Chern class of a vector bundle and the associated projective space bundle I have a very basic question regarding Chern classes. Let $X$ be a smooth projective variety and $\mathcal{E}$ a vector bundle on it. Let $\pi:\mathbb{P}(\mathcal{E})\to X$ denote the projective space bundle over $X$ associated to $\mathcal{E}$.
How are the Chern classes $c_i$ of the vector bundle $\mathcal{E}$ on $X$ related to the Chern classes $c_i'$ of the projective variety $\mathbb{P}(\mathcal{E})$, i.e. the Chern classes of the tangent bundle of $\mathbb{P}(\mathcal{E})$? My naive hope would be that we simply have $c_i'=\pi^* c_i$. Is that true? Is there a good reference?
 A: The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= \mathcal{O} \oplus \mathcal{O}$ over $\mathbb{CP}^{1}$. Then $$\mathbb{P}(E) \cong \mathbb{CP}^{1} \times \mathbb{CP}^{1}.$$
Now, note that $c_{2}(E)=0$ since $E$ is a trivial bundle, but $\int_{\mathbb{P}(E)} c_{2}(T\mathbb{P}(E)) = 4$, since this is equal to the topological Euler characteristic of $\mathbb{P}(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $\mathbb{P}(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $\mathbb{P}(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
