Correct vector space of eigenkets of angular momentum When we say an particle is in the state:
\begin{equation}
|l,m\rangle,
\end{equation}
what is the underlying state space, as a vector space? Is it a tensor product vector space, of dimension:
\begin{equation} 
l\times(2l+1)\ ?
\end{equation}
How can I find the matrix representation of the angular momentum operators that act on the $2l+1$ vector space in that tensor product? I am used to angular momentum operators taking the form of a cross-product:
\begin{equation}
x_ip_j - p_ix_j,
\end{equation}
but can we still do that for the $2l+1$ dimensional space corresponding to $m$?
 A: For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question.
When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for $L_z$). The total space of (bound) states for your system is then the infinite sum of these spaces for all possible $l$, i.e.
$$ \mathcal{H} = \bigoplus_l \mathcal{H}_l$$
A: My understanding of this limited, but this might help (too long for a comment):
The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space.
Out of convenience, we denote the eigenstates by the quantum numbers, indexing them with $ n, \ell$ and $m $ which correspond (though are not equal to) their respective eigenvalues for each of the operators.
I suppose that a state with only $\ell$ and $ m $ specified lies in the subspace of $ H $ with an orthonormal basis equal to the set of simultaneous eigenstates with those quantum numbers. 
