Can one determine the dimension of a manifold given its 1-skeleton?












14














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







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  • 1




    Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
    – Jeff Strom
    yesterday
















14














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







share|cite|improve this question


















  • 1




    Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
    – Jeff Strom
    yesterday














14












14








14


2





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Simon RoseSimon Rose

3,7712444




3,7712444








  • 1




    Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
    – Jeff Strom
    yesterday














  • 1




    Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
    – Jeff Strom
    yesterday








1




1




Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
– Jeff Strom
yesterday




Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta?
– Jeff Strom
yesterday










1 Answer
1






active

oldest

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25














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.






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  • 8




    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.
    – Richard Stanley
    yesterday













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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes









25














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.






share|cite|improve this answer

















  • 8




    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.
    – Richard Stanley
    yesterday


















25














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.






share|cite|improve this answer

















  • 8




    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.
    – Richard Stanley
    yesterday
















25












25








25






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Gjergji ZaimiGjergji Zaimi

62k4163308




62k4163308








  • 8




    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.
    – Richard Stanley
    yesterday
















  • 8




    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.
    – Richard Stanley
    yesterday










8




8




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.
– Richard Stanley
yesterday






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.
– Richard Stanley
yesterday




















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