The composition of two natural transformations is again a natural transformation. The identity id F èaè :=1 FèAè is also a natural transformation. Denition A natural transformation ' : F,!Gis called a natural isomorphism if there exists a natural transformation è : G,!Fsuch that ' è =id G and è ' =id F. The natural transformation è is uniquely determined by '. We write ',1 := è. A functor F is said to be isomorphic to a functor G if there exists a natural isomorphism ' : F,!G. Problem Let F; G : C,!Dbe functors. Show that a natural transformation ' : F,!Gis a natural isomorphism if and only if 'èaè is an isomorphism for all objects A 2C. 2. Let èa B;p A ;p B è be the product of A and B in C. Then there is a natural isomorphism Morè;A Bè ç = MorC è;aè Mor C è;bè: 3. Let C be a category with nite products. For each object A in C show that there exists a morphism A : A,! A A satisfying p 1 A =1 A = p 2 A. Show that this denes a natural transformation. What are the functors 4. Let C be a category with nite products. Show that there is a bifunctor :CC,!Csuch that è èèa; Bè is the object of a product of A and B. We denote elements in the image of this functor by A B := è èèa; Bè and similarly f g. 5. With the notation of the preceding problem show that there is a natural transformation èa; B; Cè :èa Bè C ç = A èb Cè. Show that the diagram

9 3. NATURAL TRANSFORMATIONS 9 ècoherence orconstraintsè èèa Bè Cè D èa;b;cè1 èa èb Cèè D èa;bc;dè A èèb Cè Dè èab;c;dè èa;b;cdè 1èB;C;Dè èa Bè èc Dè A èb èc Dèè commutes. 6. With the notation of the preceding problem show that there are a natural transformations çèaè :EA,! A and çèaè :AE,! A such that the diagram ècoherence orconstraintsè èa Eè B èa;e;bè Q ç ç ç QQQQs ç+ ç A B çèaè1 A èe Bè 1çèBè Denition Let C and D be categories. A covariant functor F : C,!Dis called an equivalence ofcategories if there exists a covariant functor G : D,!Cand natural isomorphisms ' : GF ç = IdC and è : FG ç = IdD. A contravariant functor F : C,!Dis called a duality of categories if there exists a contravariant functor G : D,!Cand natural isomorphisms ' : GF ç = IdC and è : FG ç = IdD. A category C is said to be equivalent to a category D if there exists an equivalence F : C,!D. A category C is said to be dual to a category D if there exists a duality F : C,!D. Problem Show that the dual category C op is dual to the category C. 2. Let D be a category dual to the category C. Show that D is equivalent to the dual category C op. 3. Let F : C,!Dbe an equivalence with respect to G : D,!C, ' : GF ç = IdC, and è : FG ç = IdD. Show that G : D,!Cis an equivalence. Show that G is uniquely determined by F up to a natural isomorphism.

10 10 8. TOOLBOX 4. Tensor Products Denition and Remark Let M R and R N be Rmodules, and let A be an abelian group. A map f : M N,! A is called Rbilinear if 1. fèm + m 0 ;nè=fèm; nè+fèm 0 ;nè; 2. fèm; n + n 0 è=fèm; nè+fèm; n 0 è; 3. fèmr;nè=fèm; rnè for all r 2 R; m; m 0 2 M; n; n 0 2 N. Let Bil R èm;n; Aè denote the set of all Rbilinear maps f : M N,! A. Bil R èm;n; Aè is an abelian group with èf + gèèm; nè :=fèm; nè+gèm; nè. Denition Let M R and R N be Rmodules. An abelian group M R N together with an Rbilinear map : M N 3 èm; nè 7! m n 2 M R N is called a tensor product of M and N over R if for each abelian group A and for each Rbilinear map f : M N,! A there exists a unique group homomorphism g : M R N,! A such that the diagram M M R N commutes. The elements of M R N are called tensors, the elements of the form m n are called decomposable tensors. Warning: If you want to dene a homomorphism f : M R N,! A with a tensor product as domain you must dene it by giving an Rbilinear map dened on M N. Lemma A tensor product èm R N;è dened bym R and R N is unique up to a unique isomorphism. Proof. Let èm R N;è and èm R N;è be tensor products. Then M H H HH Hj M R N ẖ M R N k M R N ẖ M R N implies k = h,1. Because of this fact we will henceforth talk about the tensor product of M and N over R. Proposition èrules of computation in a tensor productè Let èm R N;è be the tensor product. Then we have for all r 2 R, m; m 0 2 M, n; n 0 2 g A

11 4. TENSOR PRODUCTS M R N = f P i m i n i j m i 2 M;n i 2 Ng; 2. èm + m 0 è n = m n + m 0 n; 3. m èn + n 0 è=m n + m n 0 ; 4. mr n = m rn èobserve in particular, that : M N,! M N is not injective in generalè, 5. if f : M N,! A is an Rbilinear map and g : M R N,! A is the induced homomorphism, then gèm nè =fèm; nè: Proof. 1. Let B := hm ni ç M R N denote the subgroup of M R N generated by the decomposable tensors mn. Let j : B,! M R N be the embedding homomorphism. We get an induced map 0 : M N,! B. In the following diagram 0 M N j B M 0 id, B p j B M R N we have id B 0 = 0, p with p j 0 = p = 0 exists since 0 is Rbilinear. Because of jp = j 0 = =id M RN we get jp =id M RN, hence the embedding j is surjective and thus the identity. 2. èm + m 0 è n = èm + m 0 ;nè=èm; nè+èm 0 ;nè=mn + m 0 n. 3. and 4. analogously. 5. is precisely the denition of the induced homomorphism. Remark To construct tensor products, we use the notion of a free module. Let X be a set and R be a ring. An Rmodule RX together with a map ç : X,! RX is called a free Rmodule generated byx, if for every Rmodule M and for every map f : X,! M there exists a unique homomorphism of Rmodules g : RX,! M such that the diagram X ç M commutes. Free Rmodules exist and can be constructed as RX := f : X,! Rj for almost all x 2 X : èxè =0g. Proposition Given Rmodules M R and R N. Then there exists a tensor product èm R N;è. Proof. Dene M R N := ZfM Ng=U where ZfM Ng is a free Zmodule over M N èthe free abelian groupè and U is generated by

12 12 8. TOOLBOX çèm + m 0 ;nè, çèm; nè, çèm 0 ;nè çèm; m + n 0 è, çèm; nè, çèm; n 0 è çèmr;nè, çèm; rnè for all r 2 R, m; m 0 2 M, n; n 0 2 N. Consider M N ç ZfM Ng ç M R N = ZfM Ng=U P è PPPPPPPPPPPPq Q QQQ ç g Q Q Qs A Let è be given. Then there is a unique ç 2 HomèZfM Ng;Aè such that çç = è. Since è is Rbilinear we get çèçèm + m 0 ;nè, çèm; nè, çèm 0 nèè = èèm + m 0 ;nè, èèm; nè, èèm 0 ;nè = 0 and similarly çèçèm; n + n 0 è, çèm; nè, çèm; n 0 èè = 0 and çèçèmr;nè, çèm; rnèè = 0. So we get çèuè = 0. This implies that there is a unique g 2 HomèM R N;Aè such that gç = ç èhomomorphism theoremè. Let := ç ç. Then is bilinear since èm + m 0 è n = ç çèm + m 0 ;nè= çèçèm + m 0 ;nèè = çèçèm + m 0 ;nè, çèm; nè, çèm 0 ;nè+çèm; nè +çèm 0 ;nèè = çèçèm; nè +çèm 0 ;nèè = ç çèm; nè+ç çèm 0 ;nè=m n + m 0 n. The other two properties are obtained in an analogous way. We have toshow that èm R N;è is a tensor product. The above diagram shows that for each abelian group A and for each Rbilinear map è : M N,! A there is a g 2 HomèM R N;Aè such that g = è. Given h 2 HomèM R N;Aè with h = è. Then h ç ç = è. This implies h ç = ç = g ç hence g = h. Proposition and Denition Given two homomorphisms Then there is a unique homomorphism f 2 Hom R èm:;m 0 :è and g 2 Hom R è:n; :N 0 è: f R g 2 HomèM R N;M 0 R N 0 è such that f R gèm nè =fèmè gènè, i.e. the following diagram commutes M N f g M R N M 0 N 0 ǣ M 0 R N 0 Proof. èf gè is bilinear. f R g Notation We often write f R N := f R 1 N and M R g := 1 M R g. We have the following rule of computation: f R g =èf R N 0 è èm R gè=èm 0 R gè èf R Nè since f g =èf N 0 è èm gè =èm 0 gè èf Nè.

13 4. TENSOR PRODUCTS 13 Proposition The following dene covariant functors 1. N : ModR,! Ab; 2. M : RMod,! Ab; 3. : ModR RMod,! Ab. Proof. èf gè èf 0 g 0 è=ff 0 gg 0 implies èf R gè èf 0 R g 0 è=ff 0 gg 0. Furthermore 1 M 1 N =1 M N implies 1 M R 1 N =1 M RN. Denition Let R, S be rings and let M be a left Rmodule and a right S module. M is called an RSbimodule if èrmès = rèmsè. We dene Hom RSè:M:; :N:è := Hom R è:m; :Nè ë Hom S èm:;n:è. Remark Let M S be a right Smodule and let R M,! M a map. M is an RSbimodule if and only if 1. 8r 2 R :èm 3 m 7! rm 2 Mè 2 Hom S èm:;m:è, 2. 8r;r 0 2 R; m 2 M :èr + r 0 èm = rm + r 0 m, 3. 8r;r 0 2 R; m 2 M :èrr 0 èm = rèr 0 mè, 4. 8m 2 M :1m = m: Lemma Let R M S and S N T be bimodules. Then R èm S Nè T is a bimodule by rèm nè :=rm n and èm nèt := m nt. Proof. Obviously we have 2.4. Furthermore èr S idèèm nè =rm n = rèm nè is a homomorphism. Corollary Given bimodules R M S, S N T, R M 0 S, S N 0 T and homomorphisms f 2 Hom RSè:M:; :M 0 :è and g 2 Hom ST è:n:; :N 0 :è. Then we have f S g 2 Hom RT è:m S N:;:M 0 S N 0 :è: Proof. f S gèrm ntè =fèrmè gèntè =rèf S gèèm nèt: Remark Every module M over a commutative ring K and in particular every vector space over a eld K is a KK bimodule by çm = mç. So there is an embedding functor ç : KMod,! KModK. Observe that there are KK bimodules that do not satisfy çm = mç. Take for example an automorphism : K,! K and a left Kmodule M and dene mç := èçèm. Then M is such a KK bimodule. The tensor product M K N of two KKbimodules M and N is again a KK bimodule. If we have, however, KK bimodules M and N arising from Kmodules as above, i.e. satisfying çm = mç, then their tensor product M K N also satises this equation, so M K N comes from a module in KMod. Indeed we have çm n = mç n = m çn = m nç. Thus the following diagram of functors commutes: ç KMod KMod ç KModK KModK K KMod ç K KModK:

14 14 8. TOOLBOX So we can consider KMod as a èproperè subcategory of KModK. The tensor product over K can be restricted to KMod. We write the tensor product of two vector spaces M and N as M N. Theorem In the category KMod there are natural isomorphisms 1. Associativity Law: :èm Nè P ç = M èn P è. 2. Law of the Left Unit: ç : K M ç = M. 3. Law of the Right Unit: ç : M K ç = M. 4. Symmetry Law: ç : M N ç = N M. 5. Existence of Inner HomFunctors: HomèP M;Nè ç = HomèP; HomèM;Nèè. Proof. We only describe the corresponding homomorphisms. 1. Use è è to dene èèm nè pè :=m èn pè. 2. Dene ç : K M,! M by çèr mè :=rm. 3. Dene ç : M K,! M by çèm rè :=mr. 4. Dene çèm nè :=n m. 5. For f : P M,! N dene çèfè : P,! HomèM;Nè by çèfèèpèèmè := fèp mè. Usually one identies threefold tensor products along the map so that we use M N P =èm Nè P = M èn P è. For the notion of a monoidal or tensor category, however, this natural transformation is of central importance. Problem Give an explicit proof of M èx Y è ç = M X M Y. P n i=1 v i v i 2. Show that for every nite dimensional vector space V there is a unique element 2 V V such that the following holds 8v 2 V : X i v i èvèv i = v: èhint: Use an isomorphism EndèV è ç = V V and dual bases fv i g of V and fv i g of V.è 3. Show that the following diagrams ècoherence diagrams or constraintsè commute in KMod: èèa Bè Cè D èa;b;cè1 èa èb Cèè D èa;bc;dè A èèb Cè Dè èab;c;dè èa Bè èc Dè èa;b;cdè 1èB;C;Dè A èb èc Dèè èa Kè B èa;k;bè Q ç ç ç QQQQs ç+ ç A B çèaè1 A èk Bè 1çèBè

15 4. TENSOR PRODUCTS Write çèa; Bè :A B,! B A for çèa; Bè:a b 7! b a. Show that ç is a natural transformation èbetween which functorsè. Show that èa Bè C çèx;bè1 èb Aè C B èa Cè çèa;bcè A èb Cè èb Cè A commutes for all A; B; C 2 KMod and that 1çèA;Cè B èc Aè çèb;aèçèa; Bè=id AB for all A, B in KMod. 5. Find an example of M, N 2 KModK such that M K N 6ç = N K M.

16 16 8. TOOLBOX 5. Algebras Let K be a commutative ring. In most of our applications K will be a eld. Tensor products of Kmodules will be simply written as M N := M K N. Every such tensor product is again a Kbimodule since each Kmodule M resp. N is a Kbimodule èsee è. Denition A Kalgebra is a vector space A together with a multiplication r : A A,! A that is associative: A A A id r A A and a unit ç : K,! A: rid r A A K A ç = A ç = A K r H HHHHid A id ç A A çid H HH Hj A A A: A Kalgebra A is commutative if the following diagram commutes A A ç A A A r AA r AU A: Let A and B be Kalgebras. A homomorphism of algebras f : A,! B is a Klinear map such that the following diagrams commute: A A f f r B B r and ra A f K rb B A ça AA çb AU f A B:

17 5. ALGEBRAS 17 Remark Every Kalgebra A is a ring with the multiplication A A,! A A r,! A: The unit element is çè1è, where 1 is the unit element of K. Obviously the composition of two homomorphisms of algebras is again a homomorphism of algebras. Furthermore the identity map is a homomorphism of algebras. Hence the Kalgebras form a category KAlg. The category of commutative Kalgebras will be denoted by KcAlg. Problem Show that End K èv èisakalgebra. 2. Show that èa; r : A A,! A; ç : K,! Aè isakalgebra if and only if A with the multiplication A A,! A A r,! A and the unit çè1è is a ring and ç : K,! CentèAè is a ring homomorphism into the center of A. 3. Let V be a Kmodule. Show that DèV è:= K V with the multiplication èr 1 ;v 1 èèr 2 ;v 2 è:=èr 1 r 2 ;r 1 v 2 + r 2 v 1 è is a commutative Kalgebra. Lemma Let A and B be algebras. multiplication èa 1 b 1 èèa 2 b 2 è:=a 1 a 2 b 1 b 2. Then A B is an algebra with the Proof. Certainly the algebra properties can easily be checked by a simple calculation with elements. We prefer for later applications a diagrammatic proof. Let r A : A A,! A and r B : B B,! B denote the multiplications of the two algebras. Then the new multiplication is r AB := èr A r B èè1 A ç 1 B è: A B A B,! A B where ç : B A,! A B is the symmetry map from Theorem Now the following diagrams commute A B A B A B 1 3 ç 1 A B A A B B 1rr 1ç 1 3 A B A B 1ç 1 3 A A B A B B * 1çB;AA1 3 1ç ç 1 A A B B A B ç H HHHHHj 1 2 ç 1 2 1çBB;A1 A A A B B B 1r1r A A B B rr1 2 r1r1 rr A B A B 1ç 1 A A B B rr A B In the left upper rectangle of the diagram the quadrangle commutes by the properties of the tensor product and the two triangles commute by inner properties of ç. The right upper and left lower rectangles commute since ç is a natural transformation and the right lower rectangle commutes by the associativity of the algebras A and B.

18 18 8. TOOLBOX Furthermore we use the homomorphism ç = ç AB : K,! K K,! A B in the following commutative diagram K A B ç = A B ç = A B K A B K K 1 2 çç A B A B K K A B çç1 2 H HHHH 1ç 1 H HHHH H HH Hj H HHHH H HHHH K A K B ç1ç1 H HH Hj A K B K H HHHH 1 1ç 1 H HHHH 1ç1ç H HHHH 1ç 1 A A B B rr A B A B 1ç 1 A A B B rr Hj A B: Denition Let K be a commutative ring. Let V be a Kmodule. A K algebra T èv è together with a homomorphism of Kmodules ç : V,! T èv è is called a tensor algebra over V if for each Kalgebra A and for each homomorphism of K modules f : V,! A there exists a unique homomorphism of Kalgebras g : T èv è,! A such that the diagram V ç T èv f A commutes. Note: If you want to dene a homomorphism g : T èv è,! A with a tensor algebra as domain you should dene it by giving a homomorphism of Kmodules dened on V. Lemma A tensor algebra èt èv è;çè dened byv is unique up to a unique isomorphism. Proof. Let èt èv è;çè and èt 0 èv è;ç 0 è be tensor algebras over V. Then implies k = h,1. T èv è H HHHHç ç,, ç 0 H 0 HH Hj h T 0 èv è k T èv è h T 0 èv è g

19 5. ALGEBRAS 19 Proposition èrules of computation in a tensor algebraè Let èt èv è;çè be the tensor algebra over V. Then we have 1. ç : V,! T èv è is injective èso we may identify the elements çèvè and v for all v 2 V è, 2. T èv è=f P n; ç i v i 1 ::: v in jçi =èi 1 ;::: ;i n è multiindex of length ng; 3. if f : V,! A is a homomorphism of Kmodules, A is a Kalgebra, and g : T èv è,! A is the induced homomorphism of Kalgebras, then gè X n;ç i v i1 ::: v in è= X n;ç i fèv i1 è ::: fèv in è: Proof. 1. Use the embedding homomorphism j : V,! DèV è, where DèV è is dened as in to construct g : T èv è,! DèV è such that g ç = j. Since j is injective soisç. 2. Let B := f P n; ç i v i 1 ::: v in jçi =èi 1 ;::: ;i n èmultiindex of length ng. Obviously B is the subalgebra of T èv è generated by the elements of V. Let j : B,! T èv èbe the embedding homomorphism. Then ç : V,! T èv è factors through a linear map ç 0 : V,! B. In the following diagram ç V 0 j B T èv 0 id, B p j B T èv è we have id B ç 0 = ç 0. p with p j ç 0 = p ç = ç 0 exists since ç 0 is a homomorphism of Kmodules. Because of jp ç = j ç 0 = ç =id T èv è ç we get jp =id T èv è, hence the embedding j is surjective and thus j is the identity. 3. is precisely the denition of the induced homomorphism. Proposition Given a Kmodule V. èt èv è;çè. Then there exists a tensor algebra Proof. Dene T n èv è:=v ::: V = V n to be the nfold tensor product of V. Dene T 0 èv è:=k and T 1 èv è:=v.we dene T èv è:= M iç0 T i èv è=k V èv V è èv V V è ::: : The components T n èv èoft èv è are called homogeneous components. The canonical isomorphisms T m èv è T n èv è ç = T m+n èv è taken as multiplication r : T m èv è T n èv è,! T m+n èv è r : T èv è T èv è,! T èv è and the embedding ç : K = T 0 èv è,! T èv è induce the structure of a Kalgebra on T èv è. Furthermore we have the embedding ç : V,! T 1 èv è ç T èv è.

20 20 8. TOOLBOX We have to show that èt èv è;çè is a tensor algebra. Let f : V,! A be a homomorphism of Kmodules. Each element int èv è is a sum of decomposable tensors v 1 ::: v n. Dene g : T èv è,! A by gèv 1 ::: v n è:= fèv 1 è :::fèv n è èand èg : T 0 èv è,! Aè = èç : K,! Aèè. By induction one sees that g is a homomorphism of algebras. Since èg : T 1 èv è,! Aè = èf : V,! Aè we get g ç = f. If h : T èv è,! A is a homomorphism of algebras with h ç = f we get hèv 1 ::: v n è=hèv 1 è :::hèv n è=fèv 1 è :::fèv n è hence h = g. Proposition The construction of tensor algebras T èv è denes a functor T : KMod,! KAlg that is left adjoint to the underlying functor U : KAlg,! KMod. Proof. Follows from the universal property and Problem Let X be a set and V := KX be the free Kmodule over X. Show that X,! V,! T èv è denes a free algebra over X, i.e. for every K algebra A and every map f : X,! A there is a unique homomorphism of Kalgebras g : T èv è,! A such that the diagram T èv è commutes. We write KhXi := T èkxè and call it the polynomial ring over K in the noncommuting variables X. 2. Let T èv è and ç : V,! T èv è be a tensor algebra. Regard V as a subset of T èv èby ç. Show that there is a unique homomorphism : T èv è,! T èv è T èv è with èvè =v 1+1 v for all v 2 V. 3. Show that è 1è=è1 è : T èv è,! T èv è T èv è T èv è. 4. Show that there is a unique homomorphism of algebras " : T èv è,! K with "èvè = 0 for all v 2 V. 5. Show that è" 1è=è1 "è = id T èv è. 6. Show that there is a unique homomorphism of algebras S : T èv è,! T èv è op with Sèvè =,v. èt èv è op is the opposite algebra of T èv è with multiplication st := ts for all s; t 2 T èv è=t èv è op and where st denotes the product in T èv è.è 7. Show that the diagrams commute. T èv è " T èv è T èv è K 1S g A ç T èv è 6 r T èv è T èv è

21 5. ALGEBRAS 21 Denition Let K be a commutative ring. Let V be a Kmodule. A K algebra SèV è together with a homomorphism of Kmodules ç : V,! SèV è, such that çèvè çèv 0 è=çèv 0 è çèvè for all v;v 0 2 V, is called a symmetric algebra over V if for each Kalgebra A and for each homomorphism of Kmodules f : V,! A, such that fèvè fèv 0 è=fèv 0 è fèvè for all v;v 0 2 V, there exists a unique homomorphism of Kalgebras g : SèV è,! A such that the diagram V ç SèV commutes. Note: If you want to dene a homomorphism g : SèV è,! A with a symmetric algebra as domain you should dene it by giving a homomorphism of Kmodules f : V,! A satisfying fèvè fèv 0 è=fèv 0 è fèvè for all v;v 0 2 V. Lemma A symmetric algebra èsèv è;çè dened byv is unique up to a unique isomorphism. Proof. Let èsèv è;çè and ès 0 èv è;ç 0 è be symmetric algebras over V. Then implies k = h,1. SèV g A H HHHHç ç,, ç 0 H 0 HH Hj h S 0 èv è k SèV è h S 0 èv è Proposition èrules of computation in a symmetric algebraè Let èsèv è;çè be the symmetric algebra over V. Then we have 1. ç : V,! SèV è is injective èwe will identify the elements çèvè and v for all v 2 V è, 2. SèV è=f P n;ç i v i1 ::: v in jçi =èi 1 ;::: ;i n è multiindex of length ng; 3. if f : V,! A is a homomorphism of Kmodules satisfying fèvè fèv 0 è = fèv 0 èfèvè for all v;v 0 2 V, A is a Kalgebra, and g : SèV è,! A is the induced homomorphism Kalgebras, then gè X n;ç i v i1 ::: v in è= X n;ç i fèv i1 è ::: fèv in è: Proof. 1. Use the embedding homomorphism j : V,! DèV è, where DèV è is the commutative algebra dened in to construct g : SèV è,! DèV è such that g ç = j. Since j is injective soisç.

22 22 8. TOOLBOX 2. Let B := f P n;ç i v i1 ::: v in jç i =èi1 ;::: ;i n èmultiindex of length ng. Obviously B is the subalgebra of SèV è generated by the elements of V. Let j : B,! SèV èbe the embedding homomorphism. Then ç : V,! SèV è factors through a linear map ç 0 : V,! B. In the following diagram ç V 0 j B SèV 0 id, B p j B SèV è we have id B ç 0 = ç 0, p with p j ç 0 = p ç = ç 0 exists since ç 0 is a homomorphism of Kmodules satisfying ç 0 èvè ç 0 èv 0 è = ç 0 èv 0 è ç 0 èvè for all v;v 0 2 V. Because of jp ç = j ç 0 = ç =id SèV è ç we get jp =id SèV è, hence the embedding j is surjective and thus the identity. 3. is precisely the denition of the induced homomorphism. Proposition Let V be a Kmodule. The symmetric algebra èsèv è;çè is commutative and satises the following universal property: for each commutative Kalgebra A and for each homomorphism of Kmodules f : V,! A there exists a unique homomorphism of Kalgebras g : SèV è,! A such that the diagram commutes. V ç SèV Proof. Commutativity follows from the commutativity of the generators: vv 0 = v 0 v which carries over to the elements of the form P n; ç i v i 1 ::: v in. The universal property follows since the dening condition fèvèfèv 0 è=fèv 0 èfèvè for all v;v 0 2 V is automatically satised. Proposition Given a Kmodule V. Then there exists a symmetric algebra èsèv è;çè. Proof. Dene SèV è:=t èv è=i where I = hvv 0, v 0 vjv;v 0 2 V i is the twosided ideal generated by the elements vv 0, v 0 v. Let ç be the canonical map V,! T èv è,! SèV è. Then the universal property is easily veried by the homomorphism theorem for algebras. Proposition The construction of symmetric algebras SèV è denes a functor S : KMod,! KcAlg that is left adjoint to the underlying functor U : KcAlg,! g A

23 Proof. Follows from the universal property and ALGEBRAS 23 Problem Let X be a set and V := KX be the free Kmodule over X. Show that X,! V,! SèV è denes a free commutative algebra over X, i.e. for every commutativekalgebra A and every map f : X,! A there is a unique homomorphism of Kalgebras g : SèV è,! A such that the diagram X SèV commutes. The algebra KëXë :=SèKXè is called the polynomial ring over K in the ècommutingè variables X. 2. Let SèV è and ç : V,! SèV è be a symmetric algebra. Show that there is a unique homomorphism : SèV è,! SèV è SèV è with èvè =v 1+1 v for all v 2 V. 3. Show that è 1è=è1 è : SèV è,! SèV è SèV è SèV è. 4. Show that there is a unique homomorphism of algebras " : SèV è,! K with "èvè = 0 for all v 2 V. 5. Show that è" 1è=è1 "è = id SèV è. 6. Show that there is a unique homomorphism of algebras S : SèV è,! SèV è with Sèvè =,v. 7. Show that the diagrams commute. SèV è " SèV è SèV è K 1S g A ç SèV è 6 r SèV è SèV è Denition Let K be a commutative ring. Let V be a Kmodule. A Kalgebra EèV è together with a homomorphism of Kmodules ç : V,! EèV è, such that çèvè 2 = 0 for all v 2 V, is called an exterior algebra orgrassmann algebra over V if for each Kalgebra A and for each homomorphism of Kmodules f : V,! A, such that fèvè 2 = 0 for all v 2 V, there exists a unique homomorphism of Kalgebras g : EèV è,! A such that the diagram V ç EèV f g A

24 24 8. TOOLBOX commutes. The multiplication in EèV è is usually denoted by u ^ v. Note: If you want to dene a homomorphism g : EèV è,! A with an exterior algebra as domain you should dene it by giving a homomorphism of Kmodules dened on V satisfying fèvè 2 = 0 for all v;v 0 2 V. Problem Let f : V,! A be a linear map satisfying fèvè 2 = 0 for all v 2 V. Then fèvèfèv 0 è=,fèv 0 èfèvè for all v;v 0 2 V. 2. Let2beinvertible in K èe.g. K a eld of characteristic 6= 2è. Let f : V,! A be a linear map satisfying fèvèfèv 0 è=,fèv 0 èfèvè for all v;v 0 2 V. Then fèvè 2 =0 for all v 2 V. Lemma An exterior algebra èeèv è;çè dened byv is unique up to a unique isomorphism. Proof. Let èeèv è;çè and èe 0 èv è;ç 0 è be exterior algebras over V. Then EèV è H HHHHç ç,, ç 0 H 0 HH Hj h E 0 èv è k EèV è h E 0 èv è implies k = h,1. Proposition èrules of computation in an exterior algebraè Let èeèv è;çè be the exterior algebra over V. Then we have 1. ç : V,! EèV è is injective èwe will identify the elements çèvè and v for all v 2 V è, 2. EèV è=f P n; ç i v i 1 ^ :::^ v in jçi =èi 1 ;::: ;i n è multiindex of length ng; 3. if f : V,! A is a homomorphism of Kmodules satisfying fèvè fèv 0 è =,fèv 0 è fèvè for all v;v 0 2 V, A is a Kalgebra, and g : EèV è,! A is the induced homomorphism Kalgebras, then gè X n;ç i v i1 ^ :::^ v in è= X n;ç i fèv i1 è ::: fèv in è: Proof. 1. Use the embedding homomorphism j : V,! DèV è, where DèV è is the algebra dened in to construct g : EèV è,! DèV è such that g ç = j. Since j is injective so is ç. 2. Let B := f P n;ç i v i1^:::^v in jçi =èi 1 ;::: ;i n èmultiindex of length ng. Obviously B is the subalgebra of EèV è generated by the elements of V. Let j : B,! EèV èbe the embedding homomorphism. Then ç : V,! EèV è factors through a linear map

25 ç 0 : V,! B. In the following diagram ç V 0 j B EèV 0 id, B p j B EèV è 5. ALGEBRAS 25 we have id B ç 0 = ç 0, p with p j ç 0 = p ç = ç 0 exists since ç 0 is a homomorphism of Kmodules satisfying ç 0 èvè ç 0 èv 0 è=,ç 0 èv 0 è ç 0 èvè for all v;v 0 2 V. Because of jp ç = j ç 0 = ç =id EèV è ç we get jp =id EèV è, hence the embedding j is surjective and thus j is the identity. 3. is precisely the denition of the induced homomorphism. Proposition Given a Kmodule V. Then there exists an exterior algebra èeèv è;çè. Proof. Dene EèV è:=t èv è=i where I = hv 2 jv 2 V i is the twosided ideal generated by the elements v 2. Let ç be the canonical map V,! T èv è,! EèV è. Then the universal property is easily veried by the homomorphism theorem for algebras. Problem Let V be a nite dimensional vector space of dimension n. Show that EèV è is nite dimensional of ç dimension 2 n. èhint: The homogeneous components E i èv èhave dimension. ç n i 2. Show that the symmetric group S n operates èfrom the leftè on T n èv èby çèv 1 ::: v n è=v ç,1 è1è ::: v ç,1 ènè with ç 2 S n and v i 2 V. 3. A tensor a 2 T n èv è is called a symmetric tensor if çèaè =a for all ç 2 S n. Let ^Sn èv è be the subspace of symmetric tensors in T n èv è. aè Show that S : T n èv è 3 a 7! P ç2sn çèaè 2 T n èv è is a linear map. bè Show that S has its image in ^Sn èv è. cè Show that ImèSè = ^Sn èv èifn! isinvertible in K. dè Show that ^Sn èv è,! T n èv è ç! S n èv è is an isomorphism if n! isinvertible in K and ç : T n èv è,! S n èv è is the restriction of ç : T èv è,! SèV è, the symmetric algebra. 4. A tensor a 2 T n èv è is called an antisymmetric tensor if çèaè ="èçèa for all ç 2 S n where "èçè is the sign of the permutation ç. Let ^E n èv è be the subspace of antisymmetric tensors in T n èv è. aè Show that E : T n èv è 3 a 7! P ç2sn "èçèçèaè 2 T n èv è is a linear map. bè Show that E has its image in ^E n èv è. cè Show that ImèEè = ^E n èv èifn! isinvertible in K. dè Show that ^E n èv è,! T n èv è ç! E n èv è is an isomorphism if n! isinvertible in K and ç : T n èv è,! E n èv è is the restriction of ç : T èv è,! EèV è, the exterior algebra.

26 26 8. TOOLBOX Denition Let A be a Kalgebra. A left Amodule is a Kmodule M together with a homomorphism ç M : A M,! M, such that the diagrams A A M id ç A M and rid A M M ç = K M ç çid ç M A M H HHHH id H HH Hj commute. Let A M and A N be Amodules and let f : M,! N be a Klinear map. The map f is called a homomorphism of modules if the diagram A M çm ç M M 1f A N N çn commutes. The left Amodules and their homomorphisms form the category A M of Amodules. Problem Show that an abelian group M is a left module over the ring A if and only if M is a Kmodule and an Amodule in the sense of Denition f

27 6. COALGEBRAS Coalgebras Denition A Kcoalgebra is a Kmodule C together with a comultiplication or diagonal :C,! C C that is coassociative: C C C id and a counit or augmentation : C,! K: C H HHHHid C C id C C C C C id C C K C ç = C ç = C K: id H HH Hj A Kcoalgebra C is cocommutative if the following diagram commutes C A AA AU ç C C C C Let C and D be Kcoalgebras. A homomorphism of coalgebras f : C,! D is a Klinear map such that the following diagrams commute: C f D and C C C D D f f f C D: A C AA D AU K Remark Obviously the composition of two homomorphisms of coalgebras is again a homomorphism of coalgebras. Furthermore the identity map is a homomorphism of coalgebras. Hence the Kcoalgebras form a category KCoalg. The category of cocommutative Kcoalgebras will be denoted by KcCoalg. D

28 28 8. TOOLBOX Problem Show that V V is a coalgebra for every nite dimensional vector space V over a eld K if the comultiplication is dened by èv v è := P n i=1 v v i v i v where fv i g and fv i g are dual bases of V resp. V. 2. Show that the free Kmodules KX with the basis X and the comultiplication èxè =x x is a coalgebra. What is the counit Is the counit unique 3. Show that K V with è1è = 1 1, èvè =v 1+1 v denes a coalgebra. 4. Let C and D be coalgebras. Then C D is a coalgebra with the comultiplication CD := è1 C ç 1 D èè C D è:c D C D,! C D and counit " = " CD : C D,! K K,! K. èthe proof is analogous to the proof of Lemma è To describe the comultiplication of a Kcoalgebra in terms of elements we introduce a notation rst introduced by Sweedler similar to the notation rèa bè = ab used for algebras. Instead of ècè = P c i c 0 i we write ècè = X c è1è c è2è : Observe that only the complete expression on the right hand side makes sense, not the components c è1è or c è2è which are not considered as families of elements of C. This notation alone does not help much in the calculations we have to perform later on. So we introduce a more general notation. Denition èsweedler Notationè Let M be an arbitrary Kmodule and C be a Kcoalgebra. Then there is a bijection between all multilinear maps and all linear maps f : C ::: C,! M f 0 : C ::: C,! M: These maps are associated to each other by the formula fèc 1 ;::: ;c n è=f 0 èc 1 ::: c n è: For c 2 C we dene X fècè1è;::: ;c ènèè :=f 0 è n,1 ècèè; where n,1 denotes the n, 1fold application of, for example n,1 =è 1 ::: 1è è 1è. In particular we obtain for the bilinear map : C C 3 èc; dè 7! c d 2 C C X cè1è c è2è =ècè; and for the multilinear map 2 : C C C,! C C C X cè1è c è2è c è3è =è 1èècè =è1 èècè: With this notation one veries easily X cè1è ::: èc èiè è ::: c ènè = X c è1è ::: c èn+1è

29 6. COALGEBRAS 29 and P cè1è ::: èc èiè è ::: c ènè = P c è1è ::: 1 ::: c èn,1è = P c è1è ::: c èn,1è This notation and its application to multilinear maps will also be used in more general contexts like comodules. Proposition Let C be a coalgebra and A an algebra. Then the composition f g := r A èf gè C denes a multiplication HomèC; Aè HomèC; Aè 3 f g 7! f g 2 HomèC; Aè; such that HomèC; Aè becomes an algebra. The unit element is given by K 3 7! èc 7! çèècèèè 2 HomèC; Aè. Proof. The multiplication of HomèC; Aè obviously is a bilinear map. The multiplication is associative since èf gè h = r A èèr A èf gè C è hè C = r A èr A 1èèèf gèhèè C 1è C = r A è1r A èèf èg hèèè1 C è C = r A èf èr A èg hè C èè C = f èg hè. Furthermore it is unitary with unit 1 HomèC;Aè = ç A C since ç A C f = r A èç A C fè C = r A èç A 1 A èè1 K fèè C 1 C è C = f and similarly f ç A C = f. Denition The multiplication : HomèC; AèHomèC; Aè,! HomèC; Aè is called convolution. Corollary Let C be akcoalgebra. Then C = Hom K èc;kè is an K algebra. Proof. Use that K itself is a Kalgebra. Remark If we write the evaluation as C C 3 a c 7! ha; ci 2K then an element a 2 C is completely determined by the values of ha; ci for all c 2 C. So the product of a and b in C is uniquely determined by the formula The unit element ofc is 2 C. ha b; ci = ha b; ècèi = X aèc è1è èbèc è2è è: Lemma Let K be a eld and A be a nite dimensional Kalgebra. Then A = Hom K èa;kè is a Kcoalgebra. Proof. Dene the comultiplication on C by :A,! r èa Aè can,!,1 A A : The canonical map can : A A,! èaaè is invertible, since A is nite dimensional. By a diagrammatic proof or by calculation with elements it is easy to show that A becomes a Kcoalgebra. Remark If K is an arbitrary commutative ring, then A = Hom K èa;kè is a Kcoalgebra if A is a nitely generated projective Kmodule.

30 For m 2 M we dene X fèmè1è;::: ;m ènè;m èmèè :=f 0 è n èmèè; TOOLBOX Problem Find sucient conditions for an algebra A resp. a coalgebra C such that HomèA; Cè becomes a coalgebra with coconvolution as comultiplication. Denition Let C be a Kcoalgebra. A left Ccomodule is a Kmodule M together with a homomorphism M : M,! C M, such that the diagrams M C M and C M M id H HHHHid id C C M H HH Hj C M K M ç = M: id commute. Let C M and C N be Ccomodules and let f : M,! N be a Klinear map. The map f is called a homomorphism of comodules if the diagram M M C M f N N 1f C N commutes. The left Ccomodules and their homomorphisms form the category C M of comodules. Let N be an arbitrary Kmodule and M be a Ccomodule. Then there is a bijection between all multilinear maps and all linear maps f : C ::: M,! N f 0 : C ::: M,! N: These maps are associated to each other by the formula fèc 1 ;::: ;c n ;mè=f 0 èc 1 ::: c n mè: where n denotes the nfold application of, i.e. n =è1 ::: 1 è è1 è.

31 6. COALGEBRAS 31 In particular we obtain for the bilinear map : C M,! C M X mè1è m èmè = èmè; and for the multilinear map 2 : C C M,! C C M X mè1è m è2è m èmè =è1 èècè=è 1èèmè: Problem Show that a nite dimensional vector space V is a comodule over the coalgebra V V as dened in problem with the coaction èvè := P v v i v i 2 èv V è P V where v i v i is the dual basis of V in V V. Theorem èfundamental Theorem for Comodulesè Let K be a eld. Let M be a left Ccomodule and let m 2 M be given. Then there exists a nite dimensional subcoalgebra C 0 ç C and a nite dimensional C 0 comodule M 0 with m 2 M 0 ç M where M 0 ç M is a Ksubmodule, such that the diagram M 0 0 C 0 M 0 commutes. M C M Corollary Each element c 2 C of a coalgebra iscontained in a nite dimensional subcoalgebra of C. 2. Each element m 2 M of a comodule is contained in a nite dimensional subcomodule of M. Corollary Each nite dimensional subspace V of a coalgebra C is contained in a nite dimensional subcoalgebra C 0 of C. 2. Each nite dimensional subspace V of a comodule M is contained in a nite dimensional subcomodule M 0 of M. Corollary Each coalgebra is a union of nite dimensional subcoalgebras. 2. Each comodule is a union of nite dimensional subcomodules. Proof. èof the Theoremè We can assume that m 6= 0 for else we can use M 0 =0 and C 0 =0. Under the representations of èmè 2 C M as nite sums of decomposable tensors pick one èmè = sx i=1 c i m i

32 32 8. TOOLBOX of shortest length s. Then the families èc i ji =1;::: ;sè and èm i ji =1;::: ;sè are linearly independent. Choose coecients c ij 2 C such that èc j è= tx i=1 c i c ij ; 8j =1;::: ;s; by suitably extending the linearly independent family èc i ji =1;::: ;sè to a linearly independent family èc i ji =1;::: ;tè and t ç s. P s We rst show that we can choose t = s. By coassociativity wehave i=1 c i èm i è= P s j=1 èc jè m j = P s linearly independent we can compare coecients and get è1è èm i è= j=1 P t i=1 c i c ij m j. Since the c i and the m j are sx j=1 c ij m j ; 8i =1;::: ;s and 0 = P s j=1 c ij m j for iés. The last statement implies Hence we get t = s and c ij =0; èc j è= sx i=1 8i é s; j =1;::: ;s: c i c ij ; 8j =1;::: ;s: Dene nite dimensional subspaces C 0 = hc ij ji; j =1;::: ;siçc and M 0 = hm i ji =1;::: ;siçm. Then by è1è we get : M 0,! C 0 M 0. We show that m 2 M 0 and that the restriction of to C 0 gives a linear map : C 0,! C 0 C 0 so that the required properties of the theorem are satised. First observe that m = P "èci èm i 2 M 0 and c j = P "èc i èc ij 2 C 0. Using coassociativity we get P n i;j=1 c i èc ij è m j = P s k;j=1 èc kè c kj m j = P s i;j;k=1 c i c ik c kj m j hence è2è èc ij è= sx k=1 c ik c kj : Remark We give asketch of a second proof which is somewhat more technical. Since C is a Kcoalgebra, the dual C is an algebra. The comodule structure : M,! C M leads to a module structure by ç = èev 1èè1 è :C M,! C CM,! P M. Consider the submodule N := C m. Then N is nite dimensional, since c n m = i=1hc ;c i im i for all c 2 P C n where i=1 c i m i = èmè. Observe that C m is a subspace of the space generated by the m i. But it does not depend on the choice of the m i. Furthermore if we take èmè = P c i m i with a shortest

33 6. COALGEBRAS 33 representation then the m i are in C m since c m = P hc ;c i im i = m i for c an element of a dual basis of the c i. N is a Ccomodule since èc mè = P hc ;c i ièm i è = P hc ;c iè1è ic iè2è m i 2 C C m. Now we construct a subcoalgebra D of C such that N is a Dcomodule with the induced coaction. Let D := N N. By 8.13 N is a comodule over the coalgebra N N. Construct a linear map ç : D,! C by n n P denition of the dual basis we have n = n i hn i ;ni. Thus we get 7! P n è1è hn ;n ènè i. By èç çè D èn n è=èç çèè P n n i n i n è = P n è1è hn i ;n èn è in iè1è hn ;n iènè i = P n è1è n iè1è hn ;n ièn è ihn i ;n ènèi = P n è1è n è2è hn ;n ènè i = P C èn è1è èhn ;n ènè i = C çèn n è: Furthermore " C çèn n è = "è P n è1è hn ;n ènè i = hn ; P "èn è1è èn ènè i = hn ;ni = "ènn è. Hence ç : D,! C is a homomorphism of coalgebras, D is nite dimensional and the image C 0 := çèdè is a nite dimensional subcoalgebra of C. Clearly N is also a C 0 comodule, since it is a Dcomodule. Finally we show that the Dcomodule structure on N if lifted to the Ccomodule structure coincides with the one dened on M. Wehave C èc mè= C è P hc ;m è1è im èm è è= P hc ;m è1è im è2è m èm è = P hc ;m è1è im è2è m i hm i ;m èmè i = P hc ;m è1è im è2è hm i ;m èm è im i =èç 1èè P hc ;m è1è im èmè m i m i è=èç 1èè P c m m i m i è =èç 1è D èc mè:

34 34 8. TOOLBOX 7. Bialgebras Denition A bialgebra èb;r;ç;;è consists of an algebra èb;r;çè and a coalgebra èb;;è such that the diagrams and B B ç r B B K A AA çç AU B B B B B B B B A AA AU K r B H HHHH 1ç 1 Hj B B B B rr B B K A AA ç id AU B commute, i.e. and are homomorphisms of algebras resp. r and ç are homomorphisms of coalgebras. 2. Given bialgebras A and B. A map f : A,! B is called a homomorphism of bialgebras if it is a homomorphism of algebras and a homomorphism of coalgebras. 3. The category of bialgebras is denoted by KBialg. Problem Let èb;r;çè be an algebra and èb;;"è be a coalgebra. The following are equivalent: aè èb;r;ç;;"è is a bialgebra. bè : B,! B B and " : B,! K are homomorphisms of Kalgebras. cè r : B B,! B and ç : K,! B are homomorphisms of Kcoalgebras. 2. Let B be a nite dimensional bialgebra over eld K. Show that the dual space B is a bialgebra. One of the most important properties of bialgebras B is that the tensor product over K of two Bmodules or two Bcomodules is again a Bmodule. Proposition Let B be a bialgebra. Let M and N be left Bmodules. Then M K N is a Bmodule by the map B M N,! 1 B B M N 1ç 1,! B M B N,! çç M N: 2. Let B be a bialgebra. Let M and N be left Bcomodules. Then M K N is a Bcomodule by the map M N,! B M B N 1ç 1,! B B M N,! r1 B M N: ç K

35 " 3. K is a Bmodule by the map B K ç = B 4. K is a Bcomodule by the map K ç 7. BIALGEBRAS 35,! K.,! B ç = B K. Proof. We give a diagrammatic proof for 1. The associativity law is given by B B M N 111 B B B M N B B B B M N r11 B B B B M N ç 1 1ç ç èbb;mè1 11ç 1 B B M B N çç B B B M B N 1ç èb;bm è11 1ç1ç B B M B B N 1çç B M N 11 B B M N 1ç 1 B M B N B M N 11 rr11 B B M N The unit law is the commutativity of M N ç = K M N ç11 1ç 1 r1r1 B M B N B M N çç çç M N = ç= K K M N çç11 11 B B M N 1ç 1 M N ç = K M K N ç1ç1 1 XXXXXXXXXXX Xz 1ç 1 B M B N çç M N The corresponding properties for comodules follows from the dualized diagrams. The module and comodule properties of K are easily checked. Denition Let èb;r;ç;;è be a bialgebra. Let A be a left Bmodule with structure map ç : B A,! A. Let furthermore èa; r A ;ç A è be an algebra such that r A and ç A are homomorphisms of Bmodules. Then èa; r A ;ç A ;çè is called a Bmodule algebra. 2. Let èb;r;ç;;è be a bialgebra. Let C be a left Bmodule with structure map ç : B C,! C. Let furthermore èc; C ;" C è be a coalgebra such that C and " C are homomorphisms of B modules. Then èc; C ;" C ;çè is called a Bmodule coalgebra. 3. Let èb;r;ç;;è be a bialgebra. Let A be a left Bcomodule with structure map : A,! B A. Let furthermore èa; r A ;ç A è be an algebra such that r A and ç A are homomorphisms of Bcomodules. Then èa; r A ;ç A ;è is called a Bcomodule algebra.