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Let $X$ be a compact manifold which admits an embedding in the projective space and let $\pi: Y \to X$ a projective bundle on it. I'm trying to prove that also $Y$ is algebraic (embeddable in projective space) but I do not really have idea how to do it(without using overkill method such as GAGA).

I tried to pullback the Kahler metric on $Y$ in a concrete way but did not really make up with anything.

Tommaso Scognamiglio
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  • Duplicate of this? – user10354138 Jun 03 '19 at 08:43
  • The answer there is pretty much scheme theoretic. I was looking for a proof much more standing to complex geometry – Tommaso Scognamiglio Jun 03 '19 at 08:49
  • If $E \to X$ is the original vector bundle, the tautological line bundle $\mathcal{O}{\mathbb P(E)}(1) \to \mathbb{P}(E)$ will be positive on the fibers of the projection $\pi$ to $X$. Since $X$ is projective, it has a positive line bundle $L$, and as everything is compact a sufficiently high multiple of it will make $\mathcal O{\mathbb P(E)}(1) \otimes m \pi^* L$ be positive, thus ample. – Gunnar Þór Magnússon Jun 03 '19 at 11:19
  • I think I understood but why do we need the exponent on the pullback of $L$? To be clear, let's call $\alpha$ the curvature on $\mathcal{O}_{\mathbb{P}(E)}(1)$ which makes it positive on the fiber and let's call $\beta$ the pullback curvature of $\pi^*L$. When I take a vector in $T(\mathbb{P}(E))$ it is decomposed as $(v,w)$ with $v \in T_x(X)$ and $w \in T(\mathbb{P}(E_x))$. It should be true that $\alpha(v,u)=0$ for every other $u \in T(\mathbb{P}(E))$ while $\beta(w,u)=0$, so that we should get positivity of $\alpha +\beta$ – Tommaso Scognamiglio Jun 03 '19 at 13:10
  • Hope I've been not to confusing on the comment – Tommaso Scognamiglio Jun 03 '19 at 13:14
  • The metric on the tautological line bundle on the projective bundle is defined on the whole of its tangent space, but we only know that it is positive on the relative tangent space. For the rest, we need to offset the non-positivity by something, hence powers of $L$. – Gunnar Þór Magnússon Jun 06 '19 at 14:08
  • @Gunnar Þór Magnússon, if we generalize the fiber of $\pi$ from projective spaces to projective manifolds $P$ with $H^i(P,\mathcal O)=0, i>0$, is it still true that the total space $Y$ is projective? Someone said it is true but I don't know why, do you have any clue? – Tom Oct 26 '20 at 08:50
  • @Tom Sorry, I don't know. I don't see an obvious metric to put on the relative canonical bundle there, so the proof might proceed by more algebraic arguments. – Gunnar Þór Magnússon Oct 26 '20 at 09:37
  • @Gunnar Þór Magnússon, actually this question arise from Cao Junyan's paper 《On the approximation of Kähler manifolds by algebraic varieties》p404 remark 8, in this paper, $Y$ is a finite cover of a compact Kähler manifold $Y'$ with a semipositive Ricci curvature, so $Y$ is Kählerian, and the Kähler metric $\omega$ of $Y'$ induces a metric $\omega_Y$ of $Y$, and he claims its restrictions to the fiber $P$ satisfies $\omega_P \in H^{1,1}(P,\mathbb Q)$(I don't see why) and is strictly positive,did it provide the metric you need? – Tom Oct 26 '20 at 11:00
  • @Gunnar Þór Magnússon, and the fiber: projective manifold $P$ still satisfies $H^i(P,\mathcal O)=0,i>0$. – Tom Oct 26 '20 at 11:11
  • @Tom: I don't think it does out of the box. The metric Junyan (my old PhD brother!) gets is only semipositive on the fibers; it might be zero somewhere. There's some more work to be done. (Btw, he gets that $\omega_P$ is $(1,1)$ because the $(2,0)$ cohomology is zero by Remark 7 above; and it's positive because the restriction of a positive $(1,1)$-form to a subvariety is positive.) – Gunnar Þór Magnússon Oct 27 '20 at 10:14
  • @Gunnar Þór Magnússon, thanks a lot for the very helpful comments about your old phD brother's paper, from your comment, we can get $\omega_P$ is $(1,1)$ and positive, but in Junyan's paper, he stated "thanks to remark 7, we can suppose that $\omega_P\in H^{1,1}(P,\mathbb Q)$", did you see how to get the rational coefficients $\mathbb Q$ from remark 7? – Tom Oct 27 '20 at 14:33
  • @Tom: Oh right. Hmm. That might be subtle. Wanna continue this over email? [email protected] – Gunnar Þór Magnússon Oct 27 '20 at 17:48
  • @Gunnar Þór Magnússon, great! that's so kind of you! – Tom Oct 27 '20 at 17:52

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