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A pseudo-Riemannian manifold $M$ of dimension $n$ is said to be maximally symmetric if the space of its Killing vector fields has $n(n+1)/2$ dimensions.

If $M$ is maximally symmetric, then we have the following: for every $p\in M$ and every linear isometry $\Lambda:T_pM\to T_pM$ connected with the identity, there exists an isometry $\sigma:M\to M$ such that $\sigma(p)=p$ and $d\sigma_p=\Lambda$.

On the other hand, all maximally symmetric spaces I know of (flat spaces, sphere, hyperbolic space, de Sitter and Anti de Sitter spacetimes) have a stronger property: for every $p\in M$ and every linear isometry $\Lambda:T_pM\to T_pM$ (not necessarily connected with the identity), there exists an isometry $\sigma:M\to M$ such that $\sigma(p)=p$ and $d\sigma_p=\Lambda$.

My question is: is this stronger property a consequence of maximal symmetry? In case it is not, I would like to know of an example of a maximally symmetric space not having this stronger property.

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Careful: the existence of $n(n+1)/2$ dimensional space of Killing fields does not imply global homogeneity. Think of any open subset of Euclidean space; typically they have no global isometries.

On the other hand, maximal dimensional Killing algebra implies local isometry to a pseudoRiemannian space form. This is not hard to prove using moving frames, where you calculate the effect on the curvature, since the stabilizer of a point will still have as many dimensions as the rotations/Lorentz group.

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  • $\begingroup$ To be very specific, take as your open subset of Euclidean space the interior of a nonisoceles triangle. $\endgroup$
    – Ben McKay
    Commented Jan 20, 2017 at 22:05
  • $\begingroup$ To give more detail, break up the curvature tensor into its irreps. Then all of these vanish, except for the trivial representation (corresponding to the scalar curvature), so the traceless Ricci and the Weyl vanish. $\endgroup$
    – Ben McKay
    Commented Jan 20, 2017 at 22:09
  • $\begingroup$ Thanks Ben for your answer. What I mean by a Killing field is a vector field which satisfies the Killing equation and which is complete, so that it generates a one-parameter group of isometries. Perhaps I should have emphasized that. This excludes open subsets of Euclidean space as examples of maximally symmetric spaces. $\endgroup$
    – Guillem Pérez-Nadal
    Commented Jan 21, 2017 at 0:26
  • $\begingroup$ If your Killing vector fields are complete, then they generate a group action by a transitive group of isometries. Constant curvature (as above) implies isomorphism with a space form. So your space form examples are the only examples. $\endgroup$
    – Ben McKay
    Commented Jan 21, 2017 at 8:11

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