Talk:Positive form

Please check the signs, I tend to mix them up. Tiphareth 17:13, 11 April 2007 (UTC)

Should Condition 1 for a real positive (1,1)-form not be that $$\omega$$ is the imaginary part of a hermitian form, not $$i \omega$$? (The imaginary part of $$a+ib$$ is $$b$$, not $$ib$$.) -- Hiferator (talk) 20:11, 30 March 2015 (UTC)
 * According to Lazarsfeld's "Positivity in Algebraic Geometry I", $$\omega = - \operatorname{Im} H$$ for some positive-defninite hermitian metric $$H$$. So I will change that.
 * Lazarsfeld seems to be using a slightly stronger notion of positivity. (He also states an equivalent property to 2. with strictly positive $$\alpha_i$$.) Is there a canonical way to differentiate the two, e.g. calling one strictly positive? I think both notions should be mentioned in the article. -- Hiferator (talk) 20:52, 30 March 2015 (UTC)
 * By the way, is there a particular reason for writing $$\sqrt{-1}$$ instead of $$i$$ for the imaginary unit? To me using $$i$$ looks more clear. -- Hiferator (talk) 21:06, 30 March 2015 (UTC)

A down to earth interpretation
Is this statement true? "Suppose that a divisor $D$ on $M$ generates a positive line bundle $L$. Then if $a$ is any homology class of $M$ that can be represented by a embedded Riemannian surface (perhaps with singularities), then $Da$ is greater or equal to zero."

in case this question is true, is this the main point of positive line bundles? That is, that if it has a divisor representing it, then it will have positive intersection with any other class that is represented by a complex submfld.

ELSE: "If L is represented by a divisor, then $\langle c_1(L),[N] \rangle \geq 0$ where $N$ is a complex subvariety. Hence, if we want to know if $L = [D]$ for some divisor $D$ the first thing to check is if $L$ is positive since every $[D]$ is so."

is this the point of positive line bundles? 14:49, 27 February 2008 (UTC) — Preceding unsigned comment added by 155.198.157.118 (talk)


 * The second of the above statements is False. I capitalized "false" because there are easy counterexamples.  Blow up any complex surface, say $$CP^2$$, at a point.  Let E be the exceptional set.  Let L=[E].   Then $$\langle c_1(L),[E] \rangle =-1$$.


 * I think the usual definition of positive is that $$c_1(L)$$ is represented by a positive definite, ie Kähler form $$\omega$$, rather than just positive semi-definite. —Preceding unsigned comment added by 76.24.20.200 (talk) 08:39, 10 October 2008 (UTC)


 * '"I think the usual definition of positive is that c_1(L) is represented by a positive definite"' -- this is correct, thanks for pointing this out. My error. Fixed. Tiphareth (talk) 07:33, 12 October 2008 (UTC)

Contradiction between 2. and 3.
In the definition of a positive (1,1)-form, conditions 2 and 3 do not seem to be equivalent — they should either both use $$\sqrt{-1}$$ or they should both use $$-\sqrt{-1}$$. -- 2001:1711:FA4B:E5C0:388B:6291:6B8F:C58B (talk) 17:59, 9 January 2022 (UTC)