Can two non-equivalent polytopes of same dimension have the same graph?

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By a polytope I mean the convex hull of finitely many points. The graph of a polytope is the graph isomorphic to its 1-skeleton. By equivalence of polytopes I mean combinatorial equivalence, i.e. their face lattices are isomorphic.



I know that two polytopes can have isomorphic graphs while being non-equivalent, e.g. neighborly polytopes. However, all examples I know of are polytopes of different dimension. So I wonder:




Question: Can there be two non-equivalent poyltopes of the same dimension with the same graph?




Especially, are all $k$-neighborly polytopes of the same dimension equivalent?







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  • 3




    For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
    – j.c.
    Sep 4 at 12:54










  • In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
    – Ian Agol
    Sep 5 at 17:21















up vote
5
down vote

favorite












By a polytope I mean the convex hull of finitely many points. The graph of a polytope is the graph isomorphic to its 1-skeleton. By equivalence of polytopes I mean combinatorial equivalence, i.e. their face lattices are isomorphic.



I know that two polytopes can have isomorphic graphs while being non-equivalent, e.g. neighborly polytopes. However, all examples I know of are polytopes of different dimension. So I wonder:




Question: Can there be two non-equivalent poyltopes of the same dimension with the same graph?




Especially, are all $k$-neighborly polytopes of the same dimension equivalent?







share|cite|improve this question
















  • 3




    For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
    – j.c.
    Sep 4 at 12:54










  • In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
    – Ian Agol
    Sep 5 at 17:21













up vote
5
down vote

favorite









up vote
5
down vote

favorite











By a polytope I mean the convex hull of finitely many points. The graph of a polytope is the graph isomorphic to its 1-skeleton. By equivalence of polytopes I mean combinatorial equivalence, i.e. their face lattices are isomorphic.



I know that two polytopes can have isomorphic graphs while being non-equivalent, e.g. neighborly polytopes. However, all examples I know of are polytopes of different dimension. So I wonder:




Question: Can there be two non-equivalent poyltopes of the same dimension with the same graph?




Especially, are all $k$-neighborly polytopes of the same dimension equivalent?







share|cite|improve this question












By a polytope I mean the convex hull of finitely many points. The graph of a polytope is the graph isomorphic to its 1-skeleton. By equivalence of polytopes I mean combinatorial equivalence, i.e. their face lattices are isomorphic.



I know that two polytopes can have isomorphic graphs while being non-equivalent, e.g. neighborly polytopes. However, all examples I know of are polytopes of different dimension. So I wonder:




Question: Can there be two non-equivalent poyltopes of the same dimension with the same graph?




Especially, are all $k$-neighborly polytopes of the same dimension equivalent?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 4 at 11:31









M. Winter

554220




554220







  • 3




    For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
    – j.c.
    Sep 4 at 12:54










  • In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
    – Ian Agol
    Sep 5 at 17:21













  • 3




    For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
    – j.c.
    Sep 4 at 12:54










  • In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
    – Ian Agol
    Sep 5 at 17:21








3




3




For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
– j.c.
Sep 4 at 12:54




For $k=lfloor d/2rfloor$ your question is also answered on the wikipedia page en.wikipedia.org/wiki/Neighborly_polytope
– j.c.
Sep 4 at 12:54












In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
– Ian Agol
Sep 5 at 17:21





In 3 dimensions, the 1-skeleton determines the combinatorial type. mathoverflow.net/a/308455/1345
– Ian Agol
Sep 5 at 17:21











1 Answer
1






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5
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The 1-skeleton is usually not enough to recover the face lattice,
but under some conditions it is.
I did a quick google search, and read the abstract in this paper.






share|cite|improve this answer




















  • I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
    – M. Winter
    Sep 4 at 12:16










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote













The 1-skeleton is usually not enough to recover the face lattice,
but under some conditions it is.
I did a quick google search, and read the abstract in this paper.






share|cite|improve this answer




















  • I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
    – M. Winter
    Sep 4 at 12:16














up vote
5
down vote













The 1-skeleton is usually not enough to recover the face lattice,
but under some conditions it is.
I did a quick google search, and read the abstract in this paper.






share|cite|improve this answer




















  • I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
    – M. Winter
    Sep 4 at 12:16












up vote
5
down vote










up vote
5
down vote









The 1-skeleton is usually not enough to recover the face lattice,
but under some conditions it is.
I did a quick google search, and read the abstract in this paper.






share|cite|improve this answer












The 1-skeleton is usually not enough to recover the face lattice,
but under some conditions it is.
I did a quick google search, and read the abstract in this paper.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 4 at 12:12









Per Alexandersson

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  • I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
    – M. Winter
    Sep 4 at 12:16
















  • I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
    – M. Winter
    Sep 4 at 12:16















I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
– M. Winter
Sep 4 at 12:16




I knew the paper but I failed to remember it and to draw the conclusion from the abstract. Thank you.
– M. Winter
Sep 4 at 12:16

















 

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