Symmetry group of cube
$begingroup$
I want to carefully trace the entire sequence of interactions between the structures when getting the symmetry group of the cube. I'm just learning how to work with structures and diagrams, so please check the course of my reasoning.
So $mathbb{E}^3in Ob(Top)$ -- euclidean space. $K_3 subset mathbb{E}^3$ -- $3$-dim cube. $V$ -- group of permutations on the vertex-set. $Sym K_3 = {phiin Aut(mathbb{E}^3 mid phi(K_3) = K_3}$. All objects in the following diagram are $Mon$-category objects. $Hom(mathbb{E}^3,mathbb{E}^3)$ -- a monoid, because not all endomorphisms are invertible. (this is not monoidal-category, just category where all $Ob$ -- monoids).
$$begin{array}
A Sym(K_3) & stackrel{f}{hookrightarrow}& Aut(mathbb{E}^3)& stackrel{g}{hookrightarrow} & Hom(mathbb{E}^3,mathbb{E}^3) \
cong{H} \
V & stackrel{d}{hookrightarrow} & S_8
end{array}
$$
So $H$ -- isomorphism between $Sym(K_3)$, where elements are continuous functions, and $V$. I think, that $H$ -- functor $H: Mon rightarrow Grp$. $H$ translates those monoids, which are also groups, into these same groups.
Group $V$ action on $F_{Set}({vertices})$:
$$begin{array}
A mathbb{E}^3 & hookleftarrow & {vertices}\
downarrow{F_{Set}} & &downarrow{F_{Set}}\
X & hookleftarrow & F_{Set}({vertices}) & xleftarrow[phi]{pr_{ F_{Set}({verteces})}} & F_{Set}({vertices})times Y & xrightarrow{pr_Y} & Y & hookrightarrow &S\
& & & & & &uparrow{G_{Set}}&& uparrow{G_{Set}}\
& & & & & & V &hookrightarrow & S_8
end{array}
$$
Where $F_{Set}$ and $G_{Set}$ -- forgetful functors, $F_{Set}$ from $Top rightarrow Set$ and $G_{Set}$ from $Grprightarrow Set$, $Y$ -- group $V$ structure carrier. $phi$ -- action of group $V$ on set $X$ $phi(g, x) = g cdot x$, ${vertices}$ -- vertices (discrete subspace) of cube in euclidean space $mathbb{E}^3$.
So I want to proof, that $|Sym(K_3)| = |Vx||Vx,e||V_{x,e}| = 48$. Where $x$ -- one of vertices -- point in $mathbb{E}^3$, and $e$ -- edge exiting the vertex. Obviously, a vertex orbit consists of eight vertices: $|Vx| = 8$, from each vertex there are three edges, therefore the orbit of each of them $|Vx,e| = 3$ so $Rightarrow$ $|V_{x,e}| = 2$ -- $id$ and reflection relative to the plane passing through the edge.
And I do not really understand how to interpret the edge here? As a subspace of $mathbb{E}^3$ or like a pair of vertices $(x_1,x_2)$?
group-theory category-theory finite-groups
edited Dec 7 '18 at 12:48
user10354138
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
$endgroup$
add a comment |
$begingroup$
I want to carefully trace the entire sequence of interactions between the structures when getting the symmetry group of the cube. I'm just learning how to work with structures and diagrams, so please check the course of my reasoning.
So $mathbb{E}^3in Ob(Top)$ -- euclidean space. $K_3 subset mathbb{E}^3$ -- $3$-dim cube. $V$ -- group of permutations on the vertex-set. $Sym K_3 = {phiin Aut(mathbb{E}^3 mid phi(K_3) = K_3}$. All objects in the following diagram are $Mon$-category objects. $Hom(mathbb{E}^3,mathbb{E}^3)$ -- a monoid, because not all endomorphisms are invertible. (this is not monoidal-category, just category where all $Ob$ -- monoids).
$$begin{array}
A Sym(K_3) & stackrel{f}{hookrightarrow}& Aut(mathbb{E}^3)& stackrel{g}{hookrightarrow} & Hom(mathbb{E}^3,mathbb{E}^3) \
cong{H} \
V & stackrel{d}{hookrightarrow} & S_8
end{array}
$$
So $H$ -- isomorphism between $Sym(K_3)$, where elements are continuous functions, and $V$. I think, that $H$ -- functor $H: Mon rightarrow Grp$. $H$ translates those monoids, which are also groups, into these same groups.
Group $V$ action on $F_{Set}({vertices})$:
$$begin{array}
A mathbb{E}^3 & hookleftarrow & {vertices}\
downarrow{F_{Set}} & &downarrow{F_{Set}}\
X & hookleftarrow & F_{Set}({vertices}) & xleftarrow[phi]{pr_{ F_{Set}({verteces})}} & F_{Set}({vertices})times Y & xrightarrow{pr_Y} & Y & hookrightarrow &S\
& & & & & &uparrow{G_{Set}}&& uparrow{G_{Set}}\
& & & & & & V &hookrightarrow & S_8
end{array}
$$
Where $F_{Set}$ and $G_{Set}$ -- forgetful functors, $F_{Set}$ from $Top rightarrow Set$ and $G_{Set}$ from $Grprightarrow Set$, $Y$ -- group $V$ structure carrier. $phi$ -- action of group $V$ on set $X$ $phi(g, x) = g cdot x$, ${vertices}$ -- vertices (discrete subspace) of cube in euclidean space $mathbb{E}^3$.
So I want to proof, that $|Sym(K_3)| = |Vx||Vx,e||V_{x,e}| = 48$. Where $x$ -- one of vertices -- point in $mathbb{E}^3$, and $e$ -- edge exiting the vertex. Obviously, a vertex orbit consists of eight vertices: $|Vx| = 8$, from each vertex there are three edges, therefore the orbit of each of them $|Vx,e| = 3$ so $Rightarrow$ $|V_{x,e}| = 2$ -- $id$ and reflection relative to the plane passing through the edge.
And I do not really understand how to interpret the edge here? As a subspace of $mathbb{E}^3$ or like a pair of vertices $(x_1,x_2)$?
group-theory category-theory finite-groups
edited Dec 7 '18 at 12:48
user10354138
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
$endgroup$
1
$begingroup$
Why this question has -1 rep? Where I am incorrect?
$endgroup$
– Just do it
Dec 8 '18 at 2:24
add a comment |
$begingroup$
I want to carefully trace the entire sequence of interactions between the structures when getting the symmetry group of the cube. I'm just learning how to work with structures and diagrams, so please check the course of my reasoning.
So $mathbb{E}^3in Ob(Top)$ -- euclidean space. $K_3 subset mathbb{E}^3$ -- $3$-dim cube. $V$ -- group of permutations on the vertex-set. $Sym K_3 = {phiin Aut(mathbb{E}^3 mid phi(K_3) = K_3}$. All objects in the following diagram are $Mon$-category objects. $Hom(mathbb{E}^3,mathbb{E}^3)$ -- a monoid, because not all endomorphisms are invertible. (this is not monoidal-category, just category where all $Ob$ -- monoids).
$$begin{array}
A Sym(K_3) & stackrel{f}{hookrightarrow}& Aut(mathbb{E}^3)& stackrel{g}{hookrightarrow} & Hom(mathbb{E}^3,mathbb{E}^3) \
cong{H} \
V & stackrel{d}{hookrightarrow} & S_8
end{array}
$$
So $H$ -- isomorphism between $Sym(K_3)$, where elements are continuous functions, and $V$. I think, that $H$ -- functor $H: Mon rightarrow Grp$. $H$ translates those monoids, which are also groups, into these same groups.
Group $V$ action on $F_{Set}({vertices})$:
$$begin{array}
A mathbb{E}^3 & hookleftarrow & {vertices}\
downarrow{F_{Set}} & &downarrow{F_{Set}}\
X & hookleftarrow & F_{Set}({vertices}) & xleftarrow[phi]{pr_{ F_{Set}({verteces})}} & F_{Set}({vertices})times Y & xrightarrow{pr_Y} & Y & hookrightarrow &S\
& & & & & &uparrow{G_{Set}}&& uparrow{G_{Set}}\
& & & & & & V &hookrightarrow & S_8
end{array}
$$
Where $F_{Set}$ and $G_{Set}$ -- forgetful functors, $F_{Set}$ from $Top rightarrow Set$ and $G_{Set}$ from $Grprightarrow Set$, $Y$ -- group $V$ structure carrier. $phi$ -- action of group $V$ on set $X$ $phi(g, x) = g cdot x$, ${vertices}$ -- vertices (discrete subspace) of cube in euclidean space $mathbb{E}^3$.
So I want to proof, that $|Sym(K_3)| = |Vx||Vx,e||V_{x,e}| = 48$. Where $x$ -- one of vertices -- point in $mathbb{E}^3$, and $e$ -- edge exiting the vertex. Obviously, a vertex orbit consists of eight vertices: $|Vx| = 8$, from each vertex there are three edges, therefore the orbit of each of them $|Vx,e| = 3$ so $Rightarrow$ $|V_{x,e}| = 2$ -- $id$ and reflection relative to the plane passing through the edge.
And I do not really understand how to interpret the edge here? As a subspace of $mathbb{E}^3$ or like a pair of vertices $(x_1,x_2)$?
group-theory category-theory finite-groups
edited Dec 7 '18 at 12:48
user10354138
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
$endgroup$
I want to carefully trace the entire sequence of interactions between the structures when getting the symmetry group of the cube. I'm just learning how to work with structures and diagrams, so please check the course of my reasoning.
So $mathbb{E}^3in Ob(Top)$ -- euclidean space. $K_3 subset mathbb{E}^3$ -- $3$-dim cube. $V$ -- group of permutations on the vertex-set. $Sym K_3 = {phiin Aut(mathbb{E}^3 mid phi(K_3) = K_3}$. All objects in the following diagram are $Mon$-category objects. $Hom(mathbb{E}^3,mathbb{E}^3)$ -- a monoid, because not all endomorphisms are invertible. (this is not monoidal-category, just category where all $Ob$ -- monoids).
$$begin{array}
A Sym(K_3) & stackrel{f}{hookrightarrow}& Aut(mathbb{E}^3)& stackrel{g}{hookrightarrow} & Hom(mathbb{E}^3,mathbb{E}^3) \
cong{H} \
V & stackrel{d}{hookrightarrow} & S_8
end{array}
$$
So $H$ -- isomorphism between $Sym(K_3)$, where elements are continuous functions, and $V$. I think, that $H$ -- functor $H: Mon rightarrow Grp$. $H$ translates those monoids, which are also groups, into these same groups.
Group $V$ action on $F_{Set}({vertices})$:
$$begin{array}
A mathbb{E}^3 & hookleftarrow & {vertices}\
downarrow{F_{Set}} & &downarrow{F_{Set}}\
X & hookleftarrow & F_{Set}({vertices}) & xleftarrow[phi]{pr_{ F_{Set}({verteces})}} & F_{Set}({vertices})times Y & xrightarrow{pr_Y} & Y & hookrightarrow &S\
& & & & & &uparrow{G_{Set}}&& uparrow{G_{Set}}\
& & & & & & V &hookrightarrow & S_8
end{array}
$$
Where $F_{Set}$ and $G_{Set}$ -- forgetful functors, $F_{Set}$ from $Top rightarrow Set$ and $G_{Set}$ from $Grprightarrow Set$, $Y$ -- group $V$ structure carrier. $phi$ -- action of group $V$ on set $X$ $phi(g, x) = g cdot x$, ${vertices}$ -- vertices (discrete subspace) of cube in euclidean space $mathbb{E}^3$.
So I want to proof, that $|Sym(K_3)| = |Vx||Vx,e||V_{x,e}| = 48$. Where $x$ -- one of vertices -- point in $mathbb{E}^3$, and $e$ -- edge exiting the vertex. Obviously, a vertex orbit consists of eight vertices: $|Vx| = 8$, from each vertex there are three edges, therefore the orbit of each of them $|Vx,e| = 3$ so $Rightarrow$ $|V_{x,e}| = 2$ -- $id$ and reflection relative to the plane passing through the edge.
And I do not really understand how to interpret the edge here? As a subspace of $mathbb{E}^3$ or like a pair of vertices $(x_1,x_2)$?
group-theory category-theory finite-groups
group-theory category-theory finite-groups
edited Dec 7 '18 at 12:48
user10354138
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
edited Dec 7 '18 at 12:48
user10354138
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
edited Dec 7 '18 at 12:48
user10354138
7,3772925
edited Dec 7 '18 at 12:48
user10354138
7,3772925
edited Dec 7 '18 at 12:48
user10354138
7,3772925
7,3772925
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
asked Dec 7 '18 at 12:38
Just do itJust do it
16718
16718
1
$begingroup$
Why this question has -1 rep? Where I am incorrect?
$endgroup$
– Just do it
Dec 8 '18 at 2:24
add a comment |
1
$begingroup$
Why this question has -1 rep? Where I am incorrect?
$endgroup$
– Just do it
Dec 8 '18 at 2:24
1
1
$begingroup$
Why this question has -1 rep? Where I am incorrect?
$endgroup$
– Just do it
Dec 8 '18 at 2:24
$begingroup$
Why this question has -1 rep? Where I am incorrect?
$endgroup$
– Just do it
Dec 8 '18 at 2:24
add a comment |
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$begingroup$
Why this question has -1 rep? Where I am incorrect?
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– Just do it
Dec 8 '18 at 2:24