Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Where can I find a stricter presentation of topology of manifolds, then in section 0.4 in Griffiths-Harris? For example, they define the map $H_k \times H_{n-k}$ by presenting a cycle by a submanifold and intersection form. But it's known, that not all homology classes are representable by a submanifold. It can contain singularities in codimension 2. Griffiths and Harris wtite nothing about it. But in the classical topology all is defined strictly, but it's very difficult to calculate anything, using their definition (for example, in Hatcher, Algebraic topology). Then I want to justify the approach of Griffiths-Harris. Also they use, that we can compute homology groups buy smooth simplexes or using triangulations and simplexes, contained in triangulation. The fact of existence of triangulation, as far as I know, is quite difficult, and I don't want to use it without proof. By the wyay, Griffiths and Harris writes, that their book is quite self-contained.

share|improve this question
    
As I was once reminded on this site before, the approppriate tag for such questions is "differential-topology", not "differential-geometry". –  Piotr Pstrągowski Mar 3 '13 at 14:17

1 Answer 1

up vote 1 down vote accepted

Your question is very broad and it's not completely clear to me what you're asking, but I will try to touch the subjects you mention.

It is true that not every homology class is necessarily represented by submanifold, which is one of the things that make the definition of intersection form rather difficult. The usual way is to first define a product structure in cohomology, then prove Poincare duality to show that this also gives a certain bilinear form on homology. This is adventagous because even though the cup product is almost impossible to compute from the definition, it's easy to prove its good formal properties. Only then one proves that in the case of smooth, transversal submanifolds this "algebraic" intersection form agrees with the "geometric" one (where we simply intersect the submanifolds in question). This is done for example in these notes bu Michael Hutchings.

I have to say that I have no idea about the proof of existence of triangulations of smooth manifolds, but a quick search reveales that there are proofs (see"Simple triangulation method of smooth manifolds" by S. Cairns) that are rather short - even if dense and technical. Also, I suppose that's an entirely personal opinion, but I think this is one of the results that in many cases can just be taken on faith.

share|improve this answer

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.