I am trying to show that the cohomology ring of the Klein bottle with $\mathbb{Z}_2$ coefficients is $H^*(K,\mathbb{Z}_2) \cong \mathbb{Z}_2[x,y]/(x^3,y^2, x^2y)$.

What I know:

$H^0(K,\mathbb{Z}_2)=\mathbb{Z}_2, H^1(K,\mathbb{Z}_2)=\mathbb{Z}_2\oplus \mathbb{Z}_2, H^2(K,\mathbb{Z}_2)=\mathbb{Z}_2$

$K$ is a 2-dimensional closed manifold that is orientable over the field $\mathbb{Z}_2$, so we know that for each of the two generators of $H^1(K,\mathbb{Z}_2)=\mathbb{Z}_2\oplus \mathbb{Z}_2$ (call them $x$ and $y$), there is another element $\psi, \phi$ in $H^1(K,\mathbb{Z}_2)$ such that

$x\cup \psi$=$y\cup \phi= \omega$, where $\omega \in H^2(K,\mathbb{Z_2})\cong \mathbb{Z}_2$ is the generator.

So since $x,y,$ and $x+y$ are the only non-zero elements in $H^1(K,\mathbb{Z}_2)$, we of course have to choose from these for our $\psi$ and $\phi$. This mostly comes down to figuring out which of $x\cup x, x\cup y,y\cup y$ are equal to zero.

My Question

Is there a clear/intuitive way to do this? And assuming $x$ and $y$ correspond to the dual elements of either $A$ or $B$ in the diagram below, which one does each correspond to?

Klein Bottle

  • 1
    $\begingroup$ I think the isomorphism in the first line is incorrect. In particular $H^*(K)$ is four-dimensional, whereas a basis for the algebra given is $1,x,x^2,y,xy$. From Hatcher Example 3.8 (on page 208) the cohomology is $\mathbb{Z}/2[x,y]/(xy, x^2-y^2, x^3, y^3)$, with a similar presentation for the higher (and lower) genus non-orientable surfaces. $\endgroup$ – Ben Jul 25 '15 at 7:00

For those who want to compute it all out, if we use the delta-complex below instead of the one I gave in the question we have:

Klein Bottle

Let $\phi \in C^0(K,\mathbb{Z}_2)$ be dual to $v\in C_0(K,\mathbb{Z}_2)$,

$\alpha, \beta, \gamma \in C^1(K,\mathbb{Z}_2)$ be the dual elements to $a,b,c\in C_1(K,\mathbb{Z}_2)$, respectively, and

$\mu,\lambda\in C^2(K,\mathbb{Z}_2)$ be dual to $U,L\in C_2(K,\mathbb{Z}_2)$.

To compute cohomology we need the homology groups of the chain complex: $0 \rightarrow C^0 \rightarrow C^1\rightarrow C^2 \rightarrow 0\rightarrow ...$, so we compute the coboundary map.

$\delta(\phi)(a)=\phi(\partial(a))=\phi(v)-\phi(v)=0=\delta(\phi)(b)=\delta(\phi)(c)$, so $\delta:C^0 \rightarrow C^1$ is the zero map.

$\delta(\alpha)(U)=\alpha(\partial(U))=\alpha(a)+\alpha(b)-\alpha(c)=1+0-0=1$. Similarly, $\delta(\beta)(U)=1$ and $\delta(\gamma)(U)=-1$. And $\delta(\alpha)(L)=1$, $\delta(\beta)(L)=-1$, $\delta(\gamma)(L)=1$.

So $\delta:C^1 \rightarrow C^2$ takes $\alpha \rightarrow \mu +\lambda$, $\beta \rightarrow \mu -\lambda$, and $\gamma \rightarrow -\mu +\lambda$. Thus:

Im $\delta:C^1 \rightarrow C^2=<\mu+\lambda, \mu- \lambda>=<2\mu, \mu+\lambda>=<\mu +\lambda>$ since we are over $\mathbb{Z}_2$.

Ker$ \delta:C^1 \rightarrow C^2=<\beta+\gamma,\alpha +\beta>$

Ker$ \delta:C^2 \rightarrow C^3=<\mu,\lambda>=<\mu, \mu+\lambda>$ is all of $C^2$ since $C^3=0$.

So $H^1(K,\mathbb{Z}_2)= <\beta+\gamma,\alpha +\beta>$ (Im $\delta:C^0\rightarrow C^1=0$), and $H^2(K,\mathbb{Z}_2)= <\mu,\lambda>=<\mu, \mu+\lambda>/ <\mu +\lambda>=<\mu>$.

Now to find the ring structure we just look at products and compute them explicitly: $\beta +\gamma \cup \beta +\gamma (U)= \beta+\gamma(a)* \beta+\gamma(b)=0*1=0$

$\beta +\gamma \cup \beta +\gamma (L)= \beta+\gamma(c)* \beta+\gamma(a)=1*0=0$

$\alpha +\beta \cup \alpha +\beta (U)= \alpha+\beta(a)* \alpha+\beta(b)=1*1=1$

$\alpha +\beta \cup \alpha +\beta (L)= \alpha+\beta(c)* \alpha+\beta(a)=0*1=0$

$\alpha +\beta \cup \beta +\gamma (U)= \alpha+\beta(a)* \beta+\gamma(b)=1*1=1$

$\alpha +\beta \cup \beta +\gamma (L)= \alpha+\beta(c)* \beta+\gamma(a)=0*0=0$

Thus, $\alpha +\beta \cup \alpha +\beta= \alpha +\beta \cup \beta +\gamma=\mu$, which is generator of $H^2$. Finally, set $x=\alpha +\beta$, $y=\beta +\gamma$ to get the ring above.

Ref:UOregon Topology

  • 1
    $\begingroup$ When you begin to look at the products and compute them explicitly, why does $U$ correspond to $a$ and $b$, while $L$ corresponds to $c$ and $a$? Also, how did you determine that $\alpha + \beta \cup \alpha + \beta = \mu$? $\endgroup$ – user319128 Oct 20 '16 at 5:47

The space $H^1(K) = \mathbb{Z}^2$ is generated by the Poincaré duals $\alpha = A^*$ and $\beta = B^*$ to $A$ and $B$, respectively. (I'm working over $\mathbb{Z}_2$ throughout, so that $H^*(K)$ is actually $H^*(K, \mathbb{Z}_2)$. For reasons of dimension, the only products you need to compute in the ring $H^*(K)$ are $\alpha^2, \alpha \beta$, and $\beta^2$. But for any $a, b\in H_1(K)$, the cup product $a^* \cup b^* = (a.b) [K]$, where $a.b$ denotes the intersection product and $[K]\in H^2(K)$ denotes the fundamental class of $K$. Since we're working over $\mathbb{Z}_2$, the latter is just the number of intersection points of $a$ and $b$ (in general position); the sign coming form their orientation is irrelevant.

  • $\begingroup$ By the way, regarding this technique: In general, it's not true that every integral homology class can be represented by an embedded submanifold. There are quite a few results on situations where that (or something similar) does hold, however. It's easy to show directly in this case, but it happens to hold for $\mathbb{Z}_2$-homology in general. $\endgroup$ – anomaly Jul 23 '14 at 21:59
  • $\begingroup$ Okay, this helps. Although I haven't covered intersection numbers yet so that last sentence doesn't quite make sense to me... I'll look this up and let you know if I have any questions. Thanks $\endgroup$ – Ashley Jul 24 '14 at 0:14
  • $\begingroup$ Sure, good luck! $\endgroup$ – anomaly Jul 24 '14 at 3:09

The loops A and B determine the two Z2 cohomology classes by transverse intersections.

  • The self intersection of A is zero since you can move it parallel to itself slightly and get a disjoint loop. So its square is zero.

Its intersection with B is 1 so their product is non-zero.

  • The self intersection of B is 1 because of the half twist. So the square of B is not zero.

  • The first Stiefel Whitney class of the Klein bottle is not zero because it is non-orietable. The square of the first Stiefel-Whitney class is zero because the Klein bottle is an unoriented boundary.

Since the tangent bundle is flat with structure group, Z2, the classifying map for the tangent bundle factors through the classifying space for Z2 which is the infinite dimensional real projective space.

The loop,B, projects to the non-null homotopic loop in the projective space so A represents the first Whitney class since it is dual to B.


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