Let $\mathcal{H} \cong \mathbb{C}^2$ denote some Hilbert space. For example, this could be Hilbert space of a single spin-1/2 particle. Let $SU(2)$ be the special unitary Lie group as per usual. Let $su(2)$ be its associated algebra. I know that there exists the exponential map $exp: su(2) \rightarrow SU(2)$. So that if $g \in su(2)$, then $exp(g) \in SU(2)$.
I would like to know how one makes contact between elements of $SU(2)$ and $su(2)$ with Hilbert space $\mathcal{H}$.
I assume that you define some sort of a representation of $su(2)$ to make contact with $\mathcal{H}$, i.e. $$\pi: su(2) \times \mathcal{H} \rightarrow \mathcal{H},$$
which gives us a way to represent elements of the Lie algebra as linear transformations over the Hilbert space. I assume that you can do something similar for the Lie group itself: $$\phi: SU(2) \times \mathcal{H}\rightarrow \mathcal{H}.$$
But, another one of my confusions now comes up. Then, what is the role of the exponential map here and what does it mean to take the taylor expansion of the exponential map? In books like Sakurai and Townsend, some argument is made that unitary transformations by a finite amount of a parameter $j$ take on the form $e^{-ij\hat{O} / \hbar}$. This exponential is then taylor expanded, becoming a series of powers of $\hat{O}$, and its action on a state $\psi \in \mathcal{H}$ can then be deduced.
It then seems like
- we are implicitly using a representation of the Lie algebra containing $\hat{O}$ over $\mathcal{H}$.
- We seem to be using the exponential map to go from $su(2)$ to $SU(2)$ and then taylor expanding the exponential map to go back to $su(2)$ and then representing $su(2)$ once more as elements acting over $\mathcal{H}$.
Point 2. seems absurd. Unless the taylor expansion of the exponential together with the representation of the associated Lie algebra defines the representation of the Lie group itself. I would like some clarification about what is really mathematically going on here.
What you write is mostly correct, I only note that the representations would usually be written as $\pi: \mathfrak{su}(2) \to \mathfrak{gl}(\mathcal H)$ and $\phi: \mathrm{SU}(2) \to \mathrm{GL}(\mathcal H)$. Maybe some of your confusion stems from the fact that in the setup you are describing, there is a canonical representation $\pi = \mathrm{id}$, $\phi = \mathrm{id}$. Physics texts will therefore usually not distinguish the group / algebra and its representation.
I will very briefly (and without claiming to be 100% rigorous) recap the general theory. Beyond that, I would recommend reading an actual math book on Lie theory.
A Lie group is a differentiable group, and a Lie algebra is a vector space equipped with a Lie bracket $[.,.]$. Given a Lie group $G$, the corresponding Lie algebra $\mathfrak g$ is the tangent space on $G$ (seen as a differentiable manifold) at the identity, $\mathfrak g \equiv T_e\, G$. The group structure on $G$ induces the Lie bracket on $\mathfrak g$, but I will not go into detail how that works. The main point is that for curves $\gamma$ in $G$ with $\gamma(0) = e$, the derivatives $\gamma'(0)$ are Lie algebra elements, $\gamma'(0) \in \mathfrak g$.
Given a Lie algebra $\mathfrak g$, there is a unique simply connected Lie group $G$ such that $\mathfrak g$ is the algebra corresponding to $G$. That is Lie's third theorem.
For a vector $X \in \mathfrak g$ there is a unique 1-parameter subgroup $\gamma: \mathbb R \to G$ (that is, a curve with $\gamma(0) = e$ and $\gamma(t+s) = \gamma(t) \gamma(s)$) such that $\gamma'(0) = X$. We define the exponential map as $$ \exp(X) \equiv \gamma(1) . $$
Given a representation $\phi$ of $G$ we can construct a representation $\pi$ of $\mathfrak g$, and vice versa, in such a way that the diagram commutes: $$ \begin{array}{} \mathfrak g & \stackrel{\pi}{\longrightarrow} & \mathfrak{gl}(\mathcal H) \\ \downarrow{\exp} & & \downarrow{\exp} \\ G & \stackrel{\phi}{\longrightarrow} & \mathrm{GL}(\mathcal H) \end{array} \tag 1 $$ where $\exp$ on the right is the usual matrix exponential.
In the argument you are describing, the authors seem to start from the fact that a 1-parameter subgroup of $\mathrm{SU}(2)$ can be written as $\gamma: j \mapsto \exp(-\mathrm i j O / \hbar)$. Through the Taylor expansion, we are taking the derivative at the identity and find that the operators $O$ here are elements of the Lie algebra. (The extra factor $\mathrm i/\hbar$ is a physicist's convention but does not change the point.) Thanks to the magic in (1), the representation of $O$ is "what you would expect".