Cohomology product structures and ring structure.

The product structure of wedge forms induces a product structure on Cohomoloy classes.

If $[\omega] \in H^{k}(M)$ and $[\tau] \in H^{k}(M)$ on a manifold $M$, then natural way to define the product structure is

$[\omega] \wedge [\tau] = [\omega \wedge \tau] \in H^{k+l}(M)$.

Know that $\omega,\tau$ are closed forms. So first we need to establish that the class $[\omega \wedge \tau]$ is a closed form. Note,

$d[\omega \wedge \tau] = d\omega \wedge \tau + (-)^k \omega \wedge d \tau = 0$.

Hence, $[\omega \wedge \tau]$ is a closed form.

Since, we are dealing with classes here, we need to show that if representative $\tau$ is replaced by exact form $\tilde{\tau}= \tau + d\eta$, then we need to show that

$d[\omega \wedge \tilde{\tau}] = d\omega \wedge \tau + (-1)^{k} \omega \wedge d\eta$

Thus, $\omega \wedge \tilde{\tau}$ is equal to $d[\omega \wedge \eta]$. Hence, closed.

• For a manifold $M$ of dimension $n$, the direct sum is $H^{*}(M) = \oplus_{k=1}^{n} H^{k}(M)$
• Thus $\omega \in H^{*}(M$, can be written as $\omega=\omega_0+\omega_1+\cdots+\omega_n$ where $\omega_{i} \in H^{i}(M)$.
• Product of differential forms defined on $H^{*}(M)$ gives $H^{*}(M)$ a ring structure - called "Cohomology ring".
• Since product of differential forms is anticommutative, the ring is anticommutative.
• Direct sum gives Cohomology ring a graded algebra structure.
• Thus, $H^{*}(M)$ is anticommutative graded ring.