Prove that $\int_{\mathbb{C}\mathbb{P}^3}c_1(\mathbb{C}\mathbb{P}^3)^3=64$.

Question

I want to prove that $$\int_{\mathbb{C}\mathbb{P}^3}c_1(\mathbb{C}\mathbb{P}^3)^3=64,$$ where $c_1$ is the first Chern class. I know that for projective spaces of dimension $n$ (but maybe also in general?) it holds that $c_1(\mathbb{C}\mathbb{P}^n)=c_1(-K_{\mathbb{C}\mathbb{P}^n})=c_1(-\mathcal{O}(-n-1))$, where $K_M$ denotes the canonical bundle of $M$. So $$c_1(\mathbb{C}\mathbb{P}^3)=c_1(-K_{\mathbb{C}\mathbb{P}^3})=c_1(-\mathcal{O}(-4))=c_1(\mathcal{O}(4)),$$ and I think I am very close to the goal but still I am missing the very last step.

Answer 1

This answer might be more helpful for me then to you as im not sure we have the same background. First I will compute $c_1(\mathbb{CP}^3)$. As the total chern class is given by $$c(\mathbb{CP}^3)=(1+x)^4,$$ where $x=c_1(-K_{\mathbb{C}\mathbb{P}^3})$. I will deduce that $c_1(\mathbb{CP}^3)=4 x$. Thus we need to compute $$\int_{\mathbb{C}\mathbb{P}^3}64x^3.$$ By the whitney product formula this is the same as $$\int_{\mathbb{C}\mathbb{P}^3}64c_3\left(-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\right).$$ The top Chern class is the Euler class. Hence we can use the following neat theorem from bott and tu book differential forms in algebraic topology.

Theorem 11.17. Let $\pi: E \rightarrow M$ be an oriented rank $k$ vector bundle over a compact oriented manifold of dimension $k$. Let $s$ be a section of $E$ with a finite number of zeros. The Euler class of $E$ is Poincare dual to the zeros of $s$, counted with the appropriate multiplicities.

This theorem implies that if we can find a section of $-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}$ with only one transversal zero we are done. Finding such section $ s:\mathbb{C}\mathbb{P}^3 \to -K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}$ is easy. Let $(e_1, \dots, e_4)$ be the standart basis. We can take the section $s$ to be defined by $$[x_1, x_2, x_3, x_4]\mapsto (e_1^*,e_2^*,e_3^*), $$ Where $e^*_i$ is the dual of $e_i$. It's easy to observe that the section only zero is at the point $[0,0,0,1]$. Hence we get $$\int_{\mathbb{C}\mathbb{P}^3}64c_3\left(-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\right)=64\int_{\mathbb{C}\mathbb{P}^3}e\left(-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\oplus-K_{\mathbb{C}\mathbb{P}^3}\right)=64 \cdot 1.$$