Sheet 11. PD Dr. Ralf Holtkamp Prof. Dr. C. Schweigert Hopf algebras Winter term 2014/2015. Algebra and Number Theory Mathematics department

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1 Algebra and Number Theory Mathematics department PD Dr. Ralf Holtkamp Prof. Dr. C. Schweigert Hopf algebras Winter term 014/015 Sheet 11 Problem 1. We consider the C-algebra H generated by C and X with the relations: C = 1, X = 0 and CX + XC = Show that setting C) = C C and X) = 1 X + X C yields a well-defined Hopf-algebra. Solution. We have seen this algebra or some variation) already a few times. One should check that the definition of is compatible with the relation: C) = C C = 1 1 = 1) X) = 1 X + X CX + X XC + X C = X CX + XC) = 0 = 0) C) X) = C CX + CX 1 = C) X) The counity is given by ɛx) = 0 and ɛc) = 1. For S we set: SC) = C, SX) = CX and therefor SCX) = CXC = X. One easily checks that this gives indeed an antipode for H: SC)C = 1 = ɛc)1 S1)X + SX)C = 0 = ɛx)1 SC)CX + SCX)1 = 0 = ɛcx)1 CSC) = 1 = ɛc)1 1SX) + XSC) = 0 = ɛx)1 CSCX) + CXS1) = 0 = ɛcx)1. What is the order of S? Solution. S has of course order Show that R = C + C 1 C C) + 1 X X + X CX + CX CX CX X) is a universal R-matrix. Solution. First we need to show that R is invertible in H H. We remark that: C + C 1 C C) = 1 1 If we write a = C + C 1 C C) and t = 1 X X + X CX + CX CX CX X). We have R = a + t = a1 + at) further more, at is clearly nilpotent actually at) = 0). Hence we have: 1 at)a is the inverse of R. 1

2 We first need to show that opp C)R = R C) and opp X)R = R X). 1 opp C)R = C C) C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) = 1 C C + C C 1 1) + 1 CX CX + CX X + X X X CX) = C + C 1 C C)C C) + 1 XC XC XC X + X X + X XC) = C + C 1 C C)C C) + 1 X X X XC + XC XC + XC X)C C) = R C) opp X)R 1 = X 1 + C X) C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) 1 = X 1) C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) + C X) C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) = 1 X 1 + X C + XC 1 XC C) + 1 C X + C XC + 1 X 1 XC) = 1 X 1 + X C CX 1 + CX C) + 1 C X C CX + 1 X + 1 CX) R X) 1 = C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) 1 X + X C) 1 = C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) 1 X) C + C 1 C C) + 1 ) X X + X CX + CX CX CX X) X C) = 1 1 X + 1 CX + C X C CX) + 1 X C + X 1 + CX C CX 1) = opp X)R. Then we need to compute id )R), id)r), R 13 R 3 and R 13 R 1, but let us first rewrite R in a more convenient way: + C 1 C C C 1 C ) X X) R = C = C = 1 + C = 1 + C + C 1 C C C + X X) 1 1 C + C 1 + C ) 1 + C C + C + 1 C ) 1 X X) C + X X) 1 C C C ) 1

3 Remarking that 1+C and 1 C, are orthogonal idempotents, we obtain: R 13 R 3 = a 13 a 3 + a 13 t 3 + t 13 a 3 = C C + X 1 X) = id)a) C + + C C 1 C C C 1 C ) 1 X X) 1 C 1 + C + C 1 1 C ) C C 1 C ) 1 C 1 C C C = id)a) C + C C 1 C ) C C C 1 + C 1 X X) ) X 1 X) 1 X X) ) X C X) = id)a) C C C 1 C ) 1 X + X C) X) = id)a) + id)t) = id)r) 3

4 R 13 R 1 = a 13 a 1 + a 13 t 1 + t 13 a 1 = 1 + C C C C + C C + 1 C 1 C ) X X 1) + X 1 X) C C C 1 C ) = id )a) + C C C 1 C ) C + C C 1 + C = id )a) + C C 1 + C C ) C + C C 1 + C X X 1) ) X 1 X) X X C) ) X 1 X) = id )a) + C C 1 + C C ) X X C + X 1 X) = id )a) + C C 1 + C C ) X X C + 1 X)) This shows that R is a universal R-matrix. = id )a) + id )t) = id )R) 4. Deform R 1 into R q with q in C to obtain a one parameter family of universal R-matrices. Solution. R q = C + C 1 C C) + q X X + X CX + CX CX CX X) 5. Relate R 1 q and R q. Solution. R 1 q = τ A,A R q ) 4

5 Problem. Let H be a quasi-triangular Hopf algebra with R-matrix R = R) R 1) R ). Let X be a right H-module and define δ : X X H by v R vr 1) R ). 1. Show that X, δ) is a right H-comodule. Solution. We want to show that δ id H ) δ = id V ) δ and id V ɛ) δ = id V Let v be an element of X. We have: δ id H ) δv) = R δvr 1 )) R = R,R vr 1 R 1 R R = R,R ρ id H id H )v R 1 R 1 R R ) = R,R ρ id H id H )v R 13 R 1) = R = R ρ id H H )v R) vr 1 R ) id V ) δv) where R = R Note that we need to sum twice, that is why we take two different name for the same object) and ρ is the structural map of X as a module-h. Furthermore, we have: id V ɛ) δv) = = R vr 1 ɛr ) = v. Show that the right action and the right coaction on X fulfill the right-right) Yetter-Drinfeld condition: id X µ) τ H,X id H ) id H δρ))τ X,H id H ) id X ) =ρ µ) id X τ H,H id H ) δ ). Solution. Let v be an element of X and h an element of H. We compute: id X µ) τ H,X id H ) id H δρ))τ X,H id H ) id X )v h) = R,h) xh ) R 1 h 1) R = R,h) xr 1 h 1) R h ) = ρ µ) id X τ H,H id H ) δ )x v) 5

6 Problem 3. Let H be a bialgebra in a strict braided category C with braiding c, i.e. H is equipped with an algebra and a coalgebra structure which are compatible in the following way µ = µ µ)id c H,H id) ), η = η η, ɛµ = ɛ ɛ, ɛη = id 1. A right-right Yetter-Drinfeld module over H is an object X in C together with an associative, unital) action ρ : X H X and a coassociative, counital) coaction δ : X X H such that id X µ)c H,X id H )id H δρ))c X,H id H )id X ) =ρ µ)id X c H,H id H )δ ). 1. Assume that H is a Hopf algebra, i.e. there is a morphism S : H H such that µs id) = ηɛ = µid S). Show that X is a Yetter-Drinfeld module, if and only if δρ =id X µ)c H,X id H )id H ρ µ)s id X c H,H id H ) id H δ )c X,H id H )id X ) Solution. Graphically this is immediate: one should not forget the naturality of the braiding.. Let H be a Hopf-algebra. Show that H is a Yetter-Drinfeld module with δ := and ρ := µs µ)c H,H id)id ). Hint: The following equality holds S S) = c 1 H,H S. Solution. Graphically Problem 4. Let K be a field and let H, L be two bi-algebras over K and φ : H L a morphism of bialgebras. Denote by H-Mod resp. L-Mod the category of left modules over H resp. L and by Comod-H resp. Comod-L the category of right H resp. L comodules. 1. Show that φ induces a functor Φ : L-mod H-mod. Solution. This is clear: X, ρ) X, ρ φ id X )) f : X Y ) f : X Y ). Show that the functor Φ is strict monoidal. 6

7 Solution. We have to give natural isomorphisms between the H-modules ΦX Y ) and ΦX) ΦY ). The action on ΦX Y ) is given by The action on ΦX) ΦY ) is given by ρ X ρ Y )id τ K,X id) φ) id). ρ X ρ Y )φ id φ id)id τ H,X id) id). These coincide since φ commutes with the coproduct. We also have to give give an H-linear isomorphism ΦK) = K. But the H-modules ΦK) and K are equal, which follows since K is the ground field with the trivial action given by the counit and the counit is preserved by φ. Since the H-modules ΦX Y ) and ΦX) ΦY ) and ΦK) and K are equal we can take the identity linear maps as the needed isomorphisms. 3. Show that φ induces a functor Ψ : comod-h comod-l. Is this functor monoidal? Solution. Define Ψ : comod-h comod-k as follows X, δ) X, id X φ) δ) f : X Y ) f : X Y ) One has to check that id X φ) δ is a K-coaction on X, which follows since φ commutes with the comultiplications of H and K. Since φ also commutes with the multiplications of H and K this functor is again strict monoidal. Remember that the coaction of H resp. K on X Y involves the multiplication of H resp. K.) 4. Let H, L be quasi-triangular with R-matrices R, R. Show that in this case the functor Φ is braided, if and only if φ φ)r) = R. Solution. X, δ) X, id X φ) δ) f : X Y ) f : X Y ) One has to check that id X φ) δ is a K-coaction on X, which follows since φ commutes with the comultiplications of H and K. Since φ also commutes with the multiplications of H and K this functor is again strict monoidal. Remember that the coaction of H resp. K on X Y involves the multiplication of H resp. K.) Assume φ φ)r) = R, then Φc R X,Y ) = cr ΦX),ΦY ). Thus Φ is a braided functor. Now suppose Φ is a braided functor. We take for X and Y the regular left K-module, i.e. K with left multiplication. Since Φ is strict and braided we get the equality Φc R K,K) = c R ΦK),ΦK). If we apply these morphisms to η K η K we get the equality R = φ φ)r). Problem 5. Let H be a quasi-triangular Hopf algebra, with antipode S, R-matrix R = R 1 and Drinfeld element u = R SR ))R 1). We denote = τ. 7

8 1. Show that the following formula endow H H with a structure of module-h 4 : x y) a b c d) = Sb)xa Sd)yc. Solution. We have: x y) ) = S1)x1 S1)y1 = x y and x y) a b c d) e f g h)) = x y) ae bf cg dh) = Sbf)xae Sdh)ycg = Sf)Sb)xae Sh)Sd)ycg = Sf)Sb)xa)e Sh)Sd)yc)g = Sb)xa Sd)yc) e f g h) = x y) a b c d)) e f g h). Compute R 1 R 3 and R 1 R 3 R 13 R 1 R 14 ). Solution. R 1 R 3 = 1 1, R 1 R 3 R 13 ) = u 1 and R 1 R 3 R 13 R 1 R 14 ) = u 1 3. Prove the following equality in H 4 : R 1 )R) = R 3 R 13 R 1 R 14 R 4. Solution. We have: )R) = id H τ) id H ) id H )R) = id H τ) id H )R 13 R 1 ) = id H τ) id H )R 13 ) id H )R 1 )) = id H τ)r 14 R 4 R 13 R 3 )) = id H τ)r 13 R 3 R 14 R 4 )) Multiplying on the left by R 1, we obtain: R 1 )R) = R 1 R 13 R 3 R 14 R 4 = R 3 R 13 R 1 R 14 R 4 4. Prove that: u) = R 1 R) 1 u u) = u u)r 1 R) 1 5. Prove that g = usu 1 ) is group like, and that S 4 is an inner automorphism. 8