Lagmental Vicfred

The Torus Cohomology Ring Is an Exterior Algebra by Vicfred

tags: mathematics, cohomology, cup products, Poincare duality

Two degree-one classes generate the torus cohomology and their product is the orientation class. Keeping the exact identity in view prevents the geometric or probabilistic intuition from drifting.

Definitions first

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.

$$ a,b\in H^1(T^2;\mathbf Z) $$

A reliable calculation names domain and codomain. The notation \(\mathsf{data}\mapsto\mathsf{claim}\) is harmless only after both \(\operatorname{dom}\) and \(\operatorname{cod}\) have been fixed.

$$ H^\ast(T^2;\mathbf Z)\cong\Lambda_{\mathbf Z}(a,b) $$

A small case in full

Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.

$$ a^2=b^2=0,\qquad a\smile b=-b\smile a,\qquad\langle a\smile b,[T^2]\rangle=1 $$

The reusable statement

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

$$ \begin{aligned} \mathsf{D}\;&:\quad a,b\in H^1(T^2;\mathbf Z),\\[5pt] \mathsf{C}\;&:\quad H^\ast(T^2;\mathbf Z)\cong\Lambda_{\mathbf Z}(a,b). \end{aligned} $$

A nearby false statement

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

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] H^\ast(T^2;\mathbf Z)\cong\Lambda_{\mathbf Z}(a,b) \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Sat 12 May 2018. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.