What type of 2D lattice is depicted in the image shown below? Is it possible to determine without knowing the length of the unit vectors and the angles in between them? I'm confused because it seemingly doesn't fit with either monoclinic or orthorhombic description as $\mathrm{a \ne b}$ and $\mathrm{c\ne d}$ Bravais lattice


The picture is not a lattice, it is a periodic pattern. In a lattice, all lattice points are related by the translational symmetry given by the unit cell vectors. I can take any two lattice points and take the vector between them. Then, when I add that vector to any other lattice points, I should land on another lattice point.

This is not the case here. If I take the vector between the pair of closest points, and add the vector to the upper right point of the pair, I land on empty space.

So let's paint half of the points a different color. The points of different color are related by two-fold axis (located for instance at the center of two close pairs). The two-fold symmetry applies not only to the two closest points, but to the entire pattern, so it is called a crystallographic symmetry (as opposed to non-crystallographic symmetry).

This crystallographic symmetry imposes no restraints on the dimensions of the lattice. Other symmetry elements would impose restraints, for example in the presence of a four-fold crystallographic symmetry, the lattice would have to be tetragonal.

For the example at hand, the convention is to put the origin on one of the two-fold axes. It looks like you could have one unit cell vector along x and one along y, at approximately right angle (this is not imposed by point group symmetry, so it is "by accident" and you would have to measure carefully to see whether the angle is exactly 90 or maybe 90.0001 degrees).

To summarize, it is monoclinic with an angle that happens to be close to 90 degrees.

For specifics on the lattice types, see e.g. Wikipedia.

Thanks, @Ivan_Neretin, for the helpful comment.

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    $\begingroup$ Everything you say is true, except one thing. The points are equivalent. $\endgroup$ – Ivan Neretin Mar 7 at 16:31
  • $\begingroup$ Oh, yes I see it now. There is a two-fold in the middle of the close pairs, right? $\endgroup$ – Karsten Theis Mar 7 at 16:49
  • $\begingroup$ I updated my answer based on the comment by @Ivan_Neretin $\endgroup$ – Karsten Theis Mar 7 at 17:07
  • $\begingroup$ Now it is all right. True, they are equivalent by two-fold axis, but not by translations. $\endgroup$ – Ivan Neretin Mar 7 at 17:21

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