Question

Let $\mathcal E$ be a rank $n$ vector bundle over a curve $Y$ and let $X=\mathbb P(\mathcal E)$ and let $\pi: X \to Y$ be the projection. I would like to compute the value of the top self-intersection of the tautological line bundle $(\mathcal O_X(1))^n$.

Answer 1

Let $\xi \in A^1(X) $ be the class of $\mathcal{O}_X(1)$.

Since $\dim Y=1$, by [Hartshorne, Algebraic Geometry, p. 429] we have the equality $$\xi^n=\pi^*c_1(\mathcal{E}) \cdot \xi^{n-1}.$$

But $\mathcal{O}_X(1)$ is a relative hyperplane, so $\pi^*(\textrm{point}) \cdot \xi^{n-1}=1$. Then $$\xi^n= \deg (\mathcal{E}).$$

Answer 2

As Francesco showed you this is quite straightforward, therefore I think is a good opportunity to give a more general answer here.

If you have a rank $r$ holomorphic vector bundle $E$ over a compact complex manifold $X$ of dimension $n$, and you call $\pi\colon\mathbb P(E)\to X$ the projectivized bundle of hyperplanes of $E$ and $\mathcal O_E(1)\to\mathbb P(E)$ the tautological line bundle of one dimensional quotients of $E$, then $$ \pi_*c_1(\mathcal O_E(1))^{r-1+k}=(-1)^ks_k(E),\quad k=1,\dots,n. $$ Here, $s_k(E)$ is the $k$-th Segre class which is defined for instance by the following relation: $$ c_\bullet(E)\cdot s_\bullet(E)=1, $$ where $$ c_\bullet(E)=1+c_1(E)+\cdots+c_n(E) $$ is the total Chern class of $E$ and $$ s_\bullet(E)=1+s_1(E)+\cdots+s_n(E) $$ is the total Segre class.

Thus, the first terms for $s_\bullet$ are given in terms of the Chern classes by $$ s_1(E)=-c_1(E),\quad s_2(E)=c_1(E)^2-c_2(E),\quad\dots $$ The top Segre class $s_n(E)$ has an expansion which always contains the term $(-1)^nc_1(E)^n+$ terms involving higher Chern classes.

In particular, all such terms vanish if $X$ has dimension $1$ (or the bundle has rank one).

The answer to your question is now a particular case of all this and you get that your top self-intersection is just $c_1(\mathcal E)$, that is the degree of $\mathcal E$.