# Laplacian of the inverse of a diffeomorphism

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 $$D(\Phi^{-1})\circ \Phi = (D\Phi)^{-1}$$

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.

• Looking again at your answer after some time, I don't understand something. I understood the meaning of $*$, but which is the definition of the product of a matrix with real entries and a matrix with vector valued entries? (the product between the first two terms in (1) for example) – foo90 Jun 7 '18 at 9:50
• It's like ordinary matrix multiplication: for example, $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}$ is $\begin{pmatrix}aA+bC&aB+bD\\cA+dC&cB+dD\end{pmatrix}$ where $aA$ means multiplying vector $A$ by scalar $a$. – user357151 Jun 7 '18 at 10:10