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?

  • $\begingroup$ 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
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    $\begingroup$ @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


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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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