Existence of nonvanishing Killing field

Question

Let $(M,g)$ be a closed Riemannian manifold.

Q Is there any research about the existence of nonvanishing Killing field, especially the nontrivial example.

Answer 1

For an example just take the vector field $\frac{d}{d\theta}$ on $S^{1}$ (with the round metric inherited from $S^{1} \subset \mathbb{R}^{2}$). I don't know whether you consider this to be non-trivial. Also the product of $S^{1}$ with any Riemannian manifold will have a non-vanishing Killing vector field.

A nice general statement about the existence of Killing vector fields is as follows:

A compact Riemannian manifold $(X,g)$ has a Killing vector field if and only if it has a non-trivial isometric $S^{1}$-action.

(by the way, this fails for non-compact manifolds.)

I learnt about this statement here https://math.stackexchange.com/questions/3039085/proof-of-the-existence-of-a-nontrivial-killing-vector-field-is-equivalent-to-the/3039600#3039600. There is also a proof given as an answer there.

There are smooth manifolds which don't have any (non-trivial) smooth $S^{1}$-actions for example take a compact surface of genus $g \geq 2$ (or if you prefer simply connected examples see the answers to this question Obstruction to a general S^1-action). So we also have a non-existance result in some cases (for any Riemannian metric on these smooth manifolds).

A necessary condition for a non-vanishing killing vector field is $\chi(M) = 0$, so I should give one more example. Any compact orientable 3-manifold $M$ has $\chi(M)=0$ by Poincare duality. But the $3$-manifolds with (smooth) $S^{1}$-actions are all Seifert fibred spaces. So compact orientable hyperbolic 3-manifolds would give examples with $\chi(M)=0$ but no Killing vector fields (for any Riemannian metric on them).