Representations of simple Lie algebras

\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)So here's where most of this has been heading.

We start with a simple Lie algebra, which we'll write as \(\mf l_n\). The best thing, representation-wise, about simple Lie algebras is that all finite-dimensional representations of \(\mf l_n\) decompose into irreducible representations. The next best thing is that the only 1-dimensional representations are trivial.
The \(n\) denotes several things, but one of the things it denotes is the number of "fundamental representations" of \(\mf l_n\), which I'll write as \(V_1, V_2,\ldots,V_n\). These are irreducible representations such that if we look at all tensor products of these fundamental representations, i.e. every representation of the form \(V_1^{\otimes k_1} \otimes \cdots \otimes V_n^{\otimes k_n}\), and every irreducible representation that these representations decompose into, we get every finite-dimensional irreducible representation of \(\mf l_n\).
For instance, let's look at \(su(2)\). Also known as \(\mf a_1\), it has 1 fundamental representation \(V\) which is 2-dimensional. From there we can generate \(V\otimes V\), which is 4-dimensional, and which decomposes into the trivial 1-dimensional representation and a 3-dimensional representation. \(V\otimes V\otimes V\) is 8-dimensional and decomposes into two copies of \(V\) and a 4-dimensional representation. And so on. We can think of this in terms of particle spin: each \(V\) is a spin-1/2 particle, so that \(V\otimes V\) is 2 spin-1/2 particles. The trivial representation is the particles cancelling, the 3-dimensional representation is when they act as a single spin-1 particle.
Another example is \(su(3)\), also known as \(\mf a_2\). The fundamental representations are two 3-dimensional representations, \(V\) and \(V^\vee\). Every finite-dimensional representation of \(su(3)\) is a component of some \(V^{\otimes k} \otimes (V^\vee)^{\otimes l}\). In terms of particles we should think that each copy of \(V\) is a quark and each copy of \(V^\vee\) is an antiquark. A color-neutral configuration of quarks corresponds to the tensor product having a trivial component somewhere.

Let's look at the functions involved: a representation of \(G\) on \(\mb C^n\) gives us a map \(\rho: G \rar End(\mb C^n)\) is really a set of \(n^2\) maps from \(G\) to \(\mb C\), one for each entry in an \(n \times n\) matrix. So we can write it as \(\rho_i^j\).
We can add two such maps, or multiply them, and get more maps. We may not necessarily get more maps that show up in representations, but we nonetheless get more maps. So we get a ring of such functions.
Consider the tensor product of two representations \((V,\rho)\) and \((W,\sigma)\). The corresponding map is \(\rho \hat \otimes \sigma\), which sends \(G\) to \(End(V\otimes W)\). If we tack on indices, we get indices \(\rho_i^j\) for \(V\) and \(\sigma_k^l\) for \(W\), and so \(\rho \hat \otimes \sigma\) has indices \((\rho \hat \otimes \sigma)_{ik}^{jl}\) and factors as \(\rho_i^j \sigma_k^l\), as one would expect.
Moreover, since \(\rho \hat\otimes \sigma\) is equivalent to \(\sigma \hat \otimes \rho\) via a natural isomorphism, the corresponding functions look the same, and so the ring is commutative. There is an obvious unit function, that sends everything in \(G\) to \(1\), corresponding to the map for the trivial representation.
We also have a comultiplication, since we're looking at functions on a thing that has a multiplication. If we look at \(\rho(gh)\), we get \(\rho(g)\rho(h)\); in indices, we get \(\rho(gh)_i^j = \rho(g)_i^k \rho(h)_k^j\), so we get that \(\Delta(\rho_i^j) = \rho_i^k \otimes \rho_k^j\).
We also get a counit, which is to evaluate \(\rho_i^j\) at the identity in \(G\); since a representation always sends the identity to the identity, we get that \(\rho_i^j(e) = I_i^j\), i.e. is \(1\) if the values of \(i\) and \(j\) are equal, and 0 otherwise.
Finally, we have an antipode, with \((S\rho_i^j)(g) = \rho_i^j(g^{-1})\).
So we get a Hopf algebra \(Rep(G)\). Note that if \(G\) is a Lie group, then a representation of \(G\) becomes a representation of \(\mf g\), so we can evaluate elements of \(Rep(G)\) on \(\mf g\), and thus on \(U(\mf g)\).
In fact, if \(\mf g\) is simple, then we're in luck: \(Rep(G)\) can serve as a dual of \(U(\mf g)\), in that for any nonzero element \(X\) of \(U(\mf g)\), there is an element \(f\) of \(Rep(G)\) such that \(f(X)\) is nonzero, and vice-versa.
One would think it would be a dual to \(\mb kG\), except if \(G\) is a Lie group that isn't 0-dimensional, then \(G\) is infinite and thus \(\mb kG\) is really badly behaved, so we use \(U(\mf g)\) instead.