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My question refers to a discussion from this older thread on Neron-Severi group of a Kähler manifold. In the comments below Ted Shifrin's answer there arose a discussion when the map $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ induced by the canonical inclusion $\mathbb{Z} \subset \mathbb{C}$ may be not injective. As Ted pointed out the main reason that prevents this map being injective is the presence torsion in $H^2(X,\mathbb{Z})$.

Then he continues: "...From the line bundle side, you can see this with flat bundles (whose curvature forms vanish, and so the de Rham representative is the zero form)."

Unfortunately I not understand how his example & argument for line bundles by considering to flat bundles work. Definitely, as the curvature forms of flat bundles vanish, their de Rham representative in $H^2(X,\mathbb{C})$ is zero. That's fine.

Why does it provide an example for torsion elements in $H^2(X,\mathbb{Z})$? Does their de Rham representative not already vanish in $H^2(X,\mathbb{Z})$ as the curvature is zero? But then it not provide an example for a torsion element.

Does anybody understand what Ted had by the quoted remark in mind? I tried to ask but probably that was quite obvious point and I'm just too fool to see it.

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  • $\begingroup$ The de Rham representative is a differential form so does not give an element of $H^2(X, \mathbb{Z})$. $\endgroup$ – ulrich Feb 3 at 6:29
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    $\begingroup$ Lefschetz theorem tells you that any torsion class in $H^2(X,\mathbb{Z})$ is the first Chern class of a line bundle, then necessarily flat. Like the thread you quote, the question would be more appropriate in MSE. $\endgroup$ – abx Feb 3 at 11:11
  • $\begingroup$ @abx:The Lefschetz theorem (at least which I know) says that the first Chern map $c_1: H^1(X,\mathcal{O}_X^*)\rightarrow H^2(X,\mathbb{Z})$ (that associating to a holomorphic line bundle its Chern class $c_1$) induces isomorphism $H^1(X,\mathcal{O}_X^*)/ker(c_1) \cong H^{(1,1)}(X,\mathbb{Z})$, where the right object is given by $H^{(1,1)}(X,\mathbb{Z})=Im(H^2(X, \mathbb{Z}) \to H^2(X, \mathbb{Z})) \cap H^{1,1}(X)$. As $H^1(X,\mathcal{O}_X^*)$ classifies holom line bundles $c_1$ associates as you said first Chern classes to line bundles. (reference: D. Huybrecht's Complex Geometry page 133). $\endgroup$ – user7391733 Feb 3 at 16:24
  • $\begingroup$ One point I not understand: you say that Lefschetz theorem tells that any torsion class $H^2(X,\mathbb{Z})$ that is the first class of a line bundle (in other words is contained in the image of $c_1$,right?), is neccessary flat, right? What I don't understand is that does the asumption imply that a torsion element of $H^2(X,\mathbb{Z})$ which is contained in the image of $c_1$, is already $0$? By def $im(C_1)=H^{(1,1)}(X,\mathbb{Z})=Im(H^2(X, \mathbb{Z}) \to H^2(X, \mathbb{Z})) \cap H^{1,1}(X)$ says that image of $c_1$ has no torsion. Or do I misunderstand your point? $\endgroup$ – user7391733 Feb 3 at 16:24
  • $\begingroup$ ps: yes, you are right that this question has indeed a MSE-level. I have tried to ask it to weeks ago math.stackexchange.com/questions/3519053/… but unfortunately it doesn't get much attention. $\endgroup$ – user7391733 Feb 3 at 16:25

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