# An analytic subset as a singular homology class of a compact manifold

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 $H_{2s}(M,\mathbb{C})?$

• 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. – Donu Arapura Aug 20 '11 at 13:27
• I'm a little confused about your question: as far as I can tell $\sigma$ is an equivalence class of singular $2s$-simplicies, of which $V$ is a representative. Do you secretly have some fixed procedure in mind for producing a representative of $\sigma$ given $\omega$? If so, you should probably include an explanation of that procedure in your question. – Paul Siegel Aug 20 '11 at 13:31
• In fact, I typed my question carelessly and corrected it as above. I am so sorry about this. – vu viet Aug 20 '11 at 14:36

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.
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 $M$.