Existence of Killing field 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 ?




 A: 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 $X = 0$. $\lozenge$
