Theorem: Let $M = G/K$ be a Riemannian symmetric space. Then the curvature tensor $R$ is parallel: $\nabla R = 0$.

Proof:

Step 1: Recall symmetric space structure

By definition, for each $p \in M$, there exists geodesic symmetry $s_p: M \to M$ with:

• $s_p(p) = p$
• $ds_p|_p = -\text{id}$
• $s_p$ is an isometry

Step 2: Use isometry property

Since $s_p$ is an isometry, it preserves the Riemannian curvature tensor: $s_p^* R = R$

This means for any $q \in M$: $R_{s_p(q)} (ds_p(X), ds_p(Y), ds_p(Z), ds_p(W)) = R_q(X, Y, Z, W)$

Step 3: Apply at point $p$

At $p$ itself, we have $s_p(p) = p$ and $ds_p|_p = -\text{id}$. Therefore: $R_p(-X, -Y, -Z, -W) = R_p(X, Y, Z, W)$

Since the curvature tensor is multilinear and alternating, this is automatically satisfied. But we need more information.

Step 4: Use homogeneity

The key is that $G$ acts transitively on $M$, so we can transport the curvature from one point to any other via group elements. For any $g \in G$ and tangent vectors at $o$ (basepoint): $R_{g \cdot o} = (dg)_* R_o$

The covariant derivative of $R$ must also transform the same way.

Step 5: Compute the derivative

For a symmetric space, geodesics through the origin are of form $\gamma(t) = \exp(tX) \cdot o$ for $X \in \mathfrak{p}$. The curvature along such a geodesic is: $R_{\gamma(t)} = (d\exp(tX))_* R_o$

Taking the covariant derivative along $\gamma$: $\nabla_{\dot{\gamma}} R = \frac{D}{dt}[(d\exp(tX))_* R_o]$

Step 6: Use the Cartan decomposition

The key observation is that for symmetric spaces, the curvature at the origin is completely determined by the Lie bracket on $\mathfrak{p}$: $R_o(X, Y)Z = -[[X, Y], Z]$ for $X, Y, Z \in \mathfrak{p}$ (after identification $T_o M \cong \mathfrak{p}$).

This formula is $K$-invariant (since $K$ acts on $\mathfrak{p}$ via adjoint action), and the symmetry $s_o$ acts as $-\text{id}$ on $\mathfrak{p}$.

Step 7: Conclude parallelism

The combination of:

• Homogeneity under $G$-action
• Symmetry under $s_p$ for each $p$
• The algebraic formula for curvature

Forces the curvature tensor to be constant in the direction of any geodesic. Since $M$ is connected and geodesics fill out $M$, this implies $\nabla R = 0$ everywhere. □

Corollary: Every symmetric space is locally homogeneous and Einstein.

Proof: Parallel curvature implies the Ricci tensor is also parallel: $\nabla \text{Ric} = 0$. For a homogeneous space, the Ricci tensor must be a constant multiple of the metric: $\text{Ric} = \lambda g$ (Einstein condition). □