Subdivision

of a geometric simplicial complex $K$

A geometric simplicial complex $K_1$ such that the underlying space $|K_1|$ coincides with the underlying space $|K|$ and such that each simplex of $K_1$ is contained in some simplex of $K$. In practice, the transition to a subdivision is carried out by decomposing the simplices in $K$ into smaller simplices such that the decomposition of each simplex is matched to the decomposition of its faces. In particular, each vertex of $K$ is a vertex of $K_1$. The transition to a subdivision is usually employed to demonstrate invariance of the combinatorially defined characteristics of polyhedra (cf. Polyhedron, abstract; for example, the Euler characteristic or the homology groups, cf. Homology group), and also to obtain triangulations (cf. Triangulation) with the necessary properties (for example, sufficiently small triangulations). A stellar subdivision of a complex $K$ with centre at a point $a\in |K|$ is obtained as follows. The closed simplices of $K$ that do not contain $a$ remain unaltered. Each closed simplex $\sigma$ containing $a$ is split up into cones with their vertices at $a$ over those faces of $\sigma$ that do not contain $a$. For any two triangulations $T_1$ and $T_2$ of the same polyhedron $P$ there exists a triangulation $T_3$ of $P$ obtained not only from $T_1$ but also from $T_2$ by means of a sequence of stellar subdivisions. The concept of a stellar subdivision may be formalized in the language of abstract simplicial complexes (simplicial schemes). Any stellar subdivision of a closed subcomplex can be extended to a stellar subdivision of the entire complex. The derived complex $K^{\prime }$ of a complex $K$ is obtained as the result of a sequence of stellar subdivisions with centres in all open simplices of $K$ in the order of decreasing dimensions. For an arbitrary closed subcomplex $K$ of a complex $L$, the subcomplex $K^{\prime }\subset L^{\prime }$ is complete in the following sense: From the fact that all the vertices of a certain simplex $\sigma \in L^{\prime }$ lie in $K^{\prime }$ it follows that $\sigma \in K^{\prime }$. If one takes as the centres of the derived complex the barycentres of the simplices, one gets the barycentric subdivision. If the diameter of each simplex of an $n$-dimensional complex $K$ does not exceed $d$, the diameters of the simplices in its barycentric subdivision are bounded by $nd/(n+1)$. The diameters of the simplices in the $m$-fold barycentric subdivision of $K$ are bounded by $(n/(n+1){)}^{m}d$, and so they can be made arbitrarily small by selecting $m$ sufficiently large.

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