Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it? When I read the GTM 082, "The Splitting Principle of the complex vector bundle", I see that in the proof we split off one subbundle at a time by pulling back to the projectivization of a quotient bundle. I don't know why we need to pull this back to the projectivization of the quotient bundle. I mean if there is always a line bundle embeded in the given complex vector bundle, we can just split it on the original vector bundle, every time we take quotient of these two bundle, we get a quotient bundle with dimensional reduced by 1. So in the similar way we can conclude that every bundle can be decomposed into the direct sum of several line bundle. So is it true? Can we always find a line bundle embedded into a given complex vector bundle?
 A: I think thinking in terms of classifying spaces will help clarify the situation. We know that a rank $n$ complex vector bundle $V$ on $X$ is the same thing as a homotopy class of maps $f:X\to BU_n$. A choice of decomposition $V\cong L\oplus W$ where $L$ is a line bundle corresponds to a lift of $f$ to $B(U_1\times U_{n-1})\cong BU_1\times BU_{n-1}$. The homotopy fiber of the map $BU_1\times BU_{n-1}\to BU_n$ is $U_n/(U_1\times U_{n-1})\cong \mathbb{CP}^{n-1}$. This is a non-trivial principal fibration and so it has no section.
In particular you can take the universal vector bundle on $BU_n$ (or if you want, its restriction to some finite skeleton given by a suitable complex Grassmannian) and this will give you a vector bundle that admits no line bundle.
Note that $BU_1\times BU_{n-1}$ is precisely the projectivization of the universal vector bundle on $BU_n$, so this is just another way of phrasing the splitting principle.
More precisely you can find a sequence of obstructions in $H^{i+1}(X;\pi_i\mathbb{CP}^{n-1})$ for $i\ge2$ and you can find a sub-line bundle if and only if these obstruction vanish. For example you can always find a such a line bundle if the (cohomological) dimension $X$ is less or equal than 2
A: No, this is not true. Consider the tangent bundle to $\mathbb{CP}^2$; I claim it does not contain a sublinebundle. Let $h$ be the class of a hyperplane in $H^2(\mathbb{CP}^2)$. So $H^{\ast}(\mathbb{CP}^2) \cong \mathbb{Z}[h]/h^3$. Here are two proofs (which are really the same proof in two languages).
Proof one: Chern classes The total Chern class of the tangent bundle is $1+3h + 3 h^2$. If we had a short exact sequence $0 \to L_1 \to T \to L_2 \to 0$, then we would have $c(L_1) c(L_2) = 1+3h+3h^2$. But the polynomial $1+3x+3x^2$ is irreducible over $\mathbb{Z}$. $\square$
Proof two: a bundle with no section The projectivation of the tangent bundle is the flag manifold $\mathcal{F}\ell_3$. Choosing a sublinebundle of $T$ corresponds to choosing a (continuous) section of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$. We have $H^{\ast}(\mathcal{F}\ell_3) \cong H^{\ast}(\mathbb{CP}^2)[u]/(u^2+3hu+3h^2)$. If we had a section, this would induce a map from $H^{\ast}(\mathcal{F}\ell_3)$ to $H^{\ast}(\mathbb{CP}^2)$ which would be the identity on $H^{\ast}(\mathbb{CP}^2)$ and would have to send $u$ to $kh$ for some integer $k$. But then we would have $k^2+3k+3=0$, and this polynomial does not have integer roots. $\square$
I think of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$ as a complex analogue of the Hopf fibration, which likewise has no continuous section, although I don't know a way to make this precise. This answer is the analogue of the Hairy Ball theorem, which basically says that the tangent bundle to $S^2$ doesn't have a real sublinebundle.
