Describe a way of identifying pairs of faces of \(\Delta^3\) (the standard three simplex) to produce a \(\Delta\)-complex structure on \(S^3\) having a single 3-simplex.
Write down the chain complex corresponding to the \(\Delta\)-complex in (i). Be sure to include the differentials of the complex, but you do not need to compute the homology.
Solution
Label the vertices of \(\Delta^3\) by \(v_0,v_1,v_2,v_3\), and write \[F_0=[v_1,v_2,v_3],\quad F_1=[v_0,v_2,v_3],\quad F_2=[v_0,v_1,v_3],\quad F_3=[v_0,v_1,v_2].\]
1. (i) Identify \(F_0\) with \(F_1\) by the order-preserving affine map \(v_1\mapsto v_0\), \(v_2\mapsto v_2\), \(v_3\mapsto v_3\), and \(F_2\) with \(F_3\) by \(v_0\mapsto v_0\), \(v_1\mapsto v_1\), \(v_3\mapsto v_2\). The quotient is a \(\Delta\)-complex with one \(3\)-simplex, homeomorphic to \(S^3\).
Both maps preserve vertex order, so they are admissible \(\Delta\)-complex face identifications, and they pair the four faces of the single \(3\)-simplex. Regard the tetrahedron as the join \(\Delta^3=[v_0,v_1]*[v_2,v_3]\). Since \(F_0=\{v_1\}*[v_2,v_3]\) and \(F_1=\{v_0\}*[v_2,v_3]\), the first pairing identifies the two endpoints of \([v_0,v_1]\) and fixes the coordinate in \([v_2,v_3]\); likewise the second pairing identifies the two endpoints of \([v_2,v_3]\) and fixes the coordinate in \([v_0,v_1]\). Hence the quotient is \(([v_0,v_1]/\partial)*([v_2,v_3]/\partial)\cong S^1*S^1\cong S^3\), by the join homeomorphism \(S^p*S^q\cong S^{p+q+1}\).
2. The quotient has two vertices \(A=[v_0]=[v_1]\) and \(B=[v_2]=[v_3]\), three edges \(e=[v_0,v_1]\), \(f=[v_2,v_3]\), and \(c=[v_0,v_2]=[v_0,v_3]=[v_1,v_2]=[v_1,v_3]\), two \(2\)-simplices \(P=F_0=F_1\) and \(Q=F_2=F_3\), and one \(3\)-simplex \(\sigma=[v_0,v_1,v_2,v_3]\).
The first pairing identifies \(v_1\) with \(v_0\), \([v_1,v_2]\) with \([v_0,v_2]\), and \([v_1,v_3]\) with \([v_0,v_3]\). The second identifies \(v_3\) with \(v_2\), \([v_0,v_3]\) with \([v_0,v_2]\), and \([v_1,v_3]\) with \([v_1,v_2]\). The edges \([v_0,v_1]\) and \([v_2,v_3]\) are identified with nothing else.
3. \(\partial_3(\sigma)=0\), \(\partial_2(P)=f\), \(\partial_2(Q)=e\), \(\partial_1(e)=\partial_1(f)=0\), and \(\partial_1(c)=B-A\).
By step 2, \(\partial\sigma=F_0-F_1+F_2-F_3=P-P+Q-Q=0\). With \(P=[v_1,v_2,v_3]\) and \(Q=[v_0,v_1,v_3]\), \[\partial P=[v_2,v_3]-[v_1,v_3]+[v_1,v_2]=f-c+c=f,\qquad \partial Q=[v_1,v_3]-[v_0,v_3]+[v_0,v_1]=c-c+e=e.\] The edges \(e\) and \(f\) are loops at \(A\) and \(B\), and \(c\) runs from \(A\) to \(B\).
4. Q.E.D.
Step 1 answers (i). For (ii), in the bases \(\sigma\) of \(C_3\), \((P,Q)\) of \(C_2\), \((e,f,c)\) of \(C_1\), and \((A,B)\) of \(C_0\), step 3 gives the chain complex \[0\to\ZZ\xrightarrow{0}\ZZ^2 \xrightarrow{\begin{psmallmatrix}0&1\\1&0\\0&0\end{psmallmatrix}}\ZZ^3 \xrightarrow{\begin{psmallmatrix}0&0&-1\\0&0&1\end{psmallmatrix}}\ZZ^2\to0.\]