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Yes: Take the shrinking Gaussian $(\mathbb{R}^n, dx^2)$ with $X=\rho\nabla\rho$, where $\rho$ denotes the distance to the origin. This space is Ricci flat, so your inequality holds.

The Riemannian product $\mathbb{R}^m \times S^n$ is always of this type, where $\mathbb{R}^m$ is given the flat metric and $S^n$ the round metric of constant sectional curvature one. In this case $\l …

Here is my favorite way to do this computation. Let $$ \delta_{i_1 \dotsm i_k}^{j_1 \dotsm j_k} = \begin{cases} \mathrm{sgn}\, \sigma, & \text{if $j_k = i_{\sigma(k)}$}, \\ 0, & \text{otherwise} \end{ …

Yes, it is true. This is a consequence of the conformal invariance of the conformal Laplacian. Let $(M^n,g)$ be a Riemannian manifold. The conformal Laplacian is $$ L_2^g = -\Delta + \frac{n-2}{4(n- …

This follows easily from the fact that the Kelvin transform is a conformal diffeomorphism and the naturality of the Schouten tensor. Let $\Phi(z) := \frac{z}{\lvert z\rvert^2}$ denote the Kelvin trans …

Yes. (With the caveat that if there is a function such that $\nabla^2\varphi = \psi g$, then necessarily it is a warped product over a one-dimensional base, so your question should really require $\d …

Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equatio …

The Bach tensor is only divergence-free in dimension four. In general dimension, it holds that $\nabla^j B_{ij} = (n-4)P^{jk} ( \nabla_i P_{jk} - \nabla_j P_{ik} )$. I pulled this formula from the b …

You can also derive the lemma from the Bochner formula, which in general dimensions can be written $$ \frac{1}{2}\Delta\lvert\nabla u\rvert^2 = \lvert\nabla^2u\rvert^2 - (\Delta u)^2 + \delta\left((\ …

Theorem 1.4 in C. Frances' preprint "Rigidity at the boundary for conformal structures and other Cartan geometries" asserts that the geodesic compactification is unique (up to conformal diffeomorphism …

If you also assume that $(N,\ddot g)$ is complete, then your assumptions imply that $(N,\ddot g)$ is Ricci flat. I proved this in my article The nonexistence of quasi-Einstein metrics.

The first observation is that $Q_g(\phi)$ is conformally invariant. Define \begin{align*} L_gu & := -\Delta u + \frac{n-2}{4(n-1)}Ru , \\ B_gu & := \partial_\nu u + \frac{n-2}{2}Hu , \end{align*} wh …

Let $(M,[g])$ be a conformal manifold; i.e. $(M,g)$ is a Riemannian manifold and $[g] = \{ u^2g \mathrel{}:\mathrel{} u \in C^\infty(M), u>0 \}$ is the set of Riemannian metrics conformal to $g$. It i …

The discrepancy comes from the fact that the square of the Dirac operator is in general not conformally covariant. There are ways to modify powers of the Dirac operator to get a conformally covariant …

You are forgetting that, as part of the renormalization, the functions $\psi_s$ all satisfy $\psi_s(-P)=1$ (here $-P$ is the south pole). In particular, since $\psi$ is $C^2$ away from the north pole …