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Let $V$ be a compact connected complex analytic subvariety (possibly singular) of a complex smooth manifold. Let $n$ denote its complex dimension.

Is it true that $H_{2n}(V,\mathbb{Z})\simeq \mathbb{Z}$?

A reference would be helpful.

Remark. I believe that the answer is positive, but have no idea how to prove it (I am not a specialist.) Is it a difficult result?

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    $\begingroup$ For a smooth submanifold this is obvious (because a complex manifold is orientable). That is, the question is actually about the singular? $\endgroup$
    – Arshak Aivazian
    Commented Mar 23, 2023 at 8:46
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    $\begingroup$ @AivazianArshak: Sure, the main case is of singular variety. $\endgroup$
    – asv
    Commented Mar 23, 2023 at 9:12

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This is indeed the case. In fact in the general case (possibily non compact) an irreducible complex analytic subvariety V has a fundamental class in Borel-Moore homology (in singular homology when $V$ is compact)

$$ [V]\in H_{2n}^{BM}(V;\mathbb{Z})\cong \mathbb{Z}.[V] $$

see for example "La classe d'homologie fondamentale d'un espace analytique" by Borel and Haefliger (BSMF, 1961). Moreover if you consider a resolution of singularities $$ \pi_*:\tilde{V}\rightarrow V $$ Then you get an isomorphism $$\pi_{*}:H^{BM}_{2n}(\tilde{V};\mathbb{Z})\rightarrow H^{BM}_{2n}(V;\mathbb{Z}),\:\pi_*([\tilde{V}])=[V].$$

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  • $\begingroup$ I read "connected" not "irreducible" in the question, so the answer seems obviously "no". of course you may be assuming the word "variety" implies irreducible, but then why add "connected"? $\endgroup$
    – roy smith
    Commented Mar 29, 2023 at 0:48
  • $\begingroup$ @DavidC: Do you have a reference for the last statement (even if only in the case where $V$ is compact)? $\endgroup$
    – Michael Albanese
    Commented Apr 5, 2023 at 0:59
  • $\begingroup$ @MichaelAlbanese : chapter 19 in Fulton's book "Intersection theory". Also section 3 of Totaro's paper "Torsion algebraic cycles and complex cobordism". $\endgroup$
    – David C
    Commented Apr 5, 2023 at 14:18

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