= Solution
The <angular momentum commutation relations> are
$$
\boxed{[J_i,J_j]=i\hbar\sum_{k=1}^3\varepsilon_{ijk}J_k},\qquad
J^2=J_1^2+J_2^2+J_3^2,\qquad [J^2,J_i]=0,
$$
where $\varepsilon_{ijk}$ is the <Levi-Civita symbol>. The <angular momentum eigenstates> have
$$
\boxed{J^2|j,\mu\rangle=\hbar^2j(j+1)|j,\mu\rangle,\qquad
J_3|j,\mu\rangle=\hbar\mu|j,\mu\rangle,}
$$
with <angular momentum quantum number> $j=0,\tfrac12,1,\tfrac32,\ldots$ and <magnetic quantum number> $\mu=-j,-j+1,\ldots,j$. There are $2j+1$ such states in one <irreducible spin representation>, so the $J^2$ <eigenspace> has <dimension> $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the <magnetic quantum number> with the particle <mass> in the earlier questions.
For completeness, the <ladder operators> $J_\pm=J_1\pm iJ_2$ satisfy $[J_3,J_\pm]=\pm\hbar J_\pm$ and
$$
\|J_\pm|j,\mu\rangle\|^2=\hbar^2\bigl[j(j+1)-\mu(\mu\pm1)\bigr].
$$
The upper and lower ends are $\mu=\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet <dimension>. Pure <orbital angular momentum> permits only integer $j$, whereas general <spin> allows half-integers. If a larger <Hilbert space> contains several copies of the same <irreducible spin representation>, the full $J^2$ <eigenspace> includes all those copies: its <dimension> is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.
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