= Solution
Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U'(0)$. Differentiating $U(t)^\dagger U(t)=I$ gives $X^\dagger+X=0$; differentiating $\det U(t)=1$ gives $\operatorname{tr}X=0$. Conversely, for every <anti-Hermitian matrix> $X$ with zero <trace>, the <matrix exponential> $e^{tX}$ is unitary and has <determinant> $e^{t\operatorname{tr}X}=1$. Thus the real <Lie algebra> is
$$
\boxed{\mathfrak{su}(4)=\{X\in M_4(\mathbb C):X^\dagger=-X,\ \operatorname{tr}X=0\},\qquad [X,Y]=XY-YX.}
$$
The <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> is
$$
\boxed{A_{jk}=E_{jk}-E_{kj},\quad B_{jk}=i(E_{jk}+E_{kj})\quad(1\leq j<k\leq4),\qquad H_a=i(E_{aa}-E_{a+1,a+1})\quad(a=1,2,3).}
$$
There 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 $H_a$. Consequently
$$
\boxed{\dim_{\mathbb R}\mathfrak{su}(4)=15,\qquad\operatorname{rank}\mathfrak{su}(4)=3.}
$$
For the <rank of a semisimple Lie algebra>, the diagonal <Cartan subalgebra> consists of $i\operatorname{diag}(t_1,t_2,t_3,t_4)$ 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 $-iA_{jk},-iB_{jk},-iH_a$ and writes <group> elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.
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