# Follow up on intersection forms

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?

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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. – Rudy the Reindeer Aug 28 '12 at 11:53
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
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
Even on manifolds there are other natural bilinear pairings. For example the torsion linking form. – Ryan Budney Aug 28 '12 at 17:16

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 Poincaré 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 guarantee even if you had counterparts of dimension and fundamental classes).