Let $\mathcal E$ be an $SL(2)$-bundle on $\mathbb P^2$. Assume that the restriction of $\mathcal E$ to any $\mathbb P^1$ is non-trivial. Is it true that $\mathcal E$ is a direct sum of two line bundles?
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$\begingroup$ In positive characteristic $2$, I believe that can fail (possibly you are interested only in characteristic $0$). If you pullback the tangent bundle of $\mathbb{P}^2$ by a Frobenius map, then the degree is even. Thus, you can tensor with an invertible sheaf to make the determinant trivial. Every line in $\mathbb{P}^2$ maps under Frobenius to a line (with different coefficients). The restriction of $T_{\mathbb{P}^2}$ to a line is $\mathcal{O}(2)\oplus \mathcal{O}(1).$ $\endgroup$– Jason StarrNov 1, 2017 at 19:21
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$\begingroup$ Thanks. But what about characteristic 0? $\endgroup$– Alexander BravermanNov 2, 2017 at 2:01
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$\begingroup$ The answer by @Sasha works in all characteristics. If $E$ is a direct sum of $\mathcal{O}(-d)$ and $\mathcal{O}(d)$, then the second Chern class of $E$ equals $-d^2 c_1(\mathcal{O}(1))^2$ for the integer $d$. Thus, for a rank $2$ locally free whose first Chern class is not such a negative square, $E$ cannot be a direct sum of invertible sheaves. $\endgroup$– Jason StarrNov 2, 2017 at 2:04
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$\begingroup$ Note, that my previous comment also explains pullbacks of the tangent bundle. For every morphism $f:\mathbb{P}^2\to \mathbb{P}^2$ with $f^*\mathcal{O}(1)\cong \mathcal{O}(2b)$ for an integer $b$ (whether or not $f$ is a Frobenius morphism), for $E=\mathcal{O}(-3b)\otimes f^*T_{\mathbb{P}^2}$, then $c_2(E)$ equals $3e^2 c_1(\mathcal{O}(1))^2$, and this is not of the form $-d^2c_1(\mathcal{O}(1))^2.$ Thus, $E$ is not a direct sum of two line bundles. $\endgroup$– Jason StarrNov 2, 2017 at 2:47
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$\begingroup$ And is it obvious that $f^*T_{\mathbb P^2}$ is not a direct sum of two line bundles? $\endgroup$– Alexander BravermanNov 2, 2017 at 4:33
2 Answers
No. A simple counterexample is a nontrivial extension $$ 0 \to O(1) \to E \to I_x(-1) \to 0, $$ where $x$ is a point and $I_x$ is its ideal sheaf.
EDIT: basics of Serre's construction.
First, note that $$ Ext^1(I_x(−1),O(1)) \cong Ext^2(O_x,O(1)) \cong H^0(\mathbb{P}^2,\mathcal{E}\mathit{xt}^2(O_x,O(1))) \cong H^0(\mathbb{P}^2,O_x). $$ Next, let us rewrite the defining sequence of $E$ as $$ 0 \to O(1) \to E \to O(-1) \to O_x \to 0. $$ Applying $\mathcal{H}\mathit{om}(-,O(1))$ to it, we get $$ 0 \to O(2) \to E^\vee(1) \to O \to \mathcal{E}\mathit{xt}^2(O_x,O(1)) \to \mathcal{E}\mathit{xt}^1(E,O(1)) \to 0. $$ So, to check that $\mathcal{E}\mathit{xt}^1(E,O(1)) = 0$ and hence $E$ is locally free, it is enough to show that the map $$ O \to \mathcal{E}\mathit{xt}^2(O_x,O(1)) \cong O_x $$ is surjective. But by the above chain of isomorphisms, this map corresponds to the original extension class, hence is non-zero. And any non-zero map to $O_x$ is surjective.
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$\begingroup$ Thanks. Could you explain the notation? What is $I_x$? $\endgroup$ Nov 2, 2017 at 4:31
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$\begingroup$ @AlexanderBraverman: $I_x$ is the ideal of the point $x$. $\endgroup$– SashaNov 2, 2017 at 5:05
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1$\begingroup$ Ok, thanks. And why is such an extension locally free? (Sorry for the stupid questions) $\endgroup$ Nov 2, 2017 at 14:34
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1$\begingroup$ @AlexanderBraverman: This is a special case of Serre's correspondence. Just note that $Ext^1(I_x(-1),O(1)) \cong Ext^2(O_x,O(1)) = H^0(\mathbb{P}^2,\mathcal{E}\mathit{xt}^2(O_x,O(1))) \cong H^0(\mathbb{P}^2,O_x)$ and hence the class of the extension generates the sheaf $\mathcal{E}\mathit{xt}^2(O_x,O(1)) \cong O_x$. $\endgroup$– SashaNov 2, 2017 at 15:17
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$\begingroup$ Sorry, I am probably missing something basic: I understand why there is such a non-trivial extension, but why is it locally free? Doesn't it necessarily split in some neighbourhood of $x$? $\endgroup$ Nov 2, 2017 at 22:05
Let's look at $T_{\mathbb{P}^2}$ as Jason Starr suggests. We have the Euler exact sequence:
$$ 0 \rightarrow \mathcal{O}_{\mathbb{P}^2} \rightarrow \mathbb{C}^3 \otimes \mathcal{O}_{\mathbb{P}^2}(1) \rightarrow T_{\mathbb{P}^2} \rightarrow 0$$
Note that $\mathrm{Ext}^1(L_1,L_2) = 0$ for any two line bundles on $\mathbb{P}^2$. Hence, if $T_{\mathbb{P}^2}$ splits as direct sum of line bundles (say $L_1 \oplus L_2$), we have:
$$ \mathbb{C}^3 \otimes \mathcal{O}_{\mathbb{P}^2}(1) = \mathcal{O}_{\mathbb{P}^2} \oplus L_1 \oplus L_2,$$
which is clearly impossible. On the other hand, $T_{\mathbb{P}^2}$ is obviously $\mathrm{SL}_3$-homogeneous and by the normal exact sequence, the restriction of $T_{\mathbb{P}^2}$ to any $\mathbb{P}^1$ is $\mathcal{O}(1) \oplus \mathcal{O}(2)$, which is non-trivial.
You might be interested in the following result:
Theorem (Grauert-Mülich) : Let $E$ be a rank $2$, $\mu$-semi-stable vector bundle on $\mathbb{P}^2$ with $c_1(E) = 0.$ Then, for a generic line $l \subset \mathbb{P}^2$, the restriction $E|_{l}$ is trivial.
You could also have a look at this paper https://link.springer.com/chapter/10.1007/BFb0099972 by Maruyama. In there he studies the singularities of the curve of lines $l \in \mathbb{P}^2$ such $E|_{l}$ is non-trivial.
EDIT I misunderstood what was meant by $\mathrm{SL}_2$-bundle. By $\mathrm{SL}_2$-bundle, you mean a rank-$2$ vector bundle with trivial first Chern class. So if your vector bundle does not restrict to the trivial bundle on the generic $\mathbb{P}^1$, then the Grauert-Mülich Theorem implies that it is not $\mu$-semi-stable. In particular, it has a destabilizing sub-line bundle.
It is not obvious to me that all unstable rank $2$ vector bundle on $\mathbb{P}^2$ do split, but that is perhaps true...
EDIT(bis) By proposition 4 of this paper : http://www.seminariomatematico.unito.it/rendiconti/67-1/15.pdf, a rank $2$-vector bundle (say $E$) on $\mathbb{P}^2$ with $c_1 = 0$ splits if and only if $h^1(E(-1)) = 0$.
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$\begingroup$ You are right that $E$ should be unstable, and this precisely gives my example. The first $O(1)$ is the destabilizing subsheaf. You can replace it by $O(k)$ for any $k > 0$ (and then, of course, replace $I_x(-1)$ by $I_x(-k)$). $\endgroup$– SashaNov 4, 2017 at 21:17
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$\begingroup$ @Sasha : Ok. Sorry! I thought you had a misprint and that you wanted to write the Koszul complex twisted by $\mathcal{O}(1)$ (which would look like: $$0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O} \oplus \mathcal{O} \rightarrow I_x(1) \rightarrow 0)$$ But you are indeed considering something entirely different. $\endgroup$– LibliNov 5, 2017 at 19:26