# The Cup Product of the Klein Bottle mod 2

I didn’t want to use a brute force method, so I thought for a while about how to compute the cup product of the Klein bottle. This is what I came up with. I thought I’d share it.

Take coefficients in $$\mathbb{Z}/2$$. We compute the cup product by computing the intersection form (which defines the intersection product), then Poincare duality.

The Klein bottle is compact and $$\mathbb{Z}/2$$-orientable, so Poincare duality applies: $$a^* \smile b^* = [a.b][K]$$

where $$[K]$$ is the fundamental class of the Klein bottle, $$[a.b]$$ is the number of intersections of the cycles $$a$$ and $$b$$ (thought of as sub-manifolds embedded in the Klein bottle). Note that $$a^$$ and $$b^$$ denote the cocycles corresponding to the cycles $$a$$ and $$b$$ respectively.

We view the Klein bottle as the fiber bundle $$S^1 \to K \to S^1$$ Fix a fiber circle, $$F$$, and denote the base circle as $$B$$.

By pictoral argument: $$F$$ has zero transverse intersection with itself and intersects the base circle $$B$$ once. By wiggling the base circle by an infinitesimal amount $$\epsilon$$, we see that $$B$$ intersects transversely with itself once.

Another perspective:

We may more succinctly state the intersection product as the following matrix:

$$\begin{pmatrix} \beta \smile \beta & \beta \smile \gamma \ \gamma \smile \beta & \gamma \smile \gamma \end{pmatrix} \simeq \begin{pmatrix} [B.B] & [B.F] \ [F.B] & [F.F] \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix}$$

From the above matrix, we see that $$\beta^2 = \beta\gamma = 1[K]$$, and $$\gamma^2 = 0[K]$$. And, of course, $$\beta^3 = 0$$, because $$H^3(K) = 0$$. This gives us the ring structure of $$H^*(K; \mathbb{Z}/2)$$: $$(\mathbb{Z}/2\mathbb{Z})[\beta, \gamma]/(\beta^2 + \beta\gamma, \gamma^2, \beta^3)$$

P.S. If we wanted to be totally rigorous, we’d have to show that the cocycles $$\beta$$ and $$\gamma$$ form the basis of the $$H^1(K; \mathbb{Z}/2) \simeq \mathbb{Z}/2 \oplus \mathbb{Z}/2)$$. Until then, we don’t quite have that the intersection matrix gives us the cup product of $$H^*(K; \mathbb{Z}/2)$$.

Written on December 6, 2015