What operator lowers the total angular momentum? Assume  states $|j,m\rangle$, say $j\in\{3,2,1,0\}$, initially at $3$.
Is there any "lowering" operator I could apply such that $L_-|j,m\rangle
=
|j-1,m\rangle
$? 
How to express it in the $J_z$ basis?
 A: First you need to define the operator whose eigenvalues are j. As your avatar invites me to, I skip the fine formal fussing and just calculate instead. Such operators are routine, e.g., Curtright and Zachos (1990) PhysLettB243. For the Casimir $\vec J \cdot \vec  J$, with eigenvalues j(j+1), define
$$
L_0\equiv \frac{\sqrt{1+4\vec J \cdot \vec  J}-1}{2},
$$
with eigenvalues j,
$$
L_0| j,m\rangle = j| j,m\rangle .
$$ 
Your target operator $L_-$ should distinctly fail to commute with the Casimir, and hence $L_0$, so as to take you from a fixed j to a lower one, just as the LRL vectors do in the Hydrogen atom! That is
$$
[L_-,L_0]= L_- \\
[L_-,J_z]=0,
$$
so that 
$$
L_0(L_- |j,m\rangle)= (j-1)(L_- |j,m\rangle),\\
J_z (L_- |j,m\rangle) =m(L_- |j,m\rangle),
$$
as per your posit.
You might construct such operators, with pain, out of pieces of LRL invariants, such as $A_z$, but without context this readily slips into the slough of mootness.
A: In principle, any vector operator, i.e. any tensor operator with $\ell=1$ will shift you $\Delta \ell=0,\pm 1$.  Moreover, if your operator is parity-odd, then by Laporte’s rule or by simple properties of the CG coefficient $\Delta l=0$ is excluded. Thus, for instance, the operator $\hat z$ has the property 
\begin{align}
\langle \alpha’\ell’m’\vert z\vert \alpha \ell m\rangle =
C^{\ell’m’}_{\ell m; 10}\frac{\langle \alpha’\ell’\Vert r\Vert \alpha \ell\rangle}{\sqrt{2\ell’+1}}\delta_{m’m}\delta_{\ell’,\ell\pm 1}
\end{align}
by the Wigner-Eckart theorem, with  $C^{\ell’m’}_{\ell m;10}$ a Clebsch-Gordan coefficient.    
This is not quite what you want since it will shift up or shift down the value of $\ell$.    However, with
\begin{align}
\hat V_\pm=\mp \frac{1}{\sqrt{2}}(\hat x\pm i\hat y)\, ,\qquad \hat V_0=\hat z
\end{align}
then the sum
\begin{align}
\sum_{km} C^{\ell-1,m’}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle \tag{1}
\end{align}
will have non-zero matrix element with only $\vert\alpha’ \ell-1,m’\rangle$.  This can be seen by invoking again the Wigner-Eckart theorem:
\begin{align}
\langle \alpha’\ell’m’\vert\sum_{km} C^{\ell-1,m’}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle&=
\sum_{km}. 
\sum_{km} C^{\ell-1,m’}_{\ell m;1k} \langle\alpha’ \ell’m’\vert \hat V_k\vert\alpha \ell m\rangle\, ,\\
&=\sum_{km} C^{\ell-1,m’}_{\ell m;1k} 
C^{\ell’ m’}_{\ell m;1k} 
\frac{\langle\alpha’ \ell’ \Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ell’+1}}\, ,\\
&=\frac{\langle\alpha’ \ell’\Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ell’+1}}\delta_{\ell’,\ell-1}
\end{align}
by orthogonality of the CGs.  (1) is not a single operator but it does contain a projector to the correct $\ell’$ subspace, and you can freely choose the $m’$ value of your target state.
