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Guido Fubini, ``Sugli spazii che ammettono un gruppo continuo di movimenti,'' Annali di Mat., ser. 3, 8 (1903) 54.: Let $M$ be a Riemannian manifold of dimension $d\ge 3$. Its isometry group cannot be of dimension $d\,(d+1)/2-1$.

Does this theorem remain true in the pseudo-Riemannian setting?

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    $\begingroup$ I am not very familiar with Lie algebras but I think the following argument works: This would give a codimension 1 subgroup of $O(p,q)$. A theorem of Tits (see here) then would give a nonzero homomorphism $\mathfrak o(p,q) \to \mathfrak{sl}_2$. But $\mathfrak o(p,q)$ is simple, so this implies $d = p + q \leq 2$. $\endgroup$
    – mme
    Commented Jul 19, 2020 at 8:44
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    $\begingroup$ @MikeMiller For $d=4$ the maximal isometry groups are the anti de Sitter group $O(2,3)$, the Poincar\'e group $O(1,3)\ltimes R^4$ and the de Sitter group $O(1,4)$. $\endgroup$
    – Thomas Schucker
    Commented Jul 19, 2020 at 10:48
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    $\begingroup$ Thank you for correcting me! I will leave my comment up in case it helps anyone else who is unfamiliar. $\endgroup$
    – mme
    Commented Jul 19, 2020 at 12:42

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I just found the answer by G. S. Hall (2003) in Class. Quantum Grav. 20 3745.

Theorem 8. Let $M$ be a connected smooth paracompact manifold of dimension $n ≥ 3$ admitting a smooth metric $g$ of signature $(p,q)$ with $p ≤ q$ and $q ≥ 3$. Let $K(M)$ be the vector space of all global smooth Killing vector fields on $M$. Suppose dim$K(M)\not=1/2\,n(n+1)$. Then dim$K(M)<1/2\,n(n+1)−1$.

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