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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.

Tony
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2 Answers2

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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.

Ben
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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 $M\subset\Bbb P^n$.)

Ted Shifrin
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