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I came across this concept on this wiki page regarding killing vector field. The last sentence in section "Cartan Involution" says that "Equivalently, the curvature tensor is covariantly constant on locally symmetric spaces, and so these are locally parallelizable; this is the Cartan–Ambrose–Hicks theorem."

I'm really confused that apparently every manifold is "locally parallelizable", since you simply take a local chart and $\frac{\partial}{\partial x^i}$ automatically form a local parallelization. Therefore, I think it actually says that the manifold admits $n$ linearly independent local killing vector fields, not some general vector fields.

However, the wiki page doesn't cite anything here, so I don't know where that result come from, and of course I don't know how to prove it either. Besides, I don't understand its connection with Cartan–Ambrose–Hicks theorem.

Could someone explain that to me?

Selene
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    I never heard of this terminology: Anybody can edit Wikipedia articles, so I suggest you simply ignore this sentence. However, do read about the Cartan–Ambrose–Hicks theorem. – Moishe Kohan Jul 24 '21 at 11:55
  • @MoisheKohan So does that conjecture make sense? There locally exists a parallelization of killing vector fields? – Selene Jul 25 '21 at 03:39
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    If you mean a local parallelization by Killing fields, then it is not even a conjecture but a fact and it holds for all locally homogeneous Riemannian manifolds. However, this is not what the C-A-H theorem says. I checked on the Wikipedia page who is responsible for the offending sentence: It appears to be a physicist, https://en.wikipedia.org/wiki/User_talk:67.198.37.16. – Moishe Kohan Jul 25 '21 at 04:02
  • @MoisheKohan Thanks! Would you minding giving some references about the fact on the local parallelization by killing fields for all locally homogeneous Riemanniann manifolds? I'm still learning these things. – Selene Jul 25 '21 at 04:07
  • I will write a proof in a day or so. – Moishe Kohan Jul 25 '21 at 05:01
  • @MoisheKohan thanks – Selene Jul 25 '21 at 05:35

1 Answers1

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Recall that each locally symmetric space is locally homogeneous, and, since the question is local, I will limit myself to the discussion of homogeneous Riemannian manifolds. Namely, given a homogeneouse manifold $M$ and a point $x\in M$, I will construct a neighbourhood $U$ of $x$ and a collection of Killing fields on $M$ whose restriction to $U$ defines a parallelization of $U$.

The condition that a Riemannian manifold $M$ is homogeneous is equivalent to the assumption that this manifold is of the form $G/H$, where $G$ is a Lie group, $H< G$ is a compact Lie subgroup and $G$ is equipped with a left-$G$-invariant Riemannian metric. Since $M$ is homogeneous, this identification can be made so that $x\in M$ corresponds to the projection of the identity element $e\in G$. (The diffeomorphism $G/H\to M$ comes from the orbit map $g\mapsto gx, g\in G$.)

Now, given this description, let ${\mathfrak h}\subset {\mathfrak g}$ denote the Lie algebras of $H$ and $G$ respectively. Take any linear subspace ${\mathfrak p}\subset {\mathfrak g}$ such that we have a direct sum decomposition $$ {\mathfrak g}= {\mathfrak h}\oplus {\mathfrak p}. $$ In particular, the dimension $d$ of ${\mathfrak p}$ is the same as the dimension of $M$. Taking a basis $v_1,...,v_d$ in ${\mathfrak p}$, we see that these vectors project to Killing fields $X_1,...,X_d$ on $M$ which are linearly independent at $x$. By continuity, the same linear independence will hold at all points in a sufficiently small neighborhood $U$ of $x$ thereby, providing a "Killing" parallelization of $U$.

According to the comments, this observation appears to answer your actual question. Needless to say, this is not what the Cartan–Ambrose–Hicks theorem says. Looking at the record of a person who made the edit in the Wikipedia page claiming that "this is the Cartan–Ambrose–Hicks theorem," this person is likely to be a physicist and the quote should not be taken seriously. This is also a general remark regarding Wikipedia articles: Anybody can edit these and, while, in general, the quality of math Wikipedia articles is quite good, occasionally what's written is wrong or misleading or unclear. Reading Wikipedia is not a substitute to reading math books.

Moishe Kohan
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  • As I remember, there are manifolds where each killing vector field must have a zero. So they can't be linearly independent everywhere, i.e., not a parallelization. The projection from $\mathfrak{p}$ only implies that they are $\mathbb{R}$-linearly independent, not $C^\infty$-linearly independent. – Selene Jul 29 '21 at 07:26
  • @Selene: You are right, the statement should be local. I will correct the answer accordingly. – Moishe Kohan Jul 29 '21 at 10:12