TITLE: Tangent bundle of P^n and Euler exact sequence
QUESTION [7 upvotes]: I'm thinking about the Euler exact sequence for complex projective space, and I'm a bit confused. In the topological category, one has 
$$ T\mathbb{P}^n \oplus \mathbb{C} \cong L^{n+1} $$ 
where $L$ is the tautological line bundle and $\mathbb{C}$ is the trivial line bundle. This comes from a splitting of the vector bundle homomorphism $\mathbb{C}^n \rightarrow \mathbb{C}^n/L$ which you can get by picking a Hermitian metric. 
In algebraic geometry, you get a (non-split) exact sequence, which is the dual of the Euler sequence (http://en.wikipedia.org/wiki/Euler_sequence). 
Is the Euler sequence for the holomorphic tangent bundle of $\mathbb{P}^n$ split as above? It would seem so, since even though we are in the holomorphic setting we can run the first argument with the Fubini-Study metric, but I've read some things that seem to suggest otherwise.

REPLY [9 votes]: No. If you consider the Euler exact sequence for $\mathbb{CP}^1$, we have the sheaf of 1-form is the canonical sheaf for $\mathbb{CP}^1$ which is $O(-2)$. By Euler exact sequence ,
$$0 \rightarrow O(-2) \rightarrow O(-1)^2\rightarrow O\rightarrow 0$$.
However, $O(-2)\oplus O$ is not isomorphic to $O(-1)^2$ as algebraic vector bundle (or locally free sheaf if you like) since the former has global holomorphic section but the latter doesn't.