Deriving metric from Killing field.

Question

Finding Killing fields from the metric in a Riemannian manifold is a standard procedure written in many relevant books. My question is the reverse: How can I find a metric for a manifold whose Killing fields are known?

Answer 1

You can't. The simplest reason is that "most" metrics (in a sense that can be made precise) have no Killing fields at all, because Killing fields correspond to one-parameter groups of isometries, at least in the compact case. For example, you can start with any metric, even one that has lots of Killing fields like the round metric on $\mathbb S^n$, and perturb it slightly by putting "bumps" in finitely many places so that it has no isometries at all.

Even for metrics with plenty of Killing fields, the Killing fields don't determine the metric. Any constant multiple of the Euclidean metric on $\mathbb R^n$ will have the same Killing fields as the Euclidean metric.

If you want an example of metrics that differ by more than just a constant multiple but still have lots of Killing fields, consider the Berger metrics on $SU(2)$ (which is diffeomorphic to $\mathbb S^3$). To construct these, let $X,Y,Z$ be a basis for the Lie algebra of $SU(2)$ satisfying $[X, Y] = 2Z$, $[Y, Z] = 2X$, and $[Z, X] = 2Y$, and for each positive real number $a$ define a left-invariant metric $g_a$ by declaring $\{X,Y,aZ\}$ to be a global orthonormal frame. As long as $a\ne 1$, all of these metics have the same Killing fields; but they have very different geometric properties.

In cases with lots of Killing fields, there are a couple of things that can be said. First of all, if the group action on $M$ generated by the Killing fields is transitive (meaning you can get from any point to any other point by flowing along finitely many trajectories of the Killing fields), then the metric is determined globally provided you know it at one point. But this still allows for many different metrics, as in the case of the Berger metrics.

In the extreme case, if the space of Killing fields on a connected $n$-manifold has dimension $ n(n+1)/2$ (the maximum possible dimension), then the metric is in fact determined up to a constant multiple, and $(M,g)$ must be a quotient of a constant curvature space (Euclidean space, sphere, or hyperbolic space) by a discrete group of isometries.

For the Killing fields that generate transitive groups of isometries, this question is closely related to Klein's Erlangen program, which is concerned with describing homogeneous geometries in terms of their isometry groups. If you know the Lie algebra of Killing fields, then you know the isometry group (at least the connected component of it, up to a possible quotient by a finite subgroup). It's not the same as knowing the metric, but it gives you lots of information about the metric's properties.