Can one determine the dimension of a manifold given its 1-skeleton?
Clash Royale CLAN TAG#URR8PPP
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This may be an easy question, but I can't think of the answer at hand.
Suppose that I have a triangulated $n$-manifold $M$ (satisfying any set of conditions that you feel like). Suppose that I give to you the 1-skeleton of the triangulation. Can you tell me anything about the dimension of $M$?
(For CW-decompositions, the answer is obviously no: every sphere can be given a CW decomposition with no 1-cells. However, if it is a triangulation instead...)
gn.general-topology manifolds simplicial-complexes triangulations
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
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add a comment |
$begingroup$
This may be an easy question, but I can't think of the answer at hand.
Suppose that I have a triangulated $n$-manifold $M$ (satisfying any set of conditions that you feel like). Suppose that I give to you the 1-skeleton of the triangulation. Can you tell me anything about the dimension of $M$?
(For CW-decompositions, the answer is obviously no: every sphere can be given a CW decomposition with no 1-cells. However, if it is a triangulation instead...)
gn.general-topology manifolds simplicial-complexes triangulations
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
$endgroup$
1
$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39
add a comment |
$begingroup$
This may be an easy question, but I can't think of the answer at hand.
Suppose that I have a triangulated $n$-manifold $M$ (satisfying any set of conditions that you feel like). Suppose that I give to you the 1-skeleton of the triangulation. Can you tell me anything about the dimension of $M$?
(For CW-decompositions, the answer is obviously no: every sphere can be given a CW decomposition with no 1-cells. However, if it is a triangulation instead...)
gn.general-topology manifolds simplicial-complexes triangulations
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
$endgroup$
This may be an easy question, but I can't think of the answer at hand.
Suppose that I have a triangulated $n$-manifold $M$ (satisfying any set of conditions that you feel like). Suppose that I give to you the 1-skeleton of the triangulation. Can you tell me anything about the dimension of $M$?
(For CW-decompositions, the answer is obviously no: every sphere can be given a CW decomposition with no 1-cells. However, if it is a triangulation instead...)
gn.general-topology manifolds simplicial-complexes triangulations
gn.general-topology manifolds simplicial-complexes triangulations
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
asked Jan 6 at 19:31
Simon RoseSimon Rose
3,7812444
3,7812444
1
$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39
add a comment |
1
$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39
1
1
$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39
$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39
add a comment |
1 Answer
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oldest
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You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five dimensional sphere.
However, we can alway upperbound the dimension by one less than the connectivity of the given graph! It is a theorem of Barnette from "Decompositions of homology manifolds and their graphs" that the connectivity of the 1-skeleton of a d-dimensional manifold is at least d+1.
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
$endgroup$
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five dimensional sphere.
However, we can alway upperbound the dimension by one less than the connectivity of the given graph! It is a theorem of Barnette from "Decompositions of homology manifolds and their graphs" that the connectivity of the 1-skeleton of a d-dimensional manifold is at least d+1.
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
$endgroup$
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
add a comment |
$begingroup$
You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five dimensional sphere.
However, we can alway upperbound the dimension by one less than the connectivity of the given graph! It is a theorem of Barnette from "Decompositions of homology manifolds and their graphs" that the connectivity of the 1-skeleton of a d-dimensional manifold is at least d+1.
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
$endgroup$
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
add a comment |
$begingroup$
You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five dimensional sphere.
However, we can alway upperbound the dimension by one less than the connectivity of the given graph! It is a theorem of Barnette from "Decompositions of homology manifolds and their graphs" that the connectivity of the 1-skeleton of a d-dimensional manifold is at least d+1.
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
$endgroup$
You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five dimensional sphere.
However, we can alway upperbound the dimension by one less than the connectivity of the given graph! It is a theorem of Barnette from "Decompositions of homology manifolds and their graphs" that the connectivity of the 1-skeleton of a d-dimensional manifold is at least d+1.
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
answered Jan 6 at 20:39
Gjergji ZaimiGjergji Zaimi
62.6k4163309
62.6k4163309
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
add a comment |
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
10
10
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
$begingroup$
For every $ngeq dgeq 4$ there is a triangulation of a $(d-1)$-sphere (in fact, a simplicial polytope) with $n$ vertices, whose 1-skeleton is the complete graph $K_n$. These are the cyclic polytopes.
$endgroup$
– Richard Stanley
Jan 6 at 23:49
1
1
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
$begingroup$
Well, that kind of answers my next question as to whether or not simply connected might make a difference...
$endgroup$
– Simon Rose
Jan 9 at 21:09
add a comment |
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$begingroup$
Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
$endgroup$
– Jeff Strom
Jan 6 at 19:39