Motivation: $T\_{\mathbb P^2}$ isn't an extension of line bundles
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Here's a trick to show that the tangent bundle $T$ of $\mathbb P^2$ is not an extension of line bundles. If it were, we would have a short exact sequence
$$\def\O{\mathcal O}
0\to \O(a)\to T\to \O(b)\to 0
$$
for some integers $a$ and $b$. Then we can compute the total Chern class
$$\begin{align\*}
c(T)& =c(\O(a))\cdot c(\O(b)) \\
 &= (1+aH)(1+bH) \\
 &= 1+(a+b)H+abH^2,
\end{align\*}$$



where $H=c\_1(\O(1))$ is the class of a hyperplane.



On the other hand, we have the Euler sequence 
$$
0\to \O\to \O(1)^3\to T\to 0
$$
which tells us that
$$\begin{align\*}
c(T)&=c(T)\cdot c(\O)=c(\O(1)^3)\\
 &=c(\O(1))^3= 1+3H+3H^2.
\end{align\*}$$



Now observe that there do not exist integers $a$ and $b$ so that $a+b=ab=3$, so $T$ cannot be an extension of line bundles.




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The Question
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More generally, whenever we have an extension of vector bundles $0\to L\to E\to M\to 0$, we have $c(E)=c(L)\cdot c(M)$. So to show that $E$ has no sub-bundles (or equivalently, no quotient bundles), it suffices to show that $c(E)$ doesn't factor. The question is whether the converse is true:



> 
> Suppose $E$ is a rank $r$ vector bundle on a (smooth quasi-projective) scheme (or manifold) $X$ so that $c(E)=c(L)c(M)$ for vector bundles $L$ and $M$ of rank $i$ and $r-i$, respctively. Must $E$ have a sub-bundle or rank $i$ or $r-i$?
> 
> 
> 


**Remark 1:** The phrasing is a bit strange compared to the natural-sounding "If the total Chern class of a vector bundle factors, does it have a sub-bundle?" The point is that knowing the rank of $E$ is very important. We showed that $T\_{\mathbb P^2}$ has no sub-bundles, but $O(1)^3$ has the same total Chern class and clearly has lots of sub-bundles.


**Remark 2:** Does either $L$ or $M$ have to be a sub-bundle of $E$? NO! For example, on $\mathbb P^1$, we have that
$$
c(\O(1)\oplus \O(-1)) = (1+H)(1-H)=1 = c(\O)c(\O)
$$
but $\O(1)\oplus \O(-1)$ doesn't have a sub-bundle isomorphic to $\O$ (because it has no non-vanishing sections).


**Remark 3:** What is the answer in the case $X=\mathbb P^n$?



If the total Chern class of a vector bundle factors, does it have a sub-bundle?
The answer for projective spaces is negative. I think the simplest example are 2-bundles on $\mathbb{P}^3(\mathbb{C})$. In that case the Schwarzenberger condition is that $c\_1c\_2$ should be even. Atiyah and Rees have proved that for any pair $(c\_1,c\_2)$ satisfying this there are holomorphic vector bundles $\xi$ with $c\_1(\xi)=c\_1,c\_2(\xi)=c\_2$ (see Atiyah, Rees, Vector bundles on projective 3-space. Invent. Math. 35 (1976), 131–153.). The number of topologically distinct such bundles in 1 when $c\_1$ is odd and 2 when $c\_1$ is even. So e.g. there is a topologically nonsplit 2-bundle on $\mathbb{P}^3$ with total Chern class $(1+ka)(1-ka)$ where $a=c\_1(\mathcal{O}(1))$.


The topological classification of 2-bundles on $\mathbb{P}^3$ and the existence
of a holomorphic structure on them are also proved in Okonek, Schneider, Spindler, Vector bundles on complex projective spaces, chapter 1, 6.3.


