# Tangent bundle of P^n and Euler exact sequence

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.

No, any construction using the Hermitian metric is going to take you outside the holomorphic category! By the way, you can understand the Euler sequence very elegantly by mapping $\Bbb P^n\times \Bbb C^{n+1}\to T\Bbb P^n\otimes\mathscr L$ as follows: If $\pi\colon\Bbb C^{n+1}-\{0\}\to\Bbb P^n$ and $\pi(\tilde p) = p$, for $\xi\in T_{\tilde p}\Bbb C^{n+1}$, map $\xi$ to $\pi_{*\tilde p}\xi\otimes\tilde p$, and check this is well-defined.
You might also try to generalize the Euler sequence to a complex submanifold $M\subset\Bbb P^n$, letting $\tilde M = \pi^{-1}M$. Then you get the exact sequence $$0\to \mathscr L \to E \to TM\otimes\mathscr L\to 0\,,$$ where $E_p = T_{\tilde p}\tilde M$ for any $\tilde p\in \pi^{-1}(p)$. ($\tilde M$ is the affine cone corresponding to $M\subset\Bbb P^n$.)
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.