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Here “controllable” means density operator controllability, so all transformations on a fixed spectral orbit are available. Strict unitary operator controllability, including an independently specified global phase, is a stronger condition. Assume the usual finite-dimensional real controls and freely chosen duration.

The dimension hypothesis actually identifies the algebra, not just its size. For completeness, put the invariant inner product $⟨X,Y⟩=-\mathrm{Tr}(XY)$ on $\mathfrak{u}(N)$ and write ${\mathfrak{g}}^{\perp }=\mathbb{R}Z$. Since $\mathfrak{g}$ is closed under commutators, $[X,Z]$ lies in ${\mathfrak{g}}^{\perp }$ for every $X\in \mathfrak{g}$. The adjoint action is skew-adjoint for this inner product, so $[X,Z]$ is also orthogonal to $Z$ and must vanish. If $Z$ were nonscalar, its orthogonal eigenspaces would give a centralizer of dimension at most $(N-1{)}^2+1<N^2-1$, too small to contain $\mathfrak{g}$. Therefore $Z$ is scalar and $\mathfrak{g}=\mathfrak{su}(N)$. This proves codimension-one quantum dynamical algebra is special unitary.

a1 is true for density-state controllability; a2 is true; a3 is false. The algebra $\mathfrak{su}(N)$ generates every determinant-one unitary. For any $V\in U(N)$, choose $\phi$ with $e^{iN\phi }=\det V$; then $e^{-i\phi }V\in SU(N)$, proving that every gate is available up to global phase. But all generated Quantum Hamiltonians are traceless, andso evolution starting at the identity always has determinant one. Arbitrary determinant, requiring the additional central direction $iI$, cannot be generated. If a1 is instead read as phase-sensitive unitary-operator controllability, it is false for precisely the same reason as a3; the physical distinction is explicit here.