Consider the manifold $M := \mathbb{R}^3 \setminus B_1$ where $B_1$ is the unit ball. Equip $M$ with an asymptotically flat metric $g$ of high order. Let $\gamma$ be the induced metric on $\partial M$.
The Riesz transform is the bounded linear map $R: L^2(\partial M) \to L^2 (\Lambda^1 (T^* \partial M))$ defined by
$$R (f) = d (- \Delta_{\gamma})^{-\frac{1}{2}} f$$
Define another operator $R^* : L^2(\partial M) \to L^2 (\Lambda^1 (T^* \partial M))$ by
$$R^*(f) = d N_g^{-1} (f)$$
Where $N_g: H^1(\partial M) \to L^2(\partial M)$ is the Dirichlet to Neumann map, which is defined by the following: if $h = N_g(f)$, then $h = \nu \cdot \nabla u $ where $\nu$ is the unit normal vector field on $\partial M$ and $u$ is the unique function on $M$ that goes to $0$ at infinity and satisfies $\Delta_gu=0$ and $\left. u \right|_{\partial M} = f$.
It is well known that $N_g$ is a pesudo differential operator of order 1 with principal part $- (-\Delta_{\gamma})^{\frac{1}{2}}$. Also, $N_g$ is invertible (that's not true on bounded domains).
My question will not be very specific. I want to understand $R^*$ more as well as the relationship between $R$ and $R^*$.
Is $R^*$ also a bounded operator? If so how does $\Vert R^*\Vert$ depend on $g$? Or does it only depend on $\gamma$? What do we know about $R-R^*$? I would imagine this is a "small" operator in some sense because $N_g$ is "close" to $- (-\Delta_{\gamma})^{\frac{1}{2}}$. Maybe $R-R^*$ is a smoothing operator (which is a psuedo differential operator of order $-\infty$.)? Is there anything in the literature about $R^*$?
You can assume $g$ is the euclidean metric and so $\gamma$ is the round metric on the sphere if that makes the question easier. In that case, I think $N_g = - \sqrt{\Delta_{\gamma} + \frac{1}{4}} + \frac{3}{2}$
Any help or references will really be appreciated.
