Mixing
What does “Vertex-transitivity” tell us about an arbitrary graph?
$begingroup$
I know the formal definition, i.e. , for any arbitrary graph G,
$forall v_i,v_jin V(G), exists g in Auto(G) , s.t. g(v_i)=v_j$
But what does it mean in a broader sense?
graph-theory
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
$endgroup$
add a comment |
$begingroup$
I know the formal definition, i.e. , for any arbitrary graph G,
$forall v_i,v_jin V(G), exists g in Auto(G) , s.t. g(v_i)=v_j$
But what does it mean in a broader sense?
graph-theory
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
$endgroup$
3
$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09
add a comment |
$begingroup$
I know the formal definition, i.e. , for any arbitrary graph G,
$forall v_i,v_jin V(G), exists g in Auto(G) , s.t. g(v_i)=v_j$
But what does it mean in a broader sense?
graph-theory
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
$endgroup$
I know the formal definition, i.e. , for any arbitrary graph G,
$forall v_i,v_jin V(G), exists g in Auto(G) , s.t. g(v_i)=v_j$
But what does it mean in a broader sense?
graph-theory
graph-theory
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
asked Nov 25 '16 at 20:58
Ashkan RanjbarAshkan Ranjbar
11713
11713
3
$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09
add a comment |
3
$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09
3
3
$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09
add a comment |
3 Answers
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$begingroup$
If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.
- Graph $X$ is regular.(converse is not true Eg: Frucht Graph)
- $Aut(X)$ is non-trivial.
- Graph $X$ is symmetric.
- Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.
For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of ${1,2,3,4,5}$ of size 2. }.
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
$endgroup$
add a comment |
$begingroup$
Its very similar to what homogeneity means in topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
$endgroup$
add a comment |
$begingroup$
Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:
This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
answered Jan 26 at 18:30
alan2herealan2here
519219
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.
- Graph $X$ is regular.(converse is not true Eg: Frucht Graph)
- $Aut(X)$ is non-trivial.
- Graph $X$ is symmetric.
- Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.
For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of ${1,2,3,4,5}$ of size 2. }.
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
$endgroup$
add a comment |
$begingroup$
If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.
- Graph $X$ is regular.(converse is not true Eg: Frucht Graph)
- $Aut(X)$ is non-trivial.
- Graph $X$ is symmetric.
- Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.
For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of ${1,2,3,4,5}$ of size 2. }.
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
$endgroup$
add a comment |
$begingroup$
If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.
- Graph $X$ is regular.(converse is not true Eg: Frucht Graph)
- $Aut(X)$ is non-trivial.
- Graph $X$ is symmetric.
- Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.
For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of ${1,2,3,4,5}$ of size 2. }.
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
$endgroup$
If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.
- Graph $X$ is regular.(converse is not true Eg: Frucht Graph)
- $Aut(X)$ is non-trivial.
- Graph $X$ is symmetric.
- Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.
For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of ${1,2,3,4,5}$ of size 2. }.
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
answered Nov 26 '16 at 3:38
Ashwin KoodathilAshwin Koodathil
667
667
add a comment |
add a comment |
$begingroup$
Its very similar to what homogeneity means in topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
$endgroup$
add a comment |
$begingroup$
Its very similar to what homogeneity means in topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
$endgroup$
add a comment |
$begingroup$
Its very similar to what homogeneity means in topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
$endgroup$
Its very similar to what homogeneity means in topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
edited Jan 26 at 22:39
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
answered Nov 26 '16 at 3:42
Jorge Fernández HidalgoJorge Fernández Hidalgo
76.8k1394195
76.8k1394195
add a comment |
add a comment |
$begingroup$
Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:
This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
answered Jan 26 at 18:30
alan2herealan2here
519219
$endgroup$
add a comment |
$begingroup$
Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:
This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
answered Jan 26 at 18:30
alan2herealan2here
519219
$endgroup$
add a comment |
$begingroup$
Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:
This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
answered Jan 26 at 18:30
alan2herealan2here
519219
$endgroup$
Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:
This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
answered Jan 26 at 18:30
alan2herealan2here
519219
edited Jan 27 at 10:25
answered Jan 26 at 18:30
alan2herealan2here
519219
answered Jan 26 at 18:30
alan2herealan2here
519219
answered Jan 26 at 18:30
alan2herealan2here
519219
519219
add a comment |
add a comment |
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$begingroup$
It means there is a pleasant symmetry to the graph. Consider the graph of a soccer ball, for example, where every vertex belongs to two hexagonal faces and a pentagonal face, and the vertices are interchangeable by spinning the ball.
$endgroup$
– Théophile
Nov 25 '16 at 21:11
$begingroup$
So back to our formal definition, what you said means that for each two vertices, I could find an automorphism,namely, a rotation from one vertex exactly toward the other? @Théophile
$endgroup$
– Ashkan Ranjbar
Nov 25 '16 at 22:10
$begingroup$
Yes, that's right. So it formalizes the idea of "you can't tell the difference between the vertices". An $n$-cycle is also vertex transitive: pick any $u,v in V$, and there is an obvious automorphism sending $u$ to $v$. To put it another way, there are no "special" vertices. A star, for example, where one central vertex $w$ is connected to a number of leaves, is not vertex transitive; you can easily distinguish $w$ from the leaves (which is an informal way of saying that no automorphism sends $w$ to a leaf).
$endgroup$
– Théophile
Nov 26 '16 at 2:06
$begingroup$
Compare also to edge transitivity, which is similar but (of course) applies to edges instead of vertices. A soccer ball is not edge transitive because there are two kinds of edges: either both endpoints are on the same pentagon, or they are on different pentagons. An $n$-cycle, on the other hand, is edge transitive.
$endgroup$
– Théophile
Nov 26 '16 at 2:09