Barycentric subdivision

Remarks

The union of the simplices of $\operatorname{sd} K$ equals the union of the simplices of $K$, and each simplex of $\operatorname{sd} K$ lies in a simplex of $K$. If every simplex of $K$ has diameter at most $d$ and $K$ has dimension at most $n$, then every simplex of $\operatorname{sd} K$ has diameter at most $\frac{n}{n + 1} d$.

For an abstract simplicial complex $K$, the vertices of $\operatorname{sd} K$ are the nonempty faces of $K$. The faces of $\operatorname{sd} K$ are the finite chains of nonempty faces under inclusion.

Example: let $K$ be a $2$-simplex with vertices $v_0, v_1, v_2$ and all its faces. The maximal chains are $\set{v_i} \subsetneq \set{v_i, v_j} \subsetneq \set{v_0, v_1, v_2}$ for the $6$ ordered pairs $i \ne j$, so $\operatorname{sd} K$ has $6$ triangles, $7$ vertices and $12$ edges.

Non-example: cutting the triangle into $3$ triangles by joining its centroid to the three corners gives a subdivision, but not the barycentric one. The barycentres of the edges are not vertices of it, though each edge is a simplex of $K$.