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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 $$\Delta_{\tilde g} = e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x_j}\frac{\partial}{\partial x_i}\right),$$ 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. 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.

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    $\begingroup$ Applying (1) to the constant function 1 gives $0=e^f$, which is absurd. $\endgroup$
    – Terry Tao
    Commented Oct 28, 2020 at 18:49
  • $\begingroup$ @TerryTao: Right, that was easy. I actually oversimplified my question. What I am really interested in is not the standard Laplacian $\Delta_{\tilde g}$, but rather the conformal Laplacian $L_{\tilde g}= \Delta_{\tilde g} + C \mathrm{Scal}_{\tilde g}$. I will edit. $\endgroup$ Commented Oct 28, 2020 at 19:06
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    $\begingroup$ Even with the lower order term, a comparison of top order terms in (1) shows that $\tilde g^{ij} = e^f \delta^{ij}$ (possibly up to a normalising constant depending on your definition of conformal Laplacian), so your analysis of the conformal case applies. $\endgroup$
    – Terry Tao
    Commented Oct 28, 2020 at 19:15

1 Answer 1

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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).

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