Can every SU(4) matrix be written as a product of of these 4 matrices Consider a general $SU(4)$ matrix
$$U=e^{i a^j T^j},$$
where the generators $T^i$ are chosen in the basis
$$\{T\} = 1_2 \times \sigma^i, \quad \sigma^i \times 1_2, \quad \sigma^i \times \sigma^j, $$
for $i,j = 1,2,3$. How would I go about proving or disproving that every $U$ can be written as the following product.
$$U = e^{i(1_2 \times \sigma^j)b^j } \; e^{i(\sigma^k \times 1_2)c^k }\; e^{i((\sigma^l \times \sigma^1)d^{l} + (\sigma^m \times \sigma^2)e^{m} )}\;e^{i(\sigma^n \times \sigma^3)f^n },$$
for some $b,c,d,e,f$.
Edit: I am aware of the general Lie theory property (see for example page 3 of this paper) that if one can divide the generators $\{T\}$ into two subsets $L(K)=\{k^m\}$ and $L(P)=\{ p^n \}$ with the following commutation relations:
$$[k^{m_1},k^{m_2}]\in L(K), \quad [k^m,p^n]\in L(P), \quad [p^{n_1},p^{n_2}]\in L(K),$$
then one can decompose any $U$ as
$$ U = e^{i \alpha^m k^m} e^{i \beta^n p^n} .$$
This allows me to pull out the the first two exponentials in my desired decomposition above, by setting
$$L(K) = \{1_2 \times \sigma^i, \quad \sigma^i \times 1_2\}, \quad L(P) = \{ \sigma^i \times \sigma^j\}.$$
However, the same `trick' does not allow me to decompose any further. I don't know if this means it cannot be done at all though...
 A: The generators of su(4) in the fundamental are 4×4 traceless hermitian matrices, so 15 independent ones.
The matrices in your basis {T} are manifestly traceless and hermitian! Convince yourself of the obviousness of the fact. They are 3+3+9=15 in number, and, again, their independence should be obvious.
They are then a good basis of su(4), so the generic group element of SU(4) may be written as the exponential whose phase is their generic linear combination.
So you must convince yourself that your quadruple exponential product does not represent a mere subgroup. The exponents of the first two exponentials commute, so you may combine them into one, by the trivial limit of the CBH identity. I'm not sure how much rigor you require, but, to lower nontrivial orders of CBH-combining the 3 remaining exponentials you have a full span.
NB. Near the identity,  to linear order of your 15 ("angle") parameters, you completely cover all 15 directions of your manifold and tangent space.  Your original question asked whether higher orders could conspire to leave  "uncovered  regions" of that manifold. Presentation theory addresses these issues.
