Compute the Euler characteristic of the \(n\)-sphere \(S^n\) using the standard triangulation of an \(n\)-simplex.
Solution
The simplex dimension in the source is off by one: the boundary of an \(n\)-simplex is \(S^{n-1}\). To triangulate \(S^n\), use the boundary of an \((n+1)\)-simplex.
1. The boundary complex\[ \partial\Delta^{n+1} \]is a triangulation of \(S^n\).
The simplex \(\Delta^{n+1}\) is homeomorphic to the closed ball \(D^{n+1}\). Its topological boundary is therefore homeomorphic to \[ \partial D^{n+1}=S^n. \] The proper faces of \(\Delta^{n+1}\) form the simplicial complex \(\partial\Delta^{n+1}\), giving the standard triangulation of that boundary.
2. For each \(0\le k\le n\), the triangulation \(\partial\Delta^{n+1}\) has\[ f_k=\binom{n+2}{k+1} \]\(k\)-simplices.
The simplex \(\Delta^{n+1}\) has \(n+2\) vertices. A \(k\)-dimensional face is determined uniquely by choosing its \(k+1\) vertices. Hence there are \[ \binom{n+2}{k+1} \] such faces. Since \(k\le n\), every \(k\)-face is a proper face and therefore belongs to the boundary complex.
3. Thus\[ \chi(S^n) =\sum_{k=0}^{n}(-1)^k\binom{n+2}{k+1}. \]
For a finite simplicial complex, the Euler characteristic is the alternating sum of the numbers of simplices in each dimension. Apply this to the simplex counts from 2.
4. The alternating sum in 3 equals\[ 1+(-1)^n. \]
Set \(j=k+1\). Then \[ \chi(S^n) =-\sum_{j=1}^{n+1}(-1)^j\binom{n+2}{j}. \] The binomial theorem gives \[ 0=(1-1)^{n+2} =\sum_{j=0}^{n+2}(-1)^j\binom{n+2}{j}. \] Therefore \[ \sum_{j=1}^{n+1}(-1)^j\binom{n+2}{j} =-1-(-1)^{n+2} =-1-(-1)^n. \] Negating this identity yields \[ \chi(S^n)=1+(-1)^n. \]
5. Hence\[ \boxed{ \chi(S^n)= \begin{cases} 2,&n\text{ even},\\ 0,&n\text{ odd}. \end{cases}} \]
This is the parity form of the formula in 4.