Is there a Riemannian metric $\tilde g$ on $\mathbb R^d$ such that
$$\tag{1}
\Delta\_{\tilde g}=e^f(\Delta +1),$$
for some $f\in C^\infty(\mathbb R^d)$? Here $\Delta=\partial\_{x\_1}^2+\ldots+\partial\_{x\_d}^2.$ (Answer: no, as (1) fails on constants. See Edit below).


If there is such a $\tilde g$, it cannot be conformal to the standard Euclidean metric $g=\delta\_{ij}$. Indeed, if $\tilde g = e^{2\phi}g$, [then](https://anhngq.wordpress.com/2014/12/31/conformal-change-of-the-laplace-beltrami-operator/)
$$\Delta\_{\tilde g} = e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x\_j}\frac{\partial}{\partial x\_i}\right),$$
and either $d=2$, or the second summand in the round brackets is constant only in the trivial case $\nabla \phi=0$. In both cases (1) cannot be satisfied.


**EDIT**.
The equation (1) cannot hold verbatim, as it clearly fails on constant functions (thanks Terry Tao for this remark). Instead, let us consider
$$
\tag{1b} L\_{\tilde g} = e^f(\Delta +1), $$
where
$$
L\_{\tilde g}=\frac{d-1}{4(d-2)} \Delta\_{\tilde g} - \mathrm{Scal}\_{\tilde g}$$
is the [conformal Laplacian](https://anhngq.wordpress.com/2014/12/21/conformal-changes-of-the-green-function-for-the-conformal-laplacian/). The additive term is the scalar curvature of $\tilde g$.


In this case, the fact that $\tilde g$ cannot be conformal to the Euclidean metric is even more apparent, as $L\_{\tilde g}$ is conformally invariant.



Is there a metric on Euclidean space that turns the Helmholtz equation into the Laplace equation?
After Terry Tao's comments, I came to the conclusion that the only possible choice of a metric $\tilde g$ and of an operator
$$T\_{\tilde g}= \Delta\_{\tilde g} + \text{lower order terms} $$
that will give
$$\tag{1} T\_{\tilde g}=e^f(\Delta +1)$$
is the following, in Cartesian coordinates:
$$\tag{2}
\tilde g\_{ij}= e^{2\phi}\delta\_{ij},\qquad T\_{\tilde g}=\Delta\_{\tilde g}-e^{-2\phi}\delta^{ij}\partial\_i \phi \partial\_j + e^{-2\phi}.$$
That is, $\tilde g$ must be conformal to the standard Euclidean metric, with an arbitrary conformal factor $e^{2\phi}$. **I have no idea whether the operator $T\_{\tilde g}$ has a geometric meaning**, or if this small computation can ever be useful. In any case I am posting it here.


Proof.


We can write (1) as
$$
\tilde{g}^{ij}\partial\_i\partial\_j + a^k\partial\_k +c= e^f \delta^{ij}\partial\_i \partial\_j + e^f, $$
for some scalar fields $a\_k$ and $c$. This clearly implies $\tilde g^{ij}=e^f \delta^{ij}$, which is the first equation in (2), and also $a\_k=0, c=e^{-2\phi}$. Since
$$\Delta\_{\tilde g}= e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x\_j}\frac{\partial}{\partial x\_i}\right),$$
the only possibility is to define $T\_{\tilde g}$ like in the second equation in (2).


