Lie algebra actions, universal enveloping algebra

\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)Now that we've talked about algebra and coalgebra actions in the context of groups, we can talk about them in the context of Lie algebras.

So let's consider a Lie group \(G\) and its Lie algebra \(\mf g\). Since we'll have several identities floating around, we'll write the identity element of \(G\) as \(e\), whereas identity matrices will be denoted \(I\).
For a vector space \(V\), \(End(V)\) has a Lie bracket, where for matrices \(S\) and \(T\) we have \([S, T] = ST - TS\). A Lie algebra action of \(\mf g\) on a vector space \(V\) is a map \(\rho: \mf g \rar End(V)\) such that the bracket on \(\mf g\) becomes the bracket on \(End(V)\). In other words,
$$\rho([X,Y]) = \rho(X)\rho(Y) - \rho(Y)\rho(X).$$Given an action of \(G\) on a vector space \(V\) with map \(\rho\), we can make a Lie algebra action by saying that for \(X \in \mf g\),
$$\rho(X) = \frac{\rho(e + sX) - \rho(e)}{s}$$This looks like a derivative, and those less comfortable with infinitesimals may replace it with such.
Note that this is not an algebra action, at least not if we use the Lie bracket as the multiplication, because the bracket in the Lie algebra does not become composition in \(End(V)\).

The comultiplication on \(G\) sends \(g\) to \(g \otimes g\). In particular, the identity element \(e\) gets sent to \(e \otimes e\). We view the Lie algebra as elements \(X\) such that \(e + sX\) is in \(G\) for infinitesimal \(s\). So
$$\Delta(e + sX) = (e + sX)\otimes (e + sX) = e \otimes e + sX \otimes e + e \otimes sX$$ where we use the fact that \(s^2 = 0\).
The point of infinitesimals is that they make everything look linear, so we have the idea that
$$\Delta(e + sX) = \Delta(e) + \Delta(sX) = \Delta(e) + s\Delta(X)$$Since as mentioned, \(\Delta(e) = e \otimes e\), we get that \(\Delta(X) = X\otimes e + e \otimes X\). This comultiplication is cocommutative and coassociative.
Thus we get that for two representations \((V,\rho)\) and \((W, \sigma)\), \(\mf g\) acts on \(V\otimes W\) via
$$(\rho \hat \otimes \sigma)(X)(v \otimes w) = \rho(X)v \otimes w + v \otimes \sigma(X)w$$This should look somewhat product-rule-y, in keeping with the understanding of Lie algebras as derivatives of Lie groups.
Although the comultiplication sends \(G\) to \(G\otimes G\), we don't get that the comultiplication sends \(\mf g\) to \(\mf g \otimes \mf g\). Instead it gets sent to \((\mb ke \oplus \mf g) \otimes (\mb ke \oplus \mf g)\). Most of the time we'll leave off the \(e\).
We can make \(\mb k \oplus \mf g\) into a counital coalgebra by defining \(\epsilon(X) = 0\) for all \(X \in \mf g\) and \(\epsilon(c) = c\) for \(c \in \mb k\).

What if we want an algebra that contains the Lie algebra \(\mf g\) where Lie algebra actions of \(\mf g\) become algebra actions? We call such an object the universal enveloping algebra of \(\mf g\), denoted \(U(\mf g)\), and we create it as a quotient.
We start with the tensor algebra \(\bigotimes^* \mf g\), where we're viewing \(\mf g\) as a vector space. Now we impose the Lie algebra structure by looking at the ideal \(I = \left\langle X \otimes Y - Y \otimes X - [X, Y]\right\rangle\) where \([, ]\) is the Lie bracket, and forming the quotient \(U(\mf g) = \bigotimes^* \mf g/I\).
Recall that \(\bigotimes^* V\) has a comultiplication, given by sending \(v \in V\) to \(v \otimes 1 + 1 \otimes v\), and extending by multiplication. It also has a counit, sending \(v\) to \(0\), and an antipode that sends \(v\) to \(-v\). So \(\bigotimes^* V\) is a Hopf algebra. If \(V = \mf g\), then the ideal \(I\) defined above is both an ideal and a coideal and is closed under antipode, so the quotient \(U(\mf g)\) is not only an algebra, but a Hopf algebra.
Given a universal enveloping algebra, we can recover the Lie algebra that it is built from by looking for the elements \(X\) such that \(\Delta(X) = X \otimes 1 + 1 \otimes X\), which we call "primitive" elements. Note that \(\Delta\) respects the Lie bracket, in that
$$\Delta([X, Y]) = [X, Y] \otimes 1 + 1 \otimes [X, Y] = \Delta(X)\Delta(Y) - \Delta(Y)\Delta(X)$$Thus the primitive elements in a Hopf algebra do form a Lie algebra.
For our universal enveloping algebra, we can take a representation of \(\mf g\) and extend it to a representation of \(U(\mf g)\) as follows: any non-scalar element in \(U(\mf g)\) can be written as a sum of products of elements of \(\mf g\), so we focus on such products, writing them as \(X_1X_2\cdots X_k\). Given a representation \((V,\rho)\) of \(\mf g\), we apply \(\rho\) to \(X_1X_2\cdots X_k\) in the obvious fashion, as \(\rho(X_1)\circ \rho(X_2) \circ \cdots \circ \rho(X_k)\). Scalars map also in the obvious fashion, with \(c \in \mb k \subset U(\mf g)\) going to \(cI\in End(V)\). Since \(\rho\) respects the Lie bracket, we get a well-defined algebra representation of \(U(\mf g)\). So now we can use algebra action theorems and techniques to talk about Lie algebra actions.

If we instead look at the ideal \(I_h = \left\langle X \otimes Y - Y\otimes X - h[X,Y]\right\rangle\) for \(h \in \mb k\), we get a slightly different algebra sometimes denoted by \(U_h(\mf g)\). For \(h\) nonzero this is isomorphic to \(U(\mf g)\) just by rescaling things, but when \(h = 0\), we would get the relation \(X \otimes Y - Y \otimes X = 0\), which would leave us with things being commutative, i.e. we'd end up with the symmetric algebra \(Sym(\mf g)\) instead. So we can think of \(U(\mf g)\) as a deformed version of \(S(\mf g)\), which is not exactly commutative but where the noncommutativity is tightly controlled.