Motivation: $T_{\mathbb P^2}$ isn't an extension of line bundles

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.

The Question

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 $X=\mathbb P^n$?


2 Answers 2


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.


If you are also asking about the case of topological complex vector bundles over manifolds, consider the case $X=S^5$. There are no nontrivial rank $1$ bundles, but there is a nontrivial rank $2$ bundle, and of course its Chern class $1$ factors as $1\times 1$.


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