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A m-dimensional completed and connected (Semi-)Riemannian manifold which has $m(m+1)/2$ independent global Killing vector fields is called maximally symmetric space.

Then what are all possibilities of maximally symmetric spaces? Can simply or not simply connected maximally symmetric (Semi-)Riemannian manifold be completely classified ?

I know in the case of 3-dimensional Riemannian manifold, 3-sphere, 3-Euclidean and 3-hyperbolic space are maximally symmetric space. Besides these cases, are there any nontrivial case, such as not simply connected ?

Are there some literatures or textbooks which have solved this problem completely ?

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    $\begingroup$ I suppose you mean to require that the dimension of the vector space of global Killing vector fields is $m(m{+}1)/2$, and you probably also meant to specify that the space should be connected and complete (otherwise there are trivial examples given by taking open subsets or disjoint unions of examples with lower symmetry). With these hypotheses in place, the possibilities are much reduced. For example, most space forms do not qualify because the space of global Killing fields on a discrete quotient of a space form is generally much smaller; it can even have dimension zero. $\endgroup$
    – Robert Bryant
    Commented Nov 7, 2014 at 9:58
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    $\begingroup$ Now that you have edited your question to add the missing hypotheses, the answer is fairly straightforward. For example, in the Riemannian cases, the only examples are $S^m$, $\mathbb{RP}^m$, $\mathbb{E}^m$ and $\mathbb{H}^m$. In general, essentially, these are the flat spaces $\mathbb{E}^{p,q}$ and the connected open orbits of $\mathrm{O}(p,q{+}1)$ acting on $\mathbb{RP}^{p+q}$ for all pairs $(p,q)$ with $p+q = m$ and $p,q\ge0$ and their connected covering spaces. $\endgroup$
    – Robert Bryant
    Commented Nov 7, 2014 at 12:15
  • $\begingroup$ @RobertBryant Thanks a lot! Are there some literatures or textbooks which have solved this problem completely ? $\endgroup$
    – 346699
    Commented Nov 7, 2014 at 12:26

2 Answers 2

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Connected and complete maximally symmetric spaces have constant curvature and therefore are space forms. It is fairly easy to prove that simply connected space forms are completely characterized by their index, curvature, and dimension (see Chapter 8, proposition 23 in O'Neill Semiriemannian Geometry).

The simply connected space forms can be presented all as generalisations of Euclidean, hyperbolic, and spherical spaces in the following way (Chapter 8, Corollary 24, Ibid):

  • Euclidean $\implies$ Flat semi-Riemannian spaces of some index, including of course Minkowski spaces.
  • Sphere $\implies$ pseudo-spheres. For index $(j,k)$ (which I take to mean dimension $j+k$ and $j$ minuses and $k$ pluses), this is (the universal cover of a component of) the set $\{|x| = c^2\}$ in $\mathbb{R}^{j,k+1}$.
  • Hyperbolic $\implies$ pseudo-hyperbolic spacess. For index $(j,k)$ this is (the universal cover of a component of) the set $\{|x| = -c^2\}$ in $\mathbb{R}^{j+1,k}$.

By a covering argument all connected components of space-forms are quotients of these three guys. (For example, in 3 dimension, Riemannian case, don't forget the lens spaces.) For this you should remember that the symmetry groups are

  • pseudo-Euclidean spaces $\mathbb{R}^{j,k}$ has symmetry group $\mathbb{R}^{j,k} \rtimes SO(j,k)$
  • pseudo-spheres have the Lorentz group of the ambient space $SO(j,k+1)$
  • pseudo-hyperbolic spaces have the Lorentz group of the ambient space $SO(j+1,k)$.
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  • $\begingroup$ Actually, make that "connected maximally symmetric spaces have constant curvature". In the OP's original post (since corrected), the hypothesis of connectedness was omitted from the definition of 'maximally symmetric space', and, without it, the 'constant curvature' conclusion is false. (For example, consider the disjoint union of three tori of revolution in $3$-space. The space of Killing vector fields indeed has dimension $3$, but the surface doesn't have locally constant curvature anywhere.) $\endgroup$
    – Robert Bryant
    Commented Nov 7, 2014 at 14:29
  • $\begingroup$ @RobertBryant: oops. I included that assumption in the second sentence but forgot it in the first. Thanks. $\endgroup$
    – Willie Wong
    Commented Nov 7, 2014 at 15:51
  • $\begingroup$ Actually, once you put it in the first sentence, the second sentence simplifies, as there is only one connected component to consider. $\endgroup$
    – Robert Bryant
    Commented Nov 7, 2014 at 16:22
  • $\begingroup$ Well, now that you have modified it that way, you actually have to add the hypothesis 'complete' as well, to rule out proper open subsets of space forms (which, of course, are not space forms). $\endgroup$
    – Robert Bryant
    Commented Nov 7, 2014 at 16:26
  • $\begingroup$ @RobertBryant: how'bout I give you free permission to edit this post? :-) Thanks again. Also, note that it has converged to the revised question of the OP. $\endgroup$
    – Willie Wong
    Commented Nov 7, 2014 at 16:28
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The algebra of Killing vector fields is the Lie algebra of the isometry group. Maximally symmetric spaces are then spaces of dimension $m$ with an $m(m+1)/2$-dimensional group of isometries. Such spaces are also called spaces of maximal free mobility. Birkhoff (Extensions of Lie groups, Math. Z. 53, 1950, 226-235) has shown that a Riemannian manifold with maximal free mobility is locally isomorphic to one of the classical geometries - euclidean, spherical or hyperbolic.

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  • $\begingroup$ Watch out for completeness; Birkhoff requires it. Also, the original problem above is global, and for pseudo-Riemannian metrics. Birkhoff only considered the Riemannian case. I think the real issue is about which covering spaces you can use, not the local geometry. $\endgroup$
    – Ben McKay
    Commented Jan 31, 2019 at 11:21
  • $\begingroup$ Birkhoff's prove is very metricy, using barycentric coordinates to see that isometries are smooth, so I think it would be hard to imitate for pseudo-Riemannian manifolds. $\endgroup$
    – Ben McKay
    Commented Jan 31, 2019 at 11:33

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