Coalgebra actions

\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)I know that most everyone who has already learned group representation theory is probably thinking "why are you talking about coalgebras now? Just list a bunch of representations of groups using the usual setup". But where's the fun in that?

As we saw, an algebra action of an algebra \(A\) on a vector space \(V\) turned the multiplication in \(A\) into composition in \(End(V)\), So let's see what that says about coalgebras.
We take \(\Delta\) in a coalgebra \(C\). Does \(End(V)\) have a comultiplication (cocomposition?) that we can map \(\Delta\) to? If \(V\) is infinite-dimensional, not really.
Instead we get a multiplication operation on representations.
Take two representations \(\rho, V\) and \(\sigma, W\). Then we can make the representation \((\rho \hat \otimes \sigma, V\otimes W)\), where for \(f \in C\),
$$(\rho \hat \otimes \sigma)(f) = (\rho \otimes \sigma)\Delta(f)$$In other words, we take the comultiplication of \(f\), apply \(\rho\) to the first factor and \(\sigma\) to the second factor, getting an element of \(End(V \otimes W)\).
So the coalgebra structure, rather than restricting actions the way an algebra structure does, instead allows for the easy creation of new ones via the tensor product. Diagrammatically:

So for instance, if we have a matrix group \(G\), there's a vector space \(V\) on which it acts naturally via the defining representation \(\rho\), and \(\mb kG\) also acts on \(V\) via \(\rho\). Since \(\mb kG\) has a natural comultiplication that sends \(\hat g\) to \(\hat g \otimes \hat g\), we get that \(\mb kG\) acts on \(V\otimes V\) by
$$(\rho \hat \otimes \rho)(g)(u\otimes g) = \rho(g)u\otimes \rho(g)v$$And similarly it acts on \(V\otimes V\otimes V\) etc.

This allows us to have some fun with putting structures on \(V\).
Suppose that we have an algebra \(A\) and a coalgebra \(C\). We say that a coalgebra action \(\rho\) of \(C\) on \(A\) is compatible with the algebra structure of \(A\) if it plays well with the algebra structure of \(A\). In other words,
$$\rho(f)(m_A(a\otimes b)) = m_A((\rho \hat \otimes \rho)(f) \textbf{ }(a \otimes b))$$where \(m_A\) is the multiplication on \(A\).
We then say that \(A\) is then a \(C\)-module algebra.
So for a group \(G\), for instance, we say that it acts on an algebra \(A\) if we have that \(\rho(g)(m_A(a\otimes b)) = m_A(\rho(g)a\otimes \rho(g)b)\). This allows us to talk about the automorphism group of an algebra: defining \(GL(A)\) to be the general linear group for \(A\) treated as a vector space, we look at the subgroup of all \(g\) such that the compatibility rule for acting on an algebra holds. We call the resulting group \(Aut(A)\).

We can also talk about a coalgebra acting on a coalgebra. Suppose that \(C\) acts on \(H\), both coalgebras. Then we ask that the comultiplications and the action map play well:
$$\Delta_H(\rho(f)h) = (\rho \hat \otimes \rho)(f) \Delta_H(h)$$where \(\Delta_H\) is the comultiplication on \(H\), and we say that \(H\) is a \(C\)-module coalgebra.
Again we can define \(Aut(H)\) for the coalgebra \(H\) as the subgroup of \(GL(H)\) for which that rule holds.

By all of this, we can define a coalgebra acting on a bialgebra by combining all of these rules. By combining the coalgebra action rules with the algebra action rules, we get the rules for the action of a bialgebra.

The diagrams for a coalgebra acting on an algebra and a coalgebra look like:

if you'll forgive using the same black dot to indicate comultiplication in both \(C\) and \(H\). The lower equation should look like the multiplication-comultiplication compatibility diagram for Hopf algebras. The upper equation doesn't have such a nice analogy in general.

What about Hopf algebras, in particular the antipode map? Well, we want any action on a Hopf algebra to preserve the antipode map:
$$\rho(g)(S_Vv) = S_V(\rho(g)v)$$which diagrammatically just means that you can move \(S\)s on the dashed lines past the white boxes.

As for a Hopf algebra acting on a vector space \(V\), having \(S_H\) allows for actions on the dual of \(V\); if \(S_H\) is invertible, we actually get two actions, one using \(S_H\) and one using \(S_H^{-1}\).
Let's take \((V,\rho)\) as a representation of \(H\), and let \((V^\vee, \rho^\vee)\) be the representation of \(H\) where for \(f \in V^\vee\), \(\rho^\vee(g)(f) = f \circ \rho(S_H(g))\). Similarly, let \((^\vee V, ^\vee\rho)\) have the action \(^\vee\rho(g)(f) = f \circ \rho(S_H^{-1}(g))\). You can check that both of these are actions. For the objects that we're familiar with, groups and Lie algebras and commutative rings, \(S_H = S_H^{-1}\) and so there's only one action on the dual space, but for more general Hopf algebras they don't have to be the same.
You might also consider higher powers of \(S_H\), but we're not going to.