= Solution
A <Lie algebra representation> is a linear map $\rho:\mathfrak g\to\operatorname{End}(V)$ preserving brackets: $\rho([X,Y])=[\rho(X),\rho(Y)]$. Differentiating a smooth <unitary representation> $D$ of a <Lie group> gives $\rho(X)=\left.\frac{d}{dt}D(e^{tX})\right|_{t=0}$, an anti-Hermitian operator for each real algebra element $X$. In the physics convention one uses Hermitian generators $t_X=i\rho(X)$ and writes $D(e^{tX})=e^{-it t_X}$. Conversely, <integration of a Lie-algebra representation> gives a representation of the connected simply connected group. A representation of another group with the same algebra additionally has to be trivial on the kernel of its <covering map>.
For the <angular momentum commutation relations>, take $J_3$ Hermitian and $J_+^\dagger=J_-$ on a positive-definite inner-product space. A <highest-weight vector> has $J_3|j,j\rangle=j|j,j\rangle$ and $J_+|j,j\rangle=0$. Set $v_\ell=(J_-)^\ell|j,j\rangle$. Commuting $J_3$ through each lowering operator gives $J_3v_\ell=(j-\ell)v_\ell$. Furthermore,
$$
[J_+,(J_-)^\ell]=\sum_{r=0}^{\ell-1}(J_-)^r(2J_3)(J_-)^{\ell-1-r},
$$
so acting on the highest-weight state and summing the coefficients gives
$$
J_+v_\ell=\ell(2j-\ell+1)v_{\ell-1}.
$$
Taking the <inner product> with $v_{\ell-1}$ produces the <unitary highest-weight termination> recursion
$$
\|v_\ell\|^2=\ell(2j-\ell+1)\|v_{\ell-1}\|^2.
$$
The first factor requires $j\ge0$. If $2j$ is not an integer, the first integer $\ell>2j+1$ would make the norm negative, while every preceding norm is positive. Thus
$$
\boxed{2j\in\mathbb Z_{\ge0}.}
$$
For these values, $v_0,\ldots,v_{2j}$ have strictly positive norms and distinct <eigenvalues>, while $v_{2j+1}$ has zero norm and hence is zero. They therefore span a space of <dimension> $2j+1$, invariant under all three generators. Its weights are $j,j-1,\ldots,-j$. This proves \b[integer or half-integer <spin> and finite <dimension> $2j+1$]. The positive-definite unitary assumption matters: an unrestricted algebraic highest-weight module need not terminate. Normalizing the ladder states yields
$$
J_\pm|j,m\rangle=\sqrt{(j\mp m)(j\pm m+1)}\,|j,m\pm1\rangle.
$$
This is the <normalized highest-weight lowering formula> with the usual positive ladder phases and units $\hbar=1$.
In the ordered spin-one-half basis $(|1/2,1/2\rangle,|1/2,-1/2\rangle)$, the matrices are
$$
\boxed{J_3=\frac12\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad
J_+=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad
J_-=\begin{pmatrix}0&0\\1&0\end{pmatrix}.}
$$
Thus $J_i=\sigma_i/2$, where the $\sigma_i$ are <Pauli matrices>.
Choose the active-rotation convention. The <SU(2)> lift of an axis-angle rotation is
$$
U_{1/2}(\vartheta,\mathbf n)=e^{-i\vartheta\mathbf n\cdot\boldsymbol\sigma/2}
=\cos(\vartheta/2)I-i\sin(\vartheta/2)\mathbf n\cdot\boldsymbol\sigma.
$$
The <Pauli matrix multiplication law> shows that conjugating $\mathbf x\cdot\boldsymbol\sigma$ by this <matrix> produces $(R\mathbf x)\cdot\boldsymbol\sigma$, with
$$
R\mathbf x=\mathbf x\cos\vartheta+(\mathbf n\times\mathbf x)\sin\vartheta
+\mathbf n(\mathbf n\cdot\mathbf x)(1-\cos\vartheta).
$$
This is the ordinary three-dimensional rotation. In any spin-$j$ representation the corresponding unitary operator and its <matrix> elements are
$$
\boxed{U_j(R)=e^{-i\vartheta\mathbf n\cdot\mathbf J^{(j)}},\qquad
D^{(j)}_{mm'}(R)=\langle j,m|U_j(R)|j,m'\rangle.}
$$
For half-integer <spin>, $R$ in this notation includes a choice of lift to <SU(2)>; it does not define a single-valued representation of <SO(3)>.
For the specified Euler-angle order, the rightmost rotation acts first:
$$
\boxed{U_j(R)=e^{-i\phi J_3}e^{-i\theta J_2}e^{-i\psi J_3},\qquad
D^{(j)}_{mm'}(R)=e^{-im\phi}d^{(j)}_{mm'}(\theta)e^{-im'\psi}.}
$$
Here the <Wigner D-matrix> middle factor is $d^{(j)}_{mm'}(\theta)=\langle j,m|e^{-i\theta J_2}|j,m'\rangle$. It can be calculated explicitly by realizing <spin> $j$ as the symmetric product of $2j$ spin-one-half factors. Write $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. The fundamental rotation sends $|+\rangle$ to $c|+\rangle+s|-\rangle$ and $|-\rangle$ to $-s|+\rangle+c|-\rangle$. Expanding a normalized symmetric state with $j+m'$ plus signs gives
$$
d^{(j)}_{mm'}(\theta)=\sqrt{(j+m)!(j-m)!(j+m')!(j-m')!}
\sum_r\frac{(-1)^{m-m'+r}c^{2j+m'-m-2r}s^{m-m'+2r}}
{(j+m'-r)!\,r!\,(m-m'+r)!\,(j-m-r)!},
$$
where $\max(0,m'-m)\le r\le\min(j+m',j-m)$. The exponent $r$ counts initially positive spinors that become negative, while $m-m'+r$ counts initially negative ones that become positive; their minus signs and binomial coefficients give the displayed expression. Every factorial argument is an integer even for half-integer <spin>.
For <spin> one-half this reduces to the explicit <matrix>
$$
D^{(1/2)}(\phi,\theta,\psi)=
\begin{pmatrix}
e^{-i(\phi+\psi)/2}\cos(\theta/2)&-e^{-i(\phi-\psi)/2}\sin(\theta/2)\\
e^{i(\phi-\psi)/2}\sin(\theta/2)&e^{i(\phi+\psi)/2}\cos(\theta/2)
\end{pmatrix}.
$$
Replacing $\theta$ by $\theta+2\pi$ reverses both sine and cosine, so every entry changes sign. In particular, a pure $2\pi$ rotation gives \b[$D^{(1/2)}=-I$], while a $4\pi$ rotation gives $I$.
Finally, the <Adjoint double cover from SU(2) to SO(3)> is surjective by the axis-angle construction. Its kernel consists of matrices commuting with all Pauli matrices, hence scalar matrices; within <SU(2)> these are $\pm I$. Thus
$$
\boxed{SO(3)\cong SU(2)/\{\pm I\}.}
$$
Their <Lie algebras> are isomorphic, but their global topology and representations differ. In <spin> $j$, the central element $-I$ acts as $(-1)^{2j}I$, so exactly the integer-spin representations descend to ordinary <SO(3)> representations. This is the <descent of an SU(2) representation to SO(3)> criterion.
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