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I am stuck with the following exercise (exercise 15, chapter 9) in O'Neill's book on semi-Riemannian geometry:

"Let $M$ be a complete and connected semi-Riemannian manifold of dimension $n$. Show that the following statements are equivalent:

1) The isometry group of $M$ has dimension $n(n+1)/2$.

2) The algebra of Killing vector fields of $M$ has dimension $n(n+1)/2$.

3) Given any two points $p,q\in M$ and any linear isometry $\Lambda:T_pM\to T_q M$, there exists an isometry $\sigma:M\to M$ such that $\sigma(p)=q$ and $d\sigma_p=\Lambda$."

I have no problem in showing 1) $\Rightarrow$ 2) and 3) $\Rightarrow$ 1), but I cannot show 2) $\Rightarrow$ 3).

More specifically: given 2), it is clear to me that 3) with $p=q$ will hold whenever $\Lambda$ is connected to the identity, but I cannot see why it should hold also for $\Lambda$ not connected to the identity.

Any ideas?

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    I don't see a great way to get from $2$ to $3$. Instead, I think I would try to go from $2$ back to $1$, and then go from $1$ to $3$. (That said, I don't know much about semi-Riemannian things. For example, does the Hopf-Rinow theorem from Riemannian geometry still hold in this setting?) – 2017-01-25
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    If you could explain how to go from 1) to 3) that would be a completely satisfactory answer for me (for $M$ complete, the algebra of Killing fields is anti-isomorphic to the Lie algebra of the isometry group of $M$, so the equivalence between 1) and 2) is clear). The Hopf-Rinow theorem does not hold in semi-Riemannian geometry, as there is an example of a Lorentzian manifold, the Clifton-Pohl torus, which is compact but not complete (see https://en.wikipedia.org/wiki/Clifton–Pohl_torus). – 2017-01-25
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    Actually, the part from Hopf-Rinow I was thinking of is "complete $\implies$ any two points have a minimizing geodesic between them" - but that Clifton-Pohl torus is a cool example. Anyway, here's a sketch for the Riemannian case. Perhaps it is not too hard to adapt to the semi-Riemannian case? Let $FM\rightarrow M$ denote the frame bundle. Any isometry of $M$ induces a bundle isomorphism of $FM$. The kicker is the following statement: $f_1,f_2$ induce the same bundle isomorphism iff $f_1 = f_2$, which boils down to the following statement: If $f$ is an isometry with $f(p) = p$ and... – 2017-01-25
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    $d_p f = Id$, then $f = Id$. Once you have proven that, note that the group of isometries, acting on $M$ gives an embedding $G$ into $FM$ by first fixing a point $p\in M$ and frame $\{v_i\}$ at $p$ and then using $g\rightarrow (gp, g_\ast \{v_i\})$. By the "kicker", this map is injective. But $\dim FM = n(n+1)/2$. It follows that this embedding is a diffeo. In particular, $G$ can move any point to any other and any frame to any other. (Gotta run to class for the next while, but I will eventually respond to any comments you have!) – 2017-01-25
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    Thanks a lot! That's a very clear argument. The only part of it I don't understand is the implication "$f:G\to FM$ is injective and $\dim G=\dim FM$ $\Rightarrow$ $f$ is a diffeomorphism". For example, consider a manifold $M$ and an open subset $U$ of $M$. $U$ and $M$ have the same dimension, and the inclusion map $U\to M$ is injective, but it is not a diffeomorphism. – 2017-01-25
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    Yes, there is a gap there. I guess was implicitly assuming $G$ was compact. Then the image would be closed for free, and open by the dimension assumption, so, by conectedness, it would be everything. I will think about how to patch up the argument in the general case... – 2017-01-25
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    Kobayashi, Transformation Groups in Dif Geo, Thm 3.2 and Lemma 2 prove the image G in FM is closed,which is enough for my argument to carry through. – 2017-01-26
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    Thanks for the reference, it will probably be helpful. I think FM is in general not connected even though $M$ is (for example, if $M$ is the $n$-dimensional Euclidean space we have $FM={\mathbb R}^n\times O(n)$, which is not connected because O(n) is not connected), and hence the fact that the image of $G$ in $FM$ is both open and closed is not enough to conclude that it is equal to $FM$. – 2017-01-26
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    Another gap! If $M$ is non-orientable, then $FM$ will be connected, so the argument seems to work in that case. I will continue to think. (In the Riemannian case, what we have argued is already enough to show $M$ has constant sectional curvature. Such (complete) $M$ have been classified. I have no idea what happens in semi-Riemannian geometry) – 2017-01-26

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