For $n = 2$ or $n = 3$, the manifold $S^n \times \mathbb{R}$ does not admit any complete Ricci-flat metric.

Let $(M, g)$ be a compact connected Riemannian manifold with $\operatorname{Ric}(g) \geq 0$. Then: 1. The universal Riemannian cover $(\tilde{M}, \tilde{g})$ is isometric to a Riemannian product $X \times \mathbb{R}^q$, where $X$ is compact, simply connected, and $\operatorname{Ric}(g_X) \geq 0$. 2. If there exists a point $p \in M$ with $\operatorname{Ric}(g)_p > 0$ (strictly positive at one point), then $\pi_1(M)$ is finite. 3. The full isometry group of $\tilde{M}$ decomposes as $I(\tilde{M}) = I(X) \times E(\mathbb{R}^q)$, where $E(\mathbb{R}^q)$ denotes the group of rigid Euclidean motions $y \mapsto Ay + b$ with $A \in O(q)$ and $b \in \mathbb{R}^q$. 4. If $\tilde{M}$ is homogeneous (its isometry group acts transitively), then so is $X$.

Let $(M, g)$ be a compact connected Riemannian manifold with $\operatorname{Ric}(g) \geq 0$. If $\pi_1(M)$ is infinite, then the universal cover $\tilde{M}$ contains a line.

Let $(M, g)$ be a compact Riemannian manifold and let $(\tilde{M}, \tilde{g})$ be its universal Riemannian cover. If $\tilde{M}$ is non-compact, then $\tilde{M}$ contains a line.

Let $(M, g)$ be a complete connected Riemannian manifold with $\operatorname{Ric}(g) \geq 0$. Then $M$ is isometric to a Riemannian product \begin{align*} (M, g) \cong (X \times \mathbb{R}^q,\; g_X + g_{\mathrm{Eucl}}) \end{align*} for some $q \in \{0, 1, 2, \ldots\}$ and some complete connected Riemannian manifold $(X, g_X)$ with $\operatorname{Ric}(g_X) \geq 0$ that contains no line. The manifold $X$ and integer $q$ are determined up to isometry.

Let $(M, g)$ be a complete connected Riemannian manifold. Suppose $\operatorname{Ric}(g) \geq 0$ everywhere and $M$ contains a line $\ell : \mathbb{R} \to M$. Then $M$ is isometric to a Riemannian product \begin{align*} (M, g) \cong (N \times \mathbb{R},\; g_0 + dt^2) \end{align*} for some complete Riemannian manifold $(N, g_0)$.

If $(M, g)$ is a complete Riemannian manifold that is disconnected at infinity, then $M$ contains a line.

Let $(M, g)$ be a connected, oriented, compact Riemannian manifold, and let \begin{align*} \bigwedge^k T^* M = \bigoplus_i \Lambda_i^k \end{align*} be the decomposition into irreducible representations of $\mathrm{Hol}(g)$, so that each fiber $(\Lambda_i^k)_x \subseteq \bigwedge^k T_x^* M$ is an irreducible $\mathrm{Hol}_x(g)$-representation. Then: 1. For all $\alpha \in \Omega_i^k(M) := \Gamma(\Lambda_i^k)$, we have $\Delta \alpha \in \Omega_i^k(M)$. 2. There is a decomposition of de Rham cohomology: \begin{align*} H_{\mathrm{dR}}^k(M) = \bigoplus_i H_{i,\mathrm{dR}}^k(M), \end{align*} where $H_{i,\mathrm{dR}}^k(M) = \{[\alpha] : \alpha \in \Omega_i^k(M),\ \Delta \alpha = 0\}$. The dimensions $\dim H_{i,\mathrm{dR}}^k(M)$ are called the **refined Betti numbers** of $(M, g)$.

Let $(M, g)$ be a connected Riemannian manifold, and fix $p, q \in \mathbb{Z}_{+}$ and $x \in M$. The following are equivalent: 1. There exists a $(p, q)$-tensor field $\alpha$ on $M$ such that $\nabla \alpha = 0$. 2. There exists an element $\alpha_0 \in T_x M^{\otimes p} \otimes T_x^* M^{\otimes q}$ such that $\alpha_0$ is invariant under the action of $\mathrm{Hol}_x(M)$.

$\mathrm{Hol}^0(M) \subseteq \mathrm{SO}(n)$.