Algebra actions

\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)We started talking about groups in terms of what they do to things: rotating a cube or a peg, etc. Matrix groups have natural examples of doing stuff to vectors.
Let's formalize a little.

An action of a set \(S\) on a set \(T\) is a map \(\rho\) that assigns to each \(s \in S\) a map from \(T\) to itself, usually obeying some rules about the structures of \(S\) and \(T\). If we fix \(s \in S\), then the map \(t \mapsto \rho(s, t)\) is a map from \(T\) to \(T\). So for each \(s \in S\), \(\rho\) can be used to get an element of \(End(T)\), the set of maps from \(T\) to itself.
We say that an action is faithful if the map from \(S\) to \(End(T)\) is injective, i.e. if \(\rho(s) = \rho(s')\) then \(s = s'\).
In diagrams, we'd write the action map using a white square: the solid line takes in an element of \(S\), the dashed is for elements of \(T\).

We usually think of \(End(T)\) as respecting various structures that \(T\) may have. For instance, if \(T\) is a vector space, then \(End(T)\) is the set of linear maps from \(T\) to itself, which for finite-dimensional \(T\) manifests a square matrices. Hence we would like that \(\rho(s)\) is a linear transformation for all \(s\in S\).
In fact, we're going to nail that down right now: all of our actions are on vector spaces.
Given a vector space \(V\) and an action map \(\rho\) of \(S\) on \(V\), we often say that \((V, \rho)\) is a representation of \(S\). We call \(V\) an \(S\)-module. This actually is a bit of a mix of terminology; a representation theorist will usually say that \(V\) is a representation, or affords a representation, rather than saying that the pair \((V,\rho)\) is a representation. The term module is usually used in algebra and related disciplines like algebraic geometry and refers to slightly more general structures that don't necessarily have built-in scalar multiplication. So the distinction I'm making here isn't entirely standard convention, but I think it to be useful.

What if \(S\) has structure as well? If it's a structure that \(End(T)\) carries, then we usually want \(\rho\) to respect that structure. For instance, if \(S\) is a vector space, then we want \(\rho\) to be a linear map from \(S\) to \(End(T)\), which is also a vector space.

Given a group \(G\), we can talk about a group action or group representation \(\rho\) of \(G\) on a vector space \(V\) saying that for each \(g \in G\), \(\rho(g)\) is in \(End(V)\). But \(G\) is more than just a set, it's a group, so we need a few things.
First, we want \(\rho\) to send group composition to composition of maps, so we say that, as maps from \(V\) to itself,
$$\rho(gh) = \rho(g)\circ\rho(h)$$Also, because it makes sense to, we'll say that for the identity element \(e \in G\), we have
$$\rho(e) = id_V$$Because where else should the identity go but to the identity?
So a group representation is an action of \(G\) on \(V\) that obeys those two laws. Note that this means that \(\rho(g^{-1}) = \rho(g)^{-1}\) automatically. Furthermore, that means that \(\rho(g)\) is always invertible. Hence \(G\) ends up in the subset \(GL(V)\) contained in \(End(V)\).

What about an algebra? What about a representation \(\rho\) of an algebra \(A\) on \(V\)?
An algebra is a vector space, so we say that the map \(\rho: A\rar End(V)\) is linear. An algebra also has a multiplication map, and \(End(V)\) has a composition operation, so we say that \(\rho\) should send multiplication to composition, just as with group actions. In other words,
$$\rho(ab) = \rho(a)\circ\rho(b)$$If \(A\) is unital, then we add in that for \(c \in \mb k\),
$$\rho(\eta(c)) = cId_V$$Note that even if \(A\) is not associative, \(End(V)\) is. So \(\rho(a(bc))\) and \(\rho((ab)c)\) end up going to the same thing: \(\rho(a)\rho(b)\rho(c)\). On the other hand, \(End(V)\) isn't generally commutative, so we can still detect a difference between \(\rho(ab)\) and \(\rho(ba)\).

From a group representation \(\rho: G\rar End(V)\), we easily get an algebra representation of \(\mb kG\) just by sending \(\sum_{g \in G} a_g \hat g\) to \(\sum_{g \in G} a_g \rho(g)\).

Diagrammatically, we have

Note how similar these are to the associativity and unit diagrams for algebras, only now instead of just white circles we also have some white squares.

Examples!
The trivial representation is of \(G\) or \(\mb kG\) acting on \(\mb k\) with everything in \(G\), or correspondingly all elements of the form \(\hat g\), acting as the identity; for \(\mb kG\) we then extend by linearity. This is boring, but is important enough that we mention it.
Matrix groups are naturally subgroups of \(GL(V)\) for some \(V\) by definition, and so naturally have representations.
Given a representation \(\rho: G \rar End(V)\), there is also a representation \(\rho^\vee\) on the dual vector space \(V^\vee\). Suppose that \(f \in V^\vee\). How should \(g\) act on \(f\)? We want it to be compatible with an action of \(g\) on \(V\). The compatibility is that acting on both \(f\) and \(v\) does nothing:
$$(\rho^\vee(g)f)(\rho(g)v) = f(v)$$Note that \(f(v)\) is in \(\mb k\), so we're thinking that the combined action on \(V\) and \(V^\vee\) is the trivial action.
Anyway, this leads to saying that
$$(\rho^\vee(g)f)(v) = f(\rho(g)^{-1}v)$$Note that this an action, in that
$$\rho^\vee(g)(\rho^\vee(h)f) = \rho^\vee(gh)f$$Check this!
Basically this is just the change-of-basis law for dual spaces; if one space changes by \(\rho(g)\), the dual space changes by \(\rho(g)^{-1}\).
More examples to come when we have a bit more machinery.