Help fixing my broken example of Arf invariant

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.

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Keep in mind that the Arf-Kervaire invariant is meant to be evaluated on framed manifolds. I'm pretty sure the first few nontrivial examples come from Lie groups (which are automatically frameable). – Aaron Mazel-Gee Sep 1 '12 at 22:04
@AaronMazel-Gee Lie groups? Thanks for the hint. I had read about framed manifolds but then I thought that there had to be a simpler example for this. But good to know that there are Arf examples with Lie groups! – Rudy the Reindeer Sep 2 '12 at 6:52
What makes you want to use the Arf invariant, anyways? Of course (the isomorphism class of) the intersection form is a strictly stronger invariant, so if you have it then you should use it (instead of its Arf invariant) to distinguish your manifolds. The (Arf-)Kervaire invariant problem is a global question about manifolds, i.e. it asks about all manifolds at once. Short of using HHR's theorem as an absurdly large sledgehammer, I don't know any way of constructing a manifold with nonvanishing Arf invariant without knowing its intersection form in the first place. – Aaron Mazel-Gee Sep 2 '12 at 20:28
@AaronMazel-Gee Well, I've been reading about the Arf invariant. So far I have looked at Arf invariants of knots. Next I have to look into framed manifolds. What is HHR's theorem? – Rudy the Reindeer Sep 6 '12 at 15:38

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?

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Thank you Ryan for this great answer. Yes, I'd briefly read about Lens spaces (when searching for "spaces with same homology that aren't homeomorphic") but it looked a bit less simple than I wanted it to be, since I wasn't sure what the intersection form looks like on something that is not a surface. – Rudy the Reindeer Sep 2 '12 at 6:56