Wood's notation is used to reference the relationship between two 2D lattices when the angle between the two unit vectors $(a_1, a_2)$ of one lattice is the same as the angle between the unit vectors $(b_1, b_2)$ of the other lattice. It's written with the ratios of the corresponding unit vectors followed by the letter $R$ and then a rotation angle in degrees which is understood to be rounded:

$$\left(\frac{a_1}{b_1} \times \frac{a_2}{b_2}\right)R \theta $$

Below I show drawings of a hexagonal lattice with $a=1$ and the Wood's notation for four other commensurate hexagonal lattices.

Below I show the Wood's notation and the two new vectors themselves.

Question: I've found a trivial case and three more cases of commensurate hexagonal lattices where the second lattice is enlarged by a single scalar magnification and rotated so that its lattice points match those of the unit hexagonal lattice. I've done this by trial and error. Is there a systematic way that I can find all cases (if finite) or up to some limit in size if infinite?

Ignore the trivial rotations greater than 60 degrees and less than zero, and those that are integer multiples (e.g. $\left(2\sqrt{3} \times 2\sqrt{3}\right) \text{R} 30$)

$$\left(1 \times 1\right) \text{R} 0 \ \ \ \ \left[1, 0\right], \left[-\frac{1}{2}, \frac{\sqrt{3}}{2}\right]$$

$$\left(\sqrt{3} \times \sqrt{3}\right) \text{R} 30 \ \ \ \ \left[\frac{3}{2}, \frac{\sqrt{3}}{2}\right], \left[-\frac{3}{2}, \frac{\sqrt{3}}{2}\right]$$

$$\left(\sqrt{7} \times \sqrt{7}\right) \text{R} 19.1 \ \ \ \ \left[\frac{5}{2}, \frac{\sqrt{3}}{2}\right], \left[-2, \sqrt{3}\right]$$

$$\left(\sqrt{19} \times \sqrt{19}\right) \text{R} 23.4 \ \ \ \ \left[4, \sqrt{3}\right], \left[-\frac{7}{2}, \frac{3\sqrt{3}}{2}\right]$$

Click images for full size:

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  • $\begingroup$ I may need some help with tagging. $\endgroup$
    – uhoh
    Commented Aug 18, 2019 at 15:11
  • $\begingroup$ Wouldn't any individual cell in the 1/12-plane pizza slice (since anything past 30deg is going to have a similar reflective property) give a corresponding commensurate lattice? Like, it looks like you should also have $(2x2)R0$, $(\sqrt{13},\sqrt{13})R13.9$, using $(2,0)$ and $(7/2,\sqrt{3})$ respectively? $\endgroup$ Commented Aug 18, 2019 at 16:48
  • $\begingroup$ @DanUznanski the first one I've ruled out in the last sentence, for the second one the lengths of those vectors are 2 and $\sqrt{61/4}$, neither of which are $\sqrt{13}$. $\endgroup$
    – uhoh
    Commented Aug 18, 2019 at 22:38
  • 1
    $\begingroup$ Yeah I meant $\sqrt{3}/2$ on the y coordinate of the last one, whups $\endgroup$ Commented Aug 18, 2019 at 22:40
  • 1
    $\begingroup$ Of tangential note, every scaling factor will be the square root of one of the numbers in oeis.org/A003136 $\endgroup$ Commented Aug 18, 2019 at 22:51

1 Answer 1


Every non-origin cell in the lattice can be used to make a commensurate lattice.

We can generate all of them up to a given lattice-wise distance $d$ in this way:

Take two integers $0 < u \le v \le d-u$. Our target cell's location is $u$ cells up-right and then $v$ cells right, giving a coordinate of $\left(v + \frac12u, \frac{\sqrt{3}}{2}u\right)$, a scale of $\ell = \sqrt{u^2+v^2+uv}$ (the law of cosines to the rescue: $u$ and $v$ give distances along two vectors $1/3$ of a circle apart), and an angle of $\theta = \arctan\left(\frac{\sqrt{3}u}{2v + u}\right)$.

Each lattice then can be mirrored to the other half of the fundamental sixth of the plane by using $\frac{1}{6}\text{ circle} - \theta$ as the angle

If you need lattices that aren't simply scalings of other ones, simply insist on $\gcd(u,v)=1$.


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