scratch

§ Cup Product [TODO ]

created 2022-02-03 · last edited 2022-05-30
  • We need an ordered simplex, so there is a total ordering on the vertices. This is to split a chain apart at number kkk.
  • Can always multiply functions together. This takes a kkk chain ξ\xiξ and an lll chain η\etaη and produces ξ∪η\xi \cup \etaξ∪η which is a k+lk + lk+lcochain. The action on a (k+l)(k+l)(k+l) chain γ\gammaγ acts by (ξ∪η)(γ)≡ξ(γ≤k)⋅η(γ>k)(\xi \cup \eta)(\gamma) \equiv \xi (\gamma_{\leq k}) \cdot \eta (\gamma_{> k})(ξ∪η)(γ)≡ξ(γ≤k​)⋅η(γ>k​).
  • No way this can work for chains, can only ever work for cochains.
  • This cup product "works well" with coboundary. We have ∂(ξ∪η)≡(∂ξ∪η)+(−1)k(ξ∪∂η)\partial (\xi \cup \eta) \equiv (\partial \xi \cup \eta) + (-1)^k (\xi \cup \partial \eta)∂(ξ∪η)≡(∂ξ∪η)+(−1)k(ξ∪∂η).
  • We get cocycle cup cocyle is cocycle.
  • Similarly, coboundary cup cocycle is coboundary.
  • Simiarly, cocycle cup coboundary is coboundary.
  • The three above propositions imply that the cup product descends to cohomology groups.
  • The algebra of cohomology (cohomology plus the cup product) sees the difference between spaces of identical homology!
  • The space S1×S1S^1 \times S^1S1×S1 have the same homology as S2∩S1∩S1S^2 \cap S^1 \cap S^1S2∩S1∩S1. Both have equal homology/cohomology.
  • However, we will find that it will be zero on the torus and non-zero on other side.
  • The cup product measures how the two generators are locally product like. So if we pick two generators on the torus, we can find a triangle which gives non-zero
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