Mixing
A question on the distinguishable colorings of an object in relation to the number of fixed points in a...
$begingroup$
Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)
In class we were shown Burnsides counting theorem which states:
Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is
$$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$
This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes
$$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$
I also know that
$$operatorname{fix}(g)={xin Omega mid x^g=x}.$$
Take the following example: suppose we have a tetrahedron and we have 11 colors.
Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).
The Question
I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$
where
$tau$ is the rotation by $pi/3$ about a face,
$omega$ is the rotation by $-pi/3$ about a face, and
$delta$ is the rotation by $pi$ about an edge.
I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.
group-theory coloring
edited Jan 7 at 19:15
Amnotwhy
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
$endgroup$
add a comment |
$begingroup$
Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)
In class we were shown Burnsides counting theorem which states:
Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is
$$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$
This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes
$$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$
I also know that
$$operatorname{fix}(g)={xin Omega mid x^g=x}.$$
Take the following example: suppose we have a tetrahedron and we have 11 colors.
Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).
The Question
I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$
where
$tau$ is the rotation by $pi/3$ about a face,
$omega$ is the rotation by $-pi/3$ about a face, and
$delta$ is the rotation by $pi$ about an edge.
I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.
group-theory coloring
edited Jan 7 at 19:15
Amnotwhy
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
$endgroup$
add a comment |
$begingroup$
Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)
In class we were shown Burnsides counting theorem which states:
Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is
$$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$
This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes
$$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$
I also know that
$$operatorname{fix}(g)={xin Omega mid x^g=x}.$$
Take the following example: suppose we have a tetrahedron and we have 11 colors.
Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).
The Question
I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$
where
$tau$ is the rotation by $pi/3$ about a face,
$omega$ is the rotation by $-pi/3$ about a face, and
$delta$ is the rotation by $pi$ about an edge.
I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.
group-theory coloring
edited Jan 7 at 19:15
Amnotwhy
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
$endgroup$
Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)
In class we were shown Burnsides counting theorem which states:
Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is
$$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$
This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes
$$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$
I also know that
$$operatorname{fix}(g)={xin Omega mid x^g=x}.$$
Take the following example: suppose we have a tetrahedron and we have 11 colors.
Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).
The Question
I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$
where
$tau$ is the rotation by $pi/3$ about a face,
$omega$ is the rotation by $-pi/3$ about a face, and
$delta$ is the rotation by $pi$ about an edge.
I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.
group-theory coloring
group-theory coloring
edited Jan 7 at 19:15
Amnotwhy
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
edited Jan 7 at 19:15
Amnotwhy
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
edited Jan 7 at 19:15
Amnotwhy
283
edited Jan 7 at 19:15
Amnotwhy
283
edited Jan 7 at 19:15
Amnotwhy
283
283
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
asked Dec 8 '18 at 23:00
can'tcauchycan'tcauchy
999417
999417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll assume you're coloring the faces; when you color the vertices the same thing happens.
When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.
When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).
In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031771%2fa-question-on-the-distinguishable-colorings-of-an-object-in-relation-to-the-numb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll assume you're coloring the faces; when you color the vertices the same thing happens.
When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.
When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).
In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
$endgroup$
add a comment |
$begingroup$
I'll assume you're coloring the faces; when you color the vertices the same thing happens.
When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.
When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).
In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
$endgroup$
add a comment |
$begingroup$
I'll assume you're coloring the faces; when you color the vertices the same thing happens.
When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.
When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).
In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
$endgroup$
I'll assume you're coloring the faces; when you color the vertices the same thing happens.
When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.
When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).
In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
answered Dec 9 '18 at 0:34
Misha LavrovMisha Lavrov
45.7k656107
45.7k656107
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031771%2fa-question-on-the-distinguishable-colorings-of-an-object-in-relation-to-the-numb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
