 Solution

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.