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.


1 Answer 1


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.

  • $\begingroup$ 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 $d\phi$. $\endgroup$
    – D Ford
    Commented Nov 1, 2019 at 21:13
  • 1
    $\begingroup$ 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. $\endgroup$
    – Ivo Terek
    Commented Nov 2, 2019 at 20:18

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