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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...)

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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
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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.

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    $\begingroup$ For every $n\geq d\geq 4$ 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
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    $\begingroup$ 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

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