1
$\begingroup$

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 $\eta_2$?

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Praphulla Koushik Sep 6 '19 at 12:54
  • $\begingroup$ 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. $\endgroup$ – Grgur Palle Sep 7 '19 at 20:38
2
$\begingroup$

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 ^_^

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.