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.


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

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, $${\rm Im}(\rho)~\subsetneq~ GL(2j+1,\mathbb{R}) .$$

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  • $\begingroup$ Is the quotation from my textbook incorrect? $\endgroup$ – user148792 Jul 4 '18 at 15:30
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jul 4 '18 at 19:50

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.

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  • $\begingroup$ I also read that it is not faithful for j=/=1? Why is it the case? $\endgroup$ – user148792 Jul 3 '18 at 21:59
  • $\begingroup$ I have no idea. $\endgroup$ – ZeroTheHero Jul 3 '18 at 22:01

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