= Solution
The <carbon nanotube chiral vector> identifies equivalent sites of <graphene> that become one point on rolling:
$$
\mathbf C_h=n\mathbf a_1+m\mathbf a_2,\quad
|\mathbf a_1|=|\mathbf a_2|=a=\sqrt3b,\quad
d=\frac a\pi\sqrt{n^2+nm+m^2}.
$$
The primitive vectors meet at $60^\circ$, $b$ is the C–C bond length, and one can choose $0\le m\le n$. The <carbon nanotube chiral angle> satisfies $\tan\theta=\sqrt3m/(2n+m)$. \b[Armchair tubes have $(n,n)$ and $\theta=30^\circ$; zigzag tubes have $(n,0)$ and $\theta=0$]. Other pairs are chiral, with mirror-related left- and right-handed forms. Armchair and zigzag tubes themselves have mirror symmetry.
The circumferential <boundary condition> requires $\mathbf k\cdot\mathbf C_h=2\pi\ell$. In <nanotube zone folding>, an allowed <wavevector> line intersects one of the <Dirac valleys in graphene> precisely when \b[$n-m$ is divisible by three]. Thus all armchair tubes are metallic in this approximation. Zigzag tubes are metallic for $n$ divisible by three and otherwise semiconducting; general chiral tubes follow the same modular rule. Curvature can open small gaps in nominally metallic non-armchair tubes. Strain, interactions and broken symmetries can modify the ideal result; metallic band structure does not imply zero device resistance.
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