Permutations on vertices of cubes and hence finding volume enclosed by the vertices
Denote $C$ to be the cube $C={(x_1,x_2,x_3)|0 leq x_1,x_2,x_3 leq 1}$ and let $V={ (x_1,x_2,x_3)|x_1,x_2,x_3 in {0,1 } }$ be the set of vertices of the cube.
Let $A=$convex$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$, where convex means it is the volume enclosed within the points.
Let $sigma in S_3$ act on C by $sigma . (x_1,x_2,x_3) = (x_{phi (1)},x_{phi (2)},x_{phi (3)})$
(a) What are the sizes of the orbits?
(b) Let $A_sigma = { sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) in A }$. Determine the volume of this simplex. Show that $C = cup_{sigma in S_3} A_sigma$
(c) Show that any intersection $A_sigma$ and $A_tau$, $sigma neq tau$ cannot have any volume.
What I did:
(a) The possible sizes of the orbits are 1, 3 and 6.
Size of orbits 1: Elements that have the form $(a,a,a)$
Size of orbits 3: Elements that have the form $(a,b,b)$. In a sense, we are rearranging 3 elements but only 2 distinct.
Size of orbits 6: Elements that have the form $(a,b,c)$. We are arranging 3 distinct elements.
In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.
(b) I explained that the volume of the simplex is $frac {1}{6}$ since $|S_3|=6$ elements and that each $A_sigma, sigma in S_3$ are of equal sizes because they are just rotations/reflections of other points when acted by $sigma$.
Since volume of each simplex is $frac 1 6$ for each $sigma in S_3$, then the union of all simplexes is 1, which is the volume of the cube.
(c) Since each simplex must be of equal size, then the intersection between $A_sigma$ and $A_tau$ must have no volume for $sigma neq tau$, otherwise the union of all $A_sigma , sigma in S_3$ will not have volume 1, which happens to be the volume of the cube.
I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?
Edited: Part (a) answer
abstract-algebra group-theory finite-groups group-actions algebraic-groups
asked Nov 28 at 23:04
Icycarus
4671213
add a comment |
Denote $C$ to be the cube $C={(x_1,x_2,x_3)|0 leq x_1,x_2,x_3 leq 1}$ and let $V={ (x_1,x_2,x_3)|x_1,x_2,x_3 in {0,1 } }$ be the set of vertices of the cube.
Let $A=$convex$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$, where convex means it is the volume enclosed within the points.
Let $sigma in S_3$ act on C by $sigma . (x_1,x_2,x_3) = (x_{phi (1)},x_{phi (2)},x_{phi (3)})$
(a) What are the sizes of the orbits?
(b) Let $A_sigma = { sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) in A }$. Determine the volume of this simplex. Show that $C = cup_{sigma in S_3} A_sigma$
(c) Show that any intersection $A_sigma$ and $A_tau$, $sigma neq tau$ cannot have any volume.
What I did:
(a) The possible sizes of the orbits are 1, 3 and 6.
Size of orbits 1: Elements that have the form $(a,a,a)$
Size of orbits 3: Elements that have the form $(a,b,b)$. In a sense, we are rearranging 3 elements but only 2 distinct.
Size of orbits 6: Elements that have the form $(a,b,c)$. We are arranging 3 distinct elements.
In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.
(b) I explained that the volume of the simplex is $frac {1}{6}$ since $|S_3|=6$ elements and that each $A_sigma, sigma in S_3$ are of equal sizes because they are just rotations/reflections of other points when acted by $sigma$.
Since volume of each simplex is $frac 1 6$ for each $sigma in S_3$, then the union of all simplexes is 1, which is the volume of the cube.
(c) Since each simplex must be of equal size, then the intersection between $A_sigma$ and $A_tau$ must have no volume for $sigma neq tau$, otherwise the union of all $A_sigma , sigma in S_3$ will not have volume 1, which happens to be the volume of the cube.
I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?
Edited: Part (a) answer
abstract-algebra group-theory finite-groups group-actions algebraic-groups
asked Nov 28 at 23:04
Icycarus
4671213
1
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10
add a comment |
Denote $C$ to be the cube $C={(x_1,x_2,x_3)|0 leq x_1,x_2,x_3 leq 1}$ and let $V={ (x_1,x_2,x_3)|x_1,x_2,x_3 in {0,1 } }$ be the set of vertices of the cube.
Let $A=$convex$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$, where convex means it is the volume enclosed within the points.
Let $sigma in S_3$ act on C by $sigma . (x_1,x_2,x_3) = (x_{phi (1)},x_{phi (2)},x_{phi (3)})$
(a) What are the sizes of the orbits?
(b) Let $A_sigma = { sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) in A }$. Determine the volume of this simplex. Show that $C = cup_{sigma in S_3} A_sigma$
(c) Show that any intersection $A_sigma$ and $A_tau$, $sigma neq tau$ cannot have any volume.
What I did:
(a) The possible sizes of the orbits are 1, 3 and 6.
Size of orbits 1: Elements that have the form $(a,a,a)$
Size of orbits 3: Elements that have the form $(a,b,b)$. In a sense, we are rearranging 3 elements but only 2 distinct.
Size of orbits 6: Elements that have the form $(a,b,c)$. We are arranging 3 distinct elements.
In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.
(b) I explained that the volume of the simplex is $frac {1}{6}$ since $|S_3|=6$ elements and that each $A_sigma, sigma in S_3$ are of equal sizes because they are just rotations/reflections of other points when acted by $sigma$.
Since volume of each simplex is $frac 1 6$ for each $sigma in S_3$, then the union of all simplexes is 1, which is the volume of the cube.
(c) Since each simplex must be of equal size, then the intersection between $A_sigma$ and $A_tau$ must have no volume for $sigma neq tau$, otherwise the union of all $A_sigma , sigma in S_3$ will not have volume 1, which happens to be the volume of the cube.
I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?
Edited: Part (a) answer
abstract-algebra group-theory finite-groups group-actions algebraic-groups
asked Nov 28 at 23:04
Icycarus
4671213
Denote $C$ to be the cube $C={(x_1,x_2,x_3)|0 leq x_1,x_2,x_3 leq 1}$ and let $V={ (x_1,x_2,x_3)|x_1,x_2,x_3 in {0,1 } }$ be the set of vertices of the cube.
Let $A=$convex$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$, where convex means it is the volume enclosed within the points.
Let $sigma in S_3$ act on C by $sigma . (x_1,x_2,x_3) = (x_{phi (1)},x_{phi (2)},x_{phi (3)})$
(a) What are the sizes of the orbits?
(b) Let $A_sigma = { sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) in A }$. Determine the volume of this simplex. Show that $C = cup_{sigma in S_3} A_sigma$
(c) Show that any intersection $A_sigma$ and $A_tau$, $sigma neq tau$ cannot have any volume.
What I did:
(a) The possible sizes of the orbits are 1, 3 and 6.
Size of orbits 1: Elements that have the form $(a,a,a)$
Size of orbits 3: Elements that have the form $(a,b,b)$. In a sense, we are rearranging 3 elements but only 2 distinct.
Size of orbits 6: Elements that have the form $(a,b,c)$. We are arranging 3 distinct elements.
In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.
(b) I explained that the volume of the simplex is $frac {1}{6}$ since $|S_3|=6$ elements and that each $A_sigma, sigma in S_3$ are of equal sizes because they are just rotations/reflections of other points when acted by $sigma$.
Since volume of each simplex is $frac 1 6$ for each $sigma in S_3$, then the union of all simplexes is 1, which is the volume of the cube.
(c) Since each simplex must be of equal size, then the intersection between $A_sigma$ and $A_tau$ must have no volume for $sigma neq tau$, otherwise the union of all $A_sigma , sigma in S_3$ will not have volume 1, which happens to be the volume of the cube.
I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?
Edited: Part (a) answer
abstract-algebra group-theory finite-groups group-actions algebraic-groups
abstract-algebra group-theory finite-groups group-actions algebraic-groups
asked Nov 28 at 23:04
Icycarus
4671213
asked Nov 28 at 23:04
Icycarus
4671213
edited Nov 29 at 11:22
asked Nov 28 at 23:04
Icycarus
4671213
asked Nov 28 at 23:04
Icycarus
4671213
asked Nov 28 at 23:04
Icycarus
4671213
4671213
1
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10
add a comment |
1
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10
1
1
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10
add a comment |
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1
I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 leq a leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates.
– Pietro Gheri
Nov 29 at 2:13
I was thinking that $x_1$ gets maps to $x_{phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks!
– Icycarus
Nov 29 at 8:35
Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b).
– jgon
Nov 29 at 8:43
@jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer?
– Icycarus
Nov 29 at 10:39
Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex.
– jgon
Nov 29 at 16:10