Let $S\subset \mathbb P_\mathbb C^3$ be a cubic hypersurface. By exponential sequence we know the natrual map $$Pic(S)\cong H^1(S,\mathcal O_S^*)\to H^2(S,\mathbb Z)$$ is an isomorphism. We denote this isomorphism by $c_1$.
Now the primitive part $H^2(S,\mathbb Z)_{pr}$ is defined to be the kernel of the map $-\mapsto -\wedge \omega$ where $\omega$ is the kahler form on $S$. I want to understand this primitive part. I read from somewhere (for example here) that it seems $c_1(\mathcal O_S(1))=\omega$. I don't know why this is true?
If $\omega_{\mathbb P^3}$ is the standard Fubini-Study Kahler form on $\mathbb P^3$, then it is a standard fact that $c_1(\mathcal O_{\mathbb P^3}(1)) = \omega_{\mathbb P^3}$.
[Indeed, $H^{1,1}(\mathbb P^3)$ is one-dimensional, so $c_1(\mathcal O_{\mathbb P^3}(1))$ must be proportional to $\omega_{\mathbb P^3}$. In fact, it must be a positive multiple, since $\mathcal O_{\mathbb P^3}(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $\omega_{\mathbb P^3}$ so that $\int_{\mathbb P^3} \omega_{\mathbb P^3}^3 = 1.$ Meanwhile, we know that $\int_{\mathbb P^3} c_1(\mathcal O_{\mathbb P^3}(1))^3 = 1$ because three linearly-independent hyperplanes in $\mathbb P^3$ intersect at a single point. Hence $c_1(\mathcal O_{\mathbb P^3}(1)) = \omega_{\mathbb P^3}.$]
Now suppose $i : S \to \mathbb P^3$ is the embedding of the cubic $S$ into $\mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $\omega_S := i^\star (\omega_{\mathbb P^3})$, which is the same thing as $i^\star (c_1 (\mathcal O_{\mathbb P^3}(1)))$.
The line bundle $\mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^\star \mathcal O_{\mathbb P^3}(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^\star c_1(\mathcal O_{\mathbb P^3}(1)) = c_1 (i^\star \mathcal O_{\mathbb P^3}(1)),$$ we have $$ \omega_S=c_1(\mathcal O_S(1)).$$