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 $\mathcal H^{*\perp}_{g_t}$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.