6
$\begingroup$

This is a question related to an earlier question of mine:

I've been reading about topological invariants. Some of them are defined in terms of quadratic forms.

My current understanding is: we can turn a topological space $X$ into a module by taking any of its homologies and once we have a module we define a bilinear form on $H_i(X)$. For example, if we are taking the first homology we can define an bilinear form by defining $x \cdot y$ to be the intersection number of two representatives $x,y$ where $[x], [y] \in H_1 (X)$. Then this gives us a quadratic form $Q(x) = \frac{1}{2} x \cdot x \mod 2$. We can use this if we want to show that two topological spaces are not homeomorphic as follows: pick a symplectic basis $a_i, b_i$ for the space (we know that such a basis always exists). Then we define $A(Q) = \sum_i Q(a_i) Q(b_i)$. If this evaluates to $0$ in one space and to $1$ in the other we know that the two spaces are not homeomorphic.


Question 1: Is my current understanding correct? Or am I missing anything?

Question 2: I understand how to define a bilinear form if I take the first homology. But this might not lead anywhere because even though the two spaces might be non-homeomorphic, their first homology groups might still coincide. So I might want to take a higher homology. How do I define a bilinear form for the $k$-th homology and what is the geometric meaning?

Question 3: I found this Wikipedia article about intersection forms. Although I am quite sure that the "intersection form" is another bilinear form just like the intersection number for the first homology, I'm confused about why I could only find additional information about $4$-manifolds. What is special about $4$-manifolds? Am I wrong in assuming that I can endow any topological space with a bilinear form?

Question 4: We do this over $\mathbb Z / 2$ for convenience, right? Because it lets us ignore orientation. But we could also consider bilinear forms over any other field, is this correct?

Edit

Question 5: Poincaré duality gives us a way to define an intersection pairing without much nasty fiddling if the topological space $X$ is a closed oriented manifold. But what are the absolute minimal requirements on $X$ in order to be able to define an intersection form on it?

Thanks for your help.

  • 0
    @QiaochuYuan Though Irving Kaplansky calls symmetric bilinear forms also inner product. See [Linear Algebra and Geometry](http://books.google.com/books/about/Linear_Algebra_and_Geometry.html?id=nUcUYHrYJtgC&redir_esc=y) page 2. He provides $(x,y) = 0$ as the most trivial example of an inner product. (I wish I could link to the page, it seems to be excluded from Google Books Preview.) The definition must have evolved since the 60ies.2012-09-04

1 Answers 1

4

Q1: no. The intersection pairing is not defined on general topological spaces.

Q2: Your space will need some special structure and your question is still too general to answer. Perhaps read-up on Poincare duality?

Q3: An intersection form is a special bilinear form induced via Poincare duality. You can think of it in terms of (oriented) transverse intersections of representative cycles.

Q4: Yes, but generally you'll need your manifold to be oriented to pull this off.

  • 0
    Thank you. I had not heard of the torsion linking form. I'm reading about the Arf invariant and I wanted to find an example of a quadratic form of which I can compute the Arf invariant. The intersection form seemed like a good idea because it has a geometric meaning. Next I will look into Arf invariant and knots and Arf invariant and framed manifolds but before I do that I wanted to find a very geometrically intuitive example where I can apply it to something simple.2012-09-02