How the coproduct defines an action on $X\otimes Y$

Question

Given a bialgebra $A$ and two $A$-modules $X$ and $Y$. We can build the tensor product of the underlying vector spaces $X\otimes Y$.

What does it mean if one says 'The $A$-module structure on $X\otimes Y$ is given via the coproduct of $A$'?

Can I also define $a.(x\otimes y):=ax\otimes ay$ or, for instance, $a.(x\otimes y):=ax\otimes y$?

Answer 1

The co-product is an algebra homomorphism $\Delta:A\to A\otimes A$ (so $\Delta$ is linear and $\Delta(ab)=\Delta(a)\Delta(b)$). If we write $\Delta(a)=\sum a_{(1)}\otimes a_{(2)}$, then $$a.(x\otimes y)=\Delta(a)(x\otimes y)=\sum (a_{(1)}.x)\otimes (a_{(2)}.y).$$

For example, if $G$ is a group, then $\Delta(g)=g\otimes g$ defines a co-product on the group algebra $\mathbb{C}G$. In this case, $g.(x\otimes y)=(g.x)\otimes(g.y)$.

For a Lie algebra $\mathfrak{g}$, the co-product is given by $\Delta(a)=a\otimes 1+1\otimes a$ for $a\in \mathfrak{g}$. Therefore, $a.(x\otimes y)=(a.x)\otimes y+x\otimes(a.y)$.

Another example of a bialgebra is the ring of symmetric functions. The co-product is given by $\Delta(p_n)=p_n\otimes 1+1\otimes p_n$ for power sum symmetric functions. If you take the basis given by elementary symmetric functions, the co-product is $$\Delta(e_n)=\sum_{k=0}^n e_{n-k}\otimes e_k.$$