Existence of Killing field

Question

Definition of Killing field as picutre below, and acoording to 2.2.20, the existence of Killing field is equal to the solvability of $$ g_{ij,k}X^k+g_{kj}\frac{\partial X^k}{\partial x^j}+ g_{ik}\frac{\partial X^k}{\partial x^j} =0 $$ I don't know PDE, so I don't know whether the 1-order PDEs has solution for any given smooth Riemannian manifold $(M,g)$. Whether it must has solution ?

Answer 1

Not all Riemannian manifolds $(M,g)$ admit Killing fields. For example:

Lemma: Let $(M,g)$ be a Riemannian manifold, let $X$ be a Killing field on $(M,g)$, and let $f_X = \frac{1}{2}|X|^2$. Then $$\Delta f_X = |\nabla X|^2 - \text{Ric}(X,X).$$ Here, $\Delta$ is the Laplacian and $\nabla$ is the Levi-Civita connection of $g$.

Theorem (Bochner, 1946): Let $(M, g)$ be a compact, oriented Riemannian manifold without boundary. If $\text{Ric} < 0$, then there are no (non-trivial) Killing fields on $M$.

Proof: Let $X$ be a Killing field on $(M,g)$, and let $f_X = \frac{1}{2}|X|^2$ as above. Let $\text{vol}_g$ be the volume form on $(M,g)$. Observe that $$\Delta f_X \,\text{vol}_g = \text{div}(\text{grad } f_X)\,\text{vol}_g = d( (\text{grad}\,f_X) \,\lrcorner\,\text{vol}_g)$$ is an exact form. By Stokes' Theorem and the Lemma above: $$0 = \int_M \Delta f_X\,\text{vol}_g = \int_M (|\nabla X|^2 - \text{Ric}(X,X))\,\text{vol}_g,$$ so $$\int_M \text{Ric}(X,X)\,\text{vol}_g = \int_M |\nabla X|^2\,\text{vol}_g \geq 0.$$ Since $\text{Ric}(X,X) \leq 0$, this implies $\text{Ric}(X,X) = 0$, so $X = 0$. $\lozenge$