Find an explicit, noninductive formula for the barycentric subdivision operator \(S: C_n(X) \to C_n(X)\).
Solution
For an oriented simplex \[ \sigma=[v_0,\dots,v_n], \] and a nonempty subset \(I\subseteq\{0,\dots,n\}\), write \(b_I\) for the barycenter of the face spanned by the vertices \(v_i\) with \(i\in I\).
For a permutation \(\pi=(i_0,\dots,i_n)\in S_{n+1}\), let \[ \Delta_\pi= [b_{\{i_0\}},b_{\{i_0,i_1\}},\dots,b_{\{i_0,\dots,i_n\}}]. \] This is one of the \(n\)-simplices in the barycentric subdivision of \(\sigma\).
Then the explicit formula is \[ \boxed{ S([v_0,\dots,v_n]) = \sum_{\pi\in S_{n+1}} \operatorname{sgn}(\pi)\, [b_{\{i_0\}},b_{\{i_0,i_1\}},\dots,b_{\{i_0,\dots,i_n\}}]. } \] Extend this linearly to \(C_n(X)\).
1. The summands are exactly the \(n\)-simplices of the barycentric subdivision.
An \(n\)-simplex of the barycentric subdivision corresponds to a maximal strict chain of nonempty faces \[ F_0<F_1<\cdots<F_n=\sigma. \] Such a chain is uniquely obtained by choosing an ordering \(i_0,\dots,i_n\) of the original vertices and setting \[ F_k=[v_{i_0},\dots,v_{i_k}]. \] Thus maximal chains are in bijection with permutations of the vertices, and the associated barycentric vertices are exactly those displayed in \(\Delta_\pi\).
2. The coefficient \(\operatorname{sgn}(\pi)\) gives each subdivided simplex the orientation induced from \(\sigma\).
For the identity permutation, the ordered barycentric vertices \[ v_0,\ b_{01},\ b_{012},\dots,b_{0\cdots n} \] have the orientation of \([v_0,\dots,v_n]\). Permuting the original vertices by \(\pi\) changes orientation by \(\operatorname{sgn}(\pi)\), and the corresponding maximal face chain changes in exactly the same way. Hence the displayed coefficient is the induced orientation sign.
3. Therefore the formula agrees with the barycentric subdivision operator.
By definition, barycentric subdivision replaces an oriented simplex by the sum of all its oriented barycentric subsimplices. By 1 these are precisely the \(\Delta_\pi\), and by 2 their orientation coefficients are \(\operatorname{sgn}(\pi)\). Hence the displayed sum is exactly \(S(\sigma)\).