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Consider the projective variety $X = \mathbb{P}^2$, and the line bundle $\mathcal{O}_X(dH)$ where $H$ is a plane and $d \in \mathbb{N}$.

Let $L$ be the total space of $\mathcal{O}_X(dH)$. I know how to form this in the following way: Let $U_i$ be the standard affine opens of $X$, then $$ L = \coprod (U_i \times \mathbb{A}^1)/\sim $$ where the equivalence is given by the glueiing matrices $A_{ij}= (x_j/x_i)^d$.

I am pretty sure the total space is a quasi-projective variety. Can anybody give me an explicit embedding?

For my purposes i am happy with the cased $d \in \{1,2\}$, but a general treatment would be nice.


Edit: Projective changed to quasi projective.

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up vote 2 down vote accepted

So let me start again. If $\mathcal M$ is a coherent locally free sheaf on $X$, then the total space of the associated vector bundle $M$ is the affine $X$-scheme $$M =\mathrm{Spec} (\mathcal{Sym}_{O_X}(\mathcal M^{\vee}))\to X.$$ (the cech means dual.) This is an affine morphism, hence quasi-projective. As $X$ has an ample sheaf, this implies that $M$ admits an open immersion into some $\mathbb P^n_X$ (EGA, II.5.3.3).

This immersion can be "explicitely" constructed as follows. Let $n\ge 1$ be big enough so that $\mathcal M^{\vee}(n)$ is generated by its global sections. Write $$ O_X^m \twoheadrightarrow \mathcal M^{\vee}(n), \quad O_X^3\twoheadrightarrow O_X(n)$$ (If $T_0, T_1, T_2$ is a basis of $H^0(X, O_X(1))$, then $O_X(n)$ is generated by $T_0^n, T_1^n, T_2^n$). Then $$\mathrm{Spec} (\mathcal{Sym}_{O_X}(\mathcal M^{\vee})) \hookrightarrow \mathbb P(\mathcal M^{\vee}\oplus O_X)\simeq {\mathbb P}(\mathcal M^{\vee}(n)\oplus O_X(n)) \hookrightarrow {\mathbb P}(O_X^{m+3})={\mathbb P}^{m+2}\times X.$$

When $\mathcal M=O_X(d)$, we can take $n=d$ and $m=1$. Hence an immersion $$L \hookrightarrow {\mathbb P}^{3}\times X.$$

Now to write the map in your chart: over $U_i$, the map is $$(\lambda_i, [x_0, x_1, x_2])\mapsto ([\lambda_i x_i^d, x_0^d, x_1^d, x_2^d], [x_0, x_1, x_2]).$$

Le glueing map on $U_i\cap U_j$ is $\lambda_i x_i^d=\lambda_j x_j^d$.


Below is the embedding for the total space of $O_X(-d)$.

Let $e_1,\dots, e_n$ be a basis of $H^0(X, O_X(d))$. This means that the map $$O_X^n\to O_X(d), \quad (f_1,\dots, f_n)\mapsto \sum_i e_if_i$$ is surjective. So $L$ is a closed subvariety of $\mathbb A^n\times X$ and is therefore quasi-projective.

This holds for any quasi-projective variety $X$ and any line bundle $L$ on $X$ generated by its global sections.

EDIT Please forget the followings lines.

It can't be. Because otherwise you would have a non-constant morphism from $\mathbb P^3$ (your total spae has dimension $3$) to $\mathbb P^2$. But such a morphism doesn't exist (this is an exercice in Hartshorne: there is no non-constant morphism from $\mathbb P^n$ to $\mathbb P^m$ if $n>m$).

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Thanks @QiL. However, i don't think the argument works here. Say our space $L$ is embeddable as $L \subset \mathbb{P}^n$. We would then have a surjective morphism $\phi: L \rightarrow \mathbb{P}^2$. By your argument, this would not be extendable to a morphism on the whole of $\mathbb{P}^n$, it would be a rational map. But this is fine, and does not contradict the existence of our morphism $\phi$. Take for example a surface geometrically ruled over $\mathbb{P}^1$. – Joachim Oct 3 '12 at 4:57
@Joachim: I am really sorry. It was late and I read "projective space" instead of "projective variety", and I though "projective bundle" instead of "vector bundle". – user18119 Oct 3 '12 at 6:19
Don't be sorry @QiL it was nice to learn anyway, and thanks for your help. I do have some more to ask. I'm having trouble how "So $L$ is a closed subvariety ..." is implied by the preceding. I'm trying to work out the example $d=1$: finding the equation for the closed subset $M \subset \mathbb{P}^2 \times \mathbb{A}^3$ and the isomorphism $L \rightarrow M$ where we consider $L$ as the space obtained by glueing as above. Could you help me a bit on this? (i would be happy with hints, i think i can work it out, however if you feel like working it out completely i wouldn't mind of course =) – Joachim Oct 3 '12 at 8:19
Another problem: We know $L$ has global sections, with a zero along a hyperplane. Say $L$ is embeddable in $X \times \mathbb{A}^n$. Then the sections would correspond to functions $X \rightarrow X \times \mathbb{A}^n$ which composed with some projections would yield funtions $X \rightarrow \mathbb{A}^1$. The only such functions are constants. Hence our section must be constant, in particular nowhere vanishing, a contradiction. I fear that your procedure only works if $X$ itself is (quasi)-affine, is this true? Or is my reasoning wrong? – Joachim Oct 3 '12 at 12:40
@Joachim: sorry again. I mixed up, as usual, the total space of $O_X(d)$ with that of $O_X(-d)$. Hopefully the third tentative is correct. – user18119 Oct 3 '12 at 22:57

If $E$ is a vector bundle of rank $r$ on a projective variety $X$, then the total space of $E$ is a quasi-projective variety : this is a vast generalization of what you are asking.

And yet the proof is easy: the vector bundle $E$ has a projective completion $\bar E=\mathbb P(E\oplus \theta)$ where $\theta=X\times \mathbb A_k^1$ denotes the trivial vector bundle of rank one on $X$.
This bundle is a locally trivial bundle $p:\bar E\to X$ with fiber $\mathbb P^r$ and is a projective variety.
And finally the original vector bundle $E$ admits of an open embedding $$E\stackrel {\cong}{\to}\mathbb P(E\oplus 1) \stackrel {\text {open}}\subset \mathbb P(E\oplus \theta)=\bar E$$ into the projective variety $\bar E$, which shows that $E$ is indeed quasi-projective.

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Thanks @George, that is really nice to know. However, my question was about an explicit embedding of the bundle into some projective space (which should factor as an open immersion followed by a closed one). So can you give such an embedding for $\overline{E}$? If possible a general construction, or else just the case $\mathcal{O}_{\mathbb{P}^2}(1)$? That would be great. – Joachim Oct 3 '12 at 11:22
Add: I meant of course the case where we take the projective closure of $E= \mathcal{O}_{\mathbb{P}^2}(1)$.. – Joachim Oct 3 '12 at 11:46

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