4
$\begingroup$

Let $X$ be a complex manifold and $\mathcal{L}$ be a positive line bundle on $X$. If $E$ is any other line bundle on $X$, then is it true that for all sufficiently large $m$, $\mathcal{L}^m \otimes E$ is also positive?

When $X$ is compact, the answer is positive, and it follows by a standard compactness argument if you start with the definition that $\mathcal{L}$ is positive iff the Chern class $\omega$ of $\mathcal{L}$ satisfies: $\omega(x; v, Iv) > 0$ for all $x \in X$ and $v \in T_{\mathbb{R}, x}(X)$ (the real tangent space of $X$ at $x$) and $I: T_{\mathbb{R}, x}(X) \to T_{\mathbb{R}, x}(X)$ is the map induced by multiplication by $i$.

So my real question is: is the above question true when $X$ is not compact? What if $X$ is an affine algebraic variety?

$\endgroup$
7
  • $\begingroup$ What bundles do you call positive? For example, what is a positive bundle on C^1? $\endgroup$ Sep 6, 2010 at 9:25
  • $\begingroup$ I guess he means a bundle equipped with some hermitian metric with positive curvature. $\endgroup$
    – Henri
    Sep 6, 2010 at 9:30
  • $\begingroup$ Well - I stated one definition in the second paragraph :) In any case, it is equivalent to the following definition of Griffiths-Harris: a line bundle $\mathcal{L}$ on $X$ is positive iff there exists a hermitian metric on $\mathcal{L}$ with curvature form $\Theta$ such that $\frac{i}{2\pi}\Theta$ is a positive (1,1) form. $\endgroup$
    – pinaki
    Sep 6, 2010 at 10:17
  • $\begingroup$ @Dmitri: Could you please explain a bit? E.g. what would be a Kahler metric of positive curvature on the trivial bundle over $\mathbb{C}^2$? $\endgroup$
    – pinaki
    Sep 6, 2010 at 12:34
  • 3
    $\begingroup$ @Dmitri: It is not true that line bundles on affine varieties are trivial: if $X$ is a compact Riemann surface and $P$ is a point on $X$, then $\mathrm{Pic}(X\setminus\{P\})$ is the quotient of $\mathrm{Pic}(X)$ by the subgroup generated by $\mathcal{O}_X(P)$. $\endgroup$ Sep 6, 2010 at 15:12

1 Answer 1

2
$\begingroup$

Let us prove that for an affine variety $X$ every line bundle $E$ is "positive" according to the chosen defintion. All we need to prove is that for any hermitian metric $g$ on $E$ with curvature $w$ there is a Kahler form $w_1$ on $X$ such that $w_1>-w$. Since $X$ is affine, for any $w_1$ we have $w_1=\frac{i}{2\pi}\partial\bar\partial (f_1)$ and changing the metric $g$ on $E$ by $ge^{f_1}$ we corresponing curvature will change from $w$ to $w+w_1$, which we assume to be positive.

So we need to show the existence of arbitrary large $w_1$. Since $X$ is affine and hence admits an embedding in $\mathbb C^n$, it is enough to show this for $\mathbb C^n$. Moreover, since $\mathbb C^n=\mathbb C^1\times ...\times \mathbb C^1$ it is enought to prove the statement for $\mathbb C^1$. Now, on $\mathbb C^1$ every form of the shape $w_1=h_1dz\wedge d\bar z$ is Kahler for $h_1>0$ and we can chose $h_1$ as large as we wish.

The conclusion is that if one choses this definition, then each line bundle on an affine $X$ is positive, which sounds strange. So I am not sure what should be a reasonable definition of positivincess in non-compact case, if it exists at all.

$\endgroup$
3
  • $\begingroup$ One (probably stupid) question: why is it true that every Kahler $(1,1)$ form on an affine variety is of the form $\frac{i}{2\pi} \partial \bar \partial(f)$ for a global function $f$? Is it easy to see? $\endgroup$
    – pinaki
    Sep 8, 2010 at 11:14
  • $\begingroup$ Auniket, I should look for a refference. Notice though that for the proposed reasoning it is enough to prove this fact for any rotation invariant form on $\mathbb C^1$, this can be done by hands. $\endgroup$ Sep 8, 2010 at 12:06
  • 1
    $\begingroup$ Dmitri:Your proof goes through for any stein manifold.All you need is a strictly plurisubharmonic function whose Levi form dominates the curvature of the holomorphic vector bundle.You can do this by taking a strictly plurisubharmonic exhaustion function and compose with a suitable convex increasing function. $\endgroup$ Sep 11, 2010 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.