With \(\PP^n(\CC)\) given its classical topology, compute the integral singular cohomology ring \(H^*(\PP^n(\CC);\ZZ)\).
Solution
Put \[ X=\PP^n(\CC). \]
1. If \(n=0\), then\[ H^*(X;\ZZ)\cong\ZZ\cong\ZZ[h]/(h). \]
In this case \[ X=\PP^0(\CC) \] is a point. Its integral cohomology is \(\ZZ\) in degree \(0\) and vanishes in positive degrees, which is the displayed ring. Hence assume \(n\geq1\) below.
2. The space \(X\) has a CW decomposition with one cell in each realdimension\[ 0,2,\ldots,2n \]and no cells in odd dimensions.
For \(0\leq r\leq n\), let \[ X_r = \{[z_0:\cdots:z_n]\in X: z_{r+1}=\cdots=z_n=0\}. \] Then \[ X_r\cong\PP^r(\CC), \qquad X_0\subset X_1\subset\cdots\subset X_n=X. \] Moreover, \[ X_r\setminus X_{r-1} = \{[z_0:\cdots:z_r:0:\cdots:0]:z_r\neq0\} \cong \CC^r \cong \RR^{2r}, \] where the affine coordinates are \[ \left(\frac{z_0}{z_r},\ldots,\frac{z_{r-1}}{z_r}\right). \] Thus \(X_r\) is obtained from \(X_{r-1}\) by attaching one \(2r\)-cell, which gives the asserted CW decomposition.
3. Additively,\[ H^q(X;\ZZ) \cong \begin{cases} \ZZ,&q=0,2,\ldots,2n,\\ 0,&\text{otherwise}. \end{cases} \]
By step 2, the cellular cochain group is \[ C^q_{\mathrm{cell}}(X;\ZZ) \cong \begin{cases} \ZZ,&q=0,2,\ldots,2n,\\ 0,&\text{otherwise}. \end{cases} \] Every cellular differential has either zero source or zero target because there are no odd-dimensional cells. Hence all cellular differentials vanish, so cellular cohomology gives exactly the displayed groups.
4. Let\[ H\cong\PP^{n-1}(\CC)\subseteq X \]be a hyperplane and let\[ h=\operatorname{PD}[H]\in H^2(X;\ZZ) \]be its Poincare dual. For every \(0\leq r\leq n\),\[ h^r = \operatorname{PD}[\PP^{n-r}(\CC)], \]and \(h^r\) generates \(H^{2r}(X;\ZZ)\).
Choose \(r\) hyperplanes \[ H_1,\ldots,H_r\subseteq X \] defined by linearly independent linear forms. They meet transversely, and their intersection is a linear subspace \[ H_1\cap\cdots\cap H_r \cong \PP^{n-r}(\CC). \] All hyperplanes represent the same homology class. For oriented transverse submanifolds, cup product of Poincare duals is the Poincare dual of their intersection. With the complex orientations on the hyperplanes and their intersection, the intersection sign is positive. Therefore \[ h^r = \operatorname{PD}[H_1\cap\cdots\cap H_r] = \operatorname{PD}[\PP^{n-r}(\CC)]. \]
The standard linear \(\PP^{n-r}(\CC)\) is the closure of the unique \(2(n-r)\)-cell in the filtration from step 2, so its fundamental class generates \[ H_{2(n-r)}(X;\ZZ)\cong\ZZ. \] Poincare duality therefore shows that \(h^r\) generates \[ H^{2r}(X;\ZZ)\cong\ZZ. \]
5. The cohomology ring is\[ \boxed{ H^*(\PP^n(\CC);\ZZ) \cong \ZZ[h]/(h^{n+1}), \qquad \deg h=2. } \]
1. Assume \(n\geq1\). By step 4, the powers\[ 1,h,\ldots,h^n \]generate every nonzero cohomology group from step
3. Since \(X\) has realdimension \(2n\),\[ H^{2n+2}(X;\ZZ)=0, \]so\[ h^{n+1}=0. \]Hence the homomorphism of graded rings\[ \ZZ[h]/(h^{n+1}) \longrightarrow H^*(X;\ZZ) \]sending \(h\) to the hyperplane class is surjective. In each even degree\(0,2,\ldots,2n\), both sides are free of rank one and the image of \(h^r\) isa generator by step 4; both sides vanish in every other degree. Thereforethe homomorphism is an isomorphism.
6. Q.E.D.
Step 5 gives the requested integral singular cohomology ring.