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.

For which topological spaces $X$ can I define an intersection form $b(\cdot, \cdot)$?

I know at least one example: If $X$ is a closed orientable $2n$-manifold then one can define an intersection form by defining $b(c_1, c_2)$ to be the cup product of two representatives of two elements in the $k$-th homology group $H_k$.

But one can generalise and drop the condition of orientability and still define an intersection form: instead of taking an arbitrary ring $R$ of coefficients for the homologies $H_k$ one can consider $R = \mathbb Z / 2 \mathbb Z$ so that orientability becomes irrelevant since $\mathbb Z / 2 \mathbb Z$ has only one generator, $1$.

Reading the article about the cup product though suggests that the cup product can be defined on arbitrary topological spaces. Which would mean that I get a bilinear form for any topological space. (which would contradict this answer I got for one of my previous questions regarding this).

Question 1.1: What am I missing? Why can I not define the cup product on any topological space and therefore get a bilinear form (and therefore an intersection form) on any topological space?

Question 1.2: What property does a manifold have that a topological space does not have to get an intersection form? (surely the answer cannot be orientability since we want to choose coefficients mod two anyway)

Question 2: What I want to do with this is, I want to take my bilinear form on my space and then make it into a quadratic form $q(x) = b(x,x)$, choose a symplectic basis and then compute the Arf invariant of it. Does that give me additional requirements for my space $X$ in order to be able to do that?

share|improve this question
    
From Matt E's answer to one of my earlier questions I know that the Poincaré duality gives a geometrical meaning to my bilinear form. –  Matt N. Aug 28 '12 at 11:53
1  
intersection pairing is given by the cup product and by the fundamental class; the latter is not defined for every topological space –  user8268 Aug 28 '12 at 12:49
1  
It seems like you're using the word "intersection form" to be a synonym for "natural bilinear pairing". I think that's not very common. Intersection form usually means the form Poincare-dual to the cup product with all of its geometric properties -- that you can realize the form by taking transverse intersections of the representative cycles. –  Ryan Budney Aug 28 '12 at 17:10
1  
Even on manifolds there are other natural bilinear pairings. For example the torsion linking form. –  Ryan Budney Aug 28 '12 at 17:16
add comment

1 Answer 1

Ad 1.1. : see Ad 1.2.

Ad 1.2. : Manifolds have several properties that are used here. An easy one is the dimension. A manifold $M$ has a well-defined dimension say $n$. The second thing is that it has a "fundamental class" at least in the compact case. (let's not talk about compact cohomology for the moment). This means that there is an isomorphism $H^n(M,\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ (if the manifold is orientable you could replace $\mathbb{Z}/2\mathbb{Z}$ by $\mathbb{Z}$. And maybe a third thing is Poicaré Duality. This implies that if your manifold $M$ has even dimension $2n$ then the map $H^n(M,\mathbb{Z}/2\mathbb{Z}) \otimes H^n(M,\mathbb{Z}/2\mathbb{Z}) \to H^{2n}(M,\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ is non-degenerate, which tells you in particular that it is not zero (something which you could not garantee even if you had counterparts of dimension and fundamental classes).

So to sum this up, for an intersection product on a general space you would want to have a cup product (this you have), but you want to know that some abstract cohomology is isomorphic to a specific ring in which the product lives.. where should this come from in general?

Concerning question 2, I do not know how this should work for arbitrary spaces, due to the problems described above..

Hope this helps.

share|improve this answer
add comment

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.