TeethWhitener said:
This might be the wrong place to post this, but I was wondering about your inner automorphism example. In defining a representation ##\varphi## (equation 5), you specifically look at the action of a group ##G## on a vector space ##V##. But then in the inner automorphism example, you have the group action of ##G## on itself: ##G\times G \rightarrow G##, with the representation ##Ad: G \rightarrow GL(\mathfrak{g})##. But given the initial definition of a representation, shouldn't this be ##Ad: G \rightarrow GL(G)##? Why do we get to substitute the Lie algebra ##\mathfrak{g}## for the Lie group ##G##? Or alternatively, why doesn't the group ##G## act on the vector space ##\mathfrak{g}##?
To be able to speak of group action, all we need is - of course - a group ##G## and a set ##X## where the group acts on, resp. operates on. This is a function ##G \times X \longrightarrow X##. From here there can be various examples, depending on the nature of ##X##. As we have a group operation, ##X=G## is a natural example. So ##(g,x) \mapsto L_g(x) = g\cdot x## or ##(g,x) \mapsto \iota_g(x) = g\cdot x \cdot g^{-1}## are two possible actions of a group on itself. The latter defines an (inner) automorphism ##x \mapsto gxg^{-1}## for every ##g##, i.e. we have a function ##G \longrightarrow \operatorname{Aut}(G)##. This can always be done.
Now if ##X=V## is a vector space, we have another situation, because a group is no vector space. (And usually the groups under consideration are not the additive groups which build the vector space.) First the operation ##(g,v) \mapsto g.v \in V## has to be defined. For an operation it is required that ##1.v=v## and ##(g\cdot h).v = g.(h.v))## hold. These properties are the same as to say ##g \mapsto (v \mapsto g.v)## is a representation (of the group ##G## on the vector space ##V##) or ##G## operates on ##V## or ##G \longrightarrow GL(V)## is a group homomorphism: three languages for the same thing. If we have a matrix group (like orthogonal or Hermitian transformations) of a vector space ##V## we get a representation in a natural way: matrix times vector = vector.
As to your question about the adjoint represenation. First of all: ##GL(G)## doesn't make sense. The "L" stands for linear and "G" doesn't carry a linear structure. We can say ##\operatorname{Inn}(G)## for inner automorphisms, or ##\operatorname{Aut}(G)## for simply automorphisms. In the case of conjugations like above we have ##G \longrightarrow \operatorname{Inn}(G) \subseteq \operatorname{Aut}(G)## by the map ##g \mapsto \iota_g##.
Let's now go back to a matrix group ##G## which simply means ##G \subseteq GL(V)## for some vector space ##V##, i.e. ##G## consists of some kind of regular (here finite dimensional) matrices which apply to vectors from ##V##. As above we have an operation ##g \mapsto \iota_g## where ##\iota_g(x)=gxg^{-1}## is the normal matrix multiplication of matrices ##g,x,g^{-1}##. In the language of group homomorphisms this means ##G \longrightarrow \operatorname{Aut}(G) \subseteq \operatorname{Aut}(GL(V))##.
If ##G## is a matrix group with Lie algebra ##\mathfrak{g}##, we have the two things given needed: a group ##G## and a vector space ##\mathfrak{g}## on which the group acts. The action, resp. operation is given by ##(g,X) \mapsto gXg^{-1}## for ##g\in G\; , \;X\in \mathfrak{g}\,.## In cases where ##G## and ##\mathfrak{g}## are given, it is usually easy to see, that this defines a representation. The general case (##G## an analytic group and ##\mathfrak{g}## its left-invariant vector fields) needs a little bit more care (cp. Varadarajan [11]). This operation is induced by the conjugation in ##G##, that is the group conjugations ##\iota_g## can be extended to a conjugation of the vector fields and is called adjoint representationt of the Lie group ##G##.
... you have the group action of ##G## on itself: ##G\times G \rightarrow G##, with the representation ##Ad: G \rightarrow GL(\mathfrak{g})##
isn't precisely true. I have a group action (called conjugation, ##\iota\, : \, G \times G\longrightarrow G## ) on itself (no vector space),
and I have a group action (called adjoint, ##\operatorname{Ad}\, : \,G \times \mathfrak{g}\longrightarrow \mathfrak{g}##) on its tangent space (vector space, resp. representation space). These are two different operations, and only the second is called a representation, although it is induced by the first, but it's another operation.
Although we have matrices here, i.e. linear transformations of a vector space ##V##, these vectors don't play any role so far. And of course this is confusing, as ##\mathfrak{g}## are also matrices of the same dimension, which also apply to vectors of ##V##. However, this isn't even the end here, as ##\mathfrak{g}## also operates on the vector space ##\mathfrak{g}## by Lie left multiplication of the Lie algebra ##\mathfrak{g}##, i.e.
$$
\mathfrak{g} \times \mathfrak{g} \longrightarrow \mathfrak{g}\\
(X,Y) \longmapsto \operatorname{ad}(X)(Y) = [X,Y]
$$
is a Lie algebra homomorphism, which is also called adjoint representation, as ##\operatorname{ad}X## are inner derivations of ##\mathfrak{g}.##
All three are connected by
- Matrix group operation: conjugation induces ...
- Lie group operation: adjoint ##\operatorname{Ad}## connected via the exponential map to ...
- Lie algebra operation: adjoint ##\operatorname{ad} ##
I have tried to explain this mess earlier here:
https://www.physicsforums.com/insights/representations-precision-important/
