Are left-invariant fields mapped onto the manifold Killing vectors?

[center o bass]

Left invariant fields on a group G satisfies a lie algebra; say we have an n-dimensional Lie algebra for which the fields ${X_1, \ldots , X_n}$ is a basis. Let these satisfy the algebra $[X_a, X_b] = c_{ab}^c X_c$. Suppose now that we have a Riemannian manifold with killing vectors ${\xi_1,\ldots, \xi_n}$ and let they satisfy the same algebra $[\xi_a, \xi_b] = c_{ab}^c \xi_c$. Let $p \in M$ and the action of the group G on M be denoted $g \cdot p$. Then we have the map $F: TG \to TM$ given by

$$X_a \mapsto X_a^{*} := \left. \frac{d}{dt}\right|_{t = 0} e^{t X_a} \cdot p.$$

Is $X_a^{*}$ identical to the killing field $\xi_a$? If so, how does one prove it?

 

[quasar987]

If the action is effective, then the map X-->X* is injective. If not, then it seems to me that X-->X* may be non injective. In that case, the answer to your question is certainly negative.

Even if we assume the action is effective, then just looking at the case where n=1. Any X in Lie(G) is such that [X,X]=0 as well as any Killing field on M satisfies [K,K]=0. But as soon as there exist two or more Killing fields on M, we have a counter-example to your question.

 

[center o bass]

If the action is effective, then the map X-->X* is injective. If not, then it seems to me that X-->X* may be non injective. In that case, the answer to your question is certainly negative.

Even if we assume the action is effective, then just looking at the case where n=1. Any X in Lie(G) is such that [X,X]=0 as well as any Killing field on M satisfies [K,K]=0. But as soon as there exist two or more Killing fields on M, we have a counter-example to your question.

How does that counter-example arise? If we can't do the identification of the vector fields, does it however follow that if $X$ is a basis of the lie algebra of G, which is the same as that of the killing vectors on M, then $X^*$ is a killing vector? In other words, is $\mathcal{L}_{X^*} g = 0$ where g is the metric tensor on M?

At the start of page 182 in Arthur Besse's book "Einstein manifolds", he identifies $X^*$ with the Killing fields on M, but notes that

$$[X,Y]_{\mathcal{g}} = -[X^*,Y^*]$$

where $[ \ , \ ]_{\mathcal{g}}$ denotes the Lie bracket on the Lie algebra.

I think this would be provable if the map $X \mapsto X^*$ could formulated as a pushforward. And it seems that it would have to be related to the pushforward of the group action.

 

[quasar987]

If G acts by isometries on (M,g), then X* is Killing for every X in Lie(G). Is that what you're asking?!

 

[center o bass]

If G acts by isometries on (M,g), then X* is Killing for every X in Lie(G). Is that what you're asking?!

No, but rather: If the Lie algebra of G is 'equivalent' to the Lie algebra of the killing vectors on (M,g) will G act on M by isometries?

Take the example of spherical symmetry. According to Wald ( or Schutz spherical symmetry is defined as follows: If the Lie algebra of killing vector fields on (M,g) has a subalgebra which is the Lie algebra of SO(3), then we say that (M,g) has spherical symmetry.

Now take an basis vector of the Lie algebra of SO(3), let's call it X and map X according to $X \mapsto X^*$. Do we then have that $\mathcal{L}_{X^*} g = 0$?

 

[center o bass]

Or simply: When does G act on (M,g) by isometries? When are $X^*$ the killing vectors on (M,g)?

 

[quasar987]

So if I understand correctly the situation, you've read this abstract-looking definition and now you're trying to make sense of it. Is that what's going on?

If that is so, then I don't think you're asking the right question, because there is no a priori action of G on M in the definition of Wald. He only speak of the existence of an embedding of so(3) in k.

Here is how I would make sense of the definition however: The isometry group Isom(M,g) of a riemannian manifold is a finite dimensional Lie group with Lie algebra k (the Killing fields) (see Kobayashi-Nomizu). Therefor, by the general theory of Lie groups (see John Lee), if a is a subalgebra of k, there is a unique connected subgroup A of Isom(M,g) whose Lie algebra is a. On the other hand, since SO(3) also has Lie algebra a and is simply connected, SO(3) is the universal cover of A, and so there is a homomorphism (the covering map) SO(3)-->A\subset Isom(M,g). This, by definition, is an action of SO(3)~S² on M by isometries.

 

[center o bass]

So if I understand correctly the situation, you've read this abstract-looking definition and now you're trying to make sense of it. Is that what's going on?

If that is so, then I don't think you're asking the right question, because there is no a priori action of G on M in the definition of Wald. He only speak of the existence of an embedding of so(3) in k.

Here is how I would make sense of the definition however: The isometry group Isom(M,g) of a riemannian manifold is a finite dimensional Lie group with Lie algebra k (the Killing fields) (see Kobayashi-Nomizu). Therefor, by the general theory of Lie groups (see John Lee), if a is a subalgebra of k, there is a unique connected subgroup A of Isom(M,g) whose Lie algebra is a. On the other hand, since SO(3) also has Lie algebra a and is simply connected, SO(3) is the universal cover of A, and so there is a homomorphism (the covering map) SO(3)-->A\subset Isom(M,g). This, by definition, is an action of SO(3)~S² on M by isometries.

You are right that I am looking to understand that definition and it's consequences.

Is there no canonical action of Isom(M) on M?

From what you just argued, is it possible to argue that a a manifold with SO(3) symmetry is isometric to a sphere? Or locally isometric? What about homeomorphic?