Is there an easy way to find the cohomology ring of the complex projective plane?

Question

I am trying to find the cohomology ring of $\mathbb{C}\mathbb{P}^2$. But I don't know how. We know from the CW structure of $\mathbb{C}\mathbb{P}^2$ that the cohomology groups must be $\mathbb{Z}$ in dimensions 0,2 and 4 and 0 otherwise.So the only non obvious cup product we need to compute is the generator of the second cohomology group with it self. Is there any efficient way to do that.Or at least a way to show that it wont result to 0?

Thanks

Answer 1

I know a way using differential forms. Since $\mathbb{CP}^2$ is a manifold, most of the cohomologies of it are canonically isomorphic to each other. Because you mentioned its CW structure, I assume you are talking about the singular cohomology. You already know that,

$H^n_{\Delta}(\mathbb{CP}^2)$ = \begin{cases} \mathbb{Z} & \quad \text{if } n =0, 2, 4\\ 0 & \quad \text{otherwise.}\\ \end{cases}

Tensor it with $\mathbb{R}$, by the universal coefficient theorem you get

$H^n_{\Delta}(\mathbb{CP}^2;\mathbb{R})$ = \begin{cases} \mathbb{R} & \quad \text{if } n =0, 2, 4\\ 0 & \quad \text{otherwise.}\\ \end{cases} (This is compatible with the product structure.) By the de Rham theorem, you know $H^\ast_{\Delta}(\mathbb{CP}^2,\mathbb{R}) \cong H^\ast_{dR}(\mathbb{CP}^2)$ as $\mathbb{R}$-algebra where the multiplication on the left hand side is the cup product $\cup$ and that of the right hand side is the wedge product $\wedge$.

There is a Fubini-study form on $\mathbb{CP}^2$ which is defined by $$ \omega_{FS} = i \partial \bar{\partial} log (|u_0|^2+|u_1|^2+1)$$ where $[u_0:u_1:1] = [z_0:z_1:z_2]$ is the local coordinate.

First $\omega_{FS}$ is a closed form because $\partial \bar{\partial} f(|z|^2) = d \bar{\partial} f(|z|^2)$ for $f \in C^\infty (\mathbb{R})$. So $[\omega_{FS}] \in H^2_{dR}(\mathbb{CP}^2)$. If we can show that $ [\omega_{FS}] \wedge [\omega_{FS}] \neq 0$ in $H^2_{dR}(\mathbb{CP}^2)$, then we know the $\mathbb{R}$-algebra structure is non-trivail. We can calculate that,

\begin{align*} \omega_{FS}^2 &=\frac{-1}{(|u_0|^2+|u_1|^2+1)^3} du_0 \wedge d\bar{u_0} \wedge du_1 \wedge d\bar{u_1} \\ &=\frac{4}{(x_0^2+y_0^2+x_1^2+y_1^2+1)^3} dx_0 \wedge dy_0 \wedge dx_1 \wedge dy_1. \end{align*}

This shows $\int_{\mathbb{CP}^2} \omega_{FS}^2 \neq 0 $ and so $[\omega_{FS}^2] \neq 0$ by the Poincare duality.