Zoran Skoda
Yetter-Drinfeld module algebra

Yetter-Drinfeld modules form a monoidal category. Yetter-Drinfeld algebra is a monoid in that category. It is an algebra and a comodule with the YD condition and the property that the coaction ρ:MMH op\rho : M\to M\otimes H^{op} is an algebra map.

Let HH be a Hopf algebra. The compatibility of the left action \blacktriangleright and the right coaction ρ:XX [0]X [1]\rho : X\mapsto X_{[0]}\otimes X_{[1]} for the left-right Yetter-Drinfeld HH-module MM compatibility is

(h (1)X [0])h (2)X [1]=(h (2)X) [0](h (2)X) [1]h (1), (h_{(1)}\blacktriangleright X_{[0]})\otimes h_{(2)} X_{[1]} = (h_{(2)}\blacktriangleright X)_{[0]} \otimes (h _{(2)}\blacktriangleright X)_{[1]} h_{(1)},

for all hHh\in H and XMX\in M.

Proposition. If ρ:MMH op\rho:M\to M\otimes H^{op} in other to check the YD module property it is sufficient to check it on the generators of algebras HH and MM.

Proof. It is clear that the YD condition is linear. We therefore have to check the compatibility for products if the factors satisfy it. For the products XYX Y in MM we compute

(f (1)(XY) [0])f (2)(XY) [1] = (f (1)(X [0]Y [0]))f (2)Y [1]X [1] = (f (1)X [0])(f (2)Y [0])f (3)Y [1]X [1] = (f (1)X [0])(f (3)Y [0]) [0](f (3)Y [0]) [1]f (2)X [1] = (f (2)X) [0](f (3)Y) [0](f (3)Y [0]) [1])f (2)X) [1]f (1) = ((f (2)X)(f (3)Y)) [0]((f (2)X)(f (3)Y)) [1]f (1) = (f (2)(XY)) [0](f (2)(XY)) [1]f (1),\array{ (f_{(1)}\blacktriangleright (X Y)_{[0]})\otimes f_{(2)} (X Y)_{[1]} &=& (f_{(1)}\blacktriangleright (X_{[0]} Y_{[0]}))\otimes f_{(2)}Y_{[1]} X_{[1]} \\ &=&(f_{(1)}\blacktriangleright X_{[0]}) (f_{(2)}\blacktriangleright Y_{[0]})\otimes f_{(3)}Y_{[1]} X_{[1]} \\ &=&(f_{(1)}\blacktriangleright X_{[0]})(f_{(3)}\blacktriangleright Y_{[0]})_{[0]}\otimes (f_{(3)}\blacktriangleright Y_{[0]})_{[1]} f_{(2)} X_{[1]} \\ &=& (f_{(2)}\blacktriangleright X)_{[0]}(f_{(3)}\blacktriangleright Y)_{[0]}\otimes (f_{(3)}\blacktriangleright Y_{[0]})_{[1]})f_{(2)}\blacktriangleright X)_{[1]} f_{(1)} \\ &=& ((f_{(2)}\blacktriangleright X)(f_{(3)}\blacktriangleright Y))_{[0]}\otimes ((f_{(2)}\blacktriangleright X)(f_{(3)}\blacktriangleright Y))_{[1]} f_{(1)} \\ &=& (f_{(2)}\blacktriangleright (X Y))_{[0]}\otimes (f_{(2)}\blacktriangleright (X Y))_{[1]} f_{(1)}, }

and for the products fgf g in HH we compute

(fg) (1)X [0](fg) (2)X [1] = (f (1)(g (1)X [0]))f (2)(g (2)X [1]) = (f (2)(g (2)X) [0])(f (2)(g (2)X) [1])g (1) = (f (2)(g (2)X)) [0](f (2)(g (2)X)) [1]f (1)g (1) = ((fg) (2)X) [0]((fg) (2)X) [1](fg) (1)\array{ (f g)_{(1)} X_{[0]}\otimes (f g)_{(2)} X_{[1]} &=& (f_{(1)}\blacktriangleright (g_{(1)}\blacktriangleright X_{[0]}))\otimes f_{(2)} (g_{(2)} X_{[1]}) \\ &=& (f_{(2)}\blacktriangleright (g_{(2)}\blacktriangleright X)_{[0]})\otimes (f_{(2)} \blacktriangleright (g_{(2)}\blacktriangleright X)_{[1]}) g_{(1)} \\ &=& (f_{(2)}\blacktriangleright (g_{(2)}\blacktriangleright X))_{[0]}\otimes (f_{(2)} \blacktriangleright (g_{(2)}\blacktriangleright X))_{[1]} f_{(1)} g_{(1)} \\ &=& ((f g)_{(2)}\blacktriangleright X)_{[0]}\otimes ((f g)_{(2)}\blacktriangleright X)_{[1]} (f g)_{(1)} }

Braided commutativity:

X [0](X [1]Y)=YX X_{[0]} (X_{[1]}\blacktriangleright Y) = Y X

Products on the left

X [0]X [0](X [1](X [1]Y))=X [0](X [1]Y)X=YXX X_{[0]}X'_{[0]}(X'_{[1]}\blacktriangleright(X_{[1]}\blacktriangleright Y)) = X_{[0]}(X_{[1]}\blacktriangleright Y) X' = Y X X'

Products on the right

X [0](X [1](YY))=X [0](X [1]Y)(X [2]Y)=YX [0](X [1]Y)=YYX X_{[0]} (X_{[1]}\blacktriangleright (Y Y')) = X_{[0]} (X_{[1]}\blacktriangleright Y)(X_{[2]}\blacktriangleright Y') = Y X_{[0]} (X_{[1]}\blacktriangleright Y') =Y Y' X

Last revised on June 22, 2015 at 17:45:59. See the history of this page for a list of all contributions to it.