I need help fixing a broken example I've come up with. In particular, I wanted to use the Arf invariant to distinguish two non-homeomorphic surfaces. That's the first part that's broken since there aren't any surfaces that have the same homology groups but aren't homeomorphic. So I gain nothing because the homology groups already let me distinguish two surfaces.
Concretely, what I was trying to do was this: observe that the intersection number defines a bilinear pairing (an intersection form in this case) and then compute the Arf invariant of $q(x)=\langle x,x \rangle$. Stupidly, for orientable surfaces, this always gives me $q(x)=\langle x,x \rangle = 0$ and hence $\mathrm{Arf}(q) = 0$.
On the other hand, for $\mathbb R P^2$, there is only one basis vector hence I can't even define the Arf invariant (since I can't define an intersection form since that would need even dimension). So this is the second part that is broken.
Now I'm stuck. How can I fix this? What is the simplest example of two manifolds (or topological spaces if you like) that I can distinguish using the Arf invariant? Thanks for help.
I don't see how to fix your example, surfaces aren't rich enough.
To get something a little richer, 3-manifolds have something called a torsion linking form. This is a symmetric bilinear map:
$$\tau H_1(M,\mathbb Z) \times \tau H_1(M,\mathbb Z) \to \mathbb Q / \mathbb Z$$
where $\tau H_1(M,\mathbb Z)$ is the subgroup of torsion elements of $H_1(M,\mathbb Z)$. In the lens space $L_{p,q}$ the value of this form on $(x,x)$ where $x$ generates $H_1$ is $\pm r^2 q/p$. Where $r$ is some integer.
So you could restrict the torsion linking form of 3-manifolds to the 2-torsion subgroup and compute the Arf invariant of that.
I think the Arf invariant would distinguish
$$L_{p,q} \# L_{p,q} \text{ and } L_{p,q} \# L_{p,-q}$$
Provided $p$ is even and $L_{p,q}$ does not admit an orientation-reversing homotopy-equivalence (which I think is when $q \neq -r^2 q$ modulo $p$.
Have you looked at something like that?