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I am reading an article in which the author said that the sequence $$0 \to \mathcal {O} \to H^{ \oplus(n+1) } \to T\mathbb{P}^n \to 0$$ called the Euler sequence, where $H$ is the hyperplane line bundle and $T\mathbb{P}^n $ is the holomorphic tangent bundle.


I'm a bit confused about this, is the Euler sequence for the holomorphic tangent bundle of $\mathbb{P}^n$ split as above? I am reading page $409$ of Principles of Algebraic Geometry by Griffiths and Harris. They explain $$ H^{ \oplus(n+1) } =T\mathbb{P}^n \oplus \mathcal {O}$$ from $\mathbb{C}^{\infty}$ decomposition. Could someone please explain this ?


In Chern class at wiki, let L be a line in $ \mathbb {C} ^{n+1}$ that passes through the origin. It is elementary to see that the complex tangent space to $\mathbb {C} \mathbf {P} ^{n}$ at the point $L$ is naturally the set of linear maps from $L$ to its complement. Thus, the tangent bundle $T\mathbb {C} \mathbf {P} ^{n}$ can be identified with the $hom$ bundle $hom\left( \mathcal {O}(-1),\eta \right) $ where $\eta$ is the vector bundle such that $$\mathcal {O}(-1)\oplus \eta ={\mathcal {O}}^{\oplus (n+1)}.$$ It follows:

$$T\mathbb {C} \mathbf {P} ^{n}\oplus {\mathcal {O}=\operatorname {Hom} ({\mathcal {O}}(-1),\eta )\oplus \operatorname {Hom} ({\mathcal {O}}(-1),{\mathcal {O}}(-1))={\mathcal {O}}(1)^{\oplus (n+1)}}.$$ Could someone please explain that tangent bundle identified with the $hom$ bundle, thanks!

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    $\begingroup$ Split in what category? The Euler sequence (or any sequence of vector bundles) splits in the $C^{\infty}$ category. The Euler sequence does not split in the holomorphic (or algebraic) category. $\endgroup$ – Mohan Oct 14 '16 at 14:51
  • $\begingroup$ The Euler sequence splits in the $C^{\infty}$ category, then $H^{ \oplus(n+1) } =T\mathbb{P}^n \oplus \mathcal {O}$ ? $\endgroup$ – 李章哗啦 Oct 14 '16 at 15:25
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    $\begingroup$ Note that Griffiths and Harris give the short exact sequence (with proof) the line before they talk about the $C^\infty$ splitting. See also this discussion. $\endgroup$ – Ted Shifrin Oct 15 '16 at 0:07

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