Let $(M^3,g)$ be a closed Riemannian manifold and let $u_1, u_2 : M \to \mathbb{S}^1 = \mathbb{R}/\mathbb{Z}$ be harmonic maps and let $h_i = u_i^{*}(d \theta_i)$, where $d\theta_i$ are the "volume forms" on each $\mathbb{S}^1$. Then each $h_i$ is a harmonic $1$-form. Is it true that $h = h_1 \wedge h_2$ is a harmonic $2$-form?
My thoughts: defining $u : M \to \mathbb{S}^1 \times \mathbb{S}^1$ by $u(p) = (u_1(p), u_2(p))$, we know that $u$ is harmonic. Also, $h = u^*({d\theta_1 \wedge d\theta_2})$. I don't know if this helps.