Mixing
Different ways we can paint the faces of a tetrahedron using $4$ colors
$begingroup$
Suppose we want to paint the faces of a tetrahedron using $4$ different colors, assuming that we allow different faces to be painted with the same color.
By not taking the symmetries of tetrahedron into account, there are $|X|=4^4=256$ ways. Now, when its symmetries are introduced, from Burnside's orbit-counting theorem:
$$
r=frac{1}{|G|}sum_{g in G}|X_g|
$$
where $G$ is the symmetry group of the tetrahedron (considering only orientation preserving symmetries), $X_g$ are the elements fixed by $g$ and $r$ is the number of orbits of $X$ under $G$'s action. Therefore, we need to keep track of how many elements are fixed by every $g in G$.
$bullet space $The identity element keeps everything unchanged, so $X_e=256$
$bullet space $ Let $rho^j_i$ denote the rotations about the vertex $i$ by $j$ degrees. For the element to stay fixed, the adjacent faces to this vertex must be of the same color. So $$|X_{rho^{120}_1}|=|X_{rho_1^{240}}|=dots=|X_{rho^{240}_4}|=4 cdot4=16$$
$bullet space $ There are $3$ more symmetries left to examine, which are the $180^o$ rotations, $m_1,m_2, m_3$. Elements stay fixed only if they have two pairs of similarly colored adjacent faces. Thus
$$
|X_{m_1}|=|X_{m_2}|=|X_{m_3}|=4 cdot 4=16
$$
Taking all this into account, we yield:
$$
r=frac{1}{12}(256+8 cdot 16 + 3 cdot 16)=frac{432}{12}=36
$$
Therefore, there are $36$ different ways to color the faces of a tetrahedron using $4$ colors.
abstract-algebra combinatorics group-theory proof-verification group-actions
asked Jan 31 at 19:19
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
Suppose we want to paint the faces of a tetrahedron using $4$ different colors, assuming that we allow different faces to be painted with the same color.
By not taking the symmetries of tetrahedron into account, there are $|X|=4^4=256$ ways. Now, when its symmetries are introduced, from Burnside's orbit-counting theorem:
$$
r=frac{1}{|G|}sum_{g in G}|X_g|
$$
where $G$ is the symmetry group of the tetrahedron (considering only orientation preserving symmetries), $X_g$ are the elements fixed by $g$ and $r$ is the number of orbits of $X$ under $G$'s action. Therefore, we need to keep track of how many elements are fixed by every $g in G$.
$bullet space $The identity element keeps everything unchanged, so $X_e=256$
$bullet space $ Let $rho^j_i$ denote the rotations about the vertex $i$ by $j$ degrees. For the element to stay fixed, the adjacent faces to this vertex must be of the same color. So $$|X_{rho^{120}_1}|=|X_{rho_1^{240}}|=dots=|X_{rho^{240}_4}|=4 cdot4=16$$
$bullet space $ There are $3$ more symmetries left to examine, which are the $180^o$ rotations, $m_1,m_2, m_3$. Elements stay fixed only if they have two pairs of similarly colored adjacent faces. Thus
$$
|X_{m_1}|=|X_{m_2}|=|X_{m_3}|=4 cdot 4=16
$$
Taking all this into account, we yield:
$$
r=frac{1}{12}(256+8 cdot 16 + 3 cdot 16)=frac{432}{12}=36
$$
Therefore, there are $36$ different ways to color the faces of a tetrahedron using $4$ colors.
abstract-algebra combinatorics group-theory proof-verification group-actions
asked Jan 31 at 19:19
JevautJevaut
1,168312
$endgroup$
$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
1
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15
add a comment |
$begingroup$
Suppose we want to paint the faces of a tetrahedron using $4$ different colors, assuming that we allow different faces to be painted with the same color.
By not taking the symmetries of tetrahedron into account, there are $|X|=4^4=256$ ways. Now, when its symmetries are introduced, from Burnside's orbit-counting theorem:
$$
r=frac{1}{|G|}sum_{g in G}|X_g|
$$
where $G$ is the symmetry group of the tetrahedron (considering only orientation preserving symmetries), $X_g$ are the elements fixed by $g$ and $r$ is the number of orbits of $X$ under $G$'s action. Therefore, we need to keep track of how many elements are fixed by every $g in G$.
$bullet space $The identity element keeps everything unchanged, so $X_e=256$
$bullet space $ Let $rho^j_i$ denote the rotations about the vertex $i$ by $j$ degrees. For the element to stay fixed, the adjacent faces to this vertex must be of the same color. So $$|X_{rho^{120}_1}|=|X_{rho_1^{240}}|=dots=|X_{rho^{240}_4}|=4 cdot4=16$$
$bullet space $ There are $3$ more symmetries left to examine, which are the $180^o$ rotations, $m_1,m_2, m_3$. Elements stay fixed only if they have two pairs of similarly colored adjacent faces. Thus
$$
|X_{m_1}|=|X_{m_2}|=|X_{m_3}|=4 cdot 4=16
$$
Taking all this into account, we yield:
$$
r=frac{1}{12}(256+8 cdot 16 + 3 cdot 16)=frac{432}{12}=36
$$
Therefore, there are $36$ different ways to color the faces of a tetrahedron using $4$ colors.
abstract-algebra combinatorics group-theory proof-verification group-actions
asked Jan 31 at 19:19
JevautJevaut
1,168312
$endgroup$
Suppose we want to paint the faces of a tetrahedron using $4$ different colors, assuming that we allow different faces to be painted with the same color.
By not taking the symmetries of tetrahedron into account, there are $|X|=4^4=256$ ways. Now, when its symmetries are introduced, from Burnside's orbit-counting theorem:
$$
r=frac{1}{|G|}sum_{g in G}|X_g|
$$
where $G$ is the symmetry group of the tetrahedron (considering only orientation preserving symmetries), $X_g$ are the elements fixed by $g$ and $r$ is the number of orbits of $X$ under $G$'s action. Therefore, we need to keep track of how many elements are fixed by every $g in G$.
$bullet space $The identity element keeps everything unchanged, so $X_e=256$
$bullet space $ Let $rho^j_i$ denote the rotations about the vertex $i$ by $j$ degrees. For the element to stay fixed, the adjacent faces to this vertex must be of the same color. So $$|X_{rho^{120}_1}|=|X_{rho_1^{240}}|=dots=|X_{rho^{240}_4}|=4 cdot4=16$$
$bullet space $ There are $3$ more symmetries left to examine, which are the $180^o$ rotations, $m_1,m_2, m_3$. Elements stay fixed only if they have two pairs of similarly colored adjacent faces. Thus
$$
|X_{m_1}|=|X_{m_2}|=|X_{m_3}|=4 cdot 4=16
$$
Taking all this into account, we yield:
$$
r=frac{1}{12}(256+8 cdot 16 + 3 cdot 16)=frac{432}{12}=36
$$
Therefore, there are $36$ different ways to color the faces of a tetrahedron using $4$ colors.
abstract-algebra combinatorics group-theory proof-verification group-actions
abstract-algebra combinatorics group-theory proof-verification group-actions
asked Jan 31 at 19:19
JevautJevaut
1,168312
asked Jan 31 at 19:19
JevautJevaut
1,168312
edited Jan 31 at 20:04
Jevaut
asked Jan 31 at 19:19
JevautJevaut
1,168312
asked Jan 31 at 19:19
JevautJevaut
1,168312
asked Jan 31 at 19:19
JevautJevaut
1,168312
1,168312
$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
1
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15
add a comment |
$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
1
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15
$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
1
1
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15
add a comment |
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$begingroup$
You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries.
$endgroup$
– Lee Mosher
Jan 31 at 19:28
$begingroup$
@LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G cong A_4 leq S_4$.
$endgroup$
– Jevaut
Jan 31 at 19:31
1
$begingroup$
Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so.
$endgroup$
– Lee Mosher
Jan 31 at 19:34
$begingroup$
@LeeMosher Given this consideration, is my approach correct?
$endgroup$
– Jevaut
Jan 31 at 20:07
$begingroup$
Looks correct to me!
$endgroup$
– Mike Earnest
Jan 31 at 20:15