How to see that existence non-trivial parallel $p$-form implies $b_p\geq 1$ using De Rham cohomology? Suppose $(M,g)$ be a (for simplicity consider closed) Riemannian manifold. Because every parallel $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. How to see this using De Rham cohomology (of course without using properties of Hodge theory like $\delta$, 1-1 correspondence between De Rham and Harmonic forms, etc.)? It is well-known that parallel forms are closed. What is the next step?
Such an easy argument based on harmonic forms I think there must be similar easy argument based on De Rham cohomology or it needs somewhat difficult argument?

You should read a reference on Hodge theory.
The conclusion is not hard, that the map $\mathcal H^k \to H^k_{dR}$ is injective (that is, that harmonic forms are closed and exact harmonic forms are zero). The hard part is arguing that this map is surjective.


Lemma: A $p$-form is harmonic, $\Delta \alpha = 0$, if and only if $d \alpha = 0$ and $d^* \alpha = 0$.

Proof: Recall that $\Delta = \pm (d^* d + d d^*)$, where the sign depends on dimension of manifold and degree $p$ of the form. $$\langle \pm \Delta \alpha, \alpha \rangle = \langle d^* d \alpha, \alpha \rangle + \langle d d^* \alpha, \alpha \rangle = \langle d^* \alpha, d^*\alpha \rangle + \langle d\alpha, d\alpha \rangle = |d^*\alpha|^2 + |d\alpha|^2,$$ here using that $d^*$ is the $L^2$-adjoint of $d$; the inner product is $\langle \alpha, \beta \rangle = \int \alpha \wedge *\beta$.
The RHS can only be zero if $d^* \alpha$ and $d\alpha$ are both zero. QED.


Corollary: if $\alpha$ is an exact harmonic form, then $\alpha = 0$.

Proof: we know that $d^*\alpha = 0$. If $\alpha = d\beta$, it follows that $d^* d\beta = 0$. But then $$0 = \langle d^* d\beta, \beta \rangle = \langle d\beta, d\beta \rangle = |d\beta|^2,$$ so that $0 = d\beta = \alpha,$ as desired. The same argument shows that coexact harmonic forms ($\alpha = d^* \beta$) are also zero.
