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Maybe the simplest example of the phenomenon mentioned by Jason is the case of rank 2 vector bundles on $\mathbb{P}^3$ with $c_1$ even, studied in this paper of Atiyah and Rees. Besides $c_1$ and $c_2$ which encode the $K$-theory class, there is another topological invariant $\alpha$ with values in $\mathbb{Z}/2$, which can be $0$ or $1$ for holomorphic ...
As always, you should specify that $X$ is projective (or at least proper). Even so, this really depends on what you mean by a "connected family". Certainly this is false if you want the base of your family to be irreducible. Here is the simplest counterexample I see. On $X=\mathbb{P}^2$, let $V_1$ be $\mathcal{O}(-7)\oplus \mathcal{O}(-7)\oplus ... 2 Yes. If this map$\pi_{8t+1}U\to \pi_{8t+1}O$were trivial then$\pi_{8t+1}(O/U)$would have an element of order$2$. But$O/U$is homotopy equivalent to$\Omega O$, and$\pi_{8t+2}O$is trivial. I do not know a reference offhand. 1 Orientability of$TM|_\Sigma$is equivalent to non-triviality of normal bundle$N\Sigma$. Denote generator of$\pi_1(\Sigma)$by$\gamma$; if we have$N\Sigma$is trivial, then$TM|_{i(\gamma)}$is non-orientable, so$\gamma$cannot be nullhomotopic. 12 Yes, there is such a twistor fibration over each$S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic$n$-plane field transverse to the fibers of the mapping. Namely, one writes$S^{2n} = \mathrm{SO}(2n{+}1)/\mathrm{SO}(2n)$and then, using the inclusion$\mathrm{U}(n)\subset\mathrm{SO}(2n)$, one has the coset fibration$$... 6 One possible generalization of Penrose twistor fibration is the fibration$\pi:\mathcal{J}_M\to M$over any oriented$2n$-dimensional conformal manifold$(M,[g])$, whose fiber$\pi^{-1}(x)$is the space of orthogonal almost complex structures on$(T_xM,[g_x])$, namely the orthogonal$J_x:T_xM\to T_xM$squaring to$-\mathrm{id}\$ such that the complex ...