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It is well known (theorem of Grothendieck) that every vector bundle on $\mathbf{P}^1$ is a direct sum of line bundles. What about $\mathbf{P}^2$? I figure the answer must be no, but is the tangent sheaf $\mathscr{T}_{\mathbf{P}^2}$ a counterexample? By the Euler sequence it shares most invariants with a direct sum of line bundles.

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The total Chern class of $T_{\mathbb P^2}$ is $1 + 3h + 3h^2$ which is not a product of linear polynomials in $h$, so it cannot be the sum of two lines bundles.

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  • $\begingroup$ But doesnt the Euler sequence tell me that the total chern class of $T$ times the total chern class of $O$ (which is $1$) is the total chern class of a direct sum of $O(1)$? $\endgroup$ – Rüdiger Mar 4 '17 at 23:33
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    $\begingroup$ Yes, it is $(1+3 h + 3 h^2) = (1+h)^3 \mod h^3$, but here the point of view is different: Assuming $\mathcal{T} = \mathcal{O}_P(d) \oplus \mathcal{O}_P(e)$ one tries to write $(1 + 3 h + 3 h^2) = (1+d h) (1 + e h)$ which is impossible. $\endgroup$ – Jürgen Böhm Mar 5 '17 at 4:41
  • $\begingroup$ @Rüdiger : yes but $h^3 = 0$ since the cohomology/Chow ring of $\mathbb P^2$ is $\mathbb Z[h]/(h^3)$. $\endgroup$ – user171326 Mar 5 '17 at 7:50
  • $\begingroup$ @N.H. + Jürgen thanks for the clarification! $\endgroup$ – Rüdiger Mar 5 '17 at 10:58
  • $\begingroup$ Sure, you're welcome ! $\endgroup$ – user171326 Mar 5 '17 at 11:00
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The dimension of the space of global sections of the tangent sheaf is $8$: from the Euler sequence, we get $$ 0 \to H^0(\mathbb{P}^2, \mathcal{O}) \to H^0(\mathcal{P}^2, \mathcal{O}(1))^3 \to H^0(\mathbb{P}^2, \mathcal{T}) \to H^1(\mathbb{P}^2, \mathcal{O}) $$ and this last term is zero by standard arguments.

It follows that $\mathcal{T}$ cannot be a sum of two line bundles, since $h^0(\mathbb{P}^2, \mathcal{O}) = 1$, $h^0(\mathbb{P}^2, \mathcal{O}(1)) = 3$, $h^0(\mathbb{P}^2, \mathcal{O}(2)) = 6$, and the other positive bundles are too big.

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  • $\begingroup$ This is a really cool argument ! $\endgroup$ – user171326 Mar 5 '17 at 8:06

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