Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

up vote 7 down vote accepted

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 $M$.

share|improve this answer
1  
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. –  Thierry Zell Aug 20 '11 at 17:12
    
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. –  vu viet Jun 14 '12 at 15:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.