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12 votes
Accepted

What's the cohomology ring structure of a blow-up?

This is straightforward. Using $j^*[E]=-h$ and the projection formula $j_*(x\cdot j^*y)=j_*x\cdot y$, we get $[E]^p=j_*1\cdot [E]^{p-1}= (-1)^{p-1}j_*(h^{p-1})$. Then $$[E]^{p}\cdot \tau ^*\alpha = (-...
abx's user avatar
  • 38k
8 votes

Generalisation of Hirsch formula for the associativity of Steenrod's higher $\cup_2$ product with $\cup_1$ and cup products

If I may be allowed to interpret loosely what you're trying to do, maybe it's to understand $E_\infty$ structures in a way that's comparable to the very algebraic $A_\infty$ story, or at the very ...
Dave Benson's user avatar
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6 votes
Accepted

Cup products in the Mayer-Vietoris sequence

Here are some comments. The map $\delta^\ast$ is a composition of the form $\require{AMScd}$ \begin{CD} H^{\ast-1}(U\cap V) @> \sigma >\cong > H^{\ast}(\Sigma (U\cap V)) \to H^\ast X \end{...
John Klein's user avatar
  • 18.9k
6 votes

How to construct cup-product in a general site?

Yes. This is treated in detail in Section 8.4 of Jardine's book “Local homotopy theory”. See also the introduction to Chapter 8 there for a historical comment on cup products and Godement resolutions....
Dmitri Pavlov's user avatar
5 votes
Accepted

Calculations of cup products in Bredon cohomology

Frankly, there aren't many calculations out there. Most of the work I know of is on the calculation of the $RO(G)$-graded cohomology of a point, of a projective space, or of $B_GO(n)$. Here are some ...
Steve Costenoble's user avatar
4 votes

Calculating cup products using cellular cohomology

A recipe is given in Whitney's 1937 paper "On Products in a Complex" (Annals of Mathematics, vol. 39, no 2). You can find a copy here. In section 12 he gives a very explicit construction of ...
Nikolas Breuckmann's user avatar
4 votes
Accepted

Cup product in Tate Cohomology Ring

In Chapter XII.7 of the book Cartan, Henri; Eilenberg, Samuel, Homological algebra, Princeton Mathematical Series. 19. Princeton, New Jersey: Princeton University Press xv, 390 p. (1956). ZBL0075....
Mark Grant's user avatar
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3 votes
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Examples of a group $G$ and an $F$-representation $V$ where $\cup:H^1(G,F)\otimes H^1(G,V)\to H^2(G,V)$ is injective

$\newcommand{\bZ}{\mathbb{Z}}$Let $G$ be a finite group of order divisible by $p:=\mathrm{char}\, F$ such that $\dim_F \mathrm{Hom}(G,F)\geq 2$ (e.g. $G=\bZ/p\times\bZ/p$). Take $V$ to be the kernel ...
SashaP's user avatar
  • 7,377
3 votes

Find $a$ satisfying $x \cup_1 y = \delta a$ when $x,y \in Z^2(G,\mathbb{Z}_2)$

The $\cup_1$-product is not always a coboundary. For example, for a cocycle $x\in Z^2$, $x\cup_1 x\in Z^3$ is a representative of the class $\operatorname{Sq}^1 x$. This can be non-trivial already for ...
Achim Krause's user avatar
  • 10.8k
3 votes

Parabolic cohomology of modular groups and cup-products

It seems to me that the definition has been cooked up so that $H^0_P = H^0$, $H^1_P = \mathrm{Im}(H^1_c \to H^1)$, and $H^2_P = H^2_c$. This means that the parabolic cohomology is equal to the ...
Dan Petersen's user avatar
  • 40.3k
3 votes
Accepted

Cartan Formula for Steenrod square on cocycles

This is not necessarily true. The first warning one should give is that there are multiple definitions of "$\cup_{i}$" on cocycles, and different ones satisfy different identities. However, this is ...
Tyler Lawson's user avatar
  • 52.7k
2 votes
Accepted

Use of Steenrod's higher cup product and the graded-commutativity

For $(\delta w) \cup v - v \cup (\delta w)$ to be a coboundary, it would need to also be a cocycle: so we would have to have $$ 0 = \delta((\delta w) \cup v - v \cup (\delta w)) = (\delta w) \cup (\...
Tyler Lawson's user avatar
  • 52.7k
2 votes

Parabolic cohomology of modular groups and cup-products

Since I seem to my stuck on something myself, let me take a crack at your problem. This is just a proposal, I haven't checked if this really works. Let me start with the suggestion in Dan's answer ...
Donu Arapura's user avatar
  • 35.2k
1 vote
Accepted

Cohomology ring on non-simplicial complex

If $H^\ast(-,R)$ is cohomology with coefficients in a ring, the cup product may be defined purely via the functoriality of $H^\ast(-,-)$ and certain compatibilities of tensor products. There is a map $...
Joshua Mundinger's user avatar
1 vote

Equivalence of finiteness of $spliG$ and periodicity isomorphisms being induced by cup product

Let $P_\bullet \to \mathbb{Z}$ be a projective resolution as a $\mathbb{Z}[G]$-module, $\Delta : P_\bullet \to P_\bullet \otimes P_\bullet$ an approximation of the diagonal, and $\phi : P_\bullet \to \...
Oscar Randal-Williams's user avatar

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