Stability condition for vector bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:46:11Z http://mathoverflow.net/feeds/question/85908 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85908/stability-condition-for-vector-bundles Stability condition for vector bundles Botong Wang 2012-01-17T15:19:59Z 2012-01-18T05:02:21Z <p>Let $X$ be a smooth projective variety over $\mathbb{C}$, and fix $A$ an ample divisor as the polarization. We say a vector bundle $E$ to be (semi-)stable, if for any proper subsheaf $F$ of $E$, $\mu(F)&lt;\mu(E)$ (resp. $\leq$). </p> <p>I guess it is not sufficient to check just subbundles $F$ of $E$. But is there a counterexample? Or more precisely, is there an example of $(X, A, E, F)$ satisfying the following conditions?</p> <p>(1)$X,A,E$ are as above, and $F$ is a proper subsheaf of $E$ breaking the stability condition, i.e., $\mu(F)\geq \mu(E)$.</p> <p>(2)There exists no vector bundle which break the stability condition, i.e., $\mu(F')&lt;\mu(E)$ for any subbundle $F'$ of $E$.</p> http://mathoverflow.net/questions/85908/stability-condition-for-vector-bundles/85917#85917 Answer by Ray Hoobler for Stability condition for vector bundles Ray Hoobler 2012-01-17T17:33:16Z 2012-01-17T17:33:16Z <p>The reason for the condition to be on subsheaves is that the slope of a sheaf E on X is unchanged by removing subschemes of X of cod >1. Consequently the Harder-Narasimhan filtration, etc. must allow torsion free subsheaves in its construction. It is enough to consider 'saturated' subsheaves meaning subsheaves F such that E/F is torsion free since passing from a subsheaf F' to its saturate raises the slope.</p> http://mathoverflow.net/questions/85908/stability-condition-for-vector-bundles/85922#85922 Answer by Angelo for Stability condition for vector bundles Angelo 2012-01-17T19:33:17Z 2012-01-18T05:02:21Z <p>There are many examples of unstable bundles on a projective surface that have no non-trivial subbundles. For example, if $k$ is an integer with $k &lt; 3$ and $I$ is the sheaf of ideal of $m$ distinct points in $\mathbb P^2$, with $m > 0$, there exists an extension $$0 \longrightarrow \mathcal O \longrightarrow E \longrightarrow I(k) \longrightarrow 0$$ on $\mathbb P^2$ in which $E$ is locally free. Furthermore, the Chern classes of $E$ are $c_1(E) = k$ and $c_2(E) = m$ (for this, see page 103 of "Vector bundles on complex projective spaces", by Okonek, Schneider and Spindler). If $k &lt; 0$, this vector bundle is clearly unstable; but for most values of $k$ and $m$ it can not split as a direct sum of line bundles, hence it cannot contain a line subbundle (since every extension of line bundles on $\mathbb P^2$ splits).</p>