- #1

lavinia

Science Advisor

Gold Member

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## Main Question or Discussion Point

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.

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.