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The crystallographic restriction theorem says that you cannot have a periodic lattice with $n$-fold rotation symmetry, with $n$ different from 1,2,3,4 and 6 (for 2D and 3D).

There are many ways to prove the theorem, see the Wikipedia article. I understand some of the them, but one of the proof goes like this:

Consider a periodic lattice that is symmetric with respect to $n$-fold rotations around a given axis. The trace of the matrix associated to the spatial rotation around the given axis is either $2\cos\left(\frac{2\pi}{n}\right)$ (2D) or $1+2\cos\left(\frac{2\pi}{n}\right)$ (3D). As the rotation matrix maps lattice points into other lattice points, then the trace has to be an integer. The only solution to this is condition is $n$ to be equal to 1,2,3,4 or 6.

The solution and why the trace is like that I understand by simply writing the rotation matrix, but I would like to have more insight on why the trace has to be an integer in order to be a representation of a symmetry operation of the lattice.

In general, is there any meaning to trace=integer?

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    $\begingroup$ Fun trivia probable fact: I'm pretty sure this proof is owing to Donald Coexter, in a largely negative report on a paper he was reviewing. The subject of the paper was this theorem, and the paper was many pages long. Coexter's whole review was a short paragraph, including his proof of the same theorem. I can't seem to find the reference for this right now; the story is rather humorous (owing to Coexter's eloquent language in the matter). $\endgroup$
    – Selene Routley
    Commented Jan 13, 2018 at 22:50
  • $\begingroup$ I would gladly like to see it. He was criticizing the theorem? $\endgroup$
    – Mauricio
    Commented Jan 13, 2018 at 22:51
  • $\begingroup$ No, he was diplomatically, but sarcastically, criticizing the length of the paper author's proof. $\endgroup$
    – Selene Routley
    Commented Jan 13, 2018 at 22:53

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Consider transformation of a set of primitive translation vectors $e_a$, $a=1...d$ of a $d$-dimensional lattice under rotation $O$: $$ Oe_a = \sum_{b=1}^d k_{ab}\ e_b. $$ If rotation is a symmetry of a lattice then coefficients $k_{ab}$ are integers. Thus rotation matrix written in $e_a$ basis has integer elements and integer trace. Trace is invariant under linear transformations. Hence rotation matrix written in any basis has integer trace.

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  • $\begingroup$ Why are the k_ab integers? $\endgroup$
    – Mauricio
    Commented Jan 13, 2018 at 21:13
  • $\begingroup$ Because $Oe_a$ is translation vector and $e_a$ are primitive translation vectors. Sites of a lattice become sites of a same lattice under symmetry transformation. Hence $Oe_a$ is translation. $\endgroup$
    – Gec
    Commented Jan 13, 2018 at 21:15
  • $\begingroup$ Or said another way, if $e_a$ are primitive translation vectors, then ALL lattice points can be written as a sum of integer value * $e_a$. That is what makes it a lattice. $\endgroup$
    – PPenguin
    Commented Jan 13, 2018 at 21:17
  • $\begingroup$ @Gec isnt there a problem if your base is not orthogonal? $\endgroup$
    – Mauricio
    Commented Jan 13, 2018 at 21:19
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    $\begingroup$ @Mauricio no, there is no problem. Trace is invariant under any linear transformations, not only orthogonal. $\endgroup$
    – Gec
    Commented Jan 13, 2018 at 21:21

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