I'm trying to break the classification of locally riemannian symmetric spaces to little steps to make it more comprehensible (and s.t. the technical details can be verified without drowning completely).

The first big step which I find difficult to break to concise little pieces is how to get from general riemannian locally symmetric spaces to a product of simply connected indecomposable symmetric spaces. Here's my progress so far:

Let $(M,g,\nabla)$ be a riemannian manifold with a levi-civita connection (of dimension $n$).

For every point $p \in M$ the linear reflection on the tangent space $-Id: T_p M \to T_p M$ extends to define an involution on every normal coordinate ball $\sigma_p : N_p \to N_p$ which corresponds to reversing the geodesics through $p$.

Definition: A locally symmetric space is a riemannian manifold $(M,g)$ with the property that for every $p \in M$ the geodesic involution $\sigma_p :N_p \to N_p$ is an isometry (or equivalently affine with respect to the connection $\nabla$).

Here's how I understand the steps so far:

  • $M$ and $\nabla$ are analytic - Any manifold with a connection $\nabla$ for which both torsion and curvature are parallel is analytic and $\nabla$ is analytic. Kobayashi & Nomizu VI 7.7

  • Assume $M$ is complete - As far as I understand we can't make progress without this.

  • Universal cover - The universal cover $\pi : \tilde M \to M$ inherits a metric structure for which it is an isometry implying $\tilde M$ is again locally symmetric. We assume from now on that $M$ is simply connected.

  • Involutions extend to global isometries (?) - Kobayashi and Nomizu IV 6.3 - "Let $M \supset U \to N$ be an isometric immersion of connected analytic riemannian manifolds with $M$ complete and $N$ simply connected then $f$ extends to an isometric immersion $f: M \to N$". So the isometry group $I(M)$ acts transitively and $M = I(M)/K$ where $K$ is a compact stabilizer group. Furthermore $K \subset O(T_p M)$ at every point.

  • Holonomy is subgroup of stabilizer - Transvections $T^{\gamma}_t := \sigma_{\gamma(t/2)} \circ \sigma_{\gamma(0)}$ along geodesics $\gamma : I \to M$ form one parameter families of isometries (fixing correspondingly the points $\gamma(0)$). Every curve is $C^1$ limit of piecewise geodesic curves. Let $c : I \to M$ be a closed curve which is a limit of closed polygon geodesics. Transvections must fix the edges of the polygons and therefore define in the limit a parallel translation. $K$ Is compact in $I(M)$ in $I(M)$ (being a stabilizer subgroup). Therefore $Hol_p \subset K$.

  • De Rahm decomposition theorem (?) - There's a question about this on the site but it hasn't received very useful answers. I'd like to know where to find the statement and the proof of this theorem in the most modern language.

Adding everything together we have a product decomposition $I(M_0)/H_0 \times \dots \times I(M_k)/H_k$ where $I(M_j)$ acts irreducibly. From here on it's pretty much an algebraic road I think.

Is there something substantial missing/wrong in the above outline?

  • 2
    $\begingroup$ Already your first assertion is dangerous - take an open subset of a Riemannian symmetric space, then it becomes a locally symmetric space in your definition, because no normal coordinate ball ever extends across the (deleted) boundary. It might be safer to demand completeness from the very beginning. $\endgroup$ Commented Apr 24, 2016 at 15:54
  • $\begingroup$ Certainly your statement about extending an isometry to a local isometry is not true in general (take a metric on the sphere that's the round metric near the north pole but interestingly different elsewhere). But (as noted in Kobayashi-Nomizu's proof of the theorem this post is about) you don't need what you quote here; use Ch. VI, theorem 7.9. The de Rham decomposition theorem is also proved in KN, see Ch IV section 6. $\endgroup$
    – mme
    Commented Apr 24, 2016 at 16:16
  • $\begingroup$ @MikeMiller Ch. VI, theorem 7.9. Uses implicitly VI corollary 6.2 which does use analyticity $\endgroup$ Commented Apr 24, 2016 at 16:21
  • $\begingroup$ @SaalHardali Thanks for pointing this out. 7.7 proves that if the curvature and torsion are parallel (which the c/e I suggested above would not satisfy), the manifold has a natural analytic structure for which all the relevant things are analytic. A locally symmetric space has parallel curvature tensor. $\endgroup$
    – mme
    Commented Apr 24, 2016 at 16:29

1 Answer 1


Let me outline a different approach, namely the one É. Cartan himself took, one that doesn't depend on assuming completeness, doesn't rely on ideas about real-analyticity, and side-steps many of the issues that you would have to deal with in fleshing out the outline above.

So suppose that $(M^n,g)$ is a locally symmetric Riemannian manifold, which means that each $p\in M$ has an open neighborhood $U$ that supports a $g$-isometry $\iota_p:(U,g)\to (U,g)$ that fixes $p$ and is such that $(\iota_p)'(p):T_pU\to T_pU$ is minus the identity.

Let $B\to M$ denote the $\mathrm{O}(n)$-bundle of $g$-orthonormal coframes $u:T_pM\to\mathbb{R}^n$. Let $\omega:TB\to\mathbb{R}^n$ denote the canonical $1$-form and let $\theta:TB\to{\frak{so}}(n)$ be the unique $1$-form with values in skew-symmetric matrices that satisfies the first structure equation $$ \mathrm{d}\omega = -\theta\wedge\omega.\tag1 $$ This $\theta$ satisfies the second structure equation $$ \mathrm{d}\theta = -\theta\wedge\theta + R(\omega\wedge\omega)\tag2 $$ where $R:B\to K_n\subset \mathrm{Hom}\bigl({\frak{so}}(n),{\frak{so}}(n)\bigr)$ is the Riemann curvature function, where $K_n \subset \mathrm{Hom}\bigl({\frak{so}}(n),{\frak{so}}(n)\bigr)$ is the subspace of curvature tensors that satisfy the (first) Bianchi identity. Of course, $R$ is $\mathrm{O}(n)$-equivariant and satisfies the third structure equation $$ \mathrm{d}R = -\theta . R + R'(\omega)\tag3 $$ where $R':B\to \mathrm{Hom}\bigl(\mathbb{R}^n, K_n)$ is the curvature function that represents the first covariant derivative of the Riemann curvature tensor.

Now, the hypothesis that $(M,g)$ is locally symmetric implies that $R'$ vanishes, so let's assume this. (It's not immediately obvious that the converse holds, but it does.) Since $\mathrm{d}R = -\theta . R$, it follows that $R$ takes values in a single $\mathrm{O}(n)$-orbit in $K_n$ and, in fact, is a submersion of $B$ onto this orbit.

Let $r\in K_n$ be an element of this orbit, let $H\subset \mathrm{O}(n)$ be its stabilizer, and let $P = R^{-1}(r)\subset B$. Then $P$ is a principal right $H$-bundle over $M$, and, pulling everything back to $P$, the equation $(1)$ continues to hold, the equation $(2)$ becomes $$ \mathrm{d}\theta = -\theta\wedge\theta + r(\omega\wedge\omega)\tag{2'} $$ and $(3)$ becomes $$ 0 = -\theta.r\tag{3'}. $$ In other words, $\theta$ (pulled back to $P$) takes values in ${\frak{h}}\subset{\frak{so}}(n)$, the Lie algebra of $H\subset\mathrm{O}(n)$.

Because $r:{\frak{so}}(n)\to {\frak{so}}(n)$ is symmetric (thanks to the first Bianchi identity), its kernel is equal to its image, say $\frak{r}\subset {\frak{so}}(n)$. Differentiating $(3')$ and using $(2')$ shows that $$ 0 = \bigl(r(\omega\wedge\omega)\bigr).r, $$ implying that $\frak{r}\subseteq \frak{h}$. In particular, the right hand side of $(2')$ is a $2$-form that takes values in $\frak{h}$.

At this point, we have a coframing $$ \gamma = (\omega,\theta):TP\to \mathbb{R}^n\oplus\frak{h} = \frak{g} $$ satisfying structure equations $(1)$ and $(2')$ where the $2$-forms on the right hand sides of these equations have constant coefficients in the coframing. Since $\mathrm{d}^2 =0$ is an identity, it follows that there is a unique Lie algebra structure on $\frak{g}$ such that $(1)$ and $(2')$ are equivalent to $$ \mathrm{d}\gamma = -\tfrac12[\gamma,\gamma]. $$ The fact that the equations $(1)$ and $(2')$ would be unaffected by replacing $\omega$ by $-\omega$ implies that the automorphism of $\frak{g}$ that is minus the identity on $\mathbb{R}^n$ and the identity on $\frak{h}$ is an automorphism of the Lie algebra, i.e., the pair $(\frak{g},\frak{h})$ is an orthogonal symmetric pair.

For the converse, you need the existence of a Lie group with a given Lie algebra, etc.

N.B.: The fact that $a\in \frak{h}$ satisfies $a.r = 0$ implies the statement that $$ r\bigl([a,y]\bigr) = \bigl[a,r(y)\bigr] $$ for all $a\in \frak{h}$, so $\frak{r}$ is an ideal in $\frak{h}$. In fact, one usually (but not always) has $\frak{r}=\frak{h}$, but we don't need this to complete the argument.

N.B.: Note that $\frak{g}$ is the Lie algebra of the maximal symmetry group of $(M,g)$, while $\frak{u} = \mathbb{R}^n\oplus\frak{r}$ is the Lie algebra of the 'minimal' symmetric representation of $(M,g)$.


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