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These are probably trivial questions... (for the experts)

I'd like to get convinced (perhaps an intuitive/geometric explanation will be more effective than a formal one) of the following facts:

i. the fibre of $\pi:\mathcal{O}(1)\to\mathbb{P}^n$ over $\ell \in \mathbb{P}^n$ is canonically isomorphic to $\ell^*.$ The global coordinates $z_0,\ldots,z_n$ on $\mathbb{C}^{n+1}$ define natural sections of $\mathcal{O}(1)$ (here $\mathcal{O}(1)$ denotes the dual of the tautological line bundle)

ii. from the short exact sequence of holomorphic vector bundles $$ 0 \to E \to F \to G \to 0$$ show there exists a canonical isomorphism $\det F \simeq \det E \otimes \det G$

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i. I guess the easiest way to see that is in the following fashion. Let's denote $\mathcal O(-1)$, the tautologial bundle on $\mathbb P^n$. You can define it as the subvariety of $\mathbb P^n \times \mathbb C^{n+1}$ consisting of elements of the form $(d,x)$, where x is an element of the line $d$.

Over $\mathbb P^n$ you can consider the trivial bundle of linear forms over $\mathbb C^{n+1}$. A holomorphic section of that bundle is of course given by $f_0\phi_0+...+f_n \phi_n$ where $f_i$ are holomorphic functions over $\mathbb P^n$ which are constants and $\phi_i$ is just the projection on the i-th component on $\mathbb C^{n+1}$.

As $\mathcal O(-1)$ is a subbundle of $\mathbb P^n\times \mathbb C^{n+1}$, the dual of $\mathcal O(-1)$ is a quotient of that trivial bundle, where the arrow is given simply by restricting a (local) section $h_0\phi_0+...+h_n\phi_n$ to the fiber of the tautological bundle. This tells you that $a_0\phi_0+...+a_n\phi_n$ give you global sections of $\mathcal O(1)$. Now it doesnt prove that all the sections of $\mathcal O(1)$ arise in this manner because, as you must know, taking global sections is not a right-exact functor.

You can prove that this is indeed the case, but what i said already proves that you get a subspace of sections of $\mathcal O(1)$ by considering $\mathbb{C} \phi_0+...+\mathbb C\phi_n$, usually we just denote them by $x_i$ instead of $\phi_i$.

ii. Yes it is true, just one remark though, is you think of the bundle $E\oplus F$ it induces two obvious exact sequences and the isomorphism $\det(E\oplus F)\simeq \det E \otimes \det F$ depends on which you choose and will differ by a sign if you choose to swap $E$ and $F$.

It is the exact analog of the fact that the determinant of a bloc-diagonal matrix is the product of the determinant of the blocs.

Locally it is just defined by $(e_1\wedge ...\wedge e_p)\otimes (f_1\wedge ...\wedge f_k)\mapsto e_1\wedge ...\wedge e_p\wedge f_1\wedge ...\wedge f_k$, where strictly speaking you should take preimages of the $f_i$'s and check that the definition does not depend on this particular choice.

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  • $\begingroup$ i: Well, in my answer i talk about a surjective arrow $(\mathcal O_{\mathbb P^n})^{n+1}\to \mathcal O(1)$ whose kernel is $\Omega^1_{\mathbb P^n}\otimes \mathcal O(1)$ (i dont mention the kernel, though, you're right), so the right part of the Euler sequence. ii: I just wanted to mention that this isomorphism is not compatible with the commutativity of tensor product, but you're right it is canonical. I'll make changes accordingly, thanks. $\endgroup$ – Ahr Feb 26 '14 at 7:59

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