Talk:Positive form

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

'"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)[reply]

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)[reply]