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I was wondering if there were restrictions in what the cohomology classes corresponding to complex submanifolds of a complex manifold could be.

For example, say $T^4$ is regarded as a complex manifold as $\mathbb{C}^2/\mathbb{Z}[i] \times \mathbb{Z}[i]$, we have $H^2(T^4) = \Lambda^2(\mathbb{Z}^4) =\mathbb{Z}^6$. If $C$ is a ($1$ complex dimensional) complex submanifold, and we let $[C] \in H^2(T^4)$ be the corresponding cohomology class, could $[C]$ be any cohomology class in $H^2(T^4)$, or are there some restrictions we could impose just by knowing that it is a complex submanifold?

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    $\begingroup$ Your question is ambiguous. Are you asking about properties of cohomology classes of complex subvarieties of $(\mathbb{C}/\mathbb{Z}[i])^2$, in which case the answer is well-known (Lefschetz theorem), or of classes of submanifolds which are complex subvarieties of $T^4$ for some complex structure of $T^4$? (also well-known, but might be more interesting in general). $\endgroup$
    – abx
    Commented Apr 26, 2020 at 6:33
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    $\begingroup$ en.wikipedia.org/wiki/Hodge_conjecture $\endgroup$
    – Denis Nardin
    Commented Apr 26, 2020 at 6:52
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    $\begingroup$ This is related to the Steenrod realizability problem, though that's for the laxer question of representing classes by oriented manifolds. $\endgroup$
    – Arun Debray
    Commented Apr 26, 2020 at 18:53
  • $\begingroup$ @abx I just meant cohomology classes of complex subvarieties of $(\mathbb{C}/\mathbb{Z}[i])^2$. Thanks for letting me know about the condition that it has to be in $H^{1,1}(T^4,\mathbb{C}) \cap H^2(T^4,\mathbb{Z})$. I'm sure I'm being very stupid, though, but is it really true that any element in the subgroup generated by complex subvarieties can itself be represented by a complex subvariety? I can see that if C_1 and C_2 are subvarieties, then [C_1] + [C_2] is represented by an actual subvariety, but is the same true of [C_1] - [C_2]? $\endgroup$
    – Stella Dubois
    Commented May 4, 2020 at 19:14
  • $\begingroup$ No, of course. What I meant is that this is a much studied problem, with proven methods to decide. Look at "divisors and line bundles" in any algebraic geometry book. $\endgroup$
    – abx
    Commented May 4, 2020 at 19:40

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