Lagmental Vicfred

The Cup Product Is Graded-Commutative by Vicfred

Last updated: Thu 17 April 2025

tags: mathematics, cohomology, cup products, Poincare duality

Cohomology classes multiply with a sign determined by their degrees. The formulas are more useful when each symbol has a job rather than merely decorating the theorem.

Notation

Cohomology applies \(\operatorname{Hom}(-,R)\) to chains and reverses arrows. The cup product makes \(H^\ast(X;R)\) a graded ring rather than only a graded group.

$$ \smile:H^p(X;R)\times H^q(X;R)\to H^{p+q}(X;R) $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ \alpha\smile\beta=(-1)^{pq}\beta\smile\alpha $$

Stress the formula

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ H^\ast(S^n;R)\cong R[u]/(u^2),\qquad|u|=n $$

Interpretation

A good test for understanding is to change the presentation while keeping the invariant fixed. The aligned form makes that comparison unusually easy.

$$ \begin{aligned} \mathsf{D}\;&:\quad \smile:H^p(X;R)\times H^q(X;R)\to H^{p+q}(X;R),\\[5pt] \mathsf{C}\;&:\quad \alpha\smile\beta=(-1)^{pq}\beta\smile\alpha. \end{aligned} $$

Limit of the argument

Isomorphic cohomology groups do not guarantee isomorphic cohomology rings. Products can distinguish spaces that additive invariants cannot.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \alpha\smile\beta=(-1)^{pq}\beta\smile\alpha \end{gathered}} $$

The final box is a summary, not a new assumption; the proof still lives in the definitions and the intervening calculation. The source keeps each scope delimiter visible for later inspection.

This article was posted on Sat 26 November 2022. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.