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

A social experiment. What is the worst that can happen?

Limits and Infinite Integration by Parts

Why is so much work done on numerical verification of the Riemann Hypothesis?

Why should universal income be universal?

Can I still be respawned if I die by falling off the map?

On a tidally locked planet, would time be quantized?

Can the US President recognize Israel’s sovereignty over the Golan Heights for the USA or does that need an act of Congress?

How to explain what's wrong with this application of the chain rule?

Plot of a tornado-shaped surface

Keeping a ball lost forever

Did arcade monitors have same pixel aspect ratio as TV sets?

What is Cash Advance APR?

What features enable the Su-25 Frogfoot to operate with such a wide variety of fuels?

Fear of getting stuck on one programming language / technology that is not used in my country

What's the difference between releasing hormones and tropic hormones?

Can I say "fingers" when referring to toes?

Why does a simple loop result in ASYNC_NETWORK_IO waits?

Calculating total slots

Temporarily disable WLAN internet access for children, but allow it for adults

How can "mimic phobia" be cured or prevented?

Multiplicative persistence

What is going on with 'gets(stdin)' on the site coderbyte?

Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?

Are Captain Marvel's powers affected by Thanos' actions in Infinity War



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













6












$begingroup$


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
    Mar 13 at 16:37










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    Mar 13 at 16:40










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    Mar 13 at 16:53






  • 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
    Mar 13 at 16:53


















6












$begingroup$


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
    Mar 13 at 16:37










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    Mar 13 at 16:40










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    Mar 13 at 16:53






  • 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
    Mar 13 at 16:53
















6












6








6


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 Mar 13 at 20:24









J. W. Tanner

3,6551320




3,6551320










asked Mar 13 at 16:31









NazimJNazimJ

56619




56619












  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    Mar 13 at 16:37










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    Mar 13 at 16:40










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    Mar 13 at 16:53






  • 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
    Mar 13 at 16:53




















  • $begingroup$
    Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
    $endgroup$
    – Don Thousand
    Mar 13 at 16:37










  • $begingroup$
    That's why I suggested if anyone knows a good textbook/book then that would be good
    $endgroup$
    – NazimJ
    Mar 13 at 16:40










  • $begingroup$
    Have you tried wikipedia...
    $endgroup$
    – Don Thousand
    Mar 13 at 16:53






  • 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
    Mar 13 at 16:53


















$begingroup$
Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
$endgroup$
– Don Thousand
Mar 13 at 16:37




$begingroup$
Why do you assume it's simple to prove? Wikipedia points to a gigantic paper on its proof.
$endgroup$
– Don Thousand
Mar 13 at 16:37












$begingroup$
That's why I suggested if anyone knows a good textbook/book then that would be good
$endgroup$
– NazimJ
Mar 13 at 16:40




$begingroup$
That's why I suggested if anyone knows a good textbook/book then that would be good
$endgroup$
– NazimJ
Mar 13 at 16:40












$begingroup$
Have you tried wikipedia...
$endgroup$
– Don Thousand
Mar 13 at 16:53




$begingroup$
Have you tried wikipedia...
$endgroup$
– Don Thousand
Mar 13 at 16:53




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
Mar 13 at 16:53






$begingroup$
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
Mar 13 at 16:53












2 Answers
2






active

oldest

votes


















6












$begingroup$

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
    Mar 13 at 21:50










  • $begingroup$
    $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
    $endgroup$
    – Dietrich Burde
    Mar 13 at 22:40



















2












$begingroup$

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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146824%2fproof-that-there-exist-only-17-wallpaper-groups-tilings-of-the-plane%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    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
      Mar 13 at 21:50










    • $begingroup$
      $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
      $endgroup$
      – Dietrich Burde
      Mar 13 at 22:40
















    6












    $begingroup$

    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
      Mar 13 at 21:50










    • $begingroup$
      $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
      $endgroup$
      – Dietrich Burde
      Mar 13 at 22:40














    6












    6








    6





    $begingroup$

    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 Mar 13 at 19:34









    Dietrich BurdeDietrich Burde

    81.2k648106




    81.2k648106












    • $begingroup$
      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
      Mar 13 at 21:50










    • $begingroup$
      $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
      $endgroup$
      – Dietrich Burde
      Mar 13 at 22:40


















    • $begingroup$
      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
      Mar 13 at 21:50










    • $begingroup$
      $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
      $endgroup$
      – Dietrich Burde
      Mar 13 at 22:40
















    $begingroup$
    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
    Mar 13 at 21:50




    $begingroup$
    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
    Mar 13 at 21:50












    $begingroup$
    $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
    $endgroup$
    – Dietrich Burde
    Mar 13 at 22:40




    $begingroup$
    $phi(n)$ is Euler's totient function and $Phi_n$ the $n$-th cyclotomic polynomial. The cohomology group here is a finite group which we can compute.
    $endgroup$
    – Dietrich Burde
    Mar 13 at 22:40











    2












    $begingroup$

    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









    $endgroup$


















      2












      $begingroup$

      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









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 Mar 13 at 17:37









        Doug MDoug M

        1,663411




        1,663411






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146824%2fproof-that-there-exist-only-17-wallpaper-groups-tilings-of-the-plane%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Nidaros erkebispedøme

            Birsay

            Where did Arya get these scars? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara Favourite questions and answers from the 1st quarter of 2019Why did Arya refuse to end it?Has the pronunciation of Arya Stark's name changed?Has Arya forgiven people?Why did Arya Stark lose her vision?Why can Arya still use the faces?Has the Narrow Sea become narrower?Does Arya Stark know how to make poisons outside of the House of Black and White?Why did Nymeria leave Arya?Why did Arya not kill the Lannister soldiers she encountered in the Riverlands?What is the current canonical age of Sansa, Bran and Arya Stark?