Proof that there exist only 17 Wallpaper Groups (Tilings of the plane)elliptic functions on the 17 wallpaper...

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Proof that there exist only 17 Wallpaper Groups (Tilings of the plane)


elliptic functions on the 17 wallpaper groupsCan someone explain the math behind tessellation?Is anybody researching “ternary” groups?What reference contains the proof of the classification of the wallpaper groups?Tilings of the planeTrapezoid area proof by dividing it into two triangles?Exercises in category theory for a non-working mathematican (undergrad)What are the algebraic structures of the wallpaper groups?There are finitely many $n$-dimensional wallpaper groupsWallpaper groups for the hyperbolic plane













4












$begingroup$


I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).



The punchline is




$$There ,, are ,, exactly ,, 17 ,, wallpaper ,, groups ,,(17 ,, ways ,, to ,, tile ,, the ,, plane)$$




My question is $2$-fold:




  1. Can someone sketch out the proof or at least give some high level ideas of why this may be true?


  2. Can someone refer me to a website or a textbook that develops the proof in detail?











share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    5 hours ago










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    5 hours ago










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    5 hours ago






  • 4




    $begingroup$
    There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
    $endgroup$
    – Mark Bennet
    5 hours ago


















4












$begingroup$


I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).



The punchline is




$$There ,, are ,, exactly ,, 17 ,, wallpaper ,, groups ,,(17 ,, ways ,, to ,, tile ,, the ,, plane)$$




My question is $2$-fold:




  1. Can someone sketch out the proof or at least give some high level ideas of why this may be true?


  2. Can someone refer me to a website or a textbook that develops the proof in detail?











share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    5 hours ago










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    5 hours ago










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    5 hours ago






  • 4




    $begingroup$
    There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
    $endgroup$
    – Mark Bennet
    5 hours ago
















4












4








4


2



$begingroup$


I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).



The punchline is




$$There ,, are ,, exactly ,, 17 ,, wallpaper ,, groups ,,(17 ,, ways ,, to ,, tile ,, the ,, plane)$$




My question is $2$-fold:




  1. Can someone sketch out the proof or at least give some high level ideas of why this may be true?


  2. Can someone refer me to a website or a textbook that develops the proof in detail?











share|cite|improve this question











$endgroup$




I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).



The punchline is




$$There ,, are ,, exactly ,, 17 ,, wallpaper ,, groups ,,(17 ,, ways ,, to ,, tile ,, the ,, plane)$$




My question is $2$-fold:




  1. Can someone sketch out the proof or at least give some high level ideas of why this may be true?


  2. Can someone refer me to a website or a textbook that develops the proof in detail?








group-theory geometry reference-request symmetry tessellations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









J. W. Tanner

3,2001320




3,2001320










asked 5 hours ago









NazimJNazimJ

43218




43218












  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    5 hours ago










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    5 hours ago










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    5 hours ago






  • 4




    $begingroup$
    There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
    $endgroup$
    – Mark Bennet
    5 hours ago




















  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    5 hours ago










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    5 hours ago










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    5 hours ago






  • 4




    $begingroup$
    There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
    $endgroup$
    – Mark Bennet
    5 hours ago


















$begingroup$
Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
$endgroup$
– Don Thousand
5 hours ago




$begingroup$
Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
$endgroup$
– Don Thousand
5 hours ago












$begingroup$
That's why I suggested if anyone knows a good textbook/book then that would be good
$endgroup$
– NazimJ
5 hours ago




$begingroup$
That's why I suggested if anyone knows a good textbook/book then that would be good
$endgroup$
– NazimJ
5 hours ago












$begingroup$
Have you tried wikipedia...
$endgroup$
– Don Thousand
5 hours ago




$begingroup$
Have you tried wikipedia...
$endgroup$
– Don Thousand
5 hours ago




4




4




$begingroup$
There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
$endgroup$
– Mark Bennet
5 hours ago






$begingroup$
There is a proof in JH Conway et al's book "The Symmetries of Things" which is mentioned on the Wikipedia page. It uses techniques which prove the theorem, but also give a way of identifying and naming the various patterns involved.
$endgroup$
– Mark Bennet
5 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Sketch of the proof: Let $Gamma le {rm Iso}(Bbb R^2)$ be a wallpaper group. Then $Gamma$ has a normal subgroup isomorphic
to $Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $Bbb Z^2$ by conjugation.
We obtain a faithful representation
$$
F hookrightarrow {rm Aut}(Bbb Z^2)cong GL_2(Bbb Z).
$$

The group $GL_2(Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
called arithmetic ornament classes:
begin{align*}
C_1 & cong leftlangle begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} rightrangle,;
C_2 cong leftlangle begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix} rightrangle,;
C_3 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix} rightrangle, \
C_4 & cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} rightrangle, ;
C_6 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix} rightrangle, ;
D_1 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} rightrangle, \
D_1 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} rightrangle, ;
D_2 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
rightrangle, \
D_2 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
rightrangle, ; D_3 cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}rightrangle, \
D_3 & cong leftlangle begin{pmatrix} 0 & -1 \ -1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}
rightrangle,\
D_4 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}
rightrangle, ; D_6cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix}rightrangle.
end{align*}



This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $phi(n)=deg(Phi_n)mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
$$
1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
$$

determined by $H^2(F,Bbb Z^2)$.



By computing $H^2(F,Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
This yields $17$ different groups, because two of them turn out to be isomorphic.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
    $endgroup$
    – NazimJ
    13 mins ago



















1












$begingroup$

The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.



The members of the wallpaper group have a notation:



632 or 4*2 or *2222



It uses some sequence of numbers, and the symbols $*,circ, times$



The numbers represent rotations, the $*$ represents the presence of reflection, the $times$ represents a glide symmetry. The $circ$ indicates translations without reflections or rotations.



This notation suggests an algebra. For each digit before the star we add $frac {n-1}{n}$. The star adds 1. For each digit after the star we add $frac {n-1}{2n}$ or one half of what you otherwise would add.



$times$ adds 1, an $circ$ adds 2.



This sum must equal 2.



For the groups above: $frac {5}{6}+frac{2}{3} + frac {1}{2} = 2$ and $frac {3}{4} + 1 + frac {1}{4} = 2$ and $1+frac 14 + frac 14 + frac 14 + frac 14=2$



With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.



https://en.wikipedia.org/wiki/Orbifold_notation



However, I don't remember the proofs that associate this algebra to groups.






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    2 Answers
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    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Sketch of the proof: Let $Gamma le {rm Iso}(Bbb R^2)$ be a wallpaper group. Then $Gamma$ has a normal subgroup isomorphic
    to $Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $Bbb Z^2$ by conjugation.
    We obtain a faithful representation
    $$
    F hookrightarrow {rm Aut}(Bbb Z^2)cong GL_2(Bbb Z).
    $$

    The group $GL_2(Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
    called arithmetic ornament classes:
    begin{align*}
    C_1 & cong leftlangle begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} rightrangle,;
    C_2 cong leftlangle begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix} rightrangle,;
    C_3 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix} rightrangle, \
    C_4 & cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} rightrangle, ;
    C_6 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix} rightrangle, ;
    D_1 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} rightrangle, \
    D_1 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} rightrangle, ;
    D_2 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, \
    D_2 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, ; D_3 cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}rightrangle, \
    D_3 & cong leftlangle begin{pmatrix} 0 & -1 \ -1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}
    rightrangle,\
    D_4 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}
    rightrangle, ; D_6cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix}rightrangle.
    end{align*}



    This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $phi(n)=deg(Phi_n)mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
    $$
    1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
    $$

    determined by $H^2(F,Bbb Z^2)$.



    By computing $H^2(F,Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
    This yields $17$ different groups, because two of them turn out to be isomorphic.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
      $endgroup$
      – NazimJ
      13 mins ago
















    4












    $begingroup$

    Sketch of the proof: Let $Gamma le {rm Iso}(Bbb R^2)$ be a wallpaper group. Then $Gamma$ has a normal subgroup isomorphic
    to $Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $Bbb Z^2$ by conjugation.
    We obtain a faithful representation
    $$
    F hookrightarrow {rm Aut}(Bbb Z^2)cong GL_2(Bbb Z).
    $$

    The group $GL_2(Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
    called arithmetic ornament classes:
    begin{align*}
    C_1 & cong leftlangle begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} rightrangle,;
    C_2 cong leftlangle begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix} rightrangle,;
    C_3 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix} rightrangle, \
    C_4 & cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} rightrangle, ;
    C_6 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix} rightrangle, ;
    D_1 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} rightrangle, \
    D_1 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} rightrangle, ;
    D_2 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, \
    D_2 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, ; D_3 cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}rightrangle, \
    D_3 & cong leftlangle begin{pmatrix} 0 & -1 \ -1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}
    rightrangle,\
    D_4 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}
    rightrangle, ; D_6cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix}rightrangle.
    end{align*}



    This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $phi(n)=deg(Phi_n)mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
    $$
    1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
    $$

    determined by $H^2(F,Bbb Z^2)$.



    By computing $H^2(F,Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
    This yields $17$ different groups, because two of them turn out to be isomorphic.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
      $endgroup$
      – NazimJ
      13 mins ago














    4












    4








    4





    $begingroup$

    Sketch of the proof: Let $Gamma le {rm Iso}(Bbb R^2)$ be a wallpaper group. Then $Gamma$ has a normal subgroup isomorphic
    to $Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $Bbb Z^2$ by conjugation.
    We obtain a faithful representation
    $$
    F hookrightarrow {rm Aut}(Bbb Z^2)cong GL_2(Bbb Z).
    $$

    The group $GL_2(Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
    called arithmetic ornament classes:
    begin{align*}
    C_1 & cong leftlangle begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} rightrangle,;
    C_2 cong leftlangle begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix} rightrangle,;
    C_3 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix} rightrangle, \
    C_4 & cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} rightrangle, ;
    C_6 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix} rightrangle, ;
    D_1 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} rightrangle, \
    D_1 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} rightrangle, ;
    D_2 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, \
    D_2 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, ; D_3 cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}rightrangle, \
    D_3 & cong leftlangle begin{pmatrix} 0 & -1 \ -1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}
    rightrangle,\
    D_4 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}
    rightrangle, ; D_6cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix}rightrangle.
    end{align*}



    This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $phi(n)=deg(Phi_n)mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
    $$
    1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
    $$

    determined by $H^2(F,Bbb Z^2)$.



    By computing $H^2(F,Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
    This yields $17$ different groups, because two of them turn out to be isomorphic.






    share|cite|improve this answer









    $endgroup$



    Sketch of the proof: Let $Gamma le {rm Iso}(Bbb R^2)$ be a wallpaper group. Then $Gamma$ has a normal subgroup isomorphic
    to $Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $Bbb Z^2$ by conjugation.
    We obtain a faithful representation
    $$
    F hookrightarrow {rm Aut}(Bbb Z^2)cong GL_2(Bbb Z).
    $$

    The group $GL_2(Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
    called arithmetic ornament classes:
    begin{align*}
    C_1 & cong leftlangle begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} rightrangle,;
    C_2 cong leftlangle begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix} rightrangle,;
    C_3 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix} rightrangle, \
    C_4 & cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} rightrangle, ;
    C_6 cong leftlangle begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix} rightrangle, ;
    D_1 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} rightrangle, \
    D_1 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} rightrangle, ;
    D_2 cong leftlangle begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, \
    D_2 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}
    rightrangle, ; D_3 cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}rightrangle, \
    D_3 & cong leftlangle begin{pmatrix} 0 & -1 \ -1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & -1 end{pmatrix}
    rightrangle,\
    D_4 & cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}
    rightrangle, ; D_6cong leftlangle begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix},
    begin{pmatrix} 0 & 1 \ -1 & 1 end{pmatrix}rightrangle.
    end{align*}



    This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $phi(n)=deg(Phi_n)mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
    $$
    1rightarrow Bbb Z^2rightarrow Gammarightarrow Frightarrow 1,
    $$

    determined by $H^2(F,Bbb Z^2)$.



    By computing $H^2(F,Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
    This yields $17$ different groups, because two of them turn out to be isomorphic.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Dietrich BurdeDietrich Burde

    80.6k647104




    80.6k647104












    • $begingroup$
      Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
      $endgroup$
      – NazimJ
      13 mins ago


















    • $begingroup$
      Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
      $endgroup$
      – NazimJ
      13 mins ago
















    $begingroup$
    Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
    $endgroup$
    – NazimJ
    13 mins ago




    $begingroup$
    Thank you! I can see exactly where there are details i need to fill in and concepts to refresh on. If you wouldn't mind, before i go on a deep dive, to define or simply name for me $phi (n) $, $Phi_n $, and $H^2 (F, mathbb{R}^2 ) $?
    $endgroup$
    – NazimJ
    13 mins ago











    1












    $begingroup$

    The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.



    The members of the wallpaper group have a notation:



    632 or 4*2 or *2222



    It uses some sequence of numbers, and the symbols $*,circ, times$



    The numbers represent rotations, the $*$ represents the presence of reflection, the $times$ represents a glide symmetry. The $circ$ indicates translations without reflections or rotations.



    This notation suggests an algebra. For each digit before the star we add $frac {n-1}{n}$. The star adds 1. For each digit after the star we add $frac {n-1}{2n}$ or one half of what you otherwise would add.



    $times$ adds 1, an $circ$ adds 2.



    This sum must equal 2.



    For the groups above: $frac {5}{6}+frac{2}{3} + frac {1}{2} = 2$ and $frac {3}{4} + 1 + frac {1}{4} = 2$ and $1+frac 14 + frac 14 + frac 14 + frac 14=2$



    With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.



    https://en.wikipedia.org/wiki/Orbifold_notation



    However, I don't remember the proofs that associate this algebra to groups.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.



      The members of the wallpaper group have a notation:



      632 or 4*2 or *2222



      It uses some sequence of numbers, and the symbols $*,circ, times$



      The numbers represent rotations, the $*$ represents the presence of reflection, the $times$ represents a glide symmetry. The $circ$ indicates translations without reflections or rotations.



      This notation suggests an algebra. For each digit before the star we add $frac {n-1}{n}$. The star adds 1. For each digit after the star we add $frac {n-1}{2n}$ or one half of what you otherwise would add.



      $times$ adds 1, an $circ$ adds 2.



      This sum must equal 2.



      For the groups above: $frac {5}{6}+frac{2}{3} + frac {1}{2} = 2$ and $frac {3}{4} + 1 + frac {1}{4} = 2$ and $1+frac 14 + frac 14 + frac 14 + frac 14=2$



      With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.



      https://en.wikipedia.org/wiki/Orbifold_notation



      However, I don't remember the proofs that associate this algebra to groups.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.



        The members of the wallpaper group have a notation:



        632 or 4*2 or *2222



        It uses some sequence of numbers, and the symbols $*,circ, times$



        The numbers represent rotations, the $*$ represents the presence of reflection, the $times$ represents a glide symmetry. The $circ$ indicates translations without reflections or rotations.



        This notation suggests an algebra. For each digit before the star we add $frac {n-1}{n}$. The star adds 1. For each digit after the star we add $frac {n-1}{2n}$ or one half of what you otherwise would add.



        $times$ adds 1, an $circ$ adds 2.



        This sum must equal 2.



        For the groups above: $frac {5}{6}+frac{2}{3} + frac {1}{2} = 2$ and $frac {3}{4} + 1 + frac {1}{4} = 2$ and $1+frac 14 + frac 14 + frac 14 + frac 14=2$



        With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.



        https://en.wikipedia.org/wiki/Orbifold_notation



        However, I don't remember the proofs that associate this algebra to groups.






        share|cite|improve this answer









        $endgroup$



        The best I have, is this (and I admit it is not very good). The Euler characteristic of the infinite plane is 2.



        The members of the wallpaper group have a notation:



        632 or 4*2 or *2222



        It uses some sequence of numbers, and the symbols $*,circ, times$



        The numbers represent rotations, the $*$ represents the presence of reflection, the $times$ represents a glide symmetry. The $circ$ indicates translations without reflections or rotations.



        This notation suggests an algebra. For each digit before the star we add $frac {n-1}{n}$. The star adds 1. For each digit after the star we add $frac {n-1}{2n}$ or one half of what you otherwise would add.



        $times$ adds 1, an $circ$ adds 2.



        This sum must equal 2.



        For the groups above: $frac {5}{6}+frac{2}{3} + frac {1}{2} = 2$ and $frac {3}{4} + 1 + frac {1}{4} = 2$ and $1+frac 14 + frac 14 + frac 14 + frac 14=2$



        With this algebra, we can brute force through all possible combinations of rotations, reflections, glides, etc.



        https://en.wikipedia.org/wiki/Orbifold_notation



        However, I don't remember the proofs that associate this algebra to groups.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Doug MDoug M

        1,653411




        1,653411






























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