In two different books I found these two related statements.

  • The book by Jost defines a ``locally symmetric space" as one for which the curvature tensor is constant and which is geodesically complete.
  • Andreas' book attributes to Cartan a theorem that a space is locally symmetric if and only if its curvature tensor is constant. Where he defines a space to be locally symmetric if about every point there is a geodesic reflecting isometry.

I could not trace this theorem of Cartan anywhere else nor does Andreas' book give a reference. I would like to know some reference which would prove this stuff and resolve the apparent definition conflict between the above two statements.

Further I came across these 3 statements about classifying constant curvature spaces,

  • If the isometry group is transitive on points on the manifold, then the scalar curvature is constant.

  • If the isometry group is transitive on all one-dimensional subspaces of tangent spaces, then the Ricci curvature tensor is a scalar multiple of the metric. tensor

  • If the isometry group is transitive on all two-dimensional subspaces of tangent spaces, then the sectional curvature is constant on all two-dimensional subspaces of tangent spaces.

I have an understanding of why the first statement is true but I would like to know some reference which explains the other two statements.

Further call a space to be homogeneous if it is quotient of some Lie Group mod a closed subgroup. Then in physics literature one finds a statement of the kind that an "isotropic and homogeneous space-time is of constant curvature". I would like to know what is the precise mathematical meaning of this statement.

Further I would also like to know a reference for the fact that a maximally symmetric space always has constant curvature and how being maximally symmetric fits in with being "isotropic and homogeneous".

Many sources refer me to the volumes by Kobayashi and Nomizu or the book by Besse or Spivak for discussion on constant curvature spaces but unfortunately I don't have access to these books. It would be great if some online reference like lecture notes/expository review paper on constant curvature spaces could be linked which clarifies the above questions.

In the books on Riemannian Geometry that I have access to like the ones by Gallot et al,Andreas or Jost, I can't see much of any discussion on these topics.

{In all the above statements I suppose the Riemann Christoffel connection is being assumed. It would be interesting to know how much of the things hold for a general affine connection.}

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    $\begingroup$ The result you mention at the begining of your post should be in Helgason's book, I imagine. $\endgroup$ Commented Apr 21, 2010 at 18:08
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    $\begingroup$ By the way, when making such multiple questions, marking them with numbers or something like that so that one can answer only a subquestion and be able to identify it easily would be nice! $\endgroup$ Commented Apr 21, 2010 at 18:10
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    $\begingroup$ Anirbit, could you tell us what Jost says verbatim? Maybe something about the curvature tensor being covariant-constant? As Jose says, the definition you ascribe to Jost doesn't look right. $\endgroup$
    – Tim Perutz
    Commented Apr 21, 2010 at 19:58
  • $\begingroup$ In Jost's Riemannian Geometry and Geometric Analysis (is this the right book by Jost?), Definition 5.3.3, one reads that A complete riemannian manifold with $\nabla R \equiv 0$ is called locally symmetric. I believe this is not a correct definition: a locally symmetric space need not be complete. (Both Besse and Helgason agre with this.) $\endgroup$ Commented Apr 21, 2010 at 20:48
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    $\begingroup$ Ah, I should also add that there's an okay treatment in O'Neill's "Semi-Riemannian geometry" (chapter 8). And in a pinch you can always look up definitions and more references on the Springer EOM eom.springer.de/S/s091710.htm And since you asked about online lecture notes, you may want to try Eschenburg's math.uni-augsburg.de/~eschenbu/symspace.pdf $\endgroup$ Commented Apr 21, 2010 at 23:40

2 Answers 2


It is difficult to reconcile your first two statements, because they are actually wrong as written!

A riemannian manifold is locally symmetric if and only if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection. This condition was studied by Élie Cartan, who classified them using his classification of real semisimple Lie algebras. My favourite reference for this is Besse's Einstein manifolds, Chapter 7F, from where I quote the following theorem:

10.72 Theorem (Elie Cartan). For a Riemannian manifold (not necessarily complete) the following conditions are equivalent:

(i) $DR = 0$;

(ii) the geodesic symmetry $s_p$ around any point $p$ (which is defined only locally) is an isometry.

For a complete Riemannian manifold $(M,g)$ the two following conditions are equivalent:

(iii) for every point $p$ in $M$ the geodesic symmetry around $p$ is well defined and is an isometry;

(iv) the manifold $M$ is a homogeneous space $G/H$ where $G$ is a connected Lie group, $H$ a compact subgroup of $G$, and where there exists an involutive automorphism $\sigma$ of the group $G$ for which, if $S$ denotes the fixed point set of $\sigma$ and $S_e$ its connected component of the identity one has $S_e \subset H \subset S$. Moreover the Riemannian metric under consideration on $G/H$ is invariant under $G$.

Furthermore: if $(M, g)$ satisfies (iii) or (iv), then it satisfies (i) and (ii). If $(M, g)$ satisfies (i) or (ii) and if it is simply connected and complete then it satisfies (iii) and (iv).

10.76 Definition. A Riemannian manifold is said to be locally symmetric if it satisfies (i) or (ii) above; it is said to be symmetric if it satisfies (iii) or (iv).

I suspect the source of the confusion might be with the meaning of "constant curvature", which usually means constant sectional curvature (or perhaps parallel Riemann curvature) and not constant scalar curvature, which is a much weaker condition.

Concerning your final parenthetic question, there is a theorem of Ambrose and Singer, reformulated by Kostant, which says that a riemannian manifold $(M,g)$ is locally homogeneous if and only if it admits a metric connection with parallel torsion and parallel curvature. It is locally symmetric if (and only if) the connection is torsion-free.

  • $\begingroup$ I figured out that I have institutional access to the book by Besse on Springer website. Thanks for this precise reference. $\endgroup$
    – Anirbit
    Commented Apr 22, 2010 at 14:50

First, a clarification. When you say "the isometry group is transitive on all one-dimensional subspaces of tangent spaces", what this really means is that for any $v\in T_p M$ and $w\in T_q M$ with $|v| = |w|$, there is an isometry $f:M\rightarrow M$ with $f(p) = q$ and $d_p f v = w$.

This immediately implies that $Ric_p(v) = Ric_q(w)$ for any vectors $v$ and $w$ of the same length. This implies that the polarization of this to $Ric(\cdot, \cdot)$ are also the same (meaning $Ric_p(v,v') = Ric_q(d_p f v, d_p f v')$ for any isometry $f$ mapping $p$ to $q$).

Now, fix a $p\in M$ and let $G_p\subseteq $Iso$(M)$ be the subgroup of all isometries which fix $p$. By the assumption of the second bullet point, $G_p$ acts transitively on the sphere in $T_p M$ (i.e., $G_p$ takes any unit vector to any other unit vector). It follows that the representation of $G_p$ on $T_p M$ is irreducible.

Now, let $g$ denote the inner product at $T_p M$ and let $P:T_pM\rightarrow T_p M$ be defined by $g(P v, w) = Ric_p(v,w)$. It's easy to see that $P$ will be $G_p$ invariant, and hence, by Schur's lemma, a multiple of the identity. That is, $Ric_p$ is a multiple of the metric.

The statement about sectional curvatures is easier: Curvature is an invariant of the metric, and hence so is sectional curvature. By the assumption of bullet point 3, there is an isometry of $G$ which takes any point to any other point and moves any 2-plane to any other 2-plane. But then the sectional curvature of any 2-plane at any point is equal to the sectional curvature of any 2-plane at any other point, that is, $M$ has constant sectional curvature.

  • $\begingroup$ @Jason Thanks a lot for your great explanation. I was wondering as an extension of the third bullet point as to what can one say about the curvature if one is told that the isometry group is transitive on the set of tangent spaces (and not just on every 2-plane) If I know that the Riemannian manifold is a homogeneous space and that the isometry group is transitive on the set of tangent spaces then can one say anything about the curvature? $\endgroup$
    – Anirbit
    Commented Apr 24, 2010 at 8:55
  • $\begingroup$ First, homogeneous automatically implies the isometry group is transitive on the set of tangent spaces: for p and q in M, the fact that it's homogeneous implies there is an isometry f with f(p) = q. Then d_p f will automatically map TpM to TqM. As to your next question, there isn't much which can be said in this generality, but, for example, once you know all the curvatures of every 2plane at one point, you automatically know them everywhere. Compact homogeneous manifolds with finite fundamental group always have another metric which is still homogeneous and has sectional curvature >=0. $\endgroup$ Commented Apr 24, 2010 at 14:26
  • $\begingroup$ Probably I wasn't very clear. I was meaning homogeneous in the sense of the manifold being diffeomorphic to some G/H. In your initial explanation could you clarify why the map P with the property you are looking for should exist? Or to put it otherwise, why given any v should there exist a v' such that g(v',w)=Ric(v,w) for all w? {I can see the rest of the argument assuming the existence of the map P} $\endgroup$
    – Anirbit
    Commented Apr 24, 2010 at 20:18
  • $\begingroup$ I see - I meant homogeneous as in the (Riemannian) manifold is isometric to some G/H. To show P exists, pick an orthonormal basis $e_i$ for TpM. Create the symmetric n x n matrix $A = (A_{ij})$ where $A_{ij} = Ric(e_i, e_j)$. Then define $P$ by declaring it to be given by $A$ on the basis $e_i$. Then, for example, we can figure out $P(e_1)$ by pairing it with every e_i: we get $g(P(e_1), e_i) = A_{1i} = Ric(e_1, e_i)$ as desired. $\endgroup$ Commented Apr 24, 2010 at 22:26
  • $\begingroup$ Ah! Thanks. In connection to the first bullet point I noticed that the combination of Definition 5.3.2 and Theorem 5.3.2 in the book by Jost together implies that if the isometry group acts transitively on the manifold then the entire Riemann Curvature Tensor is constant. Then shouldn't there be a weaker criteria which makes just the scalar curvature constant? $\endgroup$
    – Anirbit
    Commented Apr 26, 2010 at 6:20

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