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

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...)

• Interesting starting point: can $S^n$ and $S^m$ be triangluated with isomorphic $1$-skeleta? – Jeff Strom Jan 6 '19 at 19:39

• 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. – Richard Stanley Jan 6 '19 at 23:49