If the total Chern class of a vector bundle factors, does it have a sub-bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:31:33Z http://mathoverflow.net/feeds/question/65086 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65086/if-the-total-chern-class-of-a-vector-bundle-factors-does-it-have-a-sub-bundle If the total Chern class of a vector bundle factors, does it have a sub-bundle? Anton Geraschenko 2011-05-15T21:56:20Z 2011-05-16T10:14:05Z <h2>Motivation: $T_{\mathbb P^2}$ isn't an extension of line bundles</h2> <p> 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*} </p> <p>where $H=c_1(\O(1))$ is the class of a hyperplane.</p> <p> 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*} </p> <p>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.</p> <hr> <h2>The Question</h2> <p>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:</p> <blockquote> <p>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 $r-i$?</p> </blockquote> <p><strong>Remark 1:</strong> 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.</p> <p><strong>Remark 2:</strong> 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).</p> <p><strong>Remark 3:</strong> What is the answer in the case $X=\mathbb P^n$?</p> http://mathoverflow.net/questions/65086/if-the-total-chern-class-of-a-vector-bundle-factors-does-it-have-a-sub-bundle/65088#65088 Answer by algori for If the total Chern class of a vector bundle factors, does it have a sub-bundle? algori 2011-05-15T22:20:26Z 2011-05-15T22:27:47Z <p>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 $a=c_1(\mathcal{O}(1))$.</p> <p>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.</p> http://mathoverflow.net/questions/65086/if-the-total-chern-class-of-a-vector-bundle-factors-does-it-have-a-sub-bundle/65100#65100 Answer by Tom Goodwillie for If the total Chern class of a vector bundle factors, does it have a sub-bundle? Tom Goodwillie 2011-05-16T01:52:54Z 2011-05-16T10:14:05Z <p>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 $1\times 1$.</p>