# Family of Hodge decomposition

It is known that a metric $$g$$ gives a Hodge decomposition: $$\Omega^*(M)=\mathcal H^*(M)\oplus d\Omega^*(M) \oplus \delta_g \Omega^*(M)$$ Note that the usual differential restricts to an isomorphism on $$\delta_g\Omega^*(M)$$, denoted by $$d_g: \delta_g\Omega^*(M) \to d\Omega^*(M)$$ (In fact, it is clearly injective and it is surjective since $$d\Omega^*(M)=d\delta_g\Omega^*(M)$$) Now let's say there is a family of Riemannian metrics $$g_t, t\in [0,1]$$, then many new operators can be created by taking derivatives, like $$\frac{d }{dt}d_{g_t}$$, $$\frac{d}{dt} d_{g_t}^{-1}$$ or $$\frac{d}{dt}\mathrm{proj}_{\mathcal H_{g_t}^*(M)}$$. In particularly, I am interested in the operator $$\frac{d}{dt} {d_{g_t}^{-1}}$$ because $$\mathrm{image} (d_{g_t}^{-1})=\delta_{g_t}\Omega^*(L)$$ gives a smooth family of subspaces in $$\Omega^*(L)$$, and this derivative is like the infinitesimal deformation of this subspace family.

Question 1: How do we describe the derivative operator $$\frac{d}{dt} {d_{g_t}^{-1}}$$? What condition on $$(g_t)$$ can infer that $$\frac{d}{dt}d_{g_t}^{-1}$$ is 'orthogonally transversal' to $$d^{-1}_{g_t}$$?

Another probably related question is that

Question 2: Can we regard $$\frac{d}{dt} \mathrm{proj}_{\mathcal H^*_{g_t}}$$ as an operator from $$\mathcal H^{*\perp}_{g_t}$$ to $$\mathcal H^{*\perp}_{g_t}$$?