Are rotation matrices faithful representations of the rotation group? I would like to use rotation matrices as representations of the rotation group.  I would like to know if these representations are faithful, i.e. isomorphic to the rotational group elements.
I read on the bottom of p. 61 in Ref. 1 that 

"Only the $j = 1$ representation is isomorphic to the rotation group itself."

Can someone explain to my why this is the case?
Note: $j=1$ means that the eigenvalue of $J^2$ is $j(j+1)$, where $J^2=J_x^2+J_y^2+J_z^2$, where $J_i$ is the generator of rotation about the $i$-axis.
References:


*

*J. Tseng, Symmetry and Relativity, lecture notes, 2017. The PDF file is available here.

 A: Given a non-negative integer $j\in\mathbb{N}_0$, the spin-$j$ group representation/homomorphism $$\rho: SO(3)~\to~ GL(2j+1,\mathbb{R}) $$
is faithful/injective iff $j>0$, but technically speaking, never a group isomorphism, since it is never surjective, $${\rm Im}(\rho)~\subsetneq~ GL(2j+1,\mathbb{R}) .$$
A: Yes there are, in the sense that 
$$
R(\Omega_1)\cdot R(\Omega_2)= R(\Omega_1\cdot \Omega_2)
\Rightarrow \sum_m D^{j}_{m_1m}(\Omega_1)D^j_{mm_2}(\Omega_2)
=D^j_{m_1m_2}(\Omega_1\cdot \Omega_2)
$$
valid for any $j$,
although of course finding analytically $\Omega_1\cdot\Omega_2$ in terms of the triples of parameters $\Omega_1$ and $\Omega_2$ is usually messy.  
The case $j=1$ is the defining representation so the matrices you get are identical as those obtained by geometrically constructing rotations in 3-space.  (Usually the $D$’s are in a spherical basis so you need to consider combinations of $\hat x\pm i\hat y$ as basis vectors.
One way to understand why this is true is that the rotation matrices are exponentials of the algebra $\{J_x,J_y,J_z\}$, that the a faithful representation of the algebra by $(2j+1)\times (2j+1)$ matrices will also exponentiate to a faithful representation (of the same dimension) of the group.
