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We know every differential manifold can be triangulable. Let $M$ be a compact complex manifold of dimension $m$ and V be an analytic subset of dimension $s$ of $M.$ If $V$ has no singularity then $V$ is a compact complex submanifold of $M.$ Hence, V can be considered as an element of $H_{2s}(M,\mathbb{C})$ (singular homology of M) for $V$ can be triangulable and compact. Now, consider the general case when V has singularity, as far as I know in general V is not triangulable.

Besides, it is well-know that the analytic set $V$ has the Poincare duality $\omega$ in $H_{DR}^{2m-2s}(M)$ (De rham cohomology of $M$), and again $\omega$ has the Poincare duality $\sigma \in H_{2s}(M,\mathbb{C}).$ That means there exists $2s^{th}$ singular homology chain $\sigma$ such that for all 2s-form $\eta$ one has $$\int_V \eta =\int_{\sigma} \eta.$$
Question: What is the geometric relation between $V$ and $\sigma$? On the other hand,

How can $V$ be considered geometrically as an element of $H_{2s}(M,\mathbb{C})?$

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  • $\begingroup$ As you already realize, $V$ determines a class $[V]$ in homology called the fundamental class of $V$. Dualizing twice gives you back $[V]$, i.e. $\sigma=[V]$. I suspect your question is "how should we understand $[V]$?" If you triangulate $M$ so that $V$ is a subcomplex, then $[V]$ is represented by a simplicial chain supported on $V$. Alternatively, the dual class $\omega$ is represented by a $2m-2s$ form compactly supported in a tubular neighbourhood of $V$. Or you can also use currents etc. $\endgroup$
    – Donu Arapura
    Commented Aug 20, 2011 at 13:27

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OK, I see. You're worried about the case when $V$ has singularities. Here are a number of things that you can do:

  1. $V$ is still triangulable. This goes back Lojasiewicz, I think. So you can still represent the fundamental class by a simplical chain as before.

  2. Use currents to represent the class (cf. Griffiths- Harris pp 366-400)

  3. Look up Atiyah-Hirzebruch, "Analytic cycles on complex manifolds", Topology 1, 1961

  4. (My personal favourite, although some consider this overkill.) Choose a resolution of singularities $\pi_:\tilde V\to V$, and push the fundamental class $[\tilde V]$ to $M$.

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    $\begingroup$ The fact that V is still triangulable can be seen as a standard consequence of Gabrielov's theorem of the complement; I do believe it was first proved by Lojasiewicz indeed. $\endgroup$
    – Thierry Zell
    Commented Aug 20, 2011 at 17:12
  • $\begingroup$ In fact, it is necessary to have some kind of triangulation, so to speak, a "differentiably" one. By that ones can intergrate in simplical chain. In my mind, that thing can be doned from Lojasiewicz's work. Just one more thing, it seems that some fundemental facts in triangulation of manifolds are rarely recalled or refered in books of AG or Algebraic topology. $\endgroup$
    – vu viet
    Commented Jun 14, 2012 at 15:43

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