In various references, we see the construction of unit cells of carbon nanotubes (CNTs) from chiral and translational vectors.
The chiral vector is given as:
$$\vec C_h = n\vec a_1 + m\vec a_2$$
While the translational vector is given by:
$$\vec T = t_1\vec a_1 + t_2\vec a_2$$
Clearly, $\vec C_h$ and $\vec T$ must be orthogonal, hence $\vec C_h \cdot \vec T = 0$. In the literature, it always appears that $t_1 = \displaystyle\frac{2m+n}{d_R}$ and $t_2 = -\displaystyle\frac{2n+m}{d_R}$, where $d_R=\mathrm{gcd}(2m+n,2n+m)$.
My questions are, how to deduce this? And how does this relate to the fact that the translational vector can determine the size of the unit cell?
I managed to replicate this idea for a hexagonal lattice, similar to graphene.
For a rectangular lattice, I successfully reproduced the chiral vector, but I haven't yet managed to understand or reproduce the translational vector (not even for CNTs, actually).
Here is a matricial approach.
The given constraints can be written :
$$\underbrace{\pmatrix{C_h & T}}_{M}=\underbrace{\pmatrix{a_1 & a_2}}_{B} \times \ \underbrace{\pmatrix{n & t_1\\m & t_2}}_{C}\tag{1}$$
where $M$ should be understood as the matrix whose columns are $C_h,T$ ; same convention for the second matrix $B$ which can also be written, for a certain basis :
$$B=\pmatrix{\sqrt{3} & \sqrt{3}\\1 & -1}\tag{2}$$
Now, let us express orthogonality of $C_h$ and $T$ in a matricial way by :
$$M^TM=\pmatrix{\ell_1^2 & 0\\0 & \ell_2^2}\tag{3}$$
where $\ell_1,\ell_2$ are the resp. length of vectors $C_h$ and $T$.
Besides, using (1) and (2), we have :
$$M^TM=C^TB^TBC=$$
$$M^TM = 2 \ \pmatrix{ 2(m^2+mn+n^2) & t_1(2n+m)+t_2(2m+n) \\ t_1(2n+m)+t_2(2m+n) & 2(t_1^2 + t_1t_2 + t_2^2)}\tag{4}$$
Identification of (3) and (4) yields the three equations :
$$\begin{cases} Eq1 : \ \ \ m^2+mn+n^2&=&\frac14 \ell_1^2\\ Eq2 : \ \ \ t_1(2n+m)+t_2(2m+n)&=&0 \\ Eq3 : \ \ \ t_1^2 + t_1t_2 + t_2^2&=& \frac14 \ell_2^2\end{cases}$$
We need only the second equation which gives, as desired, the proportionnality of $t_1$ and $t_2$ with $2m+n$ and $-(2n+m)$, with a proportionnality factor $\gcd(2m+n,2n+m)$ chosen in such a way that the results are integers.