# Locally symmetric spaces have parallel Riemannian curvature

I'm trying to prove the following result:

If $$(M,g)$$ is a locally symmetric Riemannian manifold, then the Riemannian curvature tensor is parallel: $$\nabla Rm \equiv 0$$.

By "locally symmetric", I mean that every point $$p \in M$$ has a local point reflection, i.e. a neighborhood $$U$$ and an isometry $$\phi : U \to U$$ that fixes $$p$$ and for which $$d\phi_p = -\mathrm{Id}$$. Here the Riemannian curvature tensor is $$Rm(X,Y,Z,W) = \langle R(X,Y)Z, W\rangle,$$ where $$R : \mathcal X(M) \times \mathcal X(M) \to \mathcal X(M)$$ is the curvature endomorphism $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z,$$ and $$\nabla Rm$$ is the covariant $$5$$-tensor given by $$(\nabla Rm)(X,Y,Z,W,V) = (\nabla_V Rm)(X,Y,Z,W,V)$$. I know that $$\phi^*(\nabla Rm) = -\nabla Rm$$, since $$\nabla Rm$$ has an odd number of arguments and $$d\phi_p = -\mathrm{Id}$$, and I know $$\phi^*Rm = Rm$$ by isometry-invariance of $$Rm$$. But that's about all I've been able to tease apart. I've tried expanding $$\phi^*(\nabla Rm)(X,Y,Z,W,V)$$ into \begin{align*} (\nabla_{\phi_*V} Rm)(\phi_*X,\phi_*Y,\phi_*Z,\phi_*W) &= (\phi_*V)Rm(\phi_*X,\phi_*Y,\phi_*Z,\phi_*W) \\ &\quad- Rm(\nabla_{\phi_*V}(\phi_*X),\phi_*Y,\phi_*Z,\phi_*W) \\ &\quad-Rm(\phi_*X,\nabla_{\phi_*V}(\phi_*Y),\phi_*Z,\phi_*W) \\ &\quad-Rm(\phi_*X,\phi_*Y,\nabla_{\phi_*V}(\phi_*Z),\phi_*W) \\ &\quad-Rm(\phi_*X,\phi_*Y,\phi_*Z,\nabla_{\phi_*V}(\phi_*W)) \end{align*} but this just becomes $$-(\nabla Rm)(X,Y,Z,W,V)$$ again. I have a feeling there's something fundamental I'm missing. Any suggestions?

EDIT: I'm able to answer the question assuming a kind of "naturality" of the Levi-Civita connection in the covariant tensor bundle $$T^5TM$$. I'm not certain if this naturality assumption is a good one, however. This is a question I've asked here.

If $$T$$ is a covariant $$k$$-tensor in a vector space $$V$$ which is invariant under $$-{\rm Id}_V$$, then $$T = (-1)^k T$$. Thus, if $$k$$ is odd, we necessarily must have $$T=0$$. We want to apply this observation for the Riemann tensor. Given $$p \in M$$, we want to prove that the tensor $$(\nabla R)_p$$ on $$T_pM$$ vanishes. Since $$(M,g)$$ is locally symmetric, $$-{\rm Id}_{T_pM}$$ is realized as the differential of some isometry fixing $$p$$. Since $$(\nabla R)_p$$ has rank $$5$$, which is odd, it must be zero.
• Yes, this is the crux of the algebraic argument. The only subtlety is in proving that $\phi^*(\nabla R) = \nabla R$. I was eventually able to do this, but it took a couple lines of computation to verify that the connection in $T^5TM$ is invariant under $d\phi$. Commented Nov 1, 2019 at 21:13
• Think locally: isometries preserve the Christoffel symbols associated to the connection in $TM$. The Christoffel symbols of the induced connection in $T^5TM$ are expressed in terms of the original Christoffel symbols, so they are also preserved. Commented Nov 2, 2019 at 20:18