7
$\begingroup$

What's the standard generalization and reference for the following statement:

If two oriented submanifolds $L$, $L'$ of an oriented compact manifold $M$ intersect transversally, then the Poincare dual of the fundamental class $[L\cap L']$ equals to the cup product of the Poincare duals of the fundamental classes $[L]$ and $[L']$.

It seems, in algebraic geometry $L$, $L'$ and $L\cap L'$ don't have to be smooth, it is only required that at a generic point of $L\cap L'$ we have that $L$ and $L'$ are smooth and intersect transversally. In the case of non-transversal intersection sometimes more can be said (e.g. multiplicities appear).

I am asking for a topological version of this.

To exclude proofs with differential forms, let's say we want homology/cohomology with integer or arbitrary coefficients.

In fact, I thought of a sketch of a theory along the following lines (in style of Hatcher's book and adopting an approach from Voisin's book):

First define inductively a class of "good subspaces" by saying that a closed subspace $L\subset M$ in a manifold $M$ is a good subspace of dimension $d$ if there is open $U\subset L$ such that $U$ is a submanifold of $M$ and $L\setminus U$ is a good subspace of dimension $d-1$. Let us call manifolds $U\subset L$ for which $L\setminus U$ have smaller dimension generic.

An orientation on a good space is simply an orientation on some generic subset.

Suppose $M$ is oriented and $L$ is an oriented good subspace of $M$ of codimension $n$. We probably can prove that $H^i(M, M-L)=0$ for $i<n$ and that there is at most one element $c_L\in H^n(M, M-L)$ which restricts to the correct orientation class on $H^n(M-(L-U), U)$. We call this class the cycle class if it exists.

If $L$ has a cycle class in a manifold $M$ and $L$ is compact, its fundamental class in $M$ is defined by the cap product $[L]=c_L\cap [M]$ of the cycle class of $L$ with the fundamental class $[M]\in H_N(M,M-L)$ ($N$ is the dimension of $M$, $N=d+n$). In case $L$ has a fundamental class in $H_d(L)$, e.g. if $L$ is a manifold, its pushforward to $M$ coincides with $[L]$.

Theorem. Suppose cycle classes exist for good oriented spaces $L$, $L'$ in an oriented manifold $M$. Suppose the intersection $K=L\cap L'$ is a good space, suppose there exist generic sets $U\subset L$, $U'\subset L'$, such that the intersection of $U$ and $U'$ is transverse and generic in $K$. Then cycle class of $K$ exists and $c_K=c_L\cup c_{L'}$.

In the case of compact $M$, the Poincare dual class of $L$ is the image of $c_L$ under the natural map $H^n(M,M-L)\to H^n(M)$, so the statement for Poincare dual classes follows the Theorem.

Do you see any problem with this approach, and has something like this been worked out in details in the literature?

$\endgroup$
  • 1
    $\begingroup$ Pseudo-manifolds are a standard way of defining "good" subspaces, but they are somewhat different than your definition. Roughly, a pseudo-manifold $L$ is a space such that, away from a codimension 2 set $U$, it is a manifold (a precise definition uses triangulation). There is no requirement that $U$ be a pseudo-manifold. It satisfies $H_{\operatorname{dim}(L)}(L) = \mathbb{Z}$, so that there is, up to orientation, a unique fundamental class. $\endgroup$ – Jesse Silliman Mar 15 at 16:57
  • $\begingroup$ Pseudomanifold is required to have a triangulation. In particular it has a filtration by closed sets $M_0\subset M_1\subset \cdots$ such that $M_i-M_{i-1}$ is a manifold. So pseudomanifolds satisfy my conditions. $\endgroup$ – Anton Mellit Mar 19 at 9:57
  • $\begingroup$ @JesseSilliman I think it all fits: we develop the general theory for good spaces, but then to begin intersecting we need to start with spaces for which cycle class exists. I see 2 examples when we know it exists: 1) pseudo-manifolds 2) good spaces $L$ for which $L-U$ has codimension $2$ in $L$, for instance complex analytic sets. $\endgroup$ – Anton Mellit Mar 19 at 10:09
  • $\begingroup$ Good point about triangulations. But what is the purpose of discussing a good space for which the cycle class doesn't exist? Compact complex analytic spaces are also pseudo-manifolds. $\endgroup$ – Jesse Silliman Mar 19 at 21:58
  • $\begingroup$ @JesseSilliman compact yes. Non-compact analytic spaces also have cycle classes. Maybe they can be triangulated, maybe not. In any case it is probably not so easy to prove. Suppose two pseudomanifolds intersect generically transversely. Is the intersection a pseudomanifold? $\endgroup$ – Anton Mellit Mar 20 at 15:20
4
$\begingroup$

You should look at "Lectures on algebraic topology" by A. Dold chapter VIII section 13.

A more intuitive approach, closer to H. Poincaré's viewpoint and to the intersection product of algebraic geometers, is to use singular manifolds to represent cycles. For example, one can use Kreck's stratifolds in order to get a very geometric definition of the intersection product (transversality holds in this framework), you can look at M. Kreck's book "Differential Algebraic Topology".

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ I second Dold. I think there's also a section about this in Bredon's Topology and Geometry, which is a bit more modern. $\endgroup$ – Greg Friedman Mar 15 at 19:36
  • 1
    $\begingroup$ Harper and Greenberg's book does this very carefully too. $\endgroup$ – Nicholas Kuhn Mar 16 at 2:34
  • $\begingroup$ Harper and Greenberg: only intersection numbers. Bredon: only smooth submanifolds. Kreck: you have to give in to his stratifolds. $\endgroup$ – Anton Mellit Mar 16 at 10:39
  • $\begingroup$ Dold is also not good enough: the closest is Prop. 13.23. That assumes that the intersection is a compact connected manifold. Even to understand the proof for this case one needs basically to read the whole book. I believe it's easier to do it from scratch. $\endgroup$ – Anton Mellit Mar 17 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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