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Consider a pure finite abstract simplicial complex $\Delta$. Define its diameter as the maximal distance between any two facets, i.e., between any two faces of maximal dimension $d-1$. The distance between two facets is the shortest path between them. A path of length $k$ in $\Delta$ is just a sequence of facets $(F_1, \ldots, F_{k+1})$ with $F_i \in \Delta$ such that $F_i \cap F_{i+1}$ is a ridge, i.e., a $(d-2)$-dimensional face.

For the class of simplicial complexes I am interested in, we can assume that such a path always exists.

To $\Delta$ one can associate a ring, the so-called Stanley-Reisner ring. It is defined as the quotient of a polynomial ring. Suppose $x_1, \ldots, x_n$ are the vertices of $\Delta$. Then its Stanley-Reisner ideal $I_{\Delta}$ is generated by the non-faces:

$I_{\Delta} = (x_{i_1} \cdots x_{i_s} : \lbrace x_{i_1}, \ldots, x_{i_s} \rbrace \not\in \Delta)$

Here $x_i$ denotes a vertex in $\Delta$ and at the same time also a variable in $k[x_1, \ldots, x_n]$, where $k$ is a field. The Stanley-Reisner ring is then defined as $k[\Delta] = k[x_1, \ldots, x_n]/I_{\Delta}$. These rings provide a nice bridge between combinatorics and geometry on the one hand and commutative algebra on the other.

So much for the setting. Now, what I am wondering about is if the diameter of the simplicial complex is represented by some property of $k[\Delta]$. Or is it maybe known that the diameter cannot be extracted from the ring?

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I believe the complex is determined up to isomorphism by the ring. The question is whether it has ring-theoretic meaning. –  Benjamin Steinberg Jan 31 '13 at 10:40
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@Benjamin Steinberg: Thanks for your comment. Two simplicial complexes are isomorphic if and only if their Stanley-Reisner rings are isomorphic as $k$-algebras. A proof can be found in the 1996 paper "Combinatorial invariance of Stanley-Reisner rings" by Bruns and Gubeladze. So, indeed, the question is whether the diameter has ring-theoretic meaning. –  Gregor Samsa Jan 31 '13 at 17:29
    
@Gregor Samsa: interesting question. Please, change $I_\Delta=(x_{i_1},\dots,x_{i_s}$ ... by $I_\Delta=(x_{i_1}\dots x_{i_s}$ ... –  Sasha Anan'in Sep 25 '13 at 12:09
    
@SashaAnan'in: Fixed, thanks. –  Gregor Samsa Sep 25 '13 at 13:07
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