Barycentric face chains triangulate a geometric simplex

Let a point have distinct positive coordinate levels $a_1>\cdots >a_m>0$, put $a_{m+1}=0$, and set $F_j=\{v:x_v\ge a_j\}$. These are nested nonempty faces. With $w_j=\mathrm{\#}{F}_j(a_j-a_{j+1})>0$, a vertex whose coordinate is $a_{\mathrm{\ell }}$ has coordinate ${\sum }_{j=\mathrm{\ell }}^m{w}_j\mathrm{/}\mathrm{\#}{F}_j=a_{\mathrm{\ell }}$ in ${\sum }_j{w}_j{b}_{F_j}$. Also ${\sum }_j{w}_j={\sum }_v{x}_v=1$. Thus every point lies in a chain simplex.

For any strict chain $G_1⊊\cdots ⊊G_s$, the vectors $b_{G_j}$ are linearly independent in barycentric coordinate space: in $\sum c_j{b}_{G_j}=0$, a coordinate in $G_s\setminus G_{s-1}$ gives $c_s\mathrm{/}\mathrm{\#}{G}_s=0$, and descending induction gives every $c_j=0$ (finish with any vertex of $G_1$). Hence they are affinely independent in the original simplex too, by uniqueness of its affine coordinates.

If a point is expressed in one of these chain simplices, delete zero weights. Its coordinate on the successive layers of the remaining chain is strictly decreasing, with consecutive differences equal to the corresponding positive weight divided by the face cardinality. Thus the positive-weight faces are exactly the level sets $F_j$ in the first step and the weights are exactly $w_j$. In two chain representations only common face labels can therefore carry positive weights. Conversely every convex combination of common labels belongs to both simplices. This proves precisely the intersection assertion and hence the triangulation. The empty simplex has no points; a one-vertex simplex has the sole weight $1$.