I'm trying to deal with an example of a rank two vector bundle over the complex projective plane which is non slope-stable (because the associated sheaf of sections has a coherent subsheaf of equal slope) but it does not admit subbundles with greater slope. This is the simplest example I have in mind in order to explain why in dimension greater than 1 (of the Kahler manifold) one has to deal with subsheaves and not only subbundles when checking the stability of a vector bundle. I have problems in many part of the exposition which goes like this:

Let $p\in\mathbb{CP}^2$ be a point and $\mathcal{I}_p$ its ideal sheaf. Then, by using the Koszul resolution, we have the following short exact sequence of coherent sheaves $0\rightarrow\mathcal{O}(-2)\overset{f}{\rightarrow}\mathcal{O}(-1)\oplus\mathcal{O}(-1)\rightarrow\mathcal{I}_p\rightarrow 0$. Now take any non trivial sheaf homomorphism $\mathcal{O}(-2)\overset{g}{\rightarrow}\mathcal{O}$, which exists since it can be seen as a holomorphic section of $\mathcal{O}(-2)^*\otimes\mathcal{O}\cong\mathcal{O}(2)$, and let $\mathcal{E}$ be the push-out of $f$ and $g$ in the catgeory of coherent sheaf.

Question 1: How can we prove that $\mathcal{E}$ fits in the following sequence of sheaves $0\rightarrow\mathcal{O}\rightarrow\mathcal{E}\rightarrow\mathcal{I}_p\rightarrow 0$ ? Why is it exact?

I have only understood that the first arrow is injective since $f$ it is.

By using this exact sequence we know that $\mu(\mathcal{E}):=\frac{deg(\mathcal{E})}{rk(\mathcal{E})}=\frac{0}{2}=0$ but since $\mu(\mathcal{O})=0$ we conclude that $\mathcal{E}$ is not stable.

Question 2: Why $\mathcal{E}$ does not admit subbundles of greater slope?

EDIT As Libli has wonderfully explained, this example shows that stability should be checked also on quotient sheaves and not only on quotient bundles. Moreover one has to prove that such a push-out $\mathcal{E}$ is indeed locally-free and not only coherent (as Libli has done).

EDIT 2 If you are interested in this example you can find it in the wonderful book of Huybrechts-Lehn "The Geometry of Moduli Spaces of Sheaves" (Thm 5.1.1 and Ex. 5.1.2)

  • 2
    $\begingroup$ The pushout of $f \colon A \to B$ and $g: A \to C$ is obtained by $ (B \oplus C) / A$. You have checked the map from $C$ to this is injective. Now you want to take the cokernel of this map from $C$, i.e. you want to quotient by $C$. What is it? $\endgroup$ – Will Sawin Jan 31 at 16:43
  • $\begingroup$ @WillSawin If I identify the sheaf $\mathcal{I}_p$ with Coker$(f)$, then thanks to your suggestion and using the injectivity of $\mathcal{O}\rightarrow\mathcal{E}$, in this case $(B\oplus C)/A$ should be exactly $\mathcal{I}_p$, right? $\endgroup$ – John117 Jan 31 at 17:00
  • $\begingroup$ Yes (although you might not need injectivity for this step). For Question 2, you can assume there is a line bundle $L$ of positive slope and map $L \to \mathcal E$ and show that the induced map $L \to \mathcal I_p$ must vanish, deduce that $L$ factors through $\mathcal O$, then show $L \to \mathcal O$ must vanish. $\endgroup$ – Will Sawin Jan 31 at 17:06
  • $\begingroup$ Also, you technically need $g$ to be nonzero at $p$ for this pushout to be a vector bundle (but this is easy to achieve from the holomorphic section point of view). $\endgroup$ – Will Sawin Jan 31 at 17:06
  • $\begingroup$ @WillSawin If $deg(L)>0$ it's clear for me your reasoning, sorry but I'm still a bit confused. Shouldn't I have to assume that $deg(L)\ge 0$? I want to show that checking only stability on subbundles would lead to the wrong conclusion that the vector bundle is stable, so if I assume by contradiction that exists a line subbundle with positive degree no one tells me that there cannot be one of degree zero which contradicts the stability of the bundle. $\endgroup$ – John117 Jan 31 at 17:52

This example is useful not to understand that stability should be checked on subsheaves (instead of checking only subbundles), but that it should be checked for all quotient sheaves (and not only quotient bundles). Indeed, let us show that $\mathcal{E}$ has no quotient bundle of rank $1$ with slope $0$. Assume, by absurd, it has. This means, we have an exact sequence:

$$0 \longrightarrow \mathcal{F} \longrightarrow \mathcal{E} \longrightarrow \mathcal{O} \longrightarrow 0.$$

By the local duality for locally complete intersections, we have $\mathcal{H}om(\mathcal{I}_p, \mathcal{O}) = \mathcal{O}$, $\mathcal{E}xt^1(\mathcal{I}_p, \mathcal{O}) = \mathcal{O}_p$ and $\mathcal{E}xt^i(\mathcal{I}_p, \mathcal{O}) = 0$ for all $i>1$. We deduce that $\mathcal{E}xt^i(\mathcal{E}, \mathcal{O}) = 0$ for all $i>0$. As a consequence, we have $\mathcal{E}$ is a rank $2$ vector bundle on $\mathbb{P}^2$ (you didn't ask that question, but it is certainly not obvious to see that $\mathcal{E}$ is actually a vector bundle with the painful push-out construction).

In particular, $\mathcal{F}$ would be a reflexive sheaf of rank $1$, that is a line bundle on $\mathbb{P}^2$. Since its slope is $0$, we have $\mathcal{F} = \mathcal{O}$. But $$\mathrm{Ext}^1(\mathcal{O},\mathcal{O}) = H^1(\mathbb{P}^2,\mathcal{O}) = 0,$$ so that $\mathcal{E}$ would be the direct summ $\mathcal{O} \oplus \mathcal{O}$. This is certainly not true, as they are no exact sequence:

$$ 0 \longrightarrow \mathcal{O} \longrightarrow \mathcal{O} \oplus \mathcal{O} \longrightarrow \mathcal{I}_p \longrightarrow 0$$ on $\mathbb{P}^2$. Hence, if we only look at quotient bundles of $\mathcal{E}$, one could have the wrong impression that $\mathcal{E}$ is slope-stable. This is not true as $\mathcal{E}$ has a destabilizing quotient sheaf of rank $1$, namely $\mathcal{I}_p$.

  • $\begingroup$ Thanks for your very beatiful answer, but at the beginning the quotient bundle of rank 1 and degree 0 that you assume to exist absurdly it is not necessarily isomorphic to the trivial bundle, isn't it? You made the same assumption also when you claimed that $\mathcal{F}=\mathcal{O}$. $\endgroup$ – John117 Feb 1 at 10:12
  • $\begingroup$ A line bundle of degree $0$ is necessarily $\mathcal{O}$ on $\mathbb{P}^2$. $\endgroup$ – Libli Feb 1 at 19:16
  • $\begingroup$ Yes of course, thank you! Could you please spell out how did you deduce that $\mathcal{E}$ is locally free by the vanishing of $\mathcal{Ext}^i(\mathcal{E},\mathcal{O})$ for $i>0$? $\endgroup$ – John117 Mar 4 at 16:38
  • $\begingroup$ Ah no nevermind, maybe I got it! Thanks anyway! $\endgroup$ – John117 Mar 5 at 12:53

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