Background (just for context, you can skip it if you're familiar with Lorentz representations)
A Lorentz transformation can be represented by the matrix $M(\Lambda)=exp(\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})$, where $J^{\mu\nu}$ are the 6 Lorentz generators which satisfy the Lorentz commutator algebra. From these generators we can express both boosts $K^i=J^{0i}$ and rotations $J^i=\epsilon^{ijk}J^{jk}/2$ (here $i,j=1,2,3$, while $\mu,\nu=1,2,3,4$).
In particular, we can form two independent linear combinations:
$\vec{J_I}=\frac{1}{2}(\vec{J}+i\vec{K}) \ \ \ \vec{J_D}=\frac{1}{2}(\vec{J}-i\vec{K})$
Which satisfy the SU(2) algebra (i.e. $[J^i_{I,D},J^j_{I,D}]=i\epsilon^{ijk}J^k_{I,D}/2$), and even commute between themselves.
This is extremedly useful as we can build any Lorentz representation by knowing how to represent SU(2) only. We know this from QM courses, that is, we can build the SU(2) matrices $\vec{J}^{[j]}$ of dimension $(2j+1)$ by giving their spin 0,1/2,1,etc. (i.e, for j=0, $\vec{J}^{[0]}=1$; for j=1/2, $\vec{J}^{[1/2]}=\vec{\sigma}/2$; and so on).
From above, we see we can express then the rotation generators as $\vec{J}=\vec{J_I}+\vec{J_D}$. However, for $\vec{J_I}$ and $\vec{J_D}$ we can have different spins $j_I$ and $j_D$, in general $j_I\neq j_D$, so both matrices will have a different dimensions. We can fix this by taking "reducible" representations where the $\vec{J_I}$ have the irreducible blocks $\vec{J}^{[j_I]}$ repeated $(2j_D+1)$ times on its diagonal; similarly for $\vec{J^{D}}$. Now both matrices of have dimension $(2j_I+1)(2j_D+1)$ and can be written, with $l=(l_I,l_D)$, as:
$(\vec{J_I}_{l'l})=\vec{J}^{[j_I]}_{l'_Il_I}\delta_{l'_Dl_D} \ \ \ (\vec{J_D}_{l'l})=\vec{J}^{[j_D]}_{l'_Dl_D}\delta_{l'_Il_I}$
Therefore, the rotation generators are $\vec{J}=\vec{J_I}+\vec{J_D}$, so the Lorentz representation $(j_I,j_D)$ of dimension $(2j_I+1)(2j_D+1)$ can be in general reduced with respect to the subgroup of rotations, and includes the total spins $|j_I-j_D|,...,j_I+j_D$ obtained by combining spins $j_I$ and $j_D$.
Problem
We can rebuild the $M(\Lambda)$ Lorentz transformation in terms of $J_I$ and $J_D$ (it's more convenient to write $\Lambda=exp[i\vec{\theta}\cdot\vec{J}+i\vec{\alpha}\cdot\vec{K}]$).
I want to show that $M(\Lambda)$ decomposes into a product:
$M^{l'l}=M^I_{l'_Il_I}M^D_{l'_Dl_D}$
I began by replacing $\vec{J}=\vec{J_I}+\vec{J_D}$ and $\vec{K}=i(\vec{J_I}-\vec{J_D})$ and separate the exponential into two exponentials:
$M(\Lambda)=exp[i\vec{\theta}\cdot\vec{J}+i\vec{\alpha}\cdot{K}]=exp[\vec{J_I}\cdot(i\vec{\theta}-\vec{a})+\vec{J_D}\cdot(i\vec{\theta}+\vec{a})]$
Since by definition $J_I$ and $J_D$ commute, we can indeed separate them (by the BCH theorem):
$M(\Lambda)=exp[\vec{J_I}\cdot(i\vec{\theta}-\vec{a})]exp[\vec{J_D}\cdot(i\vec{\theta}+\vec{a})]$
Substituting the definitions of $J_I$ and $J_D$ in terms of $J^{[j_I]}$ and $J^{[j_D]}$, we get:
$M(\Lambda)=exp[\vec{J}^{[j_I]}_{l'_Il_I}\delta_{l'_Dl_D}\cdot(i\vec{\theta}-\vec{a})]exp[\vec{J}^{[j_D]}_{l'_Dl_D}\delta_{l'_Il_I}\cdot(i\vec{\theta}+\vec{a})]$
And defining:
$M^I_{l'_Il_I}=exp[\vec{J}^{[j_I]}_{l'_Il_I}\delta_{l'_Dl_D}\cdot(i\vec{\theta}-\vec{a})] \ \ \ \ ; \ \ \ \ M^I_{l'_Dl_D}=exp[\vec{J}^{[j_D]}_{l'_Dl_D}\delta_{l'_Il_I}\cdot(i\vec{\theta}+\vec{a})]$
It seems the problem is complete, and both $M^I_{l'_Il_I}$ and $M^I_{l'_Dl_D}$ are square matrices of dimension $(2j_I+1)(2j_D+1)$. However, at a later exercise where I use this matrices, for it to be solvable, I must need that $M^I_{l'_Il_I}$ and $M^I_{l'_Dl_D}$ be instead of dimensions $(2j_I+1)$ and $(2j_D+1)$ respectively! So either my definitions are incorrect and the matrices have the wrong dimensions, or the other exercise is wrong (or I'm interpreting it incorrectly).
Reference: S. Weinberg (1995), The Theory of Quantum Fields Vol. I, p. 229
$\ $
$\ $
Additional information:
In case it's relevant, the next exercise (which seems to contradict my results here) asks the following:
The vectors $\phi$ on which the matrices $M$ act have components $\phi_l=\phi(l_I,l_D)$, so they can be thought as rectangular matrices $(2j_I+1)x(2j_D+1)$. We need to show that these "vectors" transform as $\phi\rightarrow M^I\phi (M^D)^T$.
Since I calculated that $M^I$ is a square matrix of size $(2j_I+1)(2j_D+1)$ and $\phi$ is a rectangular matrix of size $(2j_I+1)x(2j_D+1)$ we can't even multiply the first product.
Progress
I read the Wikipedia article proposed in one of the comments, and while the notation is a bit complex, I managed to extract the most important information about the tensor products from there and other articles.
After redoing some algebra, let's start with the previous equation:
$M(\Lambda)=exp[\vec{J_I}\cdot(i\vec{\theta}-\vec{a})]exp[\vec{J_D}\cdot(i\vec{\theta}+\vec{a})]$
While it's true the matrices $J_I$ and $J_I$ where redefined to have dimension $N=(2j_I+1)(2j_D+1)$, we're looking for a tensor product decomposition of the matrix $M$. Instead of substituting directly, however, let's treat them as two separate tensor spaces of dimensions $(2j_I+1)$ and $(2j_D+1)$. In this way, we're looking at a decomposition of the form:
$M[(2j_I+1)(2j_D+1)]=M^I[(2j_I+1)]\otimes M^D[(2j_D+1)]$
Defining now:
$M^I=exp[\vec{J}^{[j_I]}\cdot(i\vec{\theta}-\vec{a})]$
$M^D=exp[\vec{J}^{[j_D]}\cdot(i\vec{\theta}+\vec{a})]$
We recover the tensor product as:
$M=M^I\otimes M^D$
So now $M^I$ has dimension $(2j_I+1)$ and $M^D$ has dimension $(2j_D+1)$. I checked this is correct from the third reference. However, this seems to be inconsistent, as they should actually be rectangular matrices with dimension $(2j_I+1)\times (2j_D+1)$, so there seems to be a problem still.
References
http://maths.dur.ac.uk/users/philipp.b.lampe/LorentzBadDriburg.pdf http://www.math.mcgill.ca/walcher/phys580/PoincareFieldsMarcAntoine.pdf http://www.int.washington.edu/users/dbkaplan/571_12/Lorentz.pdf
