----- ATES1-9 > [!proposition] Proposition. ([[cohomology groups of complex projective space]]) > The [[complex projective space]] $\mathbb{C}P^{k}$ has [[singular homology|singular cohomology]] [[abelian group|groups]] $H^{n}(\mathbb{C}P^{k})=\begin{cases} \mathbb{Z} & n \text{ even} ;\\ 0& n \text{ odd}. \end{cases}$ ^proposition > [!proof]+ Proof. ([[cohomology groups of complex projective space]]) > Induct on $k$. > > **Base case.** In this case, $\mathbb{C} P^{1}$ is [[homeomorphism]] to $\mathbb{S}^{2}$ (CITE) and using [[(co)homology of the spheres]] we get the anticipated result. > > **Induction hypothesis.** Now assume the result holds for $\mathbb{C}P^{k-1}$. > > **Induction step.** We will use [[Mayer-Vietoris theorem|Mayer-Vietoris]]. > > First [[complex projective space|recall]] that we can embed $\mathbb{C}P^{k-1} \hookrightarrow \mathbb{C}P^{k}$ via identification with the set of points in $\mathbb{C}P^{k}$ having the form $[z_{0}: \dots : z_{k-1}: 0]$. > > Put $U=\mathbb{C}P^{k}-\mathbb{C}P^{k-1}$; clearly $U$ is open in $\mathbb{C}P^{n}$. Moreover every point in $U$ admits a unique representative of the form $[z_{0}: \dots : z_{k-1}:1]$; hence, the map $[z_{0}:\dots: z_{k-1}:1] \xmapsto{\varphi} (z_{0},\dots,z_{k-1})$ > witnesses that $U \cong \mathbb{C}^{k} \cong \mathbb{R}^{2k}$. > > Next, put $V=\mathbb{C}P^{k} - \{ [0:\dots:0: 1] \}$ (this is also clearly open). The point $[0:\dots:0:1]$ certainly belongs to $U$, and so $V \cup U=\mathbb{C}P^{n}$. Meanwhile, $V \cap U=\{ [z_{0}: \dots:z_{k}] \in \mathbb{C}P^{n}: z_{0}, \dots, z_{k} \neq 0 \}$ where now $\varphi |_{U \cap V}$ is a [[homeomorphism]] onto $\mathbb{C}^{k} - \{ 0 \} \cong \mathbb{R}^{k}-\{ 0 \}$, and is in turn [[homotopy equivalent]] to $\mathbb{S}^{2k-1}$ and we can apply [[homotopy invariance of singular homology|homotopy invariance]]. > > Finally, we claim that $V$ (strong) [[deformation retract|deformation retracts]] onto the aforementioned copy $\underline{\mathbb{C}P^{k-1}} \subset \mathbb{C}P^{k}$ of $\mathbb{C}P^{k-1}$ living inside $\mathbb{C}P^{k}$. Indeed, the map $\begin{align} > r: V &\to \underline{\mathbb{C}P^{k-1}} \\ > [z_{0}:\cdots :z_{n}] & \mapsto [z_{0} : \cdots : z_{n-1} : 0] > \end{align}$ > is evidently a [[retract|retraction]], and we can define a [[homotopy]] $\begin{align} > H : V \times I & \to V \\ > ([z_{0}:\cdots: z_{n}], t) & \mapsto [z_{0} : \cdots : z_{n-1}: t z_{n}] > \end{align}$ > where it is clear that > - $H(V \times \{ 0 \})=\iota \circ r(V)$ > - $H(V \times \{ 1 \})= \id_{V}(V)$ > - $H(\underline{\mathbb{C}P^{k-1}} \times I)= \id_{\underline{\mathbb{C}P^{k-1}}}(V)$ > and so this is indeed a [[deformation retract]]. > > [[Mayer-Vietoris theorem|Mayer-Vietoris]] now gives an [[exact sequence]] $\begin{align} > \dots \to H^{n-1}(U \cap V) \to H^{n}(\mathbb{C}P^{k}) \to H^{n}(U) \oplus H^{n}(V) \to H^{n}(U \cap V) \to \dots > \end{align}$ > after substitution this becomes $\dots \to H^{n-1}(\mathbb{S}^{2k-1}) \to H^{n}(\mathbb{C}P^{k}) \to H^{n}(\mathbb{C}P^{k-1}) \to H^{n}(\mathbb{S}^{2k-1}) \to \cdots$ > where we have used that $H^{n}(U) \cong 0$. Now, > - $H^{n-1}(\mathbb{S}^{2k-1})$ is $\mathbb{Z}$ if $n \in \{ 1, 2k \}$ and $0$ otherwise; > - $H^{n}(\mathbb{C}P^{k-1})=0$ if $n$ odd and $\mathbb{Z}$ otherwise, by the induction hypothesis; > - $H^{n}(\mathbb{S}^{2k-1})$ is $\mathbb{Z}$ if $n \in \{ 1, 2k-1 \}$ and $0$ otherwise. > In particular, it is immediate if $1<n<2k-1$ then the middle map is an [[isomorphism]]. If $n=2k-1$ then $H^{n}(\mathbb{C}P^{k-1})=0$ by the induction hypothesis, and already know that $H^{1}(\mathbb{C}P^{k})=\mathbb{Z}$. So actually we have that if $1 \leq n <2k$ then the middle map is an [[isomorphism]]. If $n=2k$ then we have an [[exact sequence]] $0 \to \mathbb{Z} \xhookrightarrow{} H^{n}(\mathbb{C}P^{k}) \xrightarrow{0} \mathbb{Z} \xrightarrow{\id} \mathbb{Z} \to 0$ > from which it follows that $\mathbb{Z} \cong H^{2k}(\mathbb{C}P^{k})$. This finishes the proof. > ^e1ddfc The closed, connected complex surface $X_{n}:=\#_{i=1}^{n} \mathbb{C}P^{2}$ satisfies $H^{1}(X_{n})=0, H^{2}(X_{n})=\mathbb{Z}^{n},$ as required. To show this, we can inductively apply [[Mayer-Vietoris theorem|Mayer-Vietoris]]. The space $X_{2}=\mathbb{C}P^{2} \# \mathbb{C}P^{2}$ is obtained by removing a small $4$-disc from each [[manifold]] and gluing along the boundary spheres. Let $U$ equal the first copy of $\mathbb{C}P^{2}$ together with some of the other copy 'past the gluing point', and symmetrically define $V$, such that $U \cup V$ provides an open cover of the space. Note that $U \cap V=\mathbb{S}^{3}$. Consider the long exact sequence on cohomology $\begin{align} \cdots \to H^{0}(X_{2}) \to H^{1}(U) \oplus H^{1}(V) \to H^{1}(A \cap B) . \end{align}$ To see this, we can realize $\mathbb{C}P^{2}$ as a [[cell complex]] (*or does this part need to use Mayer-Vietoris also?*) with one $0$-cell $p$, one $2$-cell $q$, and one $4$-cell $r$, yielding cellular [[chain complex of modules|chain complex]] $C_{\bullet}^{\text{cell}}(X_{n})= \cdots \to 0 \to \mathbb{Z}\langle r \rangle \to 0 \to \mathbb{Z}\langle q \rangle \to 0 \to \mathbb{Z}\langle p \rangle $where the differentials must all be zero. Dually, we have a [[chain complex of modules|cochain complex]] $C^{\bullet}_{\text{cell}}(X_{n})= \mathbb{Z}\langle p \rangle \to 0 \to \mathbb{Z} \langle q \rangle \to 0 \to \mathbb{Z}\langle r \rangle \to 0 \to \cdots $ where the codifferentials must all be zero. Taking cohomology, we find $H^{1}(X_{n})$ or see cell decomposition in [[complex projective space]] ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```