# Definition of the cup (wedge) product of de Rham cohomology classes

In a few places (Lemma 3.0.13 of this script, discussion after Lemma 3.2 here, Proposition 5 here, etc.) I've noticed a certain omission in the definition of a wedge product $$\wedge\colon H^k(M) \times H^{\ell}(M) \to H^{k+\ell}(M)$$ of de Rham cohomology classes. In particular, it is claimed that for closed differential forms $$\omega_1 \in Z^k(M)$$ and $$\omega_2 \in Z^{\ell}(M)$$: $$[\omega_1] \wedge [\omega_2] = [\omega_1 \wedge \omega_2]\,,$$ but only one direction of inclusion is proved.

I presume that $$[\omega_1] \wedge [\omega_2]$$ means (?): $$[\omega_1] \wedge [\omega_2] = \{(\omega_1+d\eta_1)\wedge(\omega_2+d\eta_2) \mid \eta_1 \in \Omega^{k-1}(M), \eta_2 \in \Omega^{\ell-1}(M)\}\,,$$ while $$[\omega_1 \wedge \omega_2]$$ is by definition: $$[\omega_1 \wedge \omega_2] = \{\omega_1 \wedge \omega_2 +d\eta_{12} \mid \eta_{12} \in \Omega^{k+\ell-1}(M)\}\,.$$ Then it is easy to show that for every $$\eta_1, \eta_2$$: $$(\omega_1+d\eta_1)\wedge(\omega_2+d\eta_2) = \omega_1 \wedge \omega_2 + d\eta_{12}\,, \text{with}$$ $$\eta_{12} = \eta_1 \wedge \omega_2 + (-1)^{k} \omega_1 \wedge \eta_2 + \eta_1 \wedge d\eta_2\,,$$ which proves $$[\omega_1] \wedge [\omega_2] \subseteq [\omega_1 \wedge \omega_2]$$. But the reverse is not proved!

How would one go about proving, given arbitrary $$\omega_1, \omega_2$$, that for every $$\eta_{12}$$ there exist corresponding $$\eta_1$$ and $$\eta_2$$?

• If I am not misunderstanding that is how it is defined... For closed forms $\omega_1$ and $\omega_2$, cup product of $[\omega_1]$ with $[\omega_2]$ is defined to be the equivalence class of the wedge product $\omega_1\wedge \omega_2$.. can you tell if you have other definition – Praphulla Koushik Sep 6 '19 at 12:54
• Indeed, this same though occurred to me, but then it's not clear why would they bother proving this one-sided inclusion. They mention that it makes the product "well-defined," but in what sense I do not know. – Grgur Palle Sep 7 '19 at 20:38

It is standard to define the cup product $$[\omega_1] \wedge [\omega_2]$$ to be $$[\omega_1 \wedge \omega_2]$$. The "inclusion" that is being proved in these texts is not an inclusion at all, but a check that the operation is well defined. The main source of confusion, it seems, is that $$[\omega_1] \wedge [\omega_2]$$ is not {$$(\omega_1 + d\eta_1) \wedge (\omega_2 + d\eta_2)$$}, as you claim. Instead, $$\wedge$$ is being defined, and we are reusing the name.

As a brief reminder, say we have some function $$f : A \to B$$ and some equivalence relation $$\sim$$ on $$A$$. We can view $$f$$ as a function from $$(A/\sim) \to B$$ if and only if $$f$$ does the same thing to every equivalence class. That is if and only if $$a_1 \sim a_2$$ implies $$f(a_1) = f(a_2)$$. We do this because we need to know that $$f([a])$$ (which we define to be $$f(a)$$) does not depend on the choice of representative of $$[a]$$. The overloading of $$f$$ to mean both the function $$A \to B$$ and the function $$(A/\sim) \to B$$ has been the source of confusion in mathematics since it first became common. To clarify matters, let's write $$\tilde{f} : (A/\sim) \to B$$ defined by $$\tilde{f}([a]) = f(a)$$.

Now: we have a function $$\wedge$$ defined on differential forms. We would like to define a new function $$\tilde{\wedge}$$ defined on cohomology classes (which, recall, are equivalence classes). To do this we need to show that $$[\alpha] \tilde{\wedge} [\beta]$$ (defined to be $$[\alpha \wedge \beta]$$) is well defined. Of course, $$\tilde{\wedge}$$ is well defined if and only if we get the same output regardless of which representative we use.

Now, every representative of $$[\alpha]$$ looks like $$\alpha + d\omega$$, and every representative of $$[\beta]$$ looks like $$\beta + d\eta$$. So checking well definedness means checking that $$[\alpha] \tilde{\wedge} [\beta] = [\alpha + d\omega] \tilde{\wedge} [\beta + d\eta]$$. But by definition, this amounts to checking that $$[\alpha \wedge \beta] = [(\alpha + d\omega) \wedge (\beta + d\eta)]$$.

As you showed, $$(\alpha + d\omega) \wedge (\beta + d\eta) = (\alpha \wedge \beta) + d\nu$$, but this is in the same equivalence class as $$\alpha \wedge \beta$$. So $$[\alpha \wedge \beta] = [(\alpha + d\omega) \wedge (\beta + d\eta)]$$, and the function is well defined.

Somewhat unfortunately, working mathematicians seldom distinguish between $$\wedge$$ and $$\tilde{\wedge}$$ and we write $$\wedge$$ for both. This is useful, because they really are the same operation, and it would get annoying to have to write a bunch of extra squiggles all the time, but it is also confusing to students just entering the field.

I hope this helps ^_^