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?
[Math] Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it
at.algebraic-topologydg.differential-geometry
Related Solutions
There is nothing special about the tangent bundle: for any vector bundle $E \rightarrow X$ with projectivisation $\pi : \mathbb{P}E \rightarrow X$, and tautological bundle $\mathcal{L} \rightarrow \mathbb{P}E$, we have $T\mathbb{P}E \cong (\pi^*E / \mathcal{L}) \otimes \mathcal{L}^*$. In fact the base $X$ is not really involved, so you may as well work with the case where $X$ is a point, and $E$ is just a vector space $V$. Then the result follows from the standard Euler sequence
$$ 0 \rightarrow \mathcal{O}_{\mathbb{P}V} \rightarrow V \otimes \mathcal{O}_{\mathbb{P}V}(1) \rightarrow T\mathbb{P}V \rightarrow 0, $$
where we regard $V$ as a constant sheaf on $\mathbb{P}V$.
Added in response to comment: Explicitly, given a point $l \in \mathbb{P}V$, viewed as a line in $V$, and a vector $v \in T_l \mathbb{P}V$, we can lift $v$ (under the projection $V \setminus \{0\}\rightarrow \mathbb{P}V$) to a vector in $V/l$ at each point of $l \setminus \{0\}$. Moreover, this lift is equivariant under the action of $k^\times$ on $l\setminus \{0\}$, where $k$ is our ground field (e.g. $\mathbb{C}$). In other words, it defines a linear $V/l$-valued function on $l$.
Linear functions on $l$ are precisely elements of $\mathcal{L}^*_l$, and $V/l$ is precisely $(\pi^*V/\mathcal{L})_l$, so $v$ defines an element of $((\pi^*V/\mathcal{L}) \otimes \mathcal{L}^*)_l$.
Added in response to further comments: Even more explicitly, if $e_0, e_1, \dots, e_n$ is a basis for $V$, with corresponding coordinates $w_0, w_1, \dots, w_n$, then the $w_i$ form homogeneous coordinates on $\mathbb{P}V$. On the affine patch $w_0 \neq 0$, we have local coordinates given by $z_j = w_j/w_0$, for $j=1, \dots, n$. Given a point $l = [1: l_1: \dots: l_n] \in \mathbb{P}V$, the vector
$$ \sum_{j=1}^n v_j \frac{\partial}{\partial z_j} $$
is tangent to the curve $z_j(t) = l_j + tv_j$. This lifts to the curve in $V$ given by $w_0(t)=f(t)$, $w_j(t)=f(t)(l_j+tv_j)$ for $j\geq 1$, where $f$ is any smooth $\mathbb{C}^\times$-valued function. Its tangent at $t=0$, i.e. a lift of $v$ to $V$, is
$$ f'(0) \bigg( e_0 + \sum_{j=1}^n l_je_j \bigg) + \sum_{j=1}^n v_j e_j. $$
The ambiguity coming from $f'(0)$ is exactly what is killed when we quotient by $l$, so the lift to $V/l$ is represented by $\sum v_je_j$.
Explicit computation for $V=\mathbb{C}^2$: Take coordinates $w_0, w_1$ on $\mathbb{C}^2$, and let $U_j = \{w_j \neq 0\} \subset \mathbb{P}^1$. We have
$$(\pi^*V / \mathcal{L})_{[w_0 : w_1]} = \mathbb{C}^2 \Big/ \mathbb{C} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix}.$$
On $U_0$ a complement to $\mathcal{L}$ in $\pi^*V$ is given by $\mathbb{C}e_1$; we can therefore take $[e_1]$ as a trivialising frame for the quotient bundle (where the square brackets denote the image in the quotient). Similarly on $U_1$ a frame is given by $[e_0]$. On $U_0 \cap U_1$ we have for all local holomorphic functions $f$ that
$$f e_1 = \frac{f}{w_1} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix} - \frac{fw_0}{w_1} e_0 $$
so
$$ f[e_1] = -\frac{fw_0}{w_1}[e_0].$$
The transition function $\psi_{10}$ is therefore $-w_0/w_1$, which (up to the minus sign, which we could have eliminated by choosing $[-e_0]$ as our frame on $U_1$) is the transition function for $\mathcal{O}(1)$.
Best Answer
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.