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 $\rho : M\to M\otimes H^{op}$ is an algebra map.

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

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

for all $h\in H$ and $X\in M$.

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

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 $X Y$ in $M$ we compute

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

and for the products $f g$ in $H$ we compute

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

Braided commutativity:

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

Products on the left

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

Products on the right

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