Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 21H / a

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- Solution a

Solution

For every nonempty simplex

σ

of

K

, let

b_{σ}

be its barycentre. The first barycentric subdivision

K^{'}

has the

b_{σ}

as vertices, and

[b_{σ_{0}}, \dots, b_{σ_{q}}]

is a simplex exactly when, after reordering,

σ_{0} ⊊ \dots ⊊ σ_{q}

. Its realization is the same polyhedron as that of

K

. The iterated barycentric subdivision is defined by

K^{(0)} = K, K^{(r + 1)} = (K^{(r)})^{'} .

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