Laplacian of the inverse of a diffeomorphism

Question

Let $\Phi: \mathbb{R}^n \to \mathbb{R}^n$ be a diffeomorhism. Let $\Delta$ be the componentwise Laplacian. Is it possible to wirite $$ \Delta (\Phi^{-1})\circ \Phi $$ in such a form that involves only $\Phi$ and not $\Phi^{-1}$?

To clarify my question: For the Jacobian matrix (which I will denote by $D$) it is clearly possible. Indeed $$ D(\Phi^{-1})\circ \Phi = (D\Phi)^{-1} $$

Answer 1

Yes, but as usual, working with higher order derivatives of inverse map is unpleasant. I use the notation

• $\Psi = \Phi^{-1}$
• $H\Phi$ is the Hessian matrix of $\Phi$, which is an $n$ by $n$ matrix whose entries are vectors (so, the entry $(i, j)$ is the derivative $\frac{\partial^2 \Phi}{\partial x_i\partial x_j}$).
• $A*H\Phi$ means applying matrix $A$ to each entry of $H\Phi$ (similar to multiplying an ordinary matrix by a scalar).
• $\operatorname{tr}$ is the trace of a matrix, for example $\Delta \Phi = \operatorname{tr} H\Phi$.

With this notation, $$ H \Psi = -((D\Phi)^{-1})^T ((D\Phi)^{-1}*H\Phi) (D\Phi)^{-1} \tag1 $$ hence $$ \Delta \Psi = -\operatorname{tr}(((D\Phi)^{-1})^T ((D\Phi)^{-1}*H\Phi) (D\Phi)^{-1}) \tag2 $$ which generalizes the one-dimensional formula for the second derivative of inverse function. Here and below the arguments are suppressed; it is understood that $\Phi$ and its derivatives are evaluated at some point $x$, while $\Psi$ and its derivatives are evaluated at $\Phi(x)$.

Proof: let $\phi^i$, $\psi^i$ denote the components of $\Phi$ and $\Psi$, while subscripts are used for derivatives. The relation $D\Psi = (D\Phi)^{-1}$ can be written as $$ \sum_k \psi^i_k \phi^k_j = \delta^{ij} \quad \text{(Kronecker delta)} $$ Differentiate both sides with respect to $x_\ell$ using the product and chain rules: $$ \sum_{k, m} \psi^i_{km} \phi^k_j \phi^m_\ell + \sum_k \psi^i_k \phi^k_{j\ell} = 0 $$ The first sum on the left is $(D\Phi)^TH\Psi D\Phi$ while the second is $D\Psi*H\Phi$. The claim (1) follows.