Directly from the definitions, compute the simplicial cohomology groups of \(S^1 \times S^1\) with \(\mathbb{Z}\) and \(\mathbb{Z}_2\) coefficients, using the \(\Delta\)-complex structure given in §2.1.
Do the same for \(\mathbb{RP}^2\) and the Klein bottle.
Solution
We compute directly from the cellular/simplicial cochain complexes obtained by dualizing the displayed \(\Delta\)-complex chain complexes.
1. For the torus, with one vertex, edges \(a,b,c\), and two \(2\)-simplices \(U,L\), choose orientations so that\[ \partial U=a+b-c,\qquad \partial L=a+b-c. \]Thus for coefficients in an abelian group \(G\),\[ C^0=G,\qquad C^1=G^3,\qquad C^2=G^2, \]with\[ \delta^0=0, \qquad \delta^1(x,y,z)=(x+y-z,x+y-z). \]
This is the dual of the simplicial boundary maps in the standard two-triangle \(\Delta\)-complex on the torus.
2. Hence for \(G=\mathbb Z\) or \(\mathbb Z_2\),\[ H^0(T^2;G)\cong G, \qquad H^1(T^2;G)\cong G^2, \qquad H^2(T^2;G)\cong G. \]
The kernel of \(\delta^1\) is given by one equation \(x+y-z=0\), hence is isomorphic to \(G^2\). Its image is the diagonal subgroup \(\{(t,t):t\in G\}\) of \(G^2\), whose quotient is isomorphic to \(G\).
3. For \(\mathbb{RP}^2\), choose the standard \(\Delta\)-complex with vertices \(v,w\), edges \(a,b,c\), and faces \(U,L\), with\[ \partial a=\partial b=w-v,\qquad \partial c=0, \]and\[ \partial U=-a+b+c,\qquad \partial L=a-b+c. \]Thus\[ \delta^0(s,t)=(t-s,t-s,0) \]and\[ \delta^1(p,q,r)=(-p+q+r,p-q+r). \]
These are obtained by evaluating cochains on the displayed simplicial boundaries.
4. With integral coefficients,\[ \boxed{H^0(\mathbb{RP}^2;\mathbb Z)=\mathbb Z,\quad H^1(\mathbb{RP}^2;\mathbb Z)=0,\quad H^2(\mathbb{RP}^2;\mathbb Z)=\mathbb Z_2.} \]
The kernel of \(\delta^0\) is the diagonal copy of \(\mathbb Z\), so \(H^0\cong\mathbb Z\). If \(\delta^1(p,q,r)=0\), subtracting and adding the two equations gives \(2r=0\), hence \(r=0\) and \(p=q\). Thus \[ \ker\delta^1=\mathbb Z(1,1,0)=\operatorname{im}\delta^0, \] so \(H^1=0\). The matrix of \(\delta^1\) has rank two and its \(2\times2\) minors have greatest common divisor \(2\), so its image has index \(2\) in \(\mathbb Z^2\). Hence \(H^2\cong\mathbb Z_2\).
5. With \(\mathbb Z_2\) coefficients,\[ \boxed{H^i(\mathbb{RP}^2;\mathbb Z_2)\cong\mathbb Z_2\quad(i=0,1,2).} \]
Modulo \(2\), \(\delta^0\) has rank one and \(\delta^1\) has rank one, with \[ \delta^1(p,q,r)=(p+q+r,p+q+r). \] Thus \(\dim\ker\delta^1=2\) and \(\dim\operatorname{im}\delta^0=1\), giving one-dimensional \(H^1\). The cokernel of \(\delta^1\) is also one-dimensional, giving \(H^2\cong\mathbb Z_2\).
6. For the Klein bottle, choose orientations in the standard two-triangle \(\Delta\)-complex so that all three edges are loops and\[ \partial U=a+b-c, \qquad \partial L=a-b+c. \]Then\[ \delta^0=0, \qquad \delta^1(p,q,r)=(p+q-r,p-q+r). \]
All vertices are identified, so the boundary of every edge is zero. Dualizing the two face boundaries gives the displayed coboundary.
7. Therefore\[ \boxed{H^0(K;\mathbb Z)=\mathbb Z,\quad H^1(K;\mathbb Z)=\mathbb Z,\quad H^2(K;\mathbb Z)=\mathbb Z_2.} \]
The equations \(p+q-r=0\) and \(p-q+r=0\) imply \(2p=0\), hence \(p=0\) and \(q=r\). Thus \(\ker\delta^1\cong\mathbb Z\). The \(2\times2\) minors of the matrix of \(\delta^1\) have gcd \(2\), so its image has index \(2\) in \(\mathbb Z^2\), giving \(H^2\cong\mathbb Z_2\).
8. With \(\mathbb Z_2\) coefficients,\[ \boxed{H^0(K;\mathbb Z_2)=\mathbb Z_2,\quad H^1(K;\mathbb Z_2)=(\mathbb Z_2)^2,\quad H^2(K;\mathbb Z_2)=\mathbb Z_2.} \]
Modulo \(2\), both components of \(\delta^1\) are \(p+q+r\), so \(\delta^1\) has rank one. Since \(\delta^0=0\), the kernel has dimension two and the cokernel dimension one.