The Splitting Principle for Varieties

Question

I am recently thinking of the following question (motivated by the splitting principle for vector bundles): Suppose $X$ is a projective smooth variety over an algebraically closed field $k$. Let \begin{equation} E_0\subsetneq E_1\subsetneq \cdots\subsetneq E_n \end{equation} be a sequence of vector bundles. By this I mean that each inclusion $E_i\subsetneq E_{i+1}$ realizes $E_i$ as a subbundle not only a subsheaf. Over $X$ there is no reason that $E_{i+1}\simeq E_i\oplus (E_{i+1}/E_i)$. Can we construct a projective variety $\pi: Y\to X$ such that \begin{equation} \pi^\ast E_{i+1}=\pi^\ast E_i\oplus \pi^\ast (E_{i+1}/E_i). \end{equation}

Answer 1

No, this does not exist in general. For instance, let $X = \mathbb{P}^1$ and consider the tautological embedding $$ \phi \colon \mathcal{O}(-1) \hookrightarrow \mathcal{O} \oplus \mathcal{O}. $$ Assume $\pi \colon Y \to X$ is a projective surjective morphism such that $\pi^*(\phi)$ is a split embedding. Then there is an epimorphism $$ \mathcal{O}_Y \oplus \mathcal{O}_Y \cong \pi^*(\mathcal{O} \oplus \mathcal{O}) \to \pi^*\mathcal{O}(-1). $$ In particular, the line bundle $\pi^*\mathcal{O}(-1)$ is globally generated. On the other hand, its dual $\pi^*\mathcal{O}(1)$ is also globally generated (because $\mathcal{O}(1)$ is globally generated on $X$). But this is obviously impossible.