Recognitions:

## Calculating the Z2 cohomology of the Klein Bottle using intersections

This thread asks for help calculating the Z2 cohomology ring of the Klein bottle using intersections.

This is what I think.

View the Klein bottle as a circle bundle over a circle. A fiber circle and the base circle generate the first Z2 cohomology by transverse intersection.

- The fiber cicle has zero transverse intersection with itself and intersects the base circle in a single point. This would appear to be the first Stiefel Whitney class of the tangent bundle. Since its self intersection is zero the cup product of the cohomology class that it determines with itself is zero. This seems right since the Euler characteristic of the Klein bottle is zero.

- the base circle intersects both itself and the fiber circle in a single point. so its square under the cup product is not zero.

This completely describes the cohomology ring in dimension 1.

What about the pull back of these classes under the two fold cover of the Klein bottle by a torus?

- the fiber circle class now intersects the base twice as so pulls back to zero. This makes sense because the torus is orientable so its first Stiefel Whitney class is zero.

- the base circle intersects itself twice and so has zero self intersection mod 2 - so its self cup product is now zero - but still intersects the fiber circle once. So it pulls back to one of the generators of the first cohomology of the torus.

I think this right.

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 Recognitions: Homework Help Science Advisor whats the idea here lavinia? you know the Z/2 homology and assume that poincare duality holds mod 2, so you want to find dual cocycles to the generating cycles?
 Recognitions: Science Advisor Mathwonk I was interested in computing the cohomolgy ring of the Klein bottle - not just its cohomology - using transverse circles on the boundary of its fundamental domain. I wanted to find the first Stiefel-Whitney class and see why its square is zero and was surprised that there seems to be a 1 dimensional class whose square is not zero but which is not induced from the classifying map into projective space. I just thought I may have done it wrong and wanted to make sure. But there is a broader idea. Can one compute the Z2 cohomology ring of a flat manifold of higher dimension using the same technique, intersection of hyperplanes on the boundary of a fundamental domain.? Are there cohomology classes that can not be obtained in this way? A priori without any other knowledge this could even be true of the Klein bottle so how does one see this? I was hoping that for more complicated flat manifold this technique could also be used to compute the Z2 characteristic algebra and maybe to find a manifold whose fundamental Z2 cohomology class is actually in induced from the classifying map into the holonomy group. The amazing thing to me about the Klein bottle is that its Z2 fundamental cocycle is the square of a 1 dimensional class but not the square of a class in its characteristic algebra (since the square of its 1st Stiefel Whitney class is zero). I have started trying out some simple 3 manifolds such as two way Klein bottle - a half twist along both the y and z axes. The same sort of thing happens here but now that manifold is orientable so the 1 dimensional cohomology class induced by the classifying map into projective space is not the first Stiefel Whitney class.

Recognitions:

## Calculating the Z2 cohomology of the Klein Bottle using intersections

Mathwonk

I have read you explanations of ramified covers in another thread and wonder whether a 2 fold ramified cover of the sphere by the torus can be set up to project to a 2 fold ramified cover of the projective plane by the Klein bottle.

This might be the classifying map of the tangent bundle.