Spin $\frac{3}{2}$ representation in Georgi's book? Georgi's book Lie Algebras in Particle Physics 2ed equation 3.32 lists the spin operators in the spin $\frac{3}{2}$ representation as:
$$J_1=\left(
\begin{array}{cccc}
 0 & \sqrt{\frac{3}{2}} & 0 & 0 \\
 \sqrt{\frac{3}{2}} & 0 & 2 & 0 \\
 0 & 2 & 0 & \sqrt{\frac{3}{2}} \\
 0 & 0 & \sqrt{\frac{3}{2}} & 0 \\
\end{array}
\right)$$
$$J_2=\left(
\begin{array}{cccc}
 0 & -i\sqrt{\frac{3}{2}} & 0 & 0 \\
 i\sqrt{\frac{3}{2}} & 0 & -i2 & 0 \\
 0 & i2 & 0 & -i\sqrt{\frac{3}{2}} \\
 0 & 0 & i\sqrt{\frac{3}{2}} & 0 \\
\end{array}
\right)$$
$$J_3=\left(
\begin{array}{cccc}
 \frac{3}{2} & 0 & 0 & 0 \\
 0 & \frac{1}{2} & 0 & 0 \\
 0 & 0 & -\frac{1}{2} & 0 \\
 0 & 0 & 0 & -\frac{3}{2} \\
\end{array}
\right)$$
but the commutators don't seem to work out $[J_1,J_2]\neq i J_3$. What gives?
I wrote the following Mathematica command j[n,s] to generate the n=1,2, or 3 spin s matrix. It generates the Pauli matrices and spin $1$ matrices correctly, but doesn't match the $3/2$ rep in Georgi's book.
j[3,s_/;IntegerQ[2s+1]&&s>0]:=SparseArray[Band[{1,1}]->Table[i,{i,s,-s,-1}],2s+1];
jplus[s_/;IntegerQ[2s+1]]:=SparseArray[Band[{1,2}]->Table[Sqrt[(s+1+m)(s-m)/2],{m,s-1,-s,-1}],2s+1];
jminus[s_/;IntegerQ[2s+1]]:=SparseArray[Band[{2,1}]->Table[Sqrt[(s+m)(s-m+1)/2],{m,s,1-s,-1}],2s+1];
j[1,s_/;IntegerQ[2s+1]]:=(jplus[s]+jminus[s])/Sqrt[2];
j[2,s_/;IntegerQ[2s+1]]:=(jplus[s]-jminus[s])/(I Sqrt[2]);

 A: There is a typo in the book's equation, and there doesn't appear to be an easily accessible online errata. If one follows the formulas the book gives for $J_\pm$: $$J_+=\frac{1}{\sqrt{2}}\left(J_1+i J_2\right)$$
$$J_-=\frac{1}{\sqrt{2}}\left(J_1-i J_2\right)$$
$$J_{-,m'm}\frac{\sqrt{\left(s-m\right) \left(m+s+1\right)} \delta _{m+1,m'}}{\sqrt{2}}$$
$$J_{+,m'm}\frac{\sqrt{\left(s+m\right) \left(s-m+1\right)} \delta _{m-1,m'}}{\sqrt{2}}$$
one finds:
$$J_1=\left(
\begin{array}{cccc}
 0 & \frac{\sqrt{3}}{2} & 0 & 0 \\
 \frac{\sqrt{3}}{2} & 0 & 1 & 0 \\
 0 & 1 & 0 & \frac{\sqrt{3}}{2} \\
 0 & 0 & \frac{\sqrt{3}}{2} & 0 \\
\end{array}
\right)$$
$$J_2=\left(
\begin{array}{cccc}
 0 & -\frac{1}{2} \left(i \sqrt{3}\right) & 0 & 0 \\
 \frac{i \sqrt{3}}{2} & 0 & -i & 0 \\
 0 & i & 0 & -\frac{1}{2} \left(i \sqrt{3}\right) \\
 0 & 0 & \frac{i \sqrt{3}}{2} & 0 \\
\end{array}
\right)$$
$$J_3=\left(
\begin{array}{cccc}
 \frac{3}{2} & 0 & 0 & 0 \\
 0 & \frac{1}{2} & 0 & 0 \\
 0 & 0 & -\frac{1}{2} & 0 \\
 0 & 0 & 0 & -\frac{3}{2} \\
\end{array}
\right)$$
