Show that the second barycentric subdivision of a \(\Delta\)-complex is a simplicial complex. Namely, show that the first barycentric subdivision produces a \(\Delta\)-complex with the property that each simplex has all its vertices distinct, then show that for a \(\Delta\)-complex with this property, barycentric subdivision produces a simplicial complex.
Solution
Recall that the vertices of the barycentric subdivision are barycenters of simplices, and its simplices correspond to chains of faces.
1. Every simplex in the first barycentric subdivision of a \(\Delta\)-complex has distinct vertices.
A \(k\)-simplex in the barycentric subdivision is determined inside some original simplex by a strict chain of faces \[ \sigma_0<\sigma_1<\cdots<\sigma_k. \] Its vertices are the barycenters \[ b_{\sigma_0},b_{\sigma_1},\dots,b_{\sigma_k}. \] Since the face inclusions are strict, \[ \dim\sigma_0<\dim\sigma_1<\cdots<\dim\sigma_k, \] so these simplices, and hence their barycenters as vertices of the subdivided \(\Delta\)-complex, are all distinct.
2. Suppose now that a \(\Delta\)-complex \(Y\) has the property that every simplex has distinct vertices. Then a face of a simplex of \(Y\) is determined uniquely by its set of vertices.
Let \(\tau\) be an \(m\)-simplex of \(Y\). Its characteristic simplex has \(m+1\) distinct image vertices. A face of \(\tau\) is obtained by deleting a subset of these vertices. Because the vertices are distinct, two different faces of the standard simplex have different image vertex sets. Hence within \(\tau\) a prescribed subset of its vertices determines at most one face.
3. In the barycentric subdivision \(Y'\), a finite set of vertices spans at most one simplex.
A vertex of \(Y'\) is a simplex of \(Y\). A set \[ \{\sigma_0,\dots,\sigma_k\} \] spans a simplex of \(Y'\) exactly when these simplices can be ordered into a strict face chain \[ \sigma_{i_0}<\sigma_{i_1}<\cdots<\sigma_{i_k}. \] Such an ordering is unique because dimensions strictly increase along a proper face inclusion. By 2, the individual face incidences are themselves unambiguous. Hence the given vertex set determines at most one simplex of \(Y'\).
4. Intersections of simplices in \(Y'\) are common faces.
Write two simplices as chains of simplices of \(Y\). Their common vertices are exactly the simplices appearing in both chains. Since each chain is totally ordered by face inclusion, the common vertices again form a chain, hence span a face of each subdivided simplex. By 3 there is no second simplex with the same vertex set. Therefore the intersection is precisely this common face.
Properties 3 and 4 are the defining combinatorial conditions for a simplicial complex. Applying 1 to the first barycentric subdivision \(X'\) and then <1>2–<1>4 to \(Y=X'\) shows that \[ \boxed{X''\text{ is a simplicial complex}.} \]