# Killing algebra itself enough for spherical symmetry?

Suppose that $$M$$ is a manifold with a metric $$g$$, and three independent (independent in the sense of Killing vector fields) Killing vector fields $$K_1,K_2,K_3$$ are given with commutation relations $$[K_i,K_j]=\sum_k\epsilon_{ijk}K_k.$$ Here $$\epsilon_{ijk}$$ is the standard (algebraic) Levi-Civita symbol. This setup is standard when spherical symmetry is given. Usually when spherical spacetimes are given, additional assumptions are made.

I am trying to see if those additional assumptions are necessary. In particular, does this setup imply that the orbits of the isometry group generated by the $$K_i$$ is two-dimensional?

I tried to prove by contradiction and assumed that the $$K_i$$ are pointwise independent. Then by Frobenius' theorem, $$M$$ is foliated (at least locally) by three dimensional surfaces to which the $$K_i$$ are tangent, and the $$K_i$$ then form a frame, and are also Killing fields of the induced metric. I tried to use the fact that 1) the $$K_i$$ is a frame, 2) the $$K_i$$ are Killing, 3) the $$K_i$$ satisfy the above commutation relation to arrive at a contradiction. If such a contradiction had been obtained then it would have followed that the $$K_i$$ cannot be pointwise independent, and thus the distribution generated by them would be two or one dimensional, but a one dimensional manifold cannot accomodate three independent Killing fields, so then the distribution is two-dimensional. Unfortunately I have not been able to prove this.

So, does the above setup imply that the distribution generated by the $$K_i$$ is two-dimensional? If so how? References are also welcome if not a direct proof.

I claim the answer is no: You can take the manifold $$M=SU(2)\cong S^3$$ with its nice Riemannian metric and three Killing vector fields that satisfy the commutation relation you gave, but evidently the orbit of the isometry group is $$SU(2)$$ itself.

You also see in the literature that whenever people discuss such things they have to insert a dimensionality assumption on the orbits, e.g. the paper

On the global geometry of spherically symmetric space-times by J. SZENTHE

starts with a definition of "spherical" action of $$SO(3)$$ (on a manifold by isometries) by stipulating that the maximal dimension of any orbits is $$2$$.

• I.. did not think of that. With that said I was not totally convinced, but it seems to me after taking some computations that if one takes the standard generators of $SU(2)$ to be orthonormal at the identity, and then constructs a left-invariant metric based on it, the elements of the left-invariant frame corresponding to the generators will be Killing fields of the left-invariant metric. Or not sure which metric you meant by "its nice Riemannian metric", but the construction I outlined here seems to work so +1 anyways. Jun 6, 2020 at 11:09
• @BenceRacskó: There's a unique bi-invariant (left and right) metric because it's compact which will be the same as the round $S^3$ metric, unless I'm misremembering
– user21299
Jun 6, 2020 at 14:34
• $S^3$ obviously admits an isometric action of $SO(4)\cong SU(2)\times SU(2)/\mathbb Z_2$. I don't remember which of these $SU(2)$s is the natural left action of $SU(2)\cong S^3$ on itself (and would therefore have a 3-dimensional orbit), but I'd first check whether the diagonal subgroup works.
– user21299
Jun 6, 2020 at 14:39